wavelength to the circumference 2pr n of the orbit.

Transcription

1 Exercie 131 level and (b) the n = 4 level? In each cae, compare the de Broglie energy (in electron volt) would the m = 1 intenity maximum wavelength to the circumference pr n of the orbit. occur at u = 60.6? For thi energy, i there an m = intenity maximum? Explain A CD-ROM i ued intead of a crytal in an electrondiffraction experiment. The urface of the CD-ROM ha track of (a) A nonrelativitic free particle with ma m ha kinetic energy K. Derive an expreion for the de Broglie wavelength of the particle in term of m and K. (b) What i the de Broglie wavelength of an 800-eV electron? Why Don t We Diffract? (a) Calculate the de Broglie wavelength of a typical peron walking through a doorway. Make reaonable approximation for the neceary quantitie. (b) Will the peron in part (a) exhibit wavelike behavior when walking through the ingle lit of a doorway? Why? What i the de Broglie wavelength for an electron with peed (a) v = 0.480c and (b) v = 0.960c? (Hint: Ue the correct relativitic expreion for linear momentum if neceary.) (a) If a photon and an electron each have the ame energy of 0.0 ev, find the wavelength of each. (b) If a photon and an electron each have the ame wavelength of 50 nm, find the energy of each. (c) You want to tudy an organic molecule that i about 50 nm long uing either a photon or an electron microcope. Approximately what wavelength hould you ue, and which probe, the electron or the photon, i likely to damage the molecule the leat? How fat would an electron have to move o that it de Broglie wavelength i 1.00 mm? Wavelength of a Bullet. Calculate the de Broglie wavelength of a 5.00-g bullet that i moving at 340 m>. Will the bullet exhibit wavelike propertie? Find the wavelength of a photon and an electron that have the ame energy of 5 ev. (Note: The energy of the electron i it kinetic energy.) (a) What accelerating potential i needed to produce electron of wavelength 5.00 nm? (b) What would be the energy of photon having the ame wavelength a thee electron? (c) What would be the wavelength of photon having the ame energy a the electron in part (a)? Through what potential difference mut electron be accelerated o they will have (a) the ame wavelength a an x ray of wavelength nm and (b) the ame energy a the x ray in part (a)? (a) Approximately how fat hould an electron move o it ha a wavelength that make it ueful to meaure the ditance between adjacent atom in typical crytal (about 0.10 nm)? (b) What i the kinetic energy of the electron in part (a)? (c) What would be the energy of a photon of the ame wavelength a the electron in part (b)? (d) Which would make a more effective probe of mallcale tructure: electron or photon? Why? CP A beam of electron i accelerated from ret through a potential difference of kv and then pae through a thin lit. The diffracted beam how it firt diffraction minima at 11.5 from the original direction of the beam when viewed far from the lit. (a) Do we need to ue relativity formula? How do you know? (b) How wide i the lit? A beam of neutron that all have the ame energy catter from atom that have a pacing of nm in the urface plane of a crytal. The m = 1 intenity maximum occur when the angle u in Fig. 39. i 8.6. What i the kinetic energy (in electron volt) of each neutron in the beam? A beam of 188-eV electron i directed at normal incidence onto a crytal urface a hown in Fig. 39.3b. The m = intenity maximum occur at an angle u = (a) What i the pacing between adjacent atom on the urface? (b) At what other angle or angle i there an intenity maximum? (c) For what electron tiny pit with a uniform pacing of 1.60 mm. (a) If the peed of the electron i 1.6 * 10 4 m>, at which value of u will the m = 1 and m = intenity maxima appear? (b) The cattered electron in thee maxima trike at normal incidence a piece of photographic film that i 50.0 cm from the CD-ROM. What i the pacing on the film between thee maxima? (a) In an electron microcope, what accelerating voltage i needed to produce electron with wavelength nm? (b) If proton are ued intead of electron, what accelerating voltage i needed to produce proton with wavelength nm? (Hint: In each cae the initial kinetic energy i negligible.) You want to tudy a biological pecimen by mean of a wavelength of 10.0 nm, and you have a choice of uing electromagnetic wave or an electron microcope. (a) Calculate the ratio of the energy of a 10.0-nm-wavelength photon to the kinetic energy of a 10.0-nm-wavelength electron. (b) In view of your anwer to part (a), which would be le damaging to the pecimen you are tudying: photon or electron? Section 39. The Nuclear Atom and Atomic Spectra CP A 4.78-MeV alpha particle from a 6 Ra decay make a head-on colliion with a uranium nucleu. A uranium nucleu ha 9 proton. (a) What i the ditance of cloet approach of the alpha particle to the center of the nucleu? Aume that the uranium nucleu remain at ret and that the ditance of cloet approach i much greater than the radiu of the uranium nucleu. (b) What i the force on the alpha particle at the intant when it i at the ditance of cloet approach? A beam of alpha particle i incident on a target of lead. A particular alpha particle come in head-on to a particular lead nucleu and top 6.50 * m away from the center of the nucleu. (Thi point i well outide the nucleu.) Aume that the lead nucleu, which ha 8 proton, remain at ret. The ma of the alpha particle i 6.64 * 10-7 kg. (a) Calculate the electrotatic potential energy at the intant that the alpha particle top. Expre your reult in joule and in MeV. (b) What initial kinetic energy (in joule and in MeV) did the alpha particle have? (c) What wa the initial peed of the alpha particle? Section 39.3 Energy Level and the Bohr Model of the Atom The ilicon ilicon ingle bond that form the bai of the mythical ilicon-baed creature the Horta ha a bond trength of 3.80 ev. What wavelength of photon would you need in a (mythical) phaor diintegration gun to detroy the Horta? A hydrogen atom i in a tate with energy ev. In the Bohr model, what i the angular momentum of the electron in the atom, with repect to an axi at the nucleu? A hydrogen atom initially in the ground level aborb a photon, which excite it to the n = 4 level. Determine the wavelength and frequency of the photon A triply ionized beryllium ion, Be 3+ (a beryllium atom with three electron removed), behave very much like a hydrogen atom except that the nuclear charge i four time a great. (a) What i the ground-level energy of Be 3+? How doe thi compare to the groundlevel energy of the hydrogen atom? (b) What i the ionization energy of Be 3+? How doe thi compare to the ionization energy of the

2 13 CHAPTER 39 Particle Behaving a Wave hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the n = to n = 1 tranition i 1 nm (ee Example 39.6). What i the wavelength of the photon emitted when a Be 3+ ion undergoe thi tranition? (d) For a given value of n, how doe the radiu of an orbit in Be 3+ compare to that for hydrogen? (a) Show that, a n get very large, the energy level of the hydrogen atom get cloer and cloer together in energy. (b) Do the radii of thee energy level alo get cloer together? (a) Uing the Bohr model, calculate the peed of the electron in a hydrogen atom in the n = 1,, and 3 level. (b) Calculate the orbital period in each of thee level. (c) The average lifetime of the firt excited level of a hydrogen atom i 1.0 * In the Bohr model, how many orbit doe an electron in the n = level complete before returning to the ground level? CP The energy-level Figure E39.30 cheme for the hypothetical oneelectron element Searium i hown in Fig. E The potential energy i taken to be zero for an electron at an infinite ditance from the nucleu. (a) How much energy (in electron volt) doe it take to ionize an electron from the ground level? (b) An 18-eV photon i aborbed n 5 4 n 5 3 n 5 n 5 1 ev 5 ev 10 ev 0 ev by a Searium atom in it ground level. A the atom return to it ground level, what poible energie can the emitted photon have? Aume that there can be tranition between all pair of level. (c) What will happen if a photon with an energy of 8 ev trike a Searium atom in it ground level? Why? (d) Photon emitted in the Searium tranition n = 3 S n = and n = 3 S n = 1 will eject photoelectron from an unknown metal, but the photon emitted from the tranition n = 4 S n = 3 will not. What are the limit (maximum and minimum poible value) of the work function of the metal? In a et of experiment Figure E39.31 on a hypothetical one-electron atom, you meaure the wavelength of the n 5 5 n 5 4 photon emitted from tranition ending in the ground n 5 3 tate 1n = 1, a hown in the energy-level diagram in Fig. E You alo oberve that it take ev to ionize thi atom. (a) What i the energy of the atom in each of the lev- l nm el 1n = 1, n =, etc.) n 5 1 hown in the figure? (b) If an electron made a tranition from the n = 4 to the n = level, what wavelength of light would it emit? Find the longet and hortet wavelength in the Lyman and Pachen erie for hydrogen. In what region of the electromagnetic pectrum doe each erie lie? (a) An atom initially in an energy level with E = -6.5 ev aborb a photon that ha wavelength 860 nm. What i the internal energy of the atom after it aborb the photon? (b) An atom initially in an energy level with E = -.68 ev emit a photon that ha wavelength 40 nm. What i the internal energy of the atom after it emit the photon? l nm l nm l nm n Ue Balmer formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the H g line of the Balmer erie for hydrogen. Section 39.4 The Laer BIO Laer Surgery. Uing a mixture of CO, N, and ometime He, CO laer emit a wavelength of 10.6 mm. At power output of kw, uch laer are ued for urgery. How many photon per econd doe a CO laer deliver to the tiue during it ue in an operation? BIO Removing Birthmark. Puled dye laer emit light of wavelength 585 nm in 0.45-m pule to remove kin blemihe uch a birthmark. The beam i uually focued onto a circular pot 5.0 mm in diameter. Suppoe that the output of one uch laer i 0.0 W. (a) What i the energy of each photon, in ev? (b) How many photon per quare millimeter are delivered to the blemih during each pule? How many photon per econd are emitted by a 7.50-mW CO laer that ha a wavelength of 10.6 mm? BIO PRK Surgery. Photorefractive keratectomy (PRK) i a laer-baed urgical procedure that correct near- and farightedne by removing part of the len of the eye to change it curvature and hence focal length. Thi procedure can remove layer 0.5 mm thick uing pule lating 1.0 n from a laer beam of wavelength 193 nm. Low-intenity beam can be ued becaue each individual photon ha enough energy to break the covalent bond of the tiue. (a) In what part of the electromagnetic pectrum doe thi light lie? (b) What i the energy of a ingle photon? (c) If a 1.50-mW beam i ued, how many photon are delivered to the len in each pule? A large number of neon atom are in thermal equilibrium. What i the ratio of the number of atom in a 5 tate to the number in a 3p tate at (a) 300 K; (b) 600 K; (c) 100 K? The energie of thee tate, relative to the ground tate, are E 5 = 0.66 ev and E 3p = ev. (d) At any of thee temperature, the rate at which a neon ga will pontaneouly emit 63.8-nm radiation i quite low. Explain why Figure 39.19a how the energy level of the odium atom. The two lowet excited level are hown in column labeled P and Find the ratio of the number of atom in a 3 > P 1 >. P 3 tate to the number in a > P 1 > tate for a odium ga in thermal equilibrium at 500 K. In which tate are more atom found? Section 39.5 Continuou Spectra A 100-W incandecent light bulb ha a cylindrical tungten filament 30.0 cm long, 0.40 mm in diameter, and with an emiivity of 0.6. (a) What i the temperature of the filament? (b) For what wavelength doe the pectral emittance of the bulb peak? (c) Incandecent light bulb are not very efficient ource of viible light. Explain why thi i o Determine l m, the wavelength at the peak of the Planck ditribution, and the correponding frequency ƒ, at thee temperature: (a) 3.00 K; (b) 300 K; (c) 3000 K Radiation ha been detected from pace that i characteritic of an ideal radiator at T =.78 K. (Thi radiation i a relic of the Big Bang at the beginning of the univere.) For thi temperature, at what wavelength doe the Planck ditribution peak? In what part of the electromagnetic pectrum i thi wavelength? The hortet viible wavelength i about 400 nm. What i the temperature of an ideal radiator whoe pectral emittance peak at thi wavelength?

3 Problem Two tar, both of which behave like ideal blackbodie, radiate the ame total energy per econd. The cooler one ha a urface temperature T and a diameter 3.0 time that of the hotter tar. (a) What i the temperature of the hotter tar in term of T? (b) What i the ratio of the peak-intenity wavelength of the hot tar to the peak-intenity wavelength of the cool tar? Siriu B. The brightet tar in the ky i Siriu, the Dog Star. It i actually a binary ytem of two tar, the maller one (Siriu B) being a white dwarf. Spectral analyi of Siriu B indicate that it urface temperature i 4,000 K and that it radiate energy at a total rate of 1.0 * 10 5 W. Aume that it behave like an ideal blackbody. (a) What i the total radiated intenity of Siriu B? (b) What i the peak-intenity wavelength? I thi wavelength viible to human? (c) What i the radiu of Siriu B? Expre your anwer in kilometer and a a fraction of our un radiu. (d) Which tar radiate more total energy per econd, the hot Siriu B or the (relatively) cool un with a urface temperature of 5800 K? To find out, calculate the ratio of the total power radiated by our un to the power radiated by Siriu B Blue Supergiant. A typical blue upergiant tar (the type that explode and leave behind a black hole) ha a urface temperature of 30,000 K and a viual luminoity 100,000 time that of our un. Our un radiate at the rate of 3.86 * 10 6 W. (Viual luminoity i the total power radiated at viible wavelength.) (a) Auming that thi tar behave like an ideal blackbody, what i the principal wavelength it radiate? I thi light viible? Ue your anwer to explain why thee tar are blue. (b) If we aume that the power radiated by the tar i alo 100,000 time that of our un, what i the radiu of thi tar? Compare it ize to that of our un, which ha a radiu of 6.96 * 10 5 km. (c) I it really correct to ay that the viual luminoity i proportional to the total power radiated? Explain. Section 39.6 The Uncertainty Principle Reviited A peky 1.5-mg moquito i annoying you a you attempt to tudy phyic in your room, which i 5.0 m wide and.5 m high. You decide to wat the botherome inect a it flie toward you, but you need to etimate it peed to make a ucceful hit. (a) What i the maximum uncertainty in the horizontal poition of the moquito? (b) What limit doe the Heienberg uncertainty principle place on your ability to know the horizontal velocity of thi moquito? I thi limitation a eriou impediment to your attempt to wat it? By extremely careful meaurement, you determine the x-coordinate of a car center of ma with an uncertainty of only 1.00 mm. The car ha a ma of 100 kg. (a) What i the minimum uncertainty in the x-component of the velocity of the car center of ma a precribed by the Heienberg uncertainty principle? (b) Doe the uncertainty principle impoe a practical limit on our ability to make imultaneou meaurement of the poition and velocitie of ordinary object like car, book, and people? Explain A 10.0-g marble i gently placed on a horizontal tabletop that i 1.75 m wide. (a) What i the maximum uncertainty in the horizontal poition of the marble? (b) According to the Heienberg uncertainty principle, what i the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your anwer to part (b), what i the longet time the marble could remain on the table? Compare thi time to the age of the univere, which i approximately 14 billion year. (Hint: Can you know that the horizontal velocity of the marble i exactly zero?) A cientit ha devied a new method of iolating individual particle. He claim that thi method enable him to detect imultaneouly the poition of a particle along an axi with a tandard deviation of 0.1 nm and it momentum component along thi axi with a tandard deviation of 3.0 * 10-5 kg # m>. Ue the Heienberg uncertainty principle to evaluate the validity of thi claim (a) The x-coordinate of an electron i meaured with an uncertainty of 0.0 mm. What i the x-component of the electron velocity, v x, if the minimum percentage uncertainty in a imultaneou meaurement of v x i 1.0%? (b) Repeat part (a) for a proton An atom in a metatable tate ha a lifetime of 5. m. What i the uncertainty in energy of the metatable tate? (a) The uncertainty in the y-component of a proton poition i.0 * 10-1 m. What i the minimum uncertainty in a imultaneou meaurement of the y-component of the proton velocity? (b) The uncertainty in the z-component of an electron velocity i 0.50 m>. What i the minimum uncertainty in a imultaneou meaurement of the z-coordinate of the electron? PROBLEMS The negative muon ha a charge equal to that of an electron but a ma that i 07 time a great. Conider a hydrogenlike atom coniting of a proton and a muon. (a) What i the reduced ma of the atom? (b) What i the ground-level energy (in electron volt)? (c) What i the wavelength of the radiation emitted in the tranition from the n = level to the n = 1 level? An atom with ma m emit a photon of wavelength l. (a) What i the recoil peed of the atom? (b) What i the kinetic energy K of the recoiling atom? (c) Find the ratio K>E, where E i the energy of the emitted photon. If thi ratio i much le than unity, the recoil of the atom can be neglected in the emiion proce. I the recoil of the atom more important for mall or large atomic mae? For long or hort wavelength? (d) Calculate K (in electron volt) and K>E for a hydrogen atom (ma 1.67 * 10-7 kg) that emit an ultraviolet photon of energy 10. ev. I recoil an important conideration in thi emiion proce? (a) What i the mallet amount of energy in electron volt that mut be given to a hydrogen atom initially in it ground level o that it can emit the H a line in the Balmer erie? (b) How many different poibilitie of pectral-line emiion are there for thi atom when the electron tart in the n = 3 level and eventually end up in the ground level? Calculate the wavelength of the emitted photon in each cae A large number of hydrogen atom are in thermal equilibrium. Let n >n 1 be the ratio of the number of atom in an n = excited tate to the number of atom in an n = 1 ground tate. At what temperature i n equal to (a) 10-1 (b) 10-8 (c) 10-4 >n 1 ; ;? (d) Like the un, other tar have continuou pectra with dark aborption line (ee Fig. 39.9). The aborption take place in the tar atmophere, which in all tar i compoed primarily of hydrogen. Explain why the Balmer aborption line are relatively weak in tar with low atmopheric temperature uch a the un (atmophere temperature 5800 K) but trong in tar with higher atmopheric temperature A ample of hydrogen atom i irradiated with light with wavelength 85.5 nm, and electron are oberved leaving the ga. (a) If each hydrogen atom were initially in it ground level, what would be the maximum kinetic energy in electron volt of thee photoelectron? (b) A few electron are detected with energie a much a 10. ev greater than the maximum kinetic energy calculated in part (a). How can thi be?

4 134 CHAPTER 39 Particle Behaving a Wave CP Bohr Orbit of a Satellite. A 0.0-kg atellite circle the earth once every.00 h in an orbit having a radiu of 8060 km. (a) Auming that Bohr angular-momentum reult 1L = nh>p applie to atellite jut a it doe to an electron in the hydrogen atom, find the quantum number n of the orbit of the atellite. (b) Show from Bohr angular momentum reult and Newton law of gravitation that the radiu of an earth-atellite orbit i directly proportional to the quare of the quantum number, r = kn, where k i the contant of proportionality. (c) Uing the reult from part (b), find the ditance between the orbit of the atellite in thi problem and it next allowed orbit. (Calculate a numerical value.) (d) Comment on the poibility of oberving the eparation of the two adjacent orbit. (e) Do quantized and claical orbit correpond for thi atellite? Which i the correct method for calculating the orbit? The Red Supergiant Betelgeue. The tar Betelgeue ha a urface temperature of 3000 K and i 600 time the diameter of our un. (If our un were that large, we would be inide it!) Aume that it radiate like an ideal blackbody. (a) If Betelgeue were to radiate all of it energy at the peak-intenity wavelength, how many photon per econd would it radiate? (b) Find the ratio of the power radiated by Betelgeue to the power radiated by our un (at 5800 K) CP Light from an ideal pherical blackbody 15.0 cm in diameter i analyzed uing a diffraction grating having 3850 line>cm. When you hine thi light through the grating, you oberve that the peak-intenity wavelength form a firt-order bright fringe at 11.6 from the central bright fringe. (a) What i the temperature of the blackbody? (b) How long will it take thi phere to radiate 1.0 MJ of energy? What mut be the temperature of an ideal blackbody o that photon of it radiated light having the peak-intenity wavelength can excite the electron in the Bohr-model hydrogen atom from the ground tate to the third excited tate? CP An ideal pherical blackbody 4.0 cm in diameter i maintained at 5 C by an internal electrical heater and i immered in a very large open-faced tank of water that i kept boiling by the energy radiated by the phere. You can neglect any heat tranferred by conduction and convection. Conult Table 17.4 a needed. (a) At what rate, in g>, i water evaporating from the tank? (b) If a phyic-wie thermophile organim living in the hot water i oberving thi proce, what will it meaure for the peakintenity (i) wavelength and (ii) frequency of the electromagnetic wave emitted by the phere? When a photon i emitted by an atom, the atom mut recoil to conerve momentum. Thi mean that the photon and the recoiling atom hare the tranition energy. (a) For an atom with ma m, calculate the correction l due to recoil to the wavelength of an emitted photon. Let l be the wavelength of the photon if recoil i not taken into conideration. (Hint: The correction i very mall, a Problem ugget, o ƒ l ƒ >l V 1. Ue thi fact to obtain an approximate but very accurate expreion for l. ) (b) Evaluate the correction for a hydrogen atom in which an electron in the nth level return to the ground level. How doe the anwer depend on n? An Ideal Blackbody. A large cavity with a very mall hole and maintained at a temperature T i a good approximation to an ideal radiator or blackbody. Radiation can pa into or out of the cavity only through the hole. The cavity i a perfect aborber, ince any radiation incident on the hole become trapped inide the cavity. Such a cavity at 00 C ha a hole with area 4.00 mm. How long doe it take for the cavity to radiate 100 J of energy through the hole? CALC (a) Write the Planck ditribution law in term of the frequency ƒ, rather than the wavelength l, to obtain I1ƒ. (b) Show that where I1l i the Planck ditribution formula of Eq. (39.4). (Hint: Change the integration variable from l to ƒ. You will need to ue the following tabulated integral: L0 L0 q q I1l dl = p5 k 4 15c h 3T 4 x 3 e ax - 1 dx = 1 40 a p a b 4 (c) The reult of part (b) i I and ha the form of the Stefan Boltzmann law, I = T 4 (Eq ). Evaluate the contant in part (b) to how that ha the value given in Section CP A beam of 40-eV electron traveling in the xdirection pae through a lit that i parallel to the y-axi and 5.0 mm wide. The diffraction pattern i recorded on a creen.5 m from the lit. (a) What i the de Broglie wavelength of the electron? (b) How much time doe it take the electron to travel from the lit to the creen? (c) Ue the width of the central diffraction pattern to calculate the uncertainty in the y-component of momentum of an electron jut after it ha paed through the lit. (d) Ue the reult of part (c) and the Heienberg uncertainty principle (Eq for y) to etimate the minimum uncertainty in the y-coordinate of an electron jut after it ha paed through the lit. Compare your reult to the width of the lit (a) What i the energy of a photon that ha wavelength 0.10 mm? (b) Through approximately what potential difference mut electron be accelerated o that they will exhibit wave nature in paing through a pinhole 0.10 mm in diameter? What i the peed of thee electron? (c) If proton rather than electron were ued, through what potential difference would proton have to be accelerated o they would exhibit wave nature in paing through thi pinhole? What would be the peed of thee proton? CP Electron go through a ingle lit 150 nm wide and trike a creen 4.0 cm away. You find that at angle of 0.0 from the center of the diffraction pattern, no electron hit the creen but electron hit at all point cloer to the center. (a) How fat were thee electron moving when they went through the lit? (b) What will be the next larger angle at which no electron hit the creen? CP A beam of electron i accelerated from ret and then pae through a pair of identical thin lit that are 1.5 nm apart. You oberve that the firt double-lit interference dark fringe occur at 18.0 from the original direction of the beam when viewed on a ditant creen. (a) Are thee electron relativitic? How do you know? (b) Through what potential difference were the electron accelerated? CP A beam of proton and a beam of alpha particle (of ma 6.64 * 10-7 kg and charge +e) are accelerated from ret through the ame potential difference and pa through identical circular hole in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam form it firt dark ring at 15 with repect to it original direction. When viewed imilarly, at what angle will the alpha particle form it firt dark ring? CP An electron beam and a photon beam pa through identical lit. On a ditant creen, the firt dark fringe occur at the ame angle for both of the beam. The electron peed are much

5 Problem 135 lower than that of light. (a) Expre the energy of a photon in term of the kinetic energy K of one of the electron. (b) Which i greater, the energy of a photon or the kinetic energy of an electron? CP Coherent light i paed through two narrow lit whoe eparation i 40.0 mm. The econd-order bright fringe in the interference pattern i located at an angle of rad. If electron are ued intead of light, what mut the kinetic energy (in electron volt) of the electron be if they are to produce an interference pattern for which the econd-order maximum i alo at rad? BIO What i the de Broglie wavelength of a red blood cell, with ma 1.00 * g, that i moving with a peed of cm>? Do we need to be concerned with the wave nature of the blood cell when we decribe the flow of blood in the body? Calculate the energy in electron volt of (a) an electron that ha de Broglie wavelength 400 nm and (b) a photon that ha wavelength 400 nm High-peed electron are ued to probe the interior tructure of the atomic nucleu. For uch electron the expreion l = h>p till hold, but we mut ue the relativitic expreion for momentum, p = mv> 1 - v >c. (a) Show that the peed of an electron that ha de Broglie wavelength l i c v = 1 + 1mcl>h (b) The quantity h>mc equal.46 * 10-1 m. (A we aw in Section 38.3, thi ame quantity appear in Eq. (38.7), the expreion for Compton cattering of photon by electron.) If l i mall compared to h>mc, the denominator in the expreion found in part (a) i cloe to unity and the peed v i very cloe to c. In thi cae it i convenient to write v = 11 - c and expre the peed of the electron in term of rather than v. Find an expreion for valid when l V h>mc. [Hint: Ue the binomial expanion 11 + z n = 1 + nz + n1n - 1z > + Á, valid for the cae ƒ z ƒ 6 1. ] (c) How fat mut an electron move for it de Broglie wavelength to be 1.00 * m, comparable to the ize of a proton? Expre your anwer in the form v = 11 - c, and tate the value of Suppoe that the uncertainty of poition of an electron i equal to the radiu of the n = 1 Bohr orbit for hydrogen. Calculate the imultaneou minimum uncertainty of the correponding momentum component, and compare thi with the magnitude of the momentum of the electron in the n = 1 Bohr orbit. Dicu your reult CP (a) A particle with ma m ha kinetic energy equal to three time it ret energy. What i the de Broglie wavelength of thi particle? (Hint: You mut ue the relativitic expreion for momentum and kinetic energy: E = 1pc + 1mc and K = E - mc.) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meter) if the particle in part (a) i (i) an electron and (ii) a proton Proton Energy in a Nucleu. The radii of atomic nuclei are of the order of 5.0 * m. (a) Etimate the minimum uncertainty in the momentum of a proton if it i confined within a nucleu. (b) Take thi uncertainty in momentum to be an etimate of the magnitude of the momentum. Ue the relativitic relationhip between energy and momentum, Eq. (37.39), to obtain an etimate of the kinetic energy of a proton confined within a nucleu. (c) For a proton to remain bound within a nucleu, what mut the magnitude of the (negative) potential energy for a proton be within the nucleu? Give your anwer in ev and in MeV. Compare to the potential energy for an electron in a hydrogen atom, which ha a magnitude of a few ten of ev. (Thi how why the interaction that bind the nucleu together i called the trong nuclear force. ) Electron Energy in a Nucleu. The radii of atomic nuclei are of the order of 5.0 * m. (a) Etimate the minimum uncertainty in the momentum of an electron if it i confined within a nucleu. (b) Take thi uncertainty in momentum to be an etimate of the magnitude of the momentum. Ue the relativitic relationhip between energy and momentum, Eq. (37.39), to obtain an etimate of the kinetic energy of an electron confined within a nucleu. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron eparated by 5.0 * m. On the bai of your reult, could there be electron within the nucleu? (Note: It i intereting to compare thi reult to that of Problem ) In a TV picture tube the accelerating voltage i 15.0 kv, and the electron beam pae through an aperture 0.50 mm in diameter to a creen m away. (a) Calculate the uncertainty in the component of the electron velocity perpendicular to the line between aperture and creen. (b) What i the uncertainty in poition of the point where the electron trike the creen? (c) Doe thi uncertainty affect the clarity of the picture ignificantly? (Ue nonrelativitic expreion for the motion of the electron. Thi i fairly accurate and i certainly adequate for obtaining an etimate of uncertainty effect.) The neutral pion 1p 0 i an untable particle produced in high-energy particle colliion. It ma i about 64 time that of the electron, and it exit for an average lifetime of 8.4 * before decaying into two gamma-ray photon. Uing the relationhip E = mc between ret ma and energy, find the uncertainty in the ma of the particle and expre it a a fraction of the ma Quantum Effect in Daily Life? A 1.5-mg inect flie through a 4.00-mm-diameter hole in an ordinary window creen. The thickne of the creen i mm. (a) What hould be the approximate wavelength and peed of the inect for her to how wave behavior a he goe through the hole? (b) At the peed found in part (a), how long would it take the inect to pa through the mm thickne of the hole in the creen? Compare thi time to the age of the univere (about 14 billion year). Would you expect to ee inect diffraction in daily life? Doorway Diffraction. If your wavelength were 1.0 m, you would undergo coniderable diffraction in moving through a doorway. (a) What mut your peed be for you to have thi wavelength? (Aume that your ma i 60.0 kg.) (b) At the peed calculated in part (a), how many year would it take you to move 0.80 m (one tep)? Will you notice diffraction effect a you walk through doorway? Atomic Spectra Uncertaintie. A certain atom ha an energy level.58 ev above the ground level. Once excited to thi level, the atom remain in thi level for 1.64 * 10-7 (on average) before emitting a photon and returning to the ground level. (a) What i the energy of the photon (in electron volt)? What i it wavelength (in nanometer)? (b) What i the mallet poible uncertainty in energy of the photon? Give your anwer in electron volt. (c) Show that ƒ E>E ƒ = ƒ l>l ƒ if ƒ l>l ƒ V 1. Ue thi to calculate the magnitude of the mallet poible uncertainty in the wavelength of the photon. Give your anwer in nanometer You intend to ue an electron microcope to tudy the tructure of ome crytal. For accurate reolution, you want the electron wavelength to be 1.00 nm. (a) Are thee electron relativitic? How do you know? (b) What accelerating potential i needed? (c) What i the kinetic energy of the electron you are uing? To ee if it i great enough to damage the crytal you are

6 136 CHAPTER 39 Particle Behaving a Wave tudying, compare it to the potential energy of a typical NaCl molecule, which i about 6.0 ev. (d) If you decided to ue electromagnetic wave a your probe, what energy hould their photon have to provide the ame reolution a the electron? Would thi energy damage the crytal? For x ray with wavelength nm, the m = 1 intenity maximum for a crytal occur when the angle u in Fig. 39. i At what angle u doe the m = 1 maximum occur when a beam of 4.50-keV electron i ued intead? Aume that the electron alo catter from the atom in the urface plane of thi ame crytal CP Electron diffraction can alo take place when there i interference between electron wave that catter from atom on the urface of a crytal and wave that catter from atom in the next plane below the urface, a ditance d from the urface (ee Fig. 36.3c). (a) Find an equation for the angle u at which there i an intenity maximum for electron wave of wavelength l. (b) The pacing between crytal plane in a certain metal i nm. If 71.0-eV electron are ued, find the angle at which there i an intenity maximum due to interference between cattered wave from adjacent crytal plane. The angle i meaured a hown in Fig. 36.3c. (c) The actual angle of the intenity maximum i lightly different from your reult in part (b). The reaon i the work function f of the metal (ee Section 38.1), which change the electron potential energy by -ef when it move from vacuum into the metal. If the effect of the work function i taken into account, i the angle of the intenity maximum larger or maller than the value found in part (b)? Explain A certain atom ha an energy level 3.50 ev above the ground tate. When excited to thi tate, it remain 4.0 m, on the average, before emitting a photon and returning to the ground tate. (a) What i the energy of the photon? What i it wavelength? (b) What i the mallet poible uncertainty in energy of the photon? BIO Structure of a Viru. To invetigate the tructure of extremely mall object, uch a virue, the wavelength of the probing wave hould be about one-tenth the ize of the object for harp image. But a the wavelength get horter, the energy of a photon of light get greater and could damage or detroy the object being tudied. One alternative i to ue electron matter wave intead of light. Virue vary coniderably in ize, but 50 nm i not unuual. Suppoe you want to tudy uch a viru, uing a wave of wavelength 5.00 nm. (a) If you ue light of thi wavelength, what would be the energy (in ev) of a ingle photon? (b) If you ue an electron of thi wavelength, what would be it kinetic energy (in ev)? I it now clear why matter wave (uch a in the electron microcope) are often preferable to electromagnetic wave for tudying microcopic object? CALC Zero-Point Energy. Conider a particle with ma m moving in a potential U = 1 kx, a in a ma pring ytem. The total energy of the particle i E = p >m + 1 kx. Aume that p and x are approximately related by the Heienberg uncertainty principle, o px L h. (a) Calculate the minimum poible value of the energy E, and the value of x that give thi minimum E. Thi lowet poible energy, which i not zero, i called the zero-point energy. (b) For the x calculated in part (a), what i the ratio of the kinetic to the potential energy of the particle? CALC A particle with ma m move in a potential U1x = A x, where A i a poitive contant. In a implified picture, quark (the contituent of proton, neutron, and other particle, a will be decribed in Chapter 44) have a potential energy of interaction of approximately thi form, where x repreent the eparation between a pair of quark. Becaue U1x S q a x S q, it not poible to eparate quark from each other (a phenomenon called quark confinement). (a) Claically, what i the force acting on thi particle a a function of x? (b) Uing the uncertainty principle a in Problem 39.9, determine approximately the zero-point energy of the particle Imagine another univere in which the value of Planck contant i J #, but in which the phyical law and all other phyical contant are the ame a in our univere. In thi univere, two phyic tudent are playing catch. They are 1 m apart, and one throw a 0.5-kg ball directly toward the other with a peed of 6.0 m>. (a) What i the uncertainty in the ball horizontal momentum, in a direction perpendicular to that in which it i being thrown, if the tudent throwing the ball know that it i located within a cube with volume 15 cm 3 at the time he throw it? (b) By what horizontal ditance could the ball mi the econd tudent? CHALLENGE PROBLEMS (a) Show that in the Bohr model, the frequency of revolution of an electron in it circular orbit around a tationary hydrogen nucleu i ƒ = me 4 >4P0 n 3 h 3. (b) In claical phyic, the frequency of revolution of the electron i equal to the frequency of the radiation that it emit. Show that when n i very large, the frequency of revolution doe indeed equal the radiated frequency calculated from Eq. (39.5) for a tranition from n 1 = n + 1 to n = n. (Thi illutrate Bohr correpondence principle, which i often ued a a check on quantum calculation. When n i mall, quantum phyic give reult that are very different from thoe of claical phyic. When n i large, the difference are not ignificant, and the two method then correpond. In fact, when Bohr firt tackled the hydrogen atom problem, he ought to determine ƒ a a function of n uch that it would correpond to claical reult for large n. ) CP CALC You have entered a contet in which the contetant drop a marble with ma 0.0 g from the roof of a building onto a mall target 5.0 m below. From uncertainty conideration, what i the typical ditance by which you will mi the target, given that you aim with the highet poible preciion? (Hint: The uncertainty x f in the x-coordinate of the marble when it reache the ground come in part from the uncertainty x i in the x-coordinate initially and in part from the initial uncertainty in v x. The latter give rie to an uncertainty v x in the horizontal motion of the marble a it fall. The value of x i and v x are related by the uncertainty principle. A mall x i give rie to a large v x, and vice vera. Find the value of x i that give the mallet total uncertainty in x at the ground. Ignore any effect of air reitance.)

7 Anwer 137 Anwer Chapter Opening Quetion? The mallet detail viible in an image i comparable to the wavelength ued to make the image. Electron can eaily be given a large momentum p and hence a hort wavelength l = h>p, and o can be ued to reolve extremely fine detail. (See Section 39.1.) Tet Your Undertanding Quetion 39.1 Anwer: (a) (i), (b) no From Example 39., the peed of a particle i v = h>lm and the kinetic energy i K = 1 1m>1h>lm = h >l mv = m. Thi how that for a given wavelength, the kinetic energy i inverely proportional to the ma. Hence the proton, with a maller ma, ha more kinetic energy than the neutron. For part (b), the energy of a photon i E = hf, and the frequency of a photon i ƒ = c>l. Hence E = hc>l and l = hc>e = * ev # * 10 8 m>>154 ev =.3 * 10-8 m. Thi i more than 100 time greater than the wavelength of an electron of the ame energy. While both photon and electron have wavelike propertie, they have different relationhip between their energy and momentum and hence between their frequency and wavelength. 39. Anwer: (iii) Becaue the alpha particle i more maive, it won t bounce back in even a head-on colliion with a proton that initially at ret, any more than a bowling ball would when colliding with a Ping-Pong ball at ret (ee Fig. 8.b). Thu there would be no large-angle cattering in thi cae. Rutherford aw large-angle cattering in hi experiment becaue gold nuclei are more maive than alpha particle (ee Fig. 8.a) Anwer: (iv) Figure 39.7 how that many (though not all) of the energy level of He + are the ame a thoe of H. Hence photon emitted during tranition between correponding pair of level in He + and H have the ame energy E and the ame wavelength l = hc>e. An H atom that drop from the n = level to the n = 1 level emit a photon of energy 10.0 ev and wavelength 1 nm (ee Example 39.6); a He + ion emit a photon of the ame energy and wavelength when it drop from the n = 4 level to the n = level. Inpecting Fig will how you that every evennumbered level in He + matche a level in H, while none of the odd-numbered He + level do. The firt three He + tranition given in the quetion (n to n 1, n 3 to n, and n 4 to n 3) all involve an odd-numbered level, o none of their wavelength match a wavelength emitted by H atom Anwer: (i) In a neon light fixture, a large potential difference i applied between the end of a neon-filled gla tube. Thi ionize ome of the neon atom, allowing a current of electron to flow through the ga. Some of the neon atom are truck by fatmoving electron, making them tranition to an excited level. From thi level the atom undergo pontaneou emiion, a depicted in Fig. 39.8b, and emit 63.8-nm photon in the proce. No population inverion occur and the photon are not trapped by mirror a hown in Fig. 39.9d, o there i no timulated emiion. Hence there i no laer action Anwer: (a) ye, (b) ye The Planck radiation law, Eq. (39.4), how that an ideal blackbody emit radiation at all wavelength: The pectral emittance I1l i equal to zero only for l = 0 and in the limit l S q. So a blackbody at 000 K doe indeed emit both x ray and radio wave. However, Fig how that the pectral emittance for thi temperature i very low for wavelength much horter than 1 mm (including x ray) and for wavelength much longer than a few mm (including radio wave). Hence uch a blackbody emit very little in the way of x ray or radio wave Anwer: (i) and (iii) (tie), (ii) and (iv) (tie) According to the Heienberg uncertainty principle, the maller the uncertainty x in the x-coordinate, the greater the uncertainty p x in the x-momentum. The relationhip between x and p x doe not depend on the ma of the particle, and o i the ame for a proton a for an electron. Bridging Problem Anwer: (a) 19 nm; ultraviolet (b) n 4 (c) l = nm, l 3 = nm

8 40 QUANTUM MECHANICS LEARNING GOALS By tudying thi chapter, you will learn: About the wave function that decribe the behavior of a particle and the Schrödinger equation that thi function mut atify. How to calculate the wave function and energy level for a particle confined to a box. How to analyze the quantummechanical behavior of a particle in a potential well.? Thee container hold olution of microcopic emiconductor particle of different ize. The particle glow when expoed to ultraviolet light; the mallet particle glow blue and the larget particle glow red. Why? How quantum mechanic make it poible for particle to go where Newtonian mechanic ay they cannot. How to ue quantum mechanic to analyze a harmonic ocillator. In Chapter 39 we found that particle can behave like wave. In fact, it turn out that we can ue the wave picture to completely decribe the behavior of a particle. Thi approach, called quantum mechanic, i the key to undertanding the behavior of matter on the molecular, atomic, and nuclear cale. In thi chapter we ll ee how to find the wave function of a particle by olving the Schrödinger equation, which i a fundamental to quantum mechanic a Newton law are to mechanic or a Maxwell equation are to electromagnetim. We ll begin with a quantum-mechanical analyi of a free particle that move along a traight line without being acted on by force of any kind. We ll then conider particle that are acted on by force and are trapped in bound tate, jut a electron are bound within an atom. We ll ee that olving the Schrödinger equation automatically give the poible energy level for the ytem. Beide energie, olving the Schrödinger equation give u the probabilitie of finding a particle in variou region. One urpriing reult i that there i a nonzero probability that microcopic particle will pa through thin barrier, even though uch a proce i forbidden by Newtonian mechanic. In thi chapter we ll conider the Schrödinger equation for one-dimenional motion only. In Chapter 41 we ll ee how to extend thi equation to three-dimenional problem uch a the hydrogen atom. The hydrogen-atom wave function will in turn form the foundation for our analyi of more complex atom, of the periodic table of the element, of x-ray energy level, and of other propertie of atom. PhET: Fourier: Making Wave Wave Function and the One-Dimenional Schrödinger Equation We have now een compelling evidence that on an atomic or ubatomic cale, an object uch a an electron cannot be decribed imply a a claical, Newtonian point particle. Intead, we mut take into account it wave characteritic. In the

9 40.1 Wave Function and the One-Dimenional Schrödinger Equation 139 Bohr model of the hydrogen atom (Section 39.3) we tried to have it both way: We pictured the electron a a claical particle in a circular orbit around the nucleu, and ued the de Broglie relation between particle momentum and wavelength to explain why only orbit of certain radii are allowed. A we aw in Section 39.6, however, the Heienberg uncertainty principle tell u that a hybrid decription of thi kind can t be wholly correct. In thi ection we ll explore how to decribe the tate of a particle by uing only the language of wave. Thi new decription, called quantum mechanic, will replace the claical cheme of decribing the tate of a particle by it coordinate and velocity component. Our new quantum-mechanical cheme for decribing a particle ha a lot in common with the language of claical wave motion. In Section 15.3 of Chapter 15, we decribed tranvere wave on a tring by pecifying the poition of each point in the tring at each intant of time by mean of a wave function y1x, t that repreent the diplacement from equilibrium, at time t, of a point on the tring at a ditance x from the origin (Fig. 40.1). Once we know the wave function for a particular wave motion, we know everything there i to know about the motion. For example, we can find the velocity and acceleration of any point on the tring at any time. We worked out pecific form for thee function for inuoidal wave, in which each particle undergoe imple harmonic motion. We followed a imilar pattern for ound wave in Chapter 16. The wave function p1x, t for a wave traveling along the x-direction repreented the preure variation at any point x and any time t. We ued thi language once more in Section 3.3, where we ued two wave function to decribe the electric and magnetic field in an electromagnetic wave. Thu it natural to ue a wave function a the central element of our new language of quantum mechanic. The cutomary ymbol for thi wave function i the Greek letter pi, or c. In general, we ll ue an uppercae to denote a function of all the pace coordinate and time, and a lowercae c for a function of the pace coordinate only not of time. Jut a the wave function y1x, t for mechanical wave on a tring provide a complete decription of the motion, o the wave function 1x, y, z, t for a particle contain all the information that can be known about the particle Thee children are talking over a cup-and-tring telephone. The diplacement of the tring i completely decribed by a wave function y1x, t. In an analogou way, a particle i completely decribed by a quantum-mechanical wave function 1x, y, z, t. CAUTION Particle wave v. mechanical wave Unlike for mechanical wave on a tring or ound wave in air, the wave function for a particle i not a mechanical wave that need ome material medium in order to propagate. The wave function decribe the particle, but we cannot define the function itelf in term of anything material. We can only decribe how it i related to phyically obervable effect. Wave in One Dimenion The wave function of a particle depend in general on all three dimenion of pace. For implicity, however, we ll begin our tudy of thee function by conidering one-dimenional motion, in which a particle of ma m move parallel to the x-axi and the wave function depend on the coordinate x and the time t only. (In the ame way, we tudied one-dimenional kinematic in Chapter before going on to tudy two- and three-dimenional motion in Chapter 3.) What doe a one-dimenional quantum-mechanical wave look like, and what determine it propertie? We can anwer thi quetion by firt recalling the propertie of a wave on a tring. We aw in Section 15.3 that any wave function y1x, t that decribe a wave on a tring mut atify the wave equation: 0 y1x, t 0x = 1 0 y1x, t v 0t (wave equation for wave on a tring) (40.1) In Eq. (40.1) v i the peed of the wave, which i the ame no matter what the wavelength. A an example, conider the following wave function for a

10 1330 CHAPTER 40 Quantum Mechanic wave of wavelength l and frequency ƒ moving in the poitive x-direction along a tring: (inuoidal wave y1x, t = Aco1kx - vt + Bin1kx - vt (40.) on a tring) Here k = p>l i the wave number and v = pƒ i the angular frequency. (We ued thee ame quantitie for mechanical wave in Chapter 15 and electromagnetic wave in Chapter 3.) The quantitie A and B are contant that determine the amplitude and phae of the wave. The expreion in Eq. (40.) i a valid wave function if and only if it atifie the wave equation, Eq. (40.1). To check thi, take the firt and econd derivative of y1x, t with repect to x and take the firt and econd derivative with repect to t: 0y1x, t 0x = -ka in1kx - vt + kb co1kx - vt (40.3a) 0 y1x, t 0x = -k Aco1kx - vt - k Bin1kx - vt 0y1x, t = vain1kx - vt - vbco1kx - vt 0t 0 y1x, t 0t = -v Aco1kx - vt - v Bin1kx - vt (40.3b) (40.3c) (40.3d) If we ubtitute Eq. (40.3b) and (40.3d) into the wave equation, Eq. (40.1), we get -k Aco1kx - vt - k Bin1kx - vt = 1 v 3-v Aco1kx - vt - v Bin1kx - vt4 (40.4) For Eq to be atified at all coordinate x and all time t, the coefficient of the coine function mut be the ame on both ide of the equation, and likewie for the coefficient of the ine function. You can ee that both of thee condition will be atified if k = v v or v = vk (wave on a tring) (40.5) From the definition of angular frequency v and wave number k, Eq. (40.5) i equivalent to pƒ = v p l or v = lƒ (wave on a tring) Thi equation i jut the familiar relationhip among wave peed, wavelength, and frequency for wave on a tring. So our calculation how that Eq. (40.) i a valid wave function for wave on a tring for any value of A and B, provided that v and k are related by Eq. (40.5). What we need i a quantum-mechanical verion of the wave equation, Eq. (40.1), valid for particle wave. We expect thi equation to involve partial derivative of the wave function 1x, t with repect to x and with repect to t. However, thi new equation cannot be the ame a Eq. (40.1) for wave on a tring becaue the relationhip between v and k i different. We can how thi by conidering a free particle, one that experience no force at all a it move along the x-axi. For uch a particle the potential energy U1x ha the ame value for all x (recall from Chapter 7 that F x = -du1x>dx, o zero force mean the potential energy ha zero derivative). For implicity let U = 0 for all x. Then the energy of the free particle i equal to it kinetic energy, which we can expre in term of it momentum p: E = 1 mv = m v m = 1mv m = p m (energy of a free particle) (40.6)

11 40.1 Wave Function and the One-Dimenional Schrödinger Equation 1331 The de Broglie relation that we introduced in Section 39.1 tell u that the energy E i proportional to the angular frequency v and the momentum p i proportional to the wave number: E = hf = h pf =Uv p (40.7a) p = h l = h p p l =Uk (40.7b) Remember that U=h>p. If we ubtitute Eq. (40.7) into Eq. (40.6), we find that the relationhip between v and k for a free particle i Uv = U k m (free particle) (40.8) Equation (40.8) i very different from the correponding relationhip for wave on a tring, Eq. (40.5): The angular frequency v for particle wave i proportional to the quare of the wave number, while for wave on a tring v i directly proportional to k. Our tak i therefore to contruct a quantum-mechanical verion of the wave equation whoe free-particle olution atify Eq. (40.8). We ll attack thi problem by auming a inuoidal wave function 1x, t of the ame form a Eq. (40.) for a inuoidal wave on a tring. For a wave on a tring, Eq. (40.) repreent a wave of wavelength l = p>k and frequency ƒ = v>p propagating in the poitive x-direction. By analogy, our inuoidal wave function 1x, t repreent a free particle of ma m, momentum p =Uk, and energy E =Uvmoving in the poitive x-direction: 1x, t = A co1kx - vt + B in1kx - vt (inuoidal wave function repreenting a free particle) (40.9) The wave number k and angular frequency v in Eq. (40.9) mut atify Eq. (40.8). If you look at Eq. (40.3b), you ll ee that taking the econd derivative of 1x, t in Eq. (40.9) with repect to x give u 1x, t multiplied by -k. Hence if we multiply 0 1x, t>0x by -U >m, we get - U 0 1x, t m 0x = - U m 3-k Aco1kx - vt - k Bin1kx - vt4 = U k 3A co1kx - vt + Bin1kx - vt4 m = U k 1x, t m (40.10) Equation (40.10) ugget that 1-U >m0 1x, t>0x hould be one ide of our quantum-mechanical wave equation, with the other ide equal to Uv 1x, t in order to atify Eq. (40.8). If you look at Eq. (40.3c), you ll ee that taking the firt time derivative of 1x, t in Eq. (40.9) bring out a factor of v. So we ll make the educated gue that the right-hand ide of our quantum-mechanical wave equation involve U=h>p time 0 1x, t>0t. So our tentative equation i - U 0 1x, t m 0x 0 1x, t = CU 0t (40.11) At thi point we include a contant C a a fudge factor to make ure that everything turn out right. Now let ubtitute the wave function from Eq. (40.9) into Eq. (40.11). From Eq. (40.10) and Eq. (40.3c), we get

12 133 CHAPTER 40 Quantum Mechanic U k 3Aco1kx - vt + Bin1kx - vt4 m = CUv3Ain1kx - vt - Bco1kx - vt4 (40.1) From Eq. (40.8), Uv =U k >m, o we can cancel thee factor on the two ide of Eq. (40.1). What remain i Aco1kx - vt + Bin1kx - vt = CAin1kx - vt - CBco1kx - vt (40.13) A in our dicuion above of the wave equation for wave on a tring, in order for Eq. (40.13) to be atified for all value of x and all value of t, the coefficient of the coine function mut be the ame on both ide of the equation, and likewie for the coefficient of the ine function. Hence we have the following relationhip among the coefficient A and B in Eq. (40.9) and the coefficient C in Eq. (40.11): A = -CB B = CA (40.14a) (40.14b) If we ue Eq. (40.14b) to eliminate B from Eq. (40.14a), we get A = -C A, which mean that C = -1. Thu C i equal to the imaginary number i = 1-1, and Eq. (40.11) become - U 0 1x, t m 0x 0 1x, t = iu 0t (one-dimenional Schrödinger equation for a free particle) (40.15) 40. Erwin Schrödinger ( ) developed the equation that bear hi name in 196, an accomplihment for which he hared (with the Britih phyicit P. A. M. Dirac) the 1933 Nobel Prize in phyic. Hi grave marker i adorned with a large letter c. Equation (40.15) i the one-dimenional Schrödinger equation for a free particle, developed in 196 by the Autrian phyicit Erwin Schrödinger (Fig. 40.). The preence of the imaginary number i in Eq. (40.15) mean that the olution to the Schrödinger equation are complex quantitie, with a real part and an imaginary part. (The imaginary part of 1x, t i a real function multiplied by the imaginary number i = 1-1. ) An example i our free-particle wave function from Eq. (40.9). Since we found C = i in Eq. (40.14), it follow from Eq. (40.14b) that B = ia. Then Eq. (40.9) become 1x, t = A3co1kx - vt + iin1kx - vt4 (inuoidal wave function repreenting a free particle) (40.16) The real part of 1x, t i Re 1x, t = A co1kx - vt and the imaginary part i Im 1x, t = Ain1kx - vt. Figure 40.3 graph the real and imaginary part of 1x, t at t = 0, o 1x, 0 = Aco kx + ia in kx. We can rewrite Eq. (40.16) uing Euler formula, which tate that for any angle u, e iu = co u + iin u e -iu = co1-u + iin1-u = co u - iin u (40.17) Thu our inuoidal free-particle wave function become 1x, t = Ae i1kx-vt = Ae ikx e -ivt (inuoidal wave function repreenting a free particle) (40.18)

13 40.1 Wave Function and the One-Dimenional Schrödinger Equation 1333 If k i poitive in Eq. (40.16), the wave function repreent a free particle moving in the poitive x-direction with momentum p =Ukand energy E =Uv =U k >m. If k i negative, the momentum and hence the motion are in the negative x-direction. (With a negative value of k, the wavelength i l = p> ƒ k ƒ). Interpreting the Wave Function The complex nature of the wave function for a free particle make thi function challenging to interpret. (We certainly haven t needed imaginary number before thi point to decribe real phyical phenomena.) Here how to think about thi function: 1x, t decribe the ditribution of a particle in pace, jut a the wave function for an electromagnetic wave decribe the ditribution of the electric and magnetic field. When we worked out interference and diffraction pattern in Chapter 35 and 36, we found that the intenity I of the radiation at any point in a pattern i proportional to the quare of the electric-field magnitude that i, to E. In the photon interpretation of interference and diffraction (ee Section 38.4), the intenity at each point i proportional to the number of photon triking around that point or, alternatively, to the probability that any individual photon will trike around the point. Thu the quare of the electric-field magnitude at each point i proportional to the probability of finding a photon around that point. In exactly the ame way, the quare of the wave function of a particle at each point tell u about the probability of finding the particle around that point. More preciely, we hould ay the quare of the abolute value of the wave function, ƒ ƒ. Thi i neceary becaue, a we have een, the wave function i a complex quantity with real and imaginary part. For a particle that can move only along the x-direction, the quantity ƒ 1x, tƒ dx i the probability that the particle will be found at time t at a coordinate in the range from x to x + dx. The particle i mot likely to be found in region where ƒ ƒ i large, and o on. Thi interpretation, firt made by the German phyicit Max Born (Fig. 40.4), require that the wave function be normalized. That i, the integral of ƒ 1x, tƒ dx over all poible value of x mut equal exactly 1. In other word, the probability i exactly 1, or 100%, that the particle i omewhere. CAUTION Interpreting ƒ ƒ Note that ƒ 1x, tƒ itelf i not a probability. Rather, ƒ 1x, t ƒ dx i the probability of finding the particle between poition x and poition x + dx at time t. If the length dx i made maller, it become le likely that the particle will be found within that length, o the probability decreae. A better name for ƒ 1x, tƒ i the probability ditribution function, ince it decribe how the probability of finding the particle at different location i ditributed over pace. Another common name for ƒ 1x, tƒ i the probability denity The patial wave function 1x, t = Ae i1kx-vt for a free particle of definite momentum p =Uki a complex function: It ha both a real part and an imaginary part. Thee are graphed here a function of x for t = 0. Re C(x, 0) = A co kx A O A Im C(x, 0) = A in kx A O A p/k p/k 3p/k p/k p/k 3p/k 40.4 In 196, the German phyicit Max Born ( ) devied the interpretation that ƒ ƒ i the probability ditribution function for a particle that i decribed by the wave function. He alo coined the term quantum mechanic (in the original German, Quantenmechanik). For hi contribution, Born hared (with Walther Bothe) the 1954 Nobel Prize in phyic. x x We can ue the probability interpretation of ƒ ƒ to get a better undertanding of Eq. (40.18), the wave function for a free particle. Thi function decribe a particle that ha a definite momentum p =Ukin the x-direction and no uncertainty in momentum: p x = 0. The Heienberg uncertainty principle for poition and momentum, Eq. (39.9), ay that x p x ÚU>. If p x i zero, then x mut be infinite, and we have no idea whatoever where along the x-axi the particle can be found. (We aw a imilar reult for photon in Section 38.4.) We can how thi by calculating the probability ditribution function ƒ 1x, tƒ. Thi i the product of and it complex conjugate *. To find the complex conjugate of a complex number, we imply replace all i with -i. For example, the complex conjugate of c = a + ib, where a and b are real, i c * = a - ib, o ƒ c ƒ = c * c = 1a + ib1a - ib = a + b (recall that i = -1. The complex conjugate of Eq. (40.18) i * 1x, t = A * e -i1kx-vt = A * e -ikx e ivt (We have to allow for the poibility that the coefficient A i itelf a complex number.) Hence the probability ditribution function i

14 1334 CHAPTER 40 Quantum Mechanic ƒ 1x, tƒ = * 1x, t 1x, t = 1A * e -ikx e ivt 1Ae ikx e -ivt = A * Ae 0 = ƒ Aƒ The probability ditribution function doen t depend on poition, which ay that we are equally likely to find the particle anywhere along the x-axi! Mathematically, thi i becaue the wave function 1x, t = Ae i1kx-vt = A3co1kx - vt + iin1kx - vt4 i a inuoidal function that extend all the way from x = -q to x = +q with the ame amplitude A. Thi alo mean that the wave function can t be normalized: The integral of ƒ 1x, tƒ over all pace would be infinite for any value of A. Note alo that the wave function in Eq. (40.18) decribe a particle with a definite energy E =Uv, o there i zero uncertainty in energy: E = 0. The Heienberg uncertainty principle for energy and time interval, t E ÚU [Eq. (39.30)], tell u that the time uncertainty t for thi particle i infinite. In other word, we can have no idea when the particle will pa a given point on the x-axi. That alo agree with our reult ƒ 1x, tƒ = ƒaƒ ; the probability ditribution function ha the ame value at all time. Since we alway have ome idea of where a particle i, the wave function given in Eq. (40.18) in t a realitic decription. In our tudy of light in Section 38.4, we aw that we can make a wave function that more localized in pace by uperimpoing two or more inuoidal function. (Thi would be a good time to review that ection.) A an illutration, let calculate ƒ 1x, tƒ for a wave function of thi kind. Example 40.1 A localized free-particle wave function The wave function 1x, t = Ae i1k 1x-v 1 t + Ae i1k x-v t i a uperpoition of two free-particle wave function of the form given by Eq. (40.18). Both k 1 and k are poitive. (a) Show that thi wave function atifie the Schrödinger equation for a free particle of ma m. (b) Find the probability ditribution function for 1x, t. SOLUTION IDENTIFY and SET UP: The wave function Ae i1k 1x-v 1 t and Ae i1k x-v t both repreent particle moving in the poitive x- direction, but with different momenta and kinetic energie: p 1 =Uk 1 and E 1 =Uv 1 =U k 1 >m for the firt function, p =Uk and E =Uv =U k >m for the econd function. To tet whether a uperpoition of thee i alo a valid wave function for a free particle, we ll ee whether our function 1x, t atifie the free-particle Schrödinger equation, Eq. (40.15). It ueful to remember the derivative of the exponential function: 1d>due au = ae au and 1d >du e au = a e au. The probability ditribution function ƒ 1x, tƒ i the product of 1x, t and it complex conjugate. EXECUTE: (a) If we ubtitute 1x, t into Eq. (40.15), the lefthand ide of the equation i - U 0 1x, t m 0x = - U 0 1Ae i1k 1x-v 1 t + Ae i1k x-v t m 0x = - U m 31ik 1 Ae i1k 1x-v 1 t + 1ik Ae i1k x-v t 4 = U k 1 m Aei1k 1x-v 1 t + U k m Aei1k x-v t The right-hand ide i The two ide are equal, provided that Uv 1 =U k 1 >m and Uv =U k >m. Thee are jut the relationhip that we noted above. So we conclude that 1x, t = Ae i1k 1x-v 1 t + Ae i1k x-v t i a valid free-particle wave function. In general, if we take any two wave function that are olution of the Schrödinger equation and then make a uperpoition of thee to create a third wave function 1x, t, then 1x, t i alo a olution of the Schrödinger equation. (b) The complex conjugate of 1x, t i Hence 0 1x, t iu = iu 01Aei1k 1x-v 1 t + Ae i1k x-v t 0t 0t ƒ 1x, tƒ = * 1x, t 1x, t = iu31-iv 1 Ae i1k 1x-v 1 t + 1-iv Ae i1k x-v t 4 =Uv 1 Ae i1k 1x-v 1 t +Uv Ae i1k x-v t * 1x, t = A * e -i1k 1x-v 1 t + A * e -i1k x-v t = 1A * e -i1k 1x-v 1 t + A * e -i1k x-v t 1Ae i1k 1x-v 1 t + Ae i1k x-v t = A * Ac e-i1k 1x-v 1 t e i1k 1x-v 1 t + e -i1k x-v t e i1k x-v t + e -i1k 1x-v 1 t e i1k x-v t + e -i1k x-v t e i1k 1x-v 1 t d = ƒaƒ 3e 0 + e 0 + e i31k -k 1 x-1v -v 1 t4 + e -i31k -k 1 x-1v -v 1 t4 4 To implify thi expreion, recall that e 0 = 1. From Euler formula, e iu = co u + i in u and e -iu = co u - iin u, o e iu + e -iu = co u. Hence ƒ 1x, tƒ = ƒaƒ 5 + co31k - k 1 x - 1v - v 1 t46 = ƒaƒ 51 + co31k - k 1 x - 1v - v 1 t46

15 40.1 Wave Function and the One-Dimenional Schrödinger Equation 1335 ƒ ƒ 1x, ƒ ƒ ƒ EVALUATE: Figure 40.5 i a graph of the probability ditribution tƒ v av = 1v - v 1 >1k - k 1 function 1x, tƒ at t = 0. The value of varie move at a velocity. The ubcript av remind u that v av repreent the average value of the particle between 0 and 4ƒAƒ ; probabilitie can never be negative! The particle ha become omewhat localized: The particle i mot likely to be found near a point where 1x, tƒ i maximum (where the function and Ae i1k x-v t Ae i1k 1x-v 1 t interfere contructively) and i very unlikely to be found near a point where 1x, tƒ = 0 (where and Ae i1k x-v t Ae i1k 1x-v 1 t interfere detructively). Thi i very imilar to the phenomenon of beat for ound wave (ee Section 16.7). Note alo that the probability ditribution function i not tationary, but move in the poitive x-direction like the particle that it repreent. To ee thi, recall from Section 15.3 that a inuoidal wave given by y1x, t = A co1kx - vt move in the poitive x-direction with velocity v = v>k; ince 1x, tƒ include a term velocity. The price we pay for localizing the particle omewhat i that, unlike a particle repreented by Eq. (40.18), it no longer ha either a definite momentum or a definite energy. That conitent with the Heienberg uncertainty principle: If we decreae the uncertaintie about where a particle i and when it pae a certain point, the uncertaintie in it momentum and energy mut increae. The average momentum of the particle i p av = 1Uk + Uk 1 >, the average of the momenta aociated with the free-particle wave function we added to create 1x, t. Thi correpond to the particle having an average velocity v av = p av >m = 1Uk +Uk 1 >m. Can you how that thi i equal to the expreion v av = 1v - v 1 >1k - k 1 that we found above? co31k - k 1 x - 1v - v 1 t4, the probability ditribution 40.5 The probability ditribution function at t = 0 for 1x, t = Ae i1k 1x-v 1 t + Ae i1k x-v t. C(x, 0) 3 4 A A O p 6p 4p p k k 1 k k 1 k k 1 k k 1 4p 6p x k k 1 k k 1 Wave Packet The wave function that we examined in Example 40.1 i not very well localized: The probability ditribution function till extend from x = -q to x = +q. Hence thi wave function can t be normalized, either. To make a wave function that more highly localized, imagine uperpoing two additional inuoidal wave with different wave number and amplitude o a to reinforce alternate maxima of ƒ 1x, tƒ in Fig and cancel out the in-between one. Finally, if we uperpoe wave with a very large number of different wave number, we can contruct a wave with only one maximum of ƒ 1x, tƒ (Fig. 40.6). Then, finally, we have omething that begin to look like both a particle and a wave. It i a particle in the ene that it i localized in pace; if we look from a ditance, it may look like a point. But it alo ha a periodic tructure that i characteritic of a wave. A localized wave pule like that hown in Fig i called a wave packet. We can repreent a wave packet by an expreion uch a q 1x, t = A1ke i1kx-vt dk L - q (40.19) Thi integral repreent a uperpoition of a very large number of wave, each with a different wave number k and angular frequency v =Uk >m, and each with an amplitude A1k that depend on k. There i an important relationhip between the two function 1x, t and A1k, which we how qualitatively in Fig If the function A1k i harply peaked, a in Fig. 40.7a, we are uperpoing only a narrow range of wave number. The reulting wave pule i then relatively broad (Fig. 40.7b). But if we ue

16 1336 CHAPTER 40 Quantum Mechanic 40.6 Superpoing a large number of inuoidal wave with different wave number and appropriate amplitude can produce a wave pule that ha a wavelength l av = p>k av and i localized within a region of pace of length x. Thi localized pule ha apect of both particle and wave. (a) Real part of the wave function at time t ReC(x, t) (b) Imaginary part of the wave function at time t Im C(x, t) (c) Probability ditribution function at time t O O C(x, t) Dx l av l av x x x a wider range of wave number, o that the function A1k i broader (Fig. 40.7c), then the wave pule i more narrowly localized (Fig. 40.7d). Thi i imply the uncertainty principle in action. A narrow range of k mean a narrow range of p x =Ukand thu a mall p x ; the reult i a relatively large x. A broad range of k correpond to a large p x, and the reulting x i maller. You can ee that the uncertainty principle for poition and momentum, x p x ÚU>, i really jut a conequence of the propertie of integral like Eq. (40.19). CAUTION Matter wave veru light wave in vacuum We can regard both a wave packet that repreent a particle and a hort pule of light from a laer a uperpoition of wave of different wave number and angular frequencie. An important difference i that the peed of light in vacuum i the ame for all wavelength l and hence all wave number k = p>l, but the peed of a matter wave i different for different wavelength. You can ee thi from the formula for the peed of the wave cret in a periodic wave, v = lƒ = v>k. For a matter wave, v =Uk >m, o v =Uk>m = h>ml. Hence matter wave with longer wavelength and maller wave number travel more lowly than thoe with hort wavelength and large wave number. (Thi houldn t be too urpriing. The de Broglie relation that we learned in Section 39.1 tell u that horter wavelength correpond to greater momentum and hence a greater peed.) Since the individual inuoidal wave that make up a wave packet travel at different peed, the hape of the packet change a it move. That why we ve pecified the time for which the wave packet in Fig and 40.7 are drawn; at later time, the packet become more pread out. By contrat, a pule of light wave in vacuum retain the ame hape at all time becaue all of it contituent inuoidal wave travel together at the ame peed. The One-Dimenional Schrödinger Equation with Potential Energy The one-dimenional Schrödinger equation that we preented in Eq. (40.15) i valid only for free particle, for which the potential energy function i zero: U1x = 0. But for an electron within an atom, a proton within an atomic nucleu, and many other real ituation, the potential energy play an important role. To tudy the behavior of matter wave in thee ituation, we need a verion of the Schrödinger equation that decribe a particle moving in the preence of a nonzero potential energy function U1x. Thi equation i - U 0 1x, t m 0x 0 1x, t + U1x 1x, t = iu 0t (general one-dimenional Schrödinger equation) (40.0) Note that if U1x = 0, Eq. (40.0) reduce to the free-particle Schrödinger equation given in Eq. (40.15). Here the motivation behind Eq. (40.0). If 1x, t i a inuoidal wave function for a free particle, 1x, t = Ae i1kx-vt = Ae ikx e -ivt, the derivative term in Eq. (40.0) become - U 0 1x, t m 0x 0 1x, t iu 0t = - U 0 m 0x 1Aeikx e -ivt = - U m 1ik 1Ae ikx e -ivt = U k 1x, t m = iu 0 0t 1Aeikx e -ivt = iu1-iv1ae ikx e -ivt =Uv 1x, t In thee expreion 1U k >m 1x, t i jut the kinetic energy K = p >m = U k >m multiplied by the wave function, and Uv 1x, t i the total energy E =Uvmultiplied by the wave function. So for a wave function of thi kind, Eq. (40.0) ay that kinetic energy time 1x, t plu potential energy time 1x, t equal total energy time 1x, t. That equivalent to the tatement in

17 40.1 Wave Function and the One-Dimenional Schrödinger Equation 1337 claical phyic that the um of kinetic energy and potential energy equal total mechanical energy: K + U = E. The obervation we ve jut made certainly aren t a proof that Eq. (40.0) i correct. The real reaon we know thi equation i correct i that it work: Prediction made with thi equation agree with experimental reult. In the remaining ection of thi chapter we ll apply Eq. (40.0) to everal phyical ituation, each with a different form of the function U1x. Stationary State We aw in our dicuion of wave packet that any free-particle wave function can be built up a a uperpoition of inuoidal wave function of the form 1x, t = Ae ikx e -ivt. Each uch inuoidal wave function correpond to a tate of definite energy E =Uv =U k >m and definite angular frequency v = E>U, o we can rewrite thee function a 1x, t = Ae ikx e -iet>u. If the potential energy function U1x i nonzero, thee inuoidal wave function do not atify the Schrödinger equation, Eq. (40.0), and o thee function cannot be the baic building block of more complicated wave function. However, we can till write the wave function for a tate of definite energy E in the form 1x, t = c1xe -iet>u (time-dependent wave function for a tate of definite energy) (40.1) That i, the wave function 1x, t for a tate of definite energy i the product of a time-independent wave function c1x and a factor e -iet>u. (For the free-particle inuoidal wave function, c1x = Ae ikx.) State of definite energy are of tremendou importance in quantum mechanic. For example, for each energy level in a hydrogen atom (Section 39.3) there i a pecific wave function. It i poible for an atom to be in a tate that doe not have a definite energy. The wave function for any uch tate can be written a a combination of definiteenergy wave function, in preciely the ame way that a free-particle wave packet can be written a a uperpoition of inuoidal wave function of definite energy a in Eq. (40.19). A tate of definite energy i commonly called a tationary tate. To ee where thi name come from, let multiply Eq. (40.1) by it complex conjugate to find the probabilty ditribution function ƒ ƒ : 40.7 How varying the function A1k in the wave-packet expreion, Eq. (40.19), change the character of the wave function 1x, t (hown here at a pecific time t = 0). (a) A(k) O k 0 (b) Re C(x, 0) (c) A(k) O A narrow function A(k) give a wave function C(x, 0) with a broad patial extent. k 0 O (d) Re C(x, 0) A broad function A(k) give a wave function C(x, 0) with a narrow patial extent. O k k x x ƒ 1x, tƒ = * 1x, t 1x, t = 3c * 1xe +iet>u 43c1xe -iet>u 4 = c * 1xc1xe 1+iEt>U+1-iEt>U = ƒc1xƒ e 0 = ƒc1xƒ (40.) Since ƒ c1xƒ doe not depend on time, Eq. (40.) how that the ame mut be true for the probability ditribution function ƒ 1x, tƒ. Thi jutifie the term tationary tate for a tate of definite energy. PhET: Quantum Tunneling and Wave Packet ActivPhyic 17.7: Wave Packet CAUTION A tationary tate doe not mean a tationary particle The name tationary tate may lead you to think that the particle i not in motion if it i decribed by uch a wave function. That not the cae. It the probability ditribution (that i, the relative likelihood of finding the particle at variou poition), not the particle itelf, that tationary. The Schrödinger equation, Eq. (40.0), become quite a bit impler for tationary tate. To ee thi, we ubtitute Eq. (40.1) into Eq. (40.0): - U 0 3c1xe -iet>u 4 m 0x + U1xc1xe -iet>u = iu 03c1xe -iet>u 4 0t

18 1338 CHAPTER 40 Quantum Mechanic The derivative on the firt term on the left-hand ide i with repect to x, o the factor of e -iet>u come outide of the derivative. Now we take the derivative with repect to t on the right-hand ide of the equation: - U m d c1x dx e -iet>u + U1xc1xe -iet>u = iua -ie U b3c1xe-iet>u 4 = Ec1xe If we divide both ide of thi equation by e -iet>u, we get -iet>u - U d c1x m dx + U1xc1x = Ec1x (time-independent Schrödinger equation) (40.3) Thi i called the time-independent Schrödinger equation. The time-dependent factor e -iet>u doe not appear, and Eq. (40.3) i an equation that involve only the time-independent wave function c1x. We ll devote much of thi chapter to olving thi equation to find the definite-energy, tationary-tate wave function c1x and the correponding value of E that i, the energie of the allowed level for different phyical ituation. Example 40. A tationary tate Conider the wave function c1x = A 1 e ikx + A e -ikx, where k i The reult i a contant time c1x, o thi c1x i indeed a valid poitive. I thi a valid time-independent wave function for a free tationary-tate wave function for a free particle. Comparing with particle in a tationary tate? What i the energy correponding to Eq. (40.3) how that the contant on the right-hand ide i the thi wave function? particle energy: E =U k >m. SOLUTION IDENTIFY and SET UP: A valid tationary-tate wave function for a free particle mut atify the time-independent Schrödinger equation, Eq. (40.3), with U1x = 0. To tet the given function c1x, we imply ubtitute it into the left-hand ide of the equation. If the reult i a contant time c1x, then the wave function i indeed a olution and the contant i equal to the particle energy E. EXECUTE: Subtituting c1x = A 1 e ikx + A e -ikx and U1x = 0 into Eq. (40.3), we obtain - U d c1x m dx = - U d 1A 1 e ikx + A e -ikx m dx = - U m 31ik A 1 e ikx + 1-ik A e ikx 4 = U k m 1A 1e ikx + A e -ikx = U k m c1x EVALUATE: Note that c1x i a uperpoition of two different wave function: one function 1A 1 e ikx that repreent a particle with magnitude of momentum p =Ukmoving in the poitive x-direction, and one function 1A e -ikx that repreent a particle with the ame magnitude of momentum moving in the negative x-direction. So while the combined wave function c1x repreent a tationary tate with a definite energy, thi tate doe not have a definite momentum. We ll ee in Section 40. that uch a wave function can repreent a tanding wave, and we ll explore ituation in which uch tanding matter wave can arie. Tet Your Undertanding of Section 40.1 Eq. (40.19) repreent a tationary tate? 40. Particle in a Box Doe a wave packet given by ActivPhyic 0.: Particle in a Box An important problem in quantum mechanic i how to ue the time-independent Schrödinger equation, Eq. (40.3), to determine the poible energy level and the correponding wave function for variou ytem. The fundamental problem i then the following: For a given potential energy function U1x, what are the poible tationary-tate wave function c1x, and what are the correponding energie E?

19 40. Particle in a Box 1339 In Section 40.1 we olved thi problem for the cae U1x = 0, correponding to a free particle. The allowed wave function and correponding energie are c1x = Ae ikx E = U k (free particle (40.4) m The wave number k i equal to p>l, where l i the wavelength. We found that k can have any real value, o the energy E of a free particle can have any value from zero to infinity. Furthermore, the particle can be found with equal probability at any value of x from - q to + q. Now let look at a imple model in which a particle i bound o that it cannot ecape to infinity, but rather i confined to a retricted region of pace. Our ytem conit of a particle confined between two rigid wall eparated by a ditance L (Fig. 40.8). The motion i purely one dimenional, with the particle moving along the x-axi only and the wall at x = 0 and x = L. The potential energy correponding to the rigid wall i infinite, o the particle cannot ecape; between the wall, the potential energy i zero (Fig. 40.9). Thi ituation i often decribed a a particle in a box. Thi model might repreent an electron that i free to move within a long, traight molecule or along a very thin wire. Wave Function for a Particle in a Box To olve the Schrödinger equation for thi ytem, we begin with ome retriction on the particle tationary-tate wave function c1x. Becaue the particle i confined to the region 0 x L, we expect the probability ditribution function ƒ 1x, tƒ = ƒc1xƒ and the wave function c1x to be zero outide that region. Thi agree with the Schrödinger equation: If the term U1xc1x in Eq. (40.3) i to be finite, then c1x mut be zero where U1x i infinite. Furthermore, c1x mut be a continuou function to be a mathematically wellbehaved olution to the Schrödinger equation. Thi implie that c1x mut be zero at the region boundary, x = 0 and x = L. Thee two condition erve a boundary condition for the problem. They hould look familiar, becaue they are the ame condition that we ued to find the normal mode of a vibrating tring in Section 15.8 (Fig ); you hould review that dicuion. An additional condition i that to calculate the econd derivative d c1x>dx in Eq. (40.3), the firt derivative dc1x>dx mut alo be continuou except at point where the potential energy become infinite (a it doe at the wall of the box). Thi i analogou to the requirement that a vibrating tring, like thoe hown in Fig , can t have any kink in it (which would correpond to a dicontinuity in the firt derivative of the wave function) except at the end of the tring. We now olve for the wave function in the region 0 x L ubject to the above condition. In thi region U1x = 0, o the wave function in thi region mut atify - U d c1x (40.5) m dx = Ec1x (particle in a box) Equation (40.5) i the ame Schrödinger equation a for a free particle, o it i tempting to conclude that the wave function and energie are given by Eq. (40.4). It i true that c1x = Ae ikx atifie the Schrödinger equation with U1x = 0, i continuou, and ha a continuou firt derivative dc1x>dx = ikae ikx. However, thi wave function doe not atify the boundary condition that c1x mut be zero at x = 0 and x = L: At x = 0 the wave function in Eq. (40.4) i equal to Ae 0 = A, and at x = L it i equal to Ae ikl. (Thee would be equal to zero if A = 0, but then the wave function would be zero and there would be no particle at all!) The way out of thi dilemma i to recall Example 40. (Section 40.1), in which we found that a more general tationary-tate olution to the time-independent Schrödinger equation with U1x = 0 i c1x = A 1 e ikx + A e -ikx (40.6) 40.8 The Newtonian view of a particle in a box. A particle with ma m move along a traight line at contant peed, bouncing between two rigid wall a ditance L apart. m v L 0 L 40.9 The potential-energy function for a particle in a box. The potential energy U i zero in the interval 0, x, L and i infinite everywhere outide thi interval. ` ` U(x) U L U(x) Normal mode of vibration for a tring with length L, held at both end. Each end i a node, and there are n 1 additional node between the end. L The length i an integral number of half-wavelength: L 5 nl n/. x n 5 3 n 5 n 5 1 x

20 1340 CHAPTER 40 Quantum Mechanic Thi wave function i a uperpoition of two wave: one traveling in the +x-direction of amplitude A 1, and one traveling in the -x-direction with the ame wave number but amplitude A. Thi i analogou to a tanding wave on a tring (Fig ), which we can regard a the uperpoition of two inuoidal wave propagating in oppoite direction (ee Section 15.7). The energy that correpond to Eq. (40.6) i E =U k >m, jut a for a ingle wave. To ee whether the wave function given by Eq. (40.6) can atify the boundary condition, let firt rewrite it in term of ine and coine uing Euler formula, Eq. (40.17): c1x = A 1 1co kx + iin kx + A 3co1-kx + iin1-kx4 = A 1 1co kx + iin kx + A 1co kx - iin kx = 1A 1 + A co kx + i1a 1 - A in kx (40.7) At x = 0 thi i equal to c10 = A 1 + A, which mut equal zero if we are to atify the boundary condition at that point. Hence A = -A 1, and Eq. (40.7) become c1x = ia 1 in kx = Cin kx (40.8) We have implified the expreion by introducing the contant C = ia 1. (We ll come back to thi contant later.) We can alo atify the econd boundary condition that c = 0 at x = L by chooing value of k uch that kl = np 1n = 1,, 3, Á. Hence Eq. (40.8) doe indeed give the tationary-tate wave function for a particle in a box in the region 0 x L. (Outide thi region, c1x = 0. The poible value of k and the wavelength l = p>k are k = np L and l = p k = L n 1n = 1,, 3, Á (40.9) Jut a for the tring in Fig , the length L of the region i an integral number of half-wavelength. Energy Level for a Particle in a Box The poible energy level for a particle in a box are given by E =U k >m = p >m, where p =Uk = 1h>p1p>l = h>l i the magnitude of momentum of a free particle with wave number k and wavelength l. Thi make ene, ince inide the region 0 x L the potential energy i zero and the energy i all kinetic. For each value of n, there are correponding value of p, l, and E; let call them p n, l n, and E n. Putting the piece together, we get p n = h l n = nh L and o the energy level for a particle in a box are E n = p n m = n h 8mL = n p U ml 1n = 1,, 3, Á (energy level, particle in a box) (40.30) (40.31) Each energy level ha it own value of the quantum number n and a correponding wave function, which we denote by c n. When we replace k in Eq. (40.8) by np>l from Eq. (40.9), we find c n 1x = Cin npx 1n = 1,, 3, Á (40.3) L The energy-level diagram in Fig a how the five lowet level for a particle in a box. The energy level are proportional to n, o ucceively higher level are paced farther and farther apart. There are an infinite number of level becaue the wall are perfectly rigid; even a particle of infinitely great kinetic

21 40. Particle in a Box 1341 (a) E n 5 5 n 5 4 5E 1 16E 1 (b) c(x) n 5 5 n 5 4 n (a) Energy-level diagram for a particle in a box. Each energy i n E 1, where E 1 i the ground-level energy. (b) Wave function for a particle in a box, with n = 1,, 3, 4, and 5. CAUTION: The five graph have been diplaced vertically for clarity, a in Fig Each of the horizontal dahed line repreent c = 0 for the repective wave function. n 5 3 9E 1 n 5 n 5 4E 1 n 5 1 E 5 0 E 1 n L x energy i confined within the box. Figure 40.11b how graph of the wave function c n 1x for n = 1,, 3, 4, and 5. Note that thee function look identical to thoe for a tanding wave on a tring (ee Fig ). CAUTION A particle in a box cannot have zero energy Note that the energy of a particle in a box cannot be zero. Equation (40.31) how that E = 0 would require n = 0, but ubtituting n = 0 into Eq. (40.3) give a zero wave function. Since a particle i decribed by a nonzero wave function, thi mean that there cannot be a particle with E = 0. Thi i a conequence of the Heienberg uncertainty principle: A particle in a zero-energy tate would have a definite value of momentum (preciely zero), o it poition uncertainty would be infinite and the particle could be found anywhere along the x-axi. But thi i impoible, ince a particle in a box can be found only between x 0 and x L. Hence E 0 i not allowed. By contrat, the allowed tationary-tate wave function with n 1,, 3,... do not repreent tate of definite momentum (each i an equal mixture of a tate of x-momentum +p n = nh>l and a tate of x-momentum -p n = -nh>l). Hence each tationary tate ha a nonzero momentum uncertainty, conitent with having a finite poition uncertainty. Example 40.3 Electron in an atom-ize box Find the firt two energy level for an electron confined to a onedimenional box 5.0 * m acro (about the diameter of an atom). SOLUTION IDENTIFY and SET UP: Thi problem ue what we have learned in thi ection about a particle in a box. The firt two energy level correpond to n = 1 and n = in Eq. (40.31). EXECUTE: From Eq. (40.31), E 1 = h 8mL = * J # * kg15.0 * m =.4 * J = 1.5 ev E = h 8mL = 4E 1 = 9.6 * J = 6.0 ev EVALUATE: The difference between the firt two energy level i E - E ev. An electron confined to a box i different from an electron bound in an atom, but it i reauring that thi reult i of the ame order of magnitude a the difference between actual atomic energy level. You can alo how that for a proton or neutron 1m = 1.67 * 10-7 kg confined to a box 1.1 * m acro (the width of a medium-ized atomic nucleu), the energie of the firt two level are about a million time larger: E = 1.7 * ev = 1.7 MeV, E = 4E 1 = 6.8 MeV, E - E 1 = 5.1 MeV. Thi ugget why nuclear reaction (which involve tranition between energy level in nuclei) releae o much more energy than chemical reaction (which involve tranition between energy level of electron in atom). Finally, you can how (ee Exercie 40.11) that the energy level of a billiard ball 1m = 0. kg confined to a box 1.3 m acro the width of a billard table are eparated by about 5 * J. Quantum effect won t diturb a game of billiard. Probability and Normalization Let look a bit more cloely at the wave function for a particle in a box, keeping in mind the probability interpretation of the wave function c that we dicued in Section In our one-dimenional ituation the quantity ƒc1xƒ dx i proportional

22 134 CHAPTER 40 Quantum Mechanic 40.1 Graph of (a) c1x and (b) ƒc1xƒ for the firt three wave function 1n = 1,, 3 for a particle in a box. The horizontal dahed line repreent c1x = 0 and ƒc1xƒ = 0 for each of the three level. The value of ƒc1xƒ dx at each point i the probability of finding the particle in a mall interval dx about the point. A in Fig b, the three graph in each part have been diplaced vertically for clarity. (a) n 5 3 n 5 c(x) to the probability that the particle will be found within a mall interval dx about x. For a particle in a box, ƒc1xƒ dx = C in npx L Figure 40.1 how graph of both c1x and ƒc1xƒ for n = 1,, and 3. Note that not all poition are equally likely. By contrat, in claical mechanic the particle i equally likely to be found at any poition between x = 0 and x = L. We ee from Fig. 40.1b that ƒc1xƒ = 0 at ome point, o there i zero probability of finding the particle at exactly thee point. Don t let that bother you; the uncertainty principle ha already hown u that we can t meaure poition exactly. The particle i localized only to be omewhere between x = 0 and x = L. The particle mut be omewhere on the x-axi that i, omewhere between x = -q and x = +q. So the um of the probabilitie for all the dx everywhere (the total probability of finding the particle) mut equal 1. That the normalization condition that we dicued in Section 40.1: dx n L (b) c(x) n 5 3 n 5 n L x x q L- q ƒc1xƒ dx = 1 (normalization condition) (40.33) A wave function i aid to be normalized if it ha a contant uch a C in Eq. (40.3) that i calculated to make the total probability equal 1 in Eq. (40.33). For a normalized wave function, ƒc1xƒ dx i not merely proportional to, but equal, the probability of finding the particle between the coordinate x and x + dx. That why we call ƒc1xƒ the probability ditribution function. (In Section 40.1 we called ƒ 1x, t ƒ the probability ditribution function. For the cae of a tationary-tate wave function, however, ƒ 1x, t ƒ i equal to ƒc1xƒ.) Let normalize the particle-in-a-box wave function c n 1x given by Eq. (40.3). Since c n 1x i zero except between x = 0 and x = L, Eq. (40.33) become L C in npx (40.34) L0 L dx = 1 You can evaluate thi integral uing the trigonometric identity in u = 1 co u; the reult i C 11 - L>. Thu our probability interpretation of the wave function demand that C L> = 1, or C = 1>L 1> ; the contant C i not arbitrary. (Thi i in contrat to the claical vibrating tring problem, in which C repreent an amplitude that depend on initial condition.) Thu the normalized tationary-tate wave function for a particle in a box are npx c n 1x = in A L L 1n = 1,, 3, Á (particle in a box) (40.35) Example 40.4 A noninuoidal wave function? (a) Show that c1x = Ax + B, where A and B are contant, i a olution of the Schrödinger equation for an E = 0 energy level of a particle in a box. (b) What contraint do the boundary condition at x = 0 and x = L place on the contant A and B? SOLUTION IDENTIFY and SET UP: To be phyically reaonable, a wave function mut atify both the Schrödinger equation and the appropriate boundary condition. In part (a) we ll ubtitute c1x into the Schrödinger equation for a particle in a box, Eq. (40.5), to determine whether it i a olution. In part (b) we ll ee what retriction on c1x arie from applying the boundary condition that c1x = 0 at x = 0 and x = L. EXECUTE: (a) From Eq. (40.5), the Schrödinger equation for an E 0 energy level of a particle in a box i - U d c1x m dx = Ec1x = 0

23 40.3 Potential Well 1343 in the region 0 x L. Differentiating c1x = Ax + B twice inide the box ( 0 x L) and outide: There i zero probability with repect to x give d c1x>dx = 0, o the left ide of the of finding the particle anywhere with thi wave function, and o equation i zero, and o c1x = Ax + B i a olution of thi c1x = Ax + B i not a phyically valid wave function. Schrödinger equation for E = 0. (Note that both c1x and it derivative dc1x>dx = A are continuou function, a they mut be.) (b) Applying the boundary condition at x = 0 give c10 = B = 0, and o c1x = Ax. Applying the boundary condition at x = L give c1l = AL = 0, o A = 0. Hence c1x = 0 both EVALUATE: The moral i that there are many function that atify the Schrödinger equation for a given phyical ituation, but mot of thee including the function conidered here have to be rejected becaue they don t atify the appropriate boundary condition. Time Dependence Finally, we note that the wave function c n 1x in Eq. (40.35) depend only on the patial coordinate x. Equation (40.1) how that if c(x i the wave function for a tate of definite energy E, the full time-dependent wave function i 1x, t = c1xe -iet>u. Hence the time-dependent tationary-tate wave function for a particle in a box are npx n 1x, t = ina A L L be-ie nt>u 1n = 1,, 3, Á (40.36) In thi expreion the energie E n are given by Eq. (40.31). The higher the quantum number n, the greater the angular frequency v n = E n >U at which the wave function ocillate. Note that ince ƒe -ient>u ƒ = e +ient>u e -ient>u = e 0 = 1, the probability ditribution function ƒ n 1x, tƒ = 1>Lin 1npx>L i independent of time and doe not ocillate. (Remember, thi i why we ay that thee tate of definite energy are tationary.) Tet Your Undertanding of Section 40. If a particle in a box i in the nth energy level, what i the average value of it x-component of momentum p x? (i) nh>l; (ii) 11>nh>L; (iii) 11> 1nh>L; (iv) 31>114 nh>l; (v) zero Potential Well A potential well i a potential-energy function U1x that ha a minimum. We introduced thi term in Section 7.5, and we alo ued it in our dicuion of periodic motion in Chapter 14. In Newtonian mechanic a particle trapped in a potential well can vibrate back and forth with periodic motion. Our firt application of the Schrödinger equation, the particle in a box, involved a rudimentary potential well with a function U1x that i zero within a certain interval and infinite everywhere ele. A we mentioned in Section 40., thi function correpond to a few ituation found in nature, but the correpondence i only approximate. A better approximation to everal actual phyical ituation i a finite well, which i a potential well with traight ide but finite height. Figure how a potential-energy function that i zero in the interval 0 x L and ha the value U 0 outide thi interval. Thi function i often called a quare-well potential. It could erve a a imple model of an electron within a metallic heet with thickne L, moving perpendicular to the urface of the heet. The electron can move freely inide the metal but ha to climb a potential-energy barrier with height U 0 to ecape from either urface of the metal. The energy U 0 i related to the work function that we dicued in Section 38.1 in connection with the photoelectric effect. In three dimenion, a pherical verion of a finite well give an approximate decription of the motion of proton and neutron within a nucleu. Bound State of a Square-Well Potential In Newtonian mechanic, the particle i trapped (localized) in a well if the total mechanical energy E i le than U 0. In quantum mechanic, uch a trapped tate i often called a bound tate. All tate are bound for an infinitely deep well like PhET: Double Well and Covalent Bond PhET: Quantum Bound State ActivPhyic 0.1: Potential Energy Diagram ActivPhyic 0.3: Potential Well A quare-well potential. The potential energy U i zero within the potential well (in the interval 0 # x # L) and ha the contant value U 0 outide thi interval. U(x) U 0 0 L x

24 1344 CHAPTER 40 Quantum Mechanic the one we decribed in Section 40.. For a finite well like that hown in Fig , if E i greater than U 0, the particle i not bound. Let ee how to olve the Schrödinger equation for the bound tate of a quare-well potential. Our goal i to find the energie and wave function for which E 6 U 0. The eaiet approach i to conider eparately the region where U = 0 and where U = U 0. Where U = 0, the time-independent Schrödinger equation i Thi i the ame a Eq. (40.5) from Section 40., which decribe a particle in a box. A in Section 40., we can expre the olution of thi equation a combination of co kx and in kx, where E =U k >m. We can rewrite the relationhip between E and k a k = me>u. Hence inide the quare well 10 x L we have c1x = Acoa me U (40.38) where A and B are contant. So far, thi look a lot like the particle-in-a-box analyi in Section 40.. The difference i that for the quare-well potential, the potential energy outide the well i not infinite, o the wave function c1x outide the well i not zero. For the region outide the well ( x 6 0 and x 7 L) the potential-energy function in the time-independent Schrodinger equation i U = U 0 : - U d c1x m dx - U d c1x m dx d c1x = Ec1x or (40.37) dx = - me U c1x xb + Bina me U xb (inide the well) d c1x or (40.39) dx = m1u 0 - E + U 0 c1x = Ec1x U c1x The quantity U 0 - E i poitive, o the olution of thi equation are exponential. Uing k (the Greek letter kappa) to repreent the quantity 3m1U 0 - E4 1> >U and taking k a poitive, we can write the olution a c1x = Ce kx + De -kx (outide the well) (40.40) A poible wave function for a particle in a finite potential well. The function i inuoidal inide the well 10 x L and exponential outide it. It approache zero aymptotically at large ƒ x ƒ. The function mut join moothly at x = 0 and x = L; the wave function and it derivative mut be continuou. c(x) 0 L x where C and D are contant with different value in the two region x 6 0 and x 7 L. Note that c can t be allowed to approach infinity a x S + q or x S - q. [If it did, we wouldn t be able to atify the normalization condition, Eq. (40.33).] Thi mean that in Eq. (40.40), we mut have D = 0 for x 6 0 and C = 0 for x 7 L. Our calculation o far how that the bound-tate wave function for a finite well are inuoidal inide the well [Eq. (40.38)] and exponential outide it [Eq. (40.40)]. We have to match the wave function inide and outide the well o that they atify the boundary condition that we mentioned in Section 40.: c1x and dc1x>dx mut be continuou at the boundary point x = 0 and x = L. If the wave function c1x or the lope dc1x>dx were to change dicontinuouly at a point, the econd derivative d c1x>dx would be infinite at that point. That would violate the time-independent Schrödinger equation, Eq. (40.3), which ay that at every point d c1x>dx i proportional to U - E. For a finite well U - E i finite everywhere, o d c1x>dx mut alo be finite everywhere. Matching the inuoidal and exponential function at the boundary point o that they join moothly i poible only for certain pecific value of the total energy E, o thi requirement determine the poible energy level of the finite quare well. There i no imple formula for the energy level a there wa for the infinitely deep well. Finding the level i a fairly complex mathematical problem that require olving a trancendental equation by numerical approximation; we won t go into the detail. Figure how the general hape of a poible wave function. The mot triking feature of thi wave function are the

25 40.3 Potential Well 1345 exponential tail that extend outide the well into region that are forbidden by Newtonian mechanic (becaue in thoe region the particle would have negative kinetic energy). We ee that there i ome probability for finding the particle outide the potential well, which would be impoible in claical mechanic. In Section 40.4 we ll dicu an amazing reult of thi effect. Example 40.5 Outide a finite well (a) Show that Eq. (40.40), c1x = Ce kx + De -kx, i indeed a Since from Eq. (40.40) k = m1u 0 - E>U, thi i equal to the olution of the time-independent Schrödinger equation outide a right-hand ide of the equation. The equation i atified, and c1x finite well of height U 0. (b) What happen to c1x in the limit i a olution. U 0 S q? (b) A U 0 approache infinity, k alo approache infinity. In the region x 6 0, c1x = Ce kx ; a k S q, kx S - q (ince x i SOLUTION negative) and e kx S 0, o the wave function approache zero for all x 6 0. Likewie, we can how that the wave function alo IDENTIFY and SET UP: In part (a), we try the given function c1x approache zero for all x 7 L. Thi i jut what we found in Section 40.; the wave function for a particle in a box mut be zero in the time-independent Schrödinger equation for x 6 0 and for x L, Eq. (40.39). In part (b), we note that in the limit U 0 S q 7 outide the box. the finite well become an infinite well, like thoe we conidered in Section 40. for a particle in a box. So in thi limit the wave function outide a finite well mut reduce to the wave function outide the box. EVALUATE: Our reult in part (b) how that the infinite quare well i a limiting cae of the finite well. We ve een many cae in Newtonian mechanic where it important to conider limiting EXECUTE: (a) We mut how that c1x = Ce kx + De -kx cae (uch a Example 5.11 and 5.13 in Section 5.). Limiting atifie d c1x>dx = 3m1U. We recall that 1d>due au 0 - E>U cae are no le important in quantum mechanic. 4c1x = ae au and 1d >du e au = a e au ; the left-hand ide of the Schrödinger equation i then d c1x dx = d dx 1Cekx + d dx 1De-kx = Ck e kx + D1-k e -kx = k 1Ce kx + De -kx = k c1x Comparing Finite and Infinite Square Well Let continue the comparion of the finite-depth potential well with the infinitely deep well, which we began in Example Firt, becaue the wave function for the finite well don t go to zero at x = 0 and x = L, the wavelength of the inuoidal part of each wave function i longer than it would be with an infinite well. Thi increae in l correpond to a reduced magnitude of momentum p = h>l and therefore a reduced energy. Thu each energy level, including the ground level, i lower for a finite well than for an infinitely deep well with the ame width. Second, a well with finite depth U 0 ha only a finite number of bound tate and correponding energy level, compared to the infinite number for an infinitely deep well. How many level there are depend on the magnitude of U 0 in comparion with the ground-level energy for the infinitely deep well (IDW), which we call. From Eq. (40.31), E 1-IDW E 1-IDW = p U (40.41) ml (ground-level energy, infinitely deep well) When the well i very deep o U 0 i much larger than E 1-IDW, there are many bound tate and the energie of the lowet few are nearly the ame a the energie for the infinitely deep well. When U 0 i only a few time a large a E 1-IDW there are only a few bound tate. (There i alway at leat one bound tate, no matter how hallow the well.) A with the infinitely deep well, there i no tate with E = 0; uch a tate would violate the uncertainty principle.

26 1346 CHAPTER 40 Quantum Mechanic (a) Wave function for the three bound tate for a particle in a finite potential well with depth U 0, for the cae U 0 = 6E 1-IDW. (Here E 1-IDW i the ground-level energy for an infinite well of the ame width.) The horizontal brown line for each wave function correpond to c = 0; the vertical placement of thee line indicate the energy of each bound tate (compare Fig ). (b) Energy-level diagram for thi ytem. The energie are expreed both a multiple of E 1-IDW and a fraction of U 0. All energie greater than U 0 are poible; tate with E 7 U 0 form a continuum. (a) c(x) (b) U(x) Continuum U 0 5 6E 1-IDW n 5 3 E 3 n 5 3 E E 1-IDW U 0 U 0 n 5 E n 5 E 1-IDW E 5.43E 1-IDW U 0 n L x E 1 n L x E E 1-IDW U Probability ditribution function ƒc(x)ƒ for the quare-well wave function hown in Fig The horizontal brown line for each wave function correpond to ƒcƒ = 0. n 5 3 n 5 n 5 1 c(x) 0 L x Figure how the cae U 0 = 6E 1-IDW ; for thi particular cae there are three bound tate. In the figure, we expre the energy level both a fraction of the well depth U 0 and a multiple of E 1-IDW. Note that if the well were infinitely deep, the lowet three level, a given by Eq. (40.31), would be E 1-IDW, 4E 1-IDW, and 9E 1-IDW. Figure alo how the wave function for the three bound tate. It turn out that when U 0 i le than E 1-IDW, there i only one bound tate. In the limit when U 0 i much maller than E 1-IDW (a very hallow well), the energy of thi ingle tate i approximately E = 0.68U 0. Figure how graph of the probability ditribution that i, the value of ƒcƒ for the wave function hown in Fig a. A with the infinite well, not all poition are equally likely. Unlike the infinite well, there i ome probability of finding the particle outide the well in the claically forbidden region. There are alo tate for which E i greater than U 0. In thee free-particle tate the particle i not bound but i free to move through all value of x. Any energy E greater than U 0 i poible, o the free-particle tate form a continuum rather than a dicrete et of tate with definite energy level. The free-particle wave function are inuoidal both inide and outide the well. The wavelength i horter inide the well than outide, correponding to greater kinetic energy inide the well than outide it. Figure how a graphic demontration of particle in a two-dimenional finite potential well. Example 40.6 decribe another application of the quarewell potential. Example 40.6 An electron in a finite well An electron i trapped in a quare well 0.50 nm acro (roughly five time a typical atomic diameter). (a) Find the ground-level energy E 1-IDW if the well i infinitely deep. (b) Find the energy level if the actual well depth U 0 i ix time the ground-level energy found in part (a). (c) Find the wavelength of the photon emitted when the electron make a tranition from the n level to the n 1 level. In what region of the electromagnetic pectrum doe the photon wavelength lie? (d) If the electron i in the n 1 (ground) level and aborb a photon, what i the minimum photon energy that will free the electron from the well? In what region of the pectrum doe the wavelength of thi photon lie? SOLUTION IDENTIFY and SET UP: Equation (40.41) give the ground-level energy E 1-IDW for an infinitely deep well, and Fig b how the energie for a quare well with U 0 = 6E 1-IDW. The energy of the photon emitted or aborbed in a tranition i equal to the difference

27 40.4 Potential Barrier and Tunneling 1347 in energy between two level involved in the tranition; the photon wavelength i given by E = hc>l (ee Chapter 38). EXECUTE: (a) From Eq. (40.41), E 1-IDW = p U ml = p * J # * kg10.50 * 10-9 m =.4 * J = 1.5 ev (b) We have U 0 = 6E 1-IDW = ev = 9.0 ev. We can read off the energy level from Fig b: E 1 = 0.65E 1-IDW = ev = 0.94 ev E =.43E 1-IDW = ev = 3.6 ev E 3 = 5.09E 1-IDW = ev = 7.6 ev (c) The photon energy and wavelength for the n to n 1 tranition are E - E 1 = 3.6 ev ev =.7 ev l = hc E * ev # * 10 8 m> =.7 ev = 460 nm in the blue region of the viible pectrum. (d) We ee from Fig b that the minimum energy needed to free the electron from the well from the n = 1 level i U 0 - E 1 = 9.0 ev ev = 8.1 ev, which i three time the.7-ev photon energy found in part (c). Hence the correponding photon wavelength i one-third of 460 nm, or (to two ignificant figure) 150 nm, which i in the ultraviolet region of the pectrum. EVALUATE: A a check, you can alo calculate the bound-tate energie by uing the formula E 1 = 0.104U 0, E = 0.405U 0, and E 3 = 0.848U 0 given in Fig b. A an additional check, note that the firt three energy level of an infinitely deep well of the ame width are E 1-IDW = 1.5 ev, E -IDW = 4E 1-IDW = 6.0 ev, and E 3-IDW = 9E 1-IDW = 13.5 ev. The energie we found in part (b) are le than thee value: A we mentioned earlier, the finite depth of the well lower the energy level compared to the level for an infinitely deep well. One application of thee idea i to quantum dot, which are nanometer-ized particle of a emiconductor uch a cad-mium elenide 1CdSe. An electron within a quantum dot behave much like a particle in a finite potential well of width L equal to the ize of the dot. When quantum dot are illuminated with ultraviolet light, the electron aborb the ultraviolet photon and are excited into high energy level, uch a the n = 3 level decribed in thi example. If the electron return to the ground level 1n = 1 in two or more tep (for example, from n = 3 to n = and from n = to n = 1), one of the tep will involve emitting a viible-light photon, a we have calculated here. (We decribed thi proce of fluorecence in Section 39.3.) Increaing the value of L decreae the energie of the level and hence the pacing between them, and thu decreae the energy and increae the wavelength of the emitted photon. The photograph that open thi chapter how quantum dot of different ize in olution: Each emit a characteritic wavelength that depend on the dot ize. Quantum dot can be injected into living tiue and their fluorecent glow ued a a tracer for biological reearch and for medicine. They may alo be the key to a new generation of laer and ultrafat computer. Tet Your Undertanding of Section 40.3 Suppoe that the width of the finite potential well hown in Fig i reduced by one-half. How mut the value of U 0 change o that there are till jut three bound energy level whoe energie are the fraction of U 0 hown in Fig b? U 0 mut: (i) increae by a factor of four; (ii) increae by a factor of two; (iii) remain the ame; (iv) decreae by a factor of one-half; (v) decreae by a factor of one-fourth Potential Barrier and Tunneling A potential barrier i the oppoite of a potential well; it i a potential-energy function with a maximum. Figure how an example. In claical Newtonian mechanic, if a particle (uch a a roller coater) i located to the left of the barrier (which might be a hill), and if the total mechanical energy of the ytem i E 1, the particle cannot move farther to the right than x = a. If it did, the potential energy U would be greater than the total energy E and the kinetic energy K = E - U would be negative. Thi i impoible in claical mechanic ince K = 1 mv can never be negative. A quantum-mechanical particle behave differently: If it encounter a barrier like the one in Fig and ha energy le than E, it may appear on the other ide. Thi phenomenon i called tunneling. In quantum-mechanical tunneling, unlike macrocopic, mechanical tunneling, the particle doe not actually puh through the barrier and loe no energy in the proce. Tunneling Through a Rectangular Barrier To undertand how tunneling can occur, let look at the potential-energy function U1x hown in Fig It like Fig turned upide-down; the potential energy i zero everywhere except in the range 0 x L, where it ha the value U 0. Thi might repreent a imple model for the potential energy of an To make thi image, 48 iron atom (hown a yellow peak) were placed in a circle on a copper urface. The elevation at each point inide the circle indicate the electron denity within the circle. The tanding-wave pattern i very imilar to the probability ditribution function for a particle in a one-dimenional finite potential well. (Thi image wa made with a canning tunneling microcope, dicued in Section 40.4.) PhET: Quantum Tunneling and Wave Packet ActivPhyic 0.4: Potential Barrier

28 1348 CHAPTER 40 Quantum Mechanic A potential-energy barrier. According to Newtonian mechanic, if the total energy of the ytem i E 1, a particle to the left of the barrier can go no farther than x = a. If the total energy i greater than E, the particle can pa over the barrier. E E 1 U(x) 0 a A rectangular potential-energy barrier with width L and height U 0. According to Newtonian mechanic, if the total energy E i le than U 0, a particle cannot pa over thi barrier but i confined to the ide where it tart. E U 0 U(x) 0 L x x electron in the preence of two lab of metal eparated by an air gap of thickne L. The potential energy i lower within either lab than in the gap between them. Let conider olution of the Schrödinger equation for thi potential-energy function for the cae in which E i le than U 0. We can ue our reult from Section In the region x 6 0 and x 7 L, where U = 0, the olution i inuoidal and i given by Eq. (40.38). Within the barrier 10 x L, U = U 0 and the olution i exponential a in Eq. (40.40). Jut a with the finite potential well, the function have to join moothly at the boundary point x = 0 and x = L, which mean that both c(x and dc1x>dx have to be continuou at thee point. Thee requirement lead to a wave function like the one hown in Fig The function i not zero inide the barrier (the region forbidden by Newtonian mechanic). Even more remarkable, a particle that i initially to the left of the barrier ha ome probability of being found to the right of the barrier. How great thi probability i depend on the width L of the barrier and the particle energy E in comparion with the barrier height U 0. The tunneling probability T that the particle get through the barrier i proportional to the quare of the ratio of the amplitude of the inuoidal wave function on the two ide of the barrier. Thee amplitude are determined by matching wave function and their derivative at the boundary point, a fairly involved mathematical problem. When T i much maller than unity, it i given approximately by T = Ge -kl where G = 16 E a1 - E b and k = m1u 0 - E (40.4) U 0 U 0 U (probability of tunneling) The probability decreae rapidly with increaing barrier width L. It alo depend critically on the energy difference U 0 - E, which in Newtonian phyic i the additional kinetic energy the particle would need to be able to climb over the barrier A poible wave function for a particle tunneling through the potentialenergy barrier hown in Fig The wave function i exponential within the barrier (0 # x # L)... c(x) U(x) U 0 0 L x... and inuoidal outide the barrier. The function and it derivative (lope) are continuou at x 5 0 and x 5 L o that the inuoidal and exponential function join moothly. Example 40.7 Tunneling through a barrier A.0-eV electron encounter a barrier 5.0 ev high. What i the probability that it will tunnel through the barrier if the barrier width i (a) 1.00 nm and (b) 0.50 nm? SOLUTION IDENTIFY and SET UP: Thi problem ue the idea of tunneling through a rectangular barrier, a in Fig and Our target variable i the tunneling probability T in Eq. (40.4), which we evaluate for the given value E =.0 ev (electron energy), U = 5.0 ev (barrier height), m = 9.11 * kg (ma of the electron), and L = 1.00 nm or 0.50 nm (barrier width). EXECUTE: Firt we evaluate G and k in Eq. (40.4), uing E =.0 ev:.0 ev.0 ev G = 16a ba1-5.0 ev 5.0 ev b = 3.8 U 0 - E = 5.0 ev -.0 ev = 3.0 ev = 4.8 * J k = * kg14.8 * J * J # = 8.9 * 10 9 m -1 (a) When L = 1.00 nm = 1.00 * 10-9 m, kl = 18.9 * 10 9 m * 10-9 m = 17.8 and T = Ge -kl = 3.8e = 7.1 * (b) When L = 0.50 nm, one-half of 1.00 nm, kl i one-half of 17.8, or 8.9. Hence T = 3.8e -8.9 = 5. * EVALUATE: Halving the width of thi barrier increae the tunneling probability T by a factor of 15. * 10-4 >17.1 * 10-8 = 7.3 * 10 3, or nearly ten thouand. The tunneling probability i an extremely enitive function of the barrier width.

29 40.4 Potential Barrier and Tunneling 1349 (a) y Surface electron z Specimen x Probe L (b) 40.1 (a) Schematic diagram of the probe of a canning tunneling microcope (STM). A the harp conducting probe i canned acro the urface in the x- and y-direction, it i alo moved in the z-direction to maintain a contant tunneling current. The changing poition of the probe i recorded and ued to contruct an image of the urface. (b) Thi colored STM image how quantum wire : thin trip, jut 10 atom wide, of a conductive rare-earth ilicide atop a ilicon urface. Such quantum wire may one day be the bai of ultraminiaturized circuit. Application of Tunneling Tunneling ha a number of practical application, ome of coniderable importance. When you twit two copper wire together or cloe the contact of a witch, current pae from one conductor to the other depite a thin layer of nonconducting copper oxide between them. The electron tunnel through thi thin inulating layer. A tunnel diode i a emiconductor device in which electron tunnel through a potential barrier. The current can be witched on and off very quickly (within a few picoecond) by varying the height of the barrier. A Joephon junction conit of two uperconductor eparated by an oxide layer a few atom (1 to nm) thick. Electron pair in the uperconductor can tunnel through the barrier layer, giving uch a device unuual circuit propertie. Joephon junction are ueful for etablihing precie voltage tandard and meauring tiny magnetic field, and they play a crucial role in the developing field of quantum computing. The canning tunneling microcope (STM) ue electron tunneling to create image of urface down to the cale of individual atom. An extremely harp conducting needle i brought very cloe to the urface, within 1 nm or o (Fig. 40.1a). When the needle i at a poitive potential with repect to the urface, electron can tunnel through the urface potential-energy barrier and reach the needle. A Example 40.7 how, the tunneling probability and hence the tunneling current are very enitive to change in the width L of the barrier (the ditance between the urface and the needle tip). In one mode of operation the needle i canned acro the urface and at the ame time i moved perpendicular to the urface to maintain a contant tunneling current. The needle motion i recorded, and after many parallel can, an image of the urface can be recontructed. Extremely precie control of needle motion, including iolation from vibration, i eential. Figure 40.1b how an STM image. (Figure i alo an STM image.) Tunneling i alo of great importance in nuclear phyic. A fuion reaction can occur when two nuclei tunnel through the barrier caued by their electrical repulion and approach each other cloely enough for the attractive nuclear force to caue them to fue. Fuion reaction occur in the core of tar, including the un; without tunneling, the un wouldn t hine. The emiion of alpha particle from untable nuclei uch a radium alo involve tunneling. An alpha particle i a cluter of two proton and two neutron (the ame a a nucleu of the mot common form of helium). Such cluter form naturally within larger atomic nuclei. An alpha particle trying to ecape from a nucleu encounter a potential barrier that reult from the combined effect of the attractive nuclear force and the electrical repulion of the remaining part of the nucleu (Fig. 40.). The alpha particle can ecape only by tunneling through thi barrier. Depending on the barrier height and width for a given kind of alpha-emitting nucleu, the tunneling probability can be low or high, and the alpha-emitting material will have low or high radioactivity. Recall from Section 39. that Ernet Rutherford ued alpha particle Application Electron Tunneling in Enzyme Protein molecule play eential role a enzyme in living organim. Enzyme like the one hown here are large molecule, and in many cae their function depend on the ability of electron to tunnel acro the pace that eparate one part of the molecule from another. Without tunneling, life a we know it would be impoible! 40. Approximate potential-energy function for an alpha particle interacting with a nucleu of radiu R. If an alpha particle inide the nucleu ha energy E greater than zero, it can tunnel through the barrier and ecape from the nucleu. Nucleu U(r) E R Inide the nucleu (r # R), an alpha particle encounter a quare-well potential due to the trong nuclear force. r O R Outide the nucleu (r. R), an alpha particle experience a 1/r potential due to electrotatic repulion.

30 1350 CHAPTER 40 Quantum Mechanic from a radioactive ource to dicover the atomic nucleu. Although Rutherford did not know it, tunneling by thee alpha particle made hi experiment poible! We ll learn more about alpha decay in Chapter 43. Tet Your Undertanding of Section 40.4 I it poible for a particle undergoing tunneling to be found within the barrier rather than on either ide of it? 40.5 The Harmonic Ocillator 40.3 Potential-energy function for the harmonic ocillator. In Newtonian mechanic the amplitude A i related to the total energy E by E = 1 k A, and the particle i retricted to the range from x = -A to x = A. In quantum mechanic the particle can be found at x 7 A or x 6-A. U(x) 5 1 k x E A U(x) 0 A x Sytem that ocillate are of tremendou importance in the phyical world, from the ocillation of your eardrum in repone to a ound wave to the vibration of the ground caued by an earthquake. Ocillation are equally important on the microcopic cale where quantum effect dominate. The molecule of the air around you can be et into vibration when they collide with each other, the proton and neutron in an excited atomic nucleu can ocillate in oppoite direction, and a microwave oven tranfer energy to food by making water molecule in the food flip back and forth. In thi ection we ll look at the olution of the Schrödinger equation for the implet kind of vibrating ytem, the quantummechanical harmonic ocillator. A we learned in Chapter 14, a harmonic ocillator i a particle with ma m that move along the x-axi under the influence of a conervative force F x = -k x. The contant k i called the force contant. (In Chapter 14 we ued the ymbol k for the force contant. In thi ection we ll ue the ymbol k intead to minimize confuion with the wave number k = p>l.) The force i proportional to the particle diplacement x from it equilibrium poition, x = 0. The correponding potential-energy function i U = 1 k x (Fig. 40.3). In Newtonian mechanic, when the particle i diplaced from equilibrium, it undergoe inuoidal motion with frequency ƒ = 11>p1k >m 1> and angular frequency v = pƒ = 1k >m 1>. The amplitude (that i, the maximum diplacement from equilibrium) of thee Newtonian ocillation i A, which i related to the energy E of the ocillator by E = 1 k A. Let make an enlightened gue about the energy level of a quantummechanical harmonic ocillator. In claical phyic an electron ocillating with angular frequency v emit electromagnetic radiation with that ame angular frequency. It reaonable to gue that when an excited quantum-mechanical harmonic ocillator with angular frequency v = 1k >m 1> (according to Newtonian mechanic, at leat) make a tranition from one energy level to a lower level, it would emit a photon with thi ame angular frequency v. The energy of uch a photon i hƒ = 1h>p1v>p = Uv. So we would expect that the pacing between adjacent energy level of the harmonic ocillator would be k hƒ =Uv =U (40.43) A m That the ame pacing between energy level that Planck aumed in deriving hi radiation law (ee Section 39.5). It wa a good aumption; a we ll ee, the 5 energy level are in fact half-integer A 1 3, Á B,, multiple of Uv. Wave Function, Boundary Condition, and Energy Level We ll begin our quantum-mechanical analyi of the harmonic ocillator by writing down the one-dimenional time-independent Schrödinger equation, Eq. (40.3), 1 with k x in place of U: - U d c1x m dx + 1 k x c1x = Ec1x (Schrödinger equation for the harmonic ocillator) (40.44) The olution of thi equation are wave function for the phyically poible tate of the ytem.

31 40.5 The Harmonic Ocillator 1351 In the dicuion of quare-well potential in Section 40. we found that the energy level are determined by boundary condition at the wall of the well. However, the harmonic-ocillator potential ha no wall a uch; what, then, are the appropriate boundary condition? Claically, ƒxƒ cannot be greater than the amplitude A given by E = 1 k A. Quantum mechanic doe allow ome penetration into claically forbidden region, but the probability decreae a that penetration increae. Thu the wave function mut approach zero a ƒxƒ grow large. Satifying the requirement that c1x S 0 a ƒxƒ S q i not a trivial a it may eem. To ee why thi i, let rewrite Eq. (40.44) in the form d c1x (40.45) dx = m U A 1 k x - EBc1x Equation (40.45) how that when x i large enough (either poitive or negative) to make the quantity A 1 poitive, the function and it econd derivative d c1x>dx k x - EB c1x have the ame ign. Figure 40.4 how four poible kind of behavior of c1x beginning at a point where x i greater than the claical amplitude A, o that k x - 1 k A = 1 k x - E 7 0. Let look at thee 1 four cae more cloely. Note that if c1x i poitive a hown in Fig. 40.4, Eq. (40.45) tell u that d c1x>dx i alo poitive and the function i concave upward. Note alo that d c1x>dx i the rate of change of the lope of c1x; thi will help u undertand how our four poible wave function behave. Curve a: The lope of c1x i poitive at point x. Since d c1x>dx 7 0, the function curve upward increaingly teeply and goe to infinity. Thi violate the boundary condition that c1x S 0 a ƒxƒ S q, o thi in t a viable wave function. Curve b: The lope of c1x i negative at point x, and d c1x>dx ha a large poitive value. Hence the lope change rapidly from negative to poitive and keep on increaing o, again, the wave function goe to infinity. Thi wave function in t viable either. Curve c: A for curve b, the lope i negative at point x. However, d c1x>dx now ha a mall poitive value, o the lope increae only gradually a c1x decreae to zero and croe over to negative value. Equation (40.45) tell u that once c1x become negative, d c1x>dx alo become negative. Hence the curve become concave downward and head for negative infinity. Thi wave function, too, fail to atify the requirement that c1x S 0 a ƒxƒ S q and thu in t viable. Curve d: If the lope of c1x at point x i negative, and the poitive value of d c1x>dx at thi point i neither too large nor too mall, the curve bend jut enough to glide in aymptotically to the x-axi. In thi cae c1x, dc1x>dx, and d c1x>dx all approach zero at large x. Thi cae offer the only hope of atifying the boundary condition that c1x S 0 a ƒxƒ S q, and it occur only for certain very pecial value of the energy E. Thi qualitative dicuion ugget how the boundary condition a ƒxƒ S q determine the poible energy level for the quantum-mechanical harmonic ocillator. It turn out that thee boundary condition are atiifed only if the energy E i equal to one of the value E n, given by the imple formula 40.4 Poible behavior of harmonicocillator wave function in the region 1 k x 7 E. In thi region, c1x and d c1x>dx have the ame ign. The curve i concave upward when d c1x>dx i poitive and concave downward when d c1x>dx i negative. c(x) O 1 k x, E 1 k x. E (a) (b) (d) x A (c) Only curve d, which approache the x-axi aymptotically for large x, i an acceptable wave function for thi ytem. E n = 1n + 1 U k A m = 1n + 1 Uv 1n = 0, 1,, Á (energy level, harmonic ocillator) (40.46) where n i the quantum number identifying each tate and energy level. Note that the ground level of energy E 0 = 1 Uv i denoted by n = 0, not n = 1.

32 135 CHAPTER 40 Quantum Mechanic 40.5 Energy level for the harmonic ocillator. The pacing between any two adjacent level i E =Uv. The energy of the ground level i E 0 = 1 Uv. E 5 hv U(x) O 11 E 5 5 hv 9 E 4 5 hv 7 E 3 5 hv 5 E 5 hv 3 E 1 5 hv 1 E 0 5 hv x Equation (40.46) confirm our gue [(Eq )] that adjacent energy level are eparated by a contant interval of Uv = hƒ, a Planck aumed in There are infinitely many level; thi houldn t be urpriing becaue we are dealing with an infinitely deep potential well. A ƒxƒ increae, U = 1 k x increae without bound. Figure 40.5 how the lowet ix energy level and the potential-energy function U1x). For each level n, the value of ƒxƒ at which the horizontal line repreenting the total energy E n interect U1x give the amplitude A n of the correponding Newtonian ocillator. Example 40.8 Vibration in a crytal A odium atom of ma 3.8 * 10-6 kg vibrate within a crytal. The potential energy increae by ev when the atom (a) The Newtonian angular frequency i k i diplaced nm from it equilibrium poition. Treat the v = A m = 1. N>m B atom a a harmonic ocillator. (a) Find the angular frequency of 3.8 * 10-6 kg = 1.79 * 1013 rad> the ocillation according to Newtonian mechanic. (b) Find the (b) From Eq. (40.46) and Fig. 40.5, the pacing between adjacent energy level i pacing (in electron volt) of adjacent vibrational energy level according to quantum mechanic. (c) What i the wavelength of a photon emitted a the reult of a tranition from one level to the Uv = * J # * next lower level? In what region of the electromagnetic pectrum doe thi lie? = 1.88 * ev Ja 1.60 * b = ev J (c) The energy E of the emitted photon i equal to the energy SOLUTION lot by the ocillator in the tranition, ev. Then IDENTIFY and SET UP: We ll find the force contant k from the expreion U = 1 k x for potential energy. We ll then find the l = hc * ev # * 10 8 m> = angular frequency v = 1k >m 1> E ev and ue thi in Eq. (40.46) to find the pacing between adjacent energy level. We ll calculate = 1.05 * 10-4 m = 105 mm the wavelength of the emitted photon a in Example Thi photon wavelength i in the infrared region of the pectrum. EXECUTE: We are given that U = ev = 1. * 10-1 J when x = * 10-9 m, o we can olve U = 1 k x for k : -1 J k = U 11. * 10 = x * 10-9 = 1. N>m m EVALUATE: Thi example how u that interatomic force contant are a few newton per meter, about the ame a thoe of houehold pring or pring-baed toy uch a the Slinky. It alo ugget that we can learn about the vibration of molecule by meauring the radiation that they emit in tranitioning to a lower vibrational tate. We will explore thi idea further in Chapter 4. ActivPhyic 0.1.6: Potential Energy Diagram, Quetion 6 Comparing Quantum and Newtonian Ocillator The wave function for the level n = 0, 1,, Á of the harmonic ocillator are called Hermite function; they aren t encountered in elementary calculu coure but are well known to mathematician. Each Hermite function i an exponential function multiplied by a polynomial in x. The harmonic-ocillator wave function correponding to n = 0 and E = E 0 (the ground level) i c1x = Ce - mk x >U (40.47) The contant C i choen to normalize the function that i, to make 1 q - q ƒcƒ dx = 1. (We re uing C rather than A a a normalization contant in thi ection, ince we ve already appropriated the ymbol A to denote the Newtonian amplitude of a harmonic ocillator.) You can find C uing the following reult from integral table: L q - q e -a x dx = 1p a

33 40.5 The Harmonic Ocillator The firt four wave function for the harmonic ocillator. The amplitude A of a Newtonian ocillator with the ame total energy i hown for each. Each wave function penetrate omewhat into the claically forbidden region ƒ x ƒ 7 A. The total number of finite maxima and minima for each function i n + 1, one more than the quantum number. c(x) c(x) c(x) c(x) n 5 0 n 5 1 n 5 n 5 3 O x O x O x O x A A A A A A A A To confirm that c(x a given by Eq. (40.47) really i a olution of the Schrödinger equation for the harmonic ocillator, we invite you to calculate the econd derivative of thi wave function, ubtitute it into Eq. (40.44), and verify that the equation i atified if the energy E i equal to E 0 = 1 Uv (ee Exercie 40.38). It a little mey, but the reult i atifying and worth the effort. Figure 40.6 how the the firt four harmonic-ocillator wave function. Each graph alo how the amplitude A of a Newtonian harmonic ocillator with the ame energy that i, the value of A determined from 1 k A = An + 1 BUv (40.48) In each cae there i ome penetration of the wave function into the region ƒxƒ 7 A that are forbidden by Newtonian mechanic. Thi i imilar to the effect that we noted in Section 40.3 for a particle in a finite quare well. Figure 40.7 how the probability ditribution ƒc1xƒ for thee ame tate. Each graph alo how the probability ditribution determined from Newtonian phyic, in which the probability of finding the particle near a randomly choen point i inverely proportional to the particle peed at that point. If we average out the wiggle in the quantum-mechanical probability curve, the reult for n 7 0 reemble the Newtonian prediction. Thi agreement improve with increaing n; Fig how the claical and quantum-mechanical probability function for n = 10. Notice that the pacing between zero of ƒc1xƒ in Fig increae with increaing ditance from x = 0. Thi make ene from the Newtonian perpective: A a particle move away from x = 0, it kinetic energy K and the magnitude p of it momentum both decreae. Thinking quantummechanically, thi mean that the wavelength l = h>p increae, o the pacing between zero of c1x (and hence of ƒc1xƒ ) alo increae. In the Newtonian analyi of the harmonic ocillator the minimum energy i zero, with the particle at ret at it equilibrium poition x 0. Thi i not poible in quantum mechanic; no olution of the Schrödinger equation ha E = 0 and atifie the boundary condition. Furthermore, if there were uch a tate, it 40.7 Probability ditribution function ƒc1xƒ for the harmonic-ocillator wave function hown in Fig The amplitude A of the Newtonian motion with the ame energy i hown for each. The blue line how the correponding probability ditribution for the Newtonian motion. A n increae, the averaged-out quantum-mechanical function reemble the Newtonian curve more and more. c(x) c(x) c(x) c(x) n 5 0 n 5 1 n 5 n 3 A O A x A O A x A O A x A O A x

34 1354 CHAPTER 40 Quantum Mechanic 40.8 Newtonian and quantummechanical probability ditribution function for a harmonic ocillator for the tate n = 10. The Newtonian amplitude A i alo hown. The larger the value of n, the more cloely the quantum-mechanical probability ditribution (green) matche the Newtonian probability ditribution (blue). c(x) n 5 10 A O 40.9 A potential-energy function decribing the interaction of two atom in a diatomic molecule. The ditance r i the eparation between the center of the atom, and the equilibrium eparation i r = r 0. The energy needed to diociate the molecule i U q. U` U(r) O r 0 A U U (approximation) When r i near r 0, the potential-energy curve i approximately parabolic (a hown by the red curve) and the ytem behave approximately like a harmonic ocillator. r x would violate the Heienberg uncertainty principle becaue there would be no 1 uncertainty in either poition or momentum. The energy mut be at leat Uv for the ytem to conform to the uncertainty principle. To ee qualitatively why thi 1 i o, conider a Newtonian ocillator with total energy Uv. We can find the amplitude A and the maximum velocity jut a we did in Section When the particle i at it maximum diplacement 1x = A and intantaneouly at ret, K = 0 and E = U = 1 k A. When the particle i at equilibrium 1x = 0 and moving at it maximum peed, U = 0 and E = K = 1. Setting E = 1 mv max Uv, we find E = 1 k A = 1 Uv = 1 1/ k Ua m b E = 1 mv max = 1 k A o v max = Aa k m b 1> The maximum momentum of the particle i o A = p max = mv max =U 1> k 1>4 m 1>4 = U1> k 1>4 m 3>4 Here where the Heienberg uncertainty principle come in. It turn out that the uncertaintie in the particle poition and momentum (calculated a tandard deviation) are, repectively, x = A> 1 = A> 1> and p x = p max > 1 = p max > 1>. Then the product of the two uncertaintie i U 1> bau1> k 1>4 m 1>4 x p x = a 1> k 1>4 m 1>4 1> b = U Thi product equal the minimum value allowed by Eq. (39.9), x p x ÚU>, 1 and thu atifie the uncertainty principle. If the energy had been le than Uv, the product x p x would have been le than U>, and the uncertainty principle would have been violated. Even when a potential-energy function in t preciely parabolic in hape, we may be able to approximate it by the harmonic-ocillator potential for ufficiently mall diplacement from equilibrium. Figure 40.9 how a typical potentialenergy function for an interatomic force in a molecule. At large eparation the curve of U1r veru r level off, correponding to the abence of force at great ditance. But the curve i approximately parabolic near the minimum of U1r (the equilibrium eparation of the atom). Near equilibrium the molecular vibration i approximately imple harmonic with energy level given by Eq. (40.46), a we aumed in Example U 1> k 1>4 m 1>4 Tet Your Undertanding of Section 40.5 A quantum-mechanical ytem initially in it ground level aborb a photon and end up in the firt excited tate. The ytem then aborb a econd photon and end up in the econd excited tate. For which of the following ytem doe the econd photon have a longer wavelength than the firt one? (i) a harmonic ocillator; (ii) a hydrogen atom; (iii) a particle in a box.

35 CHAPTER 40 SUMMARY Wave function: The wave function for a particle contain all of the information about that particle. If the particle move in one dimenion in the preence of a potential energy function U1x, the wave function 1x, t obey the one-dimenional Schrödinger equation. (For a free particle on which no force act, U1x = 0.) The quantity ƒ 1x, tƒ, called the probability ditribution function, determine the relative probability of finding a particle near a given poition at a given time. If the particle i in a tate of definite energy, called a tationary tate, 1x, t i a product of a function c1x that depend only on patial coordinate and a function e -iet>u that depend only on time. For a tationary tate, the probability ditribution function i independent of time. A patial tationary-tate wave function c1x for a particle that move in one dimenion in the preence of a potential-energy function U1x atifie the timeindependent Schrödinger equation. More complex wave function can be contructed by uperpoing tationarytate wave function. Thee can repreent particle that are localized in a certain region, thu repreenting both particle and wave apect. (See Example 40.1 and 40..) - U 0 1x, t m 0x + U1x 1x, t 0 1x, t = iu (40.0) 0t (general 1-D Schrödinger equation) 1x, t = c1xe -iet>u (time-dependent wave function for a tate of definite energy) - U (40.1) d c1x + U1xc1x = Ec1x m dx (time-independent Schrödinger equation) (40.3) Re C(x) = A co kx A O A Im C(x) = A in kx A O A x p/k p/k 3p/k p/k p/k 3p/k x Particle in a box: The energy level for a particle of ma m in a box (an infinitely deep quare potential well) with width L are given by Eq. (40.31). The correponding normalized tationary-tate wave function of the particle are given by Eq. (40.35). (See Example 40.3 and 40.4.) E n = p n m = n h 8mL = n p U ml 1n = 1,, 3, Á npx c n 1x = in A L L 1n = 1,, 3, Á (40.31) (40.35) c(x) n 5 3 n 5 n L x Wave function and normalization: To be a olution of the Schrödinger equation, the wave function c1x and (40.33) L- q ƒc1xƒ dx = 1 it derivative dc1x>dx mut be continuou everywhere (normalization condition)) except where the potential-energy function U1x ha an infinite dicontinuity. Wave function are uually normalized o that the total probability of finding the particle omewhere i unity. q c(x) n 5 3 n 5 n 5 1 x 0 L Finite potential well: In a potential well with finite depth U 0, the energy level are lower than thoe for an infinitely deep well with the ame width, and the number of energy level correponding to bound tate i finite. The level are obtained by matching wave function at the well wall to atify the continuity of c1x and dc1x>dx. (See Example 40.5 and 40.6.) n 5 3 n 5 n 5 1 U(x) Continuum E 1-IDW 0 L U 0 5 6E 1-IDW E E 1-IDW U 0 E 5.43E 1-IDW U 0 E E x 1-IDW U 0 Potential barrier and tunneling: There i a certain probability that a particle will penetrate a potentialenergy barrier even though it initial energy i le than the barrier height. Thi proce i called tunneling. (See Example 40.7.) c(x) U(x) U 0 0 L x 1355

36 1356 CHAPTER 40 Quantum Mechanic Quantum harmonic ocillator: The energy level for the harmonic ocillator (for which U1x = 1 k x E ) are given n = An + 1 k BU A m = An + 1 B Uv by Eq. (40.46). The pacing between any two adjacent 1n = 0, 1,, 3, Á (40.46) level i Uv, where v = k >m i the ocillation angular frequency of the correponding Newtonian harmonic ocillator. (See Example 40.8.) U(x) DE 5 hv O 11 E 5 5 hv 9 E 4 5 hv 7 E 3 5 hv 5 E 5 hv 3 E 1 5 hv 1 E 0 5 hv x BRIDGING PROBLEM A Packet in a Box A particle of ma m in an infinitely deep well ha the following wave function in the region from x 0 to x L: 1x, t = 1 c 11xe -ie 1t>U + 1 c 1xe -ie t>u Here c 1 1x and c 1x are the normalized tationary-tate wave function for the firt two level ( n = 1 and n = ), given by Eq. (40.35). E 1 and E, given by Eq. (40.31), are the energie of thee level. The wave function i zero for x 6 0 and for x 7 L. (a) Find the probability ditribution function for thi wave function. (b) Doe 1x, t repreent a tationary tate of definite energy? How can you tell? (c) Show that the wave function 1x, t i normalized. (d) Find the angular frequency of ocillation of the probability ditribution function. What i the interpretation of thi ocillation? (e) Suppoe intead that 1x, t i a combination of the wave function of the two lowet level of a finite well of length L and height U 0 equal to ix time the energy of the lowet-energy bound tate of an infinite well of length L. What would be the angular frequency of the probability ditribution function in thi cae? SOLUTION GUIDE See MateringPhyic tudy area for a Video Tutor olution. IDENTIFY and SET UP 1. In Section 40.1 we aw how to interpret a combination of two free-particle wave function of different energie. In thi problem you need to apply thee ame idea to a combination of wave function for the infinite well (Section 40.) and the finite well (Section 40.3). EXECUTE. Write down the full time-dependent wave function 1x, t and it complex conjugate * 1x, t uing the function c 1 1x and c 1x from Eq. (40.35). Ue thee to calculate the probability ditribution function, and decide whether or not thi function depend on time. 3. To check for normalization, you ll need to verify that when you integrate the probability ditribution function from tep over all value of x, the integral i equal to 1. [Hint: The trigonometric identitie in u = cou and in u in f = co1u - f - co1u + f may be helpful.] 4. To find the anwer to part (d) you ll need to identify the ocillation angular frequency v oc in your expreion from tep for the probability ditribution function. To interpret the ocillation, draw graph of the probability ditribution function at time t 0, t T> 4, t T>, and t 3T> 4, where T p>v oc i the ocillation period of the probability ditribution function. 5. For the finite well you do not have imple expreion for the firt two tationary-tate wave function c 1 1x and c 1x. However, you can till find the ocillation angular frequency v oc, which i related to the energie E 1 and E in the ame way a for the infinite-well cae. (Can you ee why?) EVALUATE 6. Why are the factor of 1> 1 in the wave function 1x, t important? 7. Why do you uppoe the ocillation angular frequency for a finite well i lower than for an infinite well of the ame width? Problem For intructor-aigned homework, go to : Problem of increaing difficulty. CP: Cumulative problem incorporating material from earlier chapter. CALC: Problem requiring calculu. BIO: Biocience problem. DISCUSSION QUESTIONS Q40.1 If quantum mechanic replace the language of Newtonian mechanic, why don t we have to ue wave function to decribe the motion of macrocopic bodie uch a baeball and car? Q40. A tudent remark that the relationhip of ray optic to the more general wave picture i analogou to the relationhip of Newtonian mechanic, with well-defined particle trajectorie, to quantum mechanic. Comment on thi remark. Q40.3 A Eq. (40.1) indicate, the time-dependent wave function for a tationary tate i a complex number having a real part and an imaginary part. How can thi function have any phyical meaning, ince part of it i imaginary?

37 Exercie 1357 Q40.4 Why mut the wave function of a particle be normalized? Q40.5 If a particle i in a tationary tate, doe that mean that the particle i not moving? If a particle move in empty pace with S contant momentum p and hence contant energy E = p >m, i it in a tationary tate? Explain your anwer. Q40.6 For the particle in a box, we choe k = np>l with n = 1,, 3, Á to fit the boundary condition that c = 0 at x = L. However, n = 0, -1, -, -3, Á alo atify that boundary condition. Why didn t we alo chooe thoe value of n? Q40.7 If c i normalized, what i the phyical ignificance of the area under a graph of ƒcƒ veru x between x 1 and x? What i the total area under the graph of ƒcƒ when all x are included? Explain. Q40.8 For a particle in a box, what would the probability ditribution function ƒcƒ look like if the particle behaved like a claical (Newtonian) particle? Do the actual probability ditribution approach thi claical form when n i very large? Explain. Q40.9 In Chapter 15 we repreented a tanding wave a a uperpoition of two wave traveling in oppoite direction. Can the wave function for a particle in a box alo be thought of a a combination of two traveling wave? Why or why not? What phyical interpretation doe thi repreentation have? Explain. Q40.10 A particle in a box i in the ground level. What i the probability of finding the particle in the right half of the box? (Refer to Fig. 40.1, but don t evaluate an integral.) I the anwer the ame if the particle i in an excited level? Explain. Q40.11 The wave function for a particle in a box (ee Fig. 40.1a) are zero at certain point. Doe thi mean that the particle can t move pat one of thee point? Explain. Q40.1 For a particle confined to an infinite quare well, i it correct to ay that each tate of definite energy i alo a tate of definite wavelength? I it alo a tate of definite momentum? Explain. (Hint: Remember that momentum i a vector.) Q40.13 For a particle in a finite potential well, i it correct to ay that each bound tate of definite energy i alo a tate of definite wavelength? I it a tate of definite momentum? Explain. Q40.14 In Fig. 40.1b, the probability function i zero at the point x = 0 and x = L, the wall of the box. Doe thi mean that the particle never trike the wall? Explain. Q40.15 A particle i confined to a finite potential well in the region 0 6 x 6 L. How doe the area under the graph of ƒcƒ in the region 0 6 x 6 L compare to the total area under the graph of ƒcƒ when including all poible x? Q40.16 Compare the wave function for the firt three energy level for a particle in a box of width L (ee Fig. 40.1a) to the correponding wave function for a finite potential well of the ame width (ee Fig a). How doe the wavelength in the interval 0 x L for the n = 1 level of the particle in a box compare to the correponding wavelength for the n = 1 level of the finite potential well? Ue thi to explain why E 1 i le than E 1-IDW in the ituation depicted in Fig b. Q40.17 It i tated in Section 40.3 that a finite potential well alway ha at leat one bound level, no matter how hallow the well. Doe thi mean that a U 0 S 0, E 1 S 0? Doe thi violate the Heienberg uncertainty principle? Explain. Q40.18 Figure 40.15a how that the higher the energy of a bound tate for a finite potential well, the more the wave function extend outide the well (into the interval x 6 0 and x 7 L). Explain why thi happen. Q40.19 In claical (Newtonian) mechanic, the total energy E of a particle can never be le than the potential energy U becaue the kinetic energy K cannot be negative. Yet in barrier tunneling (ee Section 40.4) a particle pae through region where E i le than U. I thi a contradiction? Explain. Q40.0 Figure how the canning tunneling microcope image of 48 iron atom placed on a copper urface, the pattern indicating the denity of electron on the copper urface. What can you infer about the potential-energy function inide the circle of iron atom? Q40.1 Qualitatively, how would you expect the probability for a particle to tunnel through a potential barrier to depend on the height of the barrier? Explain. Q40. The wave function hown in Fig i nonzero for both x 6 0 and x 7 L. Doe thi mean that the particle plit into two part when it trike the barrier, with one part tunneling through the barrier and the other part bouncing off the barrier? Explain. Q40.3 The probability ditribution for the harmonic ocillator wave function (ee Fig and 40.8) begin to reemble the claical (Newtonian) probability ditribution when the quantum number n become large. Would the ditribution become the ame a in the claical cae in the limit of very large n? Explain. Q40.4 In Fig. 40.8, how doe the probability of finding a particle in the center half of the region -A 6 x 6 A compare to the probability of finding the particle in the outer half of the region? I thi conitent with the phyical interpretation of the ituation? Q40.5 Compare the allowed energy level for the hydrogen atom, the particle in a box, and the harmonic ocillator. What are the value of the quantum number n for the ground level and the econd excited level of each ytem? Q40.6 Sketch the wave function for the potential-energy well hown in Fig. Q40.6 when E 1 i le than U 0 and when i greater than U 0. E 3 Figure Q40.6 EXERCISES Section 40.1 Wave Function and the One-Dimenional Schrödinger Equation An electron i moving a a free particle in the - x-direction with momentum that ha magnitude 4.50 * 10-4 kg # m>. What i the one-dimenional time-dependent wave function of the electron? 40.. A free particle moving in one dimenion ha wave function 1x, t = A3e i1kx-vt - e i1kx-4vt 4 where k and v are poitive real contant. (a) At t = 0 what are the two mallet poitive value of x for which the probability function ƒ 1x, tƒ i a maximum? (b) Repeat part (a) for time t = p>v. (c) Calculate v av a the ditance the maxima have moved divided by the elaped time. Compare your reult to the expreion v av = 1v - v 1 >1k - k 1 from Example Conider the free-particle wave function of Example Let k = 3k 1 = 3k. At t = the probability ditribution function ƒ 1x, tƒ ha a maximum at 0x = 0. (a) What i the mallet poitive value of x for which the probability ditribution function ha a maximum at time t = p>v, where v =Uk >m. (b) From your reult in part (a), what i the average peed with which the probability ditribution i moving in the + x-direction? Compare your reult to the expreion v av = 1v - v 1 >1k - k 1 from Example Conider the free particle of Example Show that v av = 1v - v 1 >1k - k 1 can be written a v av = p av >m, where p av = 1Uk +Uk 1 >. ` U 0 U(x) O A B ` x

38 1358 CHAPTER 40 Quantum Mechanic Conider a wave function given by c1x = Ain kx, where k = p>l and A i a real contant. (a) For what value of x i there the highet probability of finding the particle decribed by thi wave function? Explain. (b) For which value of x i the probability zero? Explain Compute ƒ ƒ for =cin vt, where c i time independent and v i a real contant. I thi a wave function for a tationary tate? Why or why not? CALC Let c 1 and c be two olution of Eq. (40.3) with energie E 1 and E, repectively, where E 1 Z E. I c = Ac 1 + Bc, where A and B are nonzero contant, a olution to Eq. (40.3)? Explain your anwer A particle i decribed by a wave function c1x = Ae -ax, where A and a are real, poitive contant. If the value of a i increaed, what effect doe thi have on (a) the particle uncertainty in poition and (b) the particle uncertainty in momentum? Explain your anwer CALC Linear Combination of Wave Function. Let c 1 and c be two olution of Eq. (40.3) with the ame energy E. Show that c = Bc 1 + Cc i alo a olution with energy E, for any value of the contant B and C. Section 40. Particle in a Box CALC A particle moving in one dimenion (the x-axi) i decribed by the wave function c1x = e Ae-bx, for x Ú 0 Ae bx, for x 6 0 where b =.00 m -1, A 7 0, and the +x-axi point toward the right. (a) Determine A o that the wave function i normalized. (b) Sketch the graph of the wave function. (c) Find the probability of finding thi particle in each of the following region: (i) within 50.0 cm of the origin, (ii) on the left ide of the origin (can you firt gue the anwer by looking at the graph of the wave function?), (iii) between x = m and x = 1.00 m Ground-Level Billiard. (a) Find the lowet energy level for a particle in a box if the particle i a billiard ball 1m = 0.0 kg and the box ha a width of 1.3 m, the ize of a billiard table. (Aume that the billiard ball lide without friction rather than roll; that i, ignore the rotational kinetic energy.) (b) Since the energy in part (a) i all kinetic, to what peed doe thi correpond? How much time would it take at thi peed for the ball to move from one ide of the table to the other? (c) What i the difference in energy between the n = and n = 1 level? (d) Are quantum-mechanical effect important for the game of billiard? A proton i in a box of width L. What mut the width of the box be for the ground-level energy to be 5.0 MeV, a typical value for the energy with which the particle in a nucleu are bound? Compare your reult to the ize of a nucleu that i, on the order of m Find the width L of a one-dimenional box for which the ground-tate energy of an electron in the box equal the abolute value of the ground tate of a hydrogen atom When a hydrogen atom undergoe a tranition from the n = to the n = 1 level, a photon with l = 1 nm i emitted. (a) If the atom i modeled a an electron in a one-dimenional box, what i the width of the box in order for the n = to n = 1 tranition to correpond to emiion of a photon of thi energy? (b) For a box with the width calculated in part (a), what i the ground-tate energy? How doe thi correpond to the ground-tate energy of a hydrogen atom? (c) Do you think a one-dimenional box i a good model for a hydrogen atom? Explain. (Hint: Compare the pacing between adjacent energy level a a function of n.) A certain atom require 3.0 ev of energy to excite an electron from the ground level to the firt excited level. Model the atom a an electron in a box and find the width L of the box An electron in a one-dimenional box ha ground-tate energy 1.00 ev. What i the wavelength of the photon aborbed when the electron make a tranition to the econd excited tate? CALC Show that the time-dependent wave function given by Eq. (40.35) i a olution to the one-dimenional Schrödinger equation, Eq. (40.3) Recall that ƒcƒ dx i the probability of finding the particle that ha normalized wave function c1x in the interval x to x + dx. Conider a particle in a box with rigid wall at x = 0 and x = L. Let the particle be in the ground level and ue c n a given in Eq. (40.35). (a) For which value of x, if any, in the range from 0 to L i the probability of finding the particle zero? (b) For which value of x i the probability highet? (c) In part (a) and (b) are your anwer conitent with Fig. 40.1? Explain Repeat Exercie for the particle in the firt excited level CALC (a) Show that c = A in kx i a olution to Eq. (40.5) if k = 1mE>U. (b) Explain why thi i an acceptable wave function for a particle in a box with rigid wall at x = 0 and x = L only if k i an integer multiple of p>l CALC (a) Repeat Exercie 40.0 for c = A co kx. (b) Explain why thi cannot be an acceptable wave function for a particle in a box with rigid wall at x = 0 and x = L no matter what the value of k (a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that ha a width of 0.15 nm. (b) The electron make a tranition from the n = 1 to n = 4 level by aborbing a photon. Calculate the wavelength of thi photon An electron i in a box of width 3.0 * m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it i in (a) the n = 1 level; (b) the n = level; (c) the n = 3 level? In each cae how doe the wavelength compare to the width of the box? CALC Normalization of the Wave Function. Conider a particle moving in one dimenion, which we hall call the x-axi. (a) What doe it mean for the wave function of thi particle to be normalized? (b) I the wave function c1x = e ax, where a i a poitive real number, normalized? Could thi be a valid wave function? (c) If the particle decribed by the wave function c1x = Ae bx, where A and b are poitive real number, i confined to the range x Ú 0, determine A (including it unit) o that the wave function i normalized. Section 40.3 Potential Well CALC (a) Show that c = Ain kx, where k i a real (not complex) contant, i not a olution of Eq. (40.3) for U = U 0 and E 6 U 0. (b) I thi c a olution for E 7 U 0? An electron i moving pat the quare well hown in Fig The electron ha energy E = 3U 0. What i the ratio of the de Broglie wavelength of the electron in the region x 7 L to the wavelength for 0 6 x 6 L? An electron i bound in a quare well of depth U 0 = 6E 1-IDW. What i the width of the well if it ground-tate energy i.00 ev? An electron i bound in a quare well of width 1.50 nm and depth U 0 = 6E 1-IDW. If the electron i initially in the ground

39 Problem 1359 level and aborb a photon, what maximum wavelength can the photon have and till liberate the electron from the well? CALC Calculate d c>dx for the wave function of Eq. (40.38), and how that the function i a olution of Eq. (40.37) An electron i bound in a quare well with a depth equal to ix time the ground-level energy E 1-IDW of an infinite well of the ame width. The longet-wavelength photon that i aborbed by the electron ha a wavelength of nm. Determine the width of the well A proton i bound in a quare well of width 4.0 fm = 4.0 * m. The depth of the well i ix time the ground-level energy E 1-IDW of the correponding infinite well. If the proton make a tranition from the level with energy E 1 to the level with energy E 3 by aborbing a photon, find the wavelength of the photon. Section 40.4 Potential Barrier and Tunneling Alpha Decay. In a Figure E40.3 imple model for a radioactive nucleu, an alpha particle 1m = U(r) 6.64 * 10-7 kg i trapped by a U 0 quare barrier that ha width.0 fm 1.0 MeV E and height 30.0 MeV. (a) What i the tunneling probability when the alpha particle encounter the barrier if it kinetic energy i 1.0 MeV below the top of the barrier (Fig. E40.3)? (b) What i the O r tunneling probability if the energy.0 fm of the alpha particle i 10.0 MeV below the top of the barrier? An electron with initial kinetic energy 6.0 ev encounter a barrier with height 11.0 ev. What i the probability of tunneling if the width of the barrier i (a) 0.80 nm and (b) 0.40 nm? An electron with initial kinetic energy 5.0 ev encounter a barrier with height U 0 and width 0.60 nm. What i the tranmiion coefficient if (a) U 0 = 7.0 ev; (b) U 0 = 9.0 ev; (c) U 0 = 13.0 ev? An electron i moving pat the quare barrier hown in Fig , but the energy of the electron i greater than the barrier height. If E = U 0, what i the ratio of the de Broglie wavelength of the electron in the region x 7 L to the wavelength for 0 6 x 6 L? A proton with initial kinetic energy 50.0 ev encounter a barrier of height 70.0 ev. What i the width of the barrier if the probability of tunneling i 3.0 * 10-3? How doe thi compare with the barrier width for an electron with the ame energy tunneling through a barrier of the ame height with the ame probability? (a) An electron with initial kinetic energy 3 ev encounter a quare barrier with height 41 ev and width 0.5 nm. What i the probability that the electron will tunnel through the barrier? (b) A proton with the ame kinetic energy encounter the ame barrier. What i the probability that the proton will tunnel through the barrier? Section 40.5 The Harmonic Ocillator CALC Show that c1x given by Eq. (40.47) i a olution to Eq. (40.44) with energy E 0 =Uv> A wooden block with ma 0.50 kg i ocillating on the end of a pring that ha force contant 110 N>m. Calculate the ground-level energy and the energy eparation between adjacent level. Expre your reult in joule and in electron volt. Are quantum effect important? A harmonic ocillator aborb a photon of wavelength 8.65 * 10-6 m when it undergoe a tranition from the ground tate to the firt excited tate. What i the ground-tate energy, in electron volt, of the ocillator? Chemit ue infrared aborption pectra to identify chemical in a ample. In one ample, a chemit find that light of wavelength 5.8 mm i aborbed when a molecule make a tranition from it ground harmonic ocillator level to it firt excited level. (a) Find the energy of thi tranition. (b) If the molecule can be treated a a harmonic ocillator with ma 5.6 * 10-6 kg, find the force contant The ground-tate energy of a harmonic ocillator i 5.60 ev. If the ocillator undergoe a tranition from it n = 3 to n = level by emitting a photon, what i the wavelength of the photon? In Section 40.5 it i hown that for the ground level of a harmonic ocillator, x p x =U>. Do a imilar analyi for an excited level that ha quantum number n. How doe the uncertainty product x p x depend on n? For the ground-level harmonic ocillator wave function c(x given in Eq. (40.47), ƒcƒ ha a maximum at x = 0. (a) Compute the ratio of ƒcƒ at x = +A to ƒcƒ at x = 0, where A i given by Eq. (40.48) with n = 0 for the ground level. (b) Compute the ratio of ƒcƒ at x = +A to ƒcƒ at x = 0. In each cae i your reult conitent with what i hown in Fig. 40.7? For the odium atom of Example 40.8, find (a) the ground-tate energy, (b) the wavelength of a photon emitted when the n = 4 to n = 3 tranition occur; (c) the energy difference for any n = 1 tranition. PROBLEMS The dicuion in Section 40.1 how that the wave function =ce -ivt i a tationary tate, where c i time independent and v i a real (not complex) contant. Conider the wave function =c 1 e -iv 1t + c e -ivt, where c 1 and c are different timeindependent function and v 1 and v are different real contant. Aume that c and are real-valued function, o that c * 1 c and c * 1 = c 1 = c. I thi a wave function for a tationary tate? Why or why not? A particle of ma m in a one-dimenional box ha the following wave function in the region x = 0 to x = L: 1x, t = 1 1 c 11xe -ie 1t>U c 31xe -ie 3t>U Here c 1 1x and c 3 1x are the normalized tationary-tate wave function for the n = 1 and n = 3 level, and E 1 and E 3 are the energie of thee level. The wave function i zero for x 6 0 and for x 7 L. (a) Find the value of the probability ditribution function at x = L> a a function of time. (b) Find the angular frequency at which the probability ditribution function ocillate CALC Conider the wave packet defined by q c1x = B1k co kx dk L 0 Let B1k = e -a k. (a) The function B1k ha it maximum value at k = 0. Let k h be the value of k at which B1k ha fallen to half it maximum value, and define the width of B1k a w k = k h. In term of a, what i w k? (b) Ue integral table to evaluate the integral that give c1x. For what value of x i c1x maximum? (c) Define the width of c1x a w x = x h, where i the poitive x h

40 1360 CHAPTER 40 Quantum Mechanic value of x at which ha fallen to half it maximum value. Calculate in term of a. (d) The momentum p i equal to hk>p, o the width of B in momentum i w p = hw k >p. Calculate the product w p w x and compare to the Heienberg uncertainty principle CALC (a) Uing the integral in Problem 40.48, determine the wave function c1x for a function B1k given by w x c1x CALC What i the probability of finding a particle in a 0 k 6 0 B1k = c 1>k 0, 0 k k 0 0, k 7 k 0 Thi repreent an equal combination of all wave number between 0 and k 0. Thu c1x repreent a particle with average wave number k 0 >, with a total pread or uncertainty in wave number of k 0. We will call thi pread the width w k of B1k, o w k = k 0. (b) Graph B1k veru k and c1x veru x for the cae k 0 = p>l, where L i a length. Locate the point where c1x ha it maximum value and label thi point on your graph. Locate the two point cloet to thi maximum (one on each ide of it) where c1x = 0, and define the ditance along the x-axi between thee two point a w x, the width of c1x. Indicate the ditance w x on your graph. What i the value of w x if k 0 = p>l? (c) Repeat part (b) for the cae k 0 = p>l. (d) The momentum p i equal to hk>p, o the width of B in momentum i w p = hw k >p. Calculate the product w p w x for each of the cae k 0 = p>l and k 0 = p>l. Dicu your reult in light of the Heienberg uncertainty principle CALC Show that the wave function c1x = Ae ikx i a olution of Eq. (40.3) for a particle of ma m, in a region where the potential energy i a contant U 0 6 E. Find an expreion for k, and relate it to the particle momentum and to it de Broglie wavelength CALC Wave function like the one in Problem can repreent free particle moving with velocity v = p>m in the x-direction. Conider a beam of uch particle incident on a potential-energy tep U1x = 0, for x 6 0, and U1x = U 0 6 E, for x 7 0. The wave function for x 6 0 i c1x = Ae ik1x + Be -ik1x, repreenting incident and reflected particle, and for x 7 0 i c1x = Ce ikx, repreenting tranmitted particle. Ue the condition that both c and it firt derivative mut be continuou at x = 0 to find the contant B and C in term of k 1, k, and A Let E n be the energy difference between the adjacent energy level E n and E n+1 for a particle in a box. The ratio R n = E n >E n compare the energy of a level to the energy eparation of the next higher energy level. (a) For what value of n i R n larget, and what i thi larget R n? (b) What doe R n approach a n become very large? How doe thi reult compare to the claical value for thi quantity? Photon in a Dye Laer. An electron in a long, organic molecule ued in a dye laer behave approximately like a particle in a box with width 4.18 nm. What i the wavelength of the photon emitted when the electron undergoe a tranition (a) from the firt excited level to the ground level and (b) from the econd excited level to the firt excited level? CALC A particle i in the ground level of a box that extend from x = 0 to x = L. (a) What i the probability of finding the particle in the region between 0 and L>4? Calculate thi by integrating ƒc1x ƒ dx, where c i normalized, from x = 0 to x = L>4. (b) What i the probability of finding the particle in the region x = L>4 to x = L>? (c) How do the reult of part (a) and (b) compare? Explain. (d) Add the probabilitie calculated in part (a) and (b). (e) Are your reult in part (a), (b), and (d) conitent with Fig. 40.1b? Explain. box of length L in the region between x = L>4 and x = 3L>4 when the particle i in (a) the ground level and (b) the firt excited level? (Hint: Integrate ƒc1x ƒ dx, where c i normalized, between L>4 and 3L>4. ) (c) Are your reult in part (a) and (b) conitent with Fig. 40.1b? Explain Conider a particle in a box with rigid wall at x = 0 and x = L. Let the particle be in the ground level. Calculate the probability ƒcƒ dx that the particle will be found in the interval x to x + dx for (a) x = L>4; (b) x = L>; (c) x = 3L> Repeat Problem for a particle in the firt excited level CP A particle i confined within a box with perfectly rigid wall at x = 0 and x = L. Although the magnitude of the intantaneou force exerted on the particle by the wall i infinite and the time over which it act i zero, the impule (that involve a product of force and time) i both finite and quantized. Show that the impule exerted by the wall at x = 0 i 1nh>Lın and that the impule exerted by the wall at x = L i -1nh>Lın. (Hint: You may wih to review Section 8.1.) CALC A fellow tudent propoe that a poible wave function for a free particle with ma m (one for which the potential-energy function U1x i zero) i c1x = e e+kx, x 6 0 e -kx, x Ú 0 where k i a poitive contant. (a) Graph thi propoed wave function. (b) Show that the propoed wave function atifie the Schrödinger equation for x 6 0 if the energy i E = -U k >m that i, if the energy of the particle i negative. (c) Show that the propoed wave function alo atifie the Schrödinger equation for x Ú 0 with the ame energy a in part (b). (d) Explain why the propoed wave function i nonethele not an acceptable olution of the Schrödinger equation for a free particle. (Hint: What i the behavior of the function at x = 0?) It i in fact impoible for a free particle (one for which U1x = 0 to have an energy le than zero The penetration ditance h in a finite potential well i the ditance at which the wave function ha decreaed to 1>e of the wave function at the claical turning point: c1x = L + h = 1 e c1l The penetration ditance can be hown to be h = U m1u 0 - E The probability of finding the particle beyond the penetration ditance i nearly zero. (a) Find h for an electron having a kinetic energy of 13 ev in a potential well with U 0 = 0 ev. (b) Find h for a 0.0-MeV proton trapped in a 30.0-MeV-deep potential well CALC (a) For the finite potential well of Fig , what relationhip among the contant A and B of Eq. (40.38) and C and D of Eq. (40.40) are obtained by applying the boundary condition that c be continuou at x = 0 and at x = L? (b) What relationhip among A, B, C, and D are obtained by applying the boundary condition that dc>dx be continuou at x = 0 and at x = L? An electron with initial kinetic energy 5.5 ev encounter a quare potential barrier with height 10.0 ev. What i the width of

41 Challenge Problem 1361 the barrier if the electron ha a 0.10% probability of tunneling through the barrier? A particle with ma m and total energy E tunnel through a quare barrier of height U 0 and width L. When the tranmiion coefficient i not much le than unity, it i given by T = c1 + 1U 0 inhkl 4E1U 0 - E d -1 where inh kl = 1e kl - e -kl > i the hyperbolic ine of kl. (a) Show that if kl W 1, thi expreion for T approache Eq. (40.4). (b) Explain why the retriction kl W 1 in part (a) implie either that the barrier i relatively wide or that the energy E i relatively low compared to U 0. (c) Show that a the particle incident kinetic energy E approache the barrier height U 0, T approache kL> 4-1, where k = me>u i the wave number of the incident particle. (Hint: If ƒzƒ V 1, then inh z L z CP A harmonic ocillator conit of a 0.00-kg ma on a pring. It frequency i 1.50 Hz, and the ma ha a peed of m> a it pae the equilibrium poition. (a) What i the value of the quantum number n for it energy level? (b) What i the difference in energy between the level E n and E n+1? I thi difference detectable? For mall amplitude of ocillation the motion of a pendulum i imple harmonic. For a pendulum with a period of 0.500, find the ground-level energy and the energy difference between adjacent energy level. Expre your reult in joule and in electron volt. Are thee value detectable? Some nm photon are emitted in a n = 1 tranition within a olid-tate lattice. The lattice i modeled a electron in a box having length nm. What tranition correpond to the emitted light? CALC Show that for c1x given by Eq. (40.47), the probability ditribution function ha a maximum at x = CALC (a) Show by direct ubtitution in the Schrödinger equation for the one-dimenional harmonic ocillator that the wave function c where a 1 1x = A 1 xe -a x >, = mv>u, i a olution with energy correponding to n = 1 in Eq. (40.46). (b) Find the normalization contant A 1. (c) Show that the probability denity ha a minimum at x = 0 and maxima at x = 1>a, correponding to the claical turning point for the ground tate n = CP (a) The wave nature of particle reult in the quantum-mechanical ituation that a particle confined in a box can aume only wavelength that reult in tanding wave in the box, with node at the box wall. Ue thi to how that an electron confined in a one-dimenional box of length L will have energy level given by E n = n h 8mL (Hint: Recall that the relationhip between the de Broglie wavelength and the peed of a nonrelativitic particle i mv = h>l. The 1 energy of the particle i mv.) (b) If a hydrogen atom i modeled a a one-dimenional box with length equal to the Bohr radiu, what i the energy (in electron volt) of the lowet energy level of the electron? Conider a potential well defined a U1x = q for x 6 0, U1x = 0 for 0 6 x 6 L, and U1x = U for x 7 L (Fig. P40.70). Conider a particle with ma m and kinetic energy E 6 U 0 that i trapped in the well. (a) The boundary condition at the infinite wall 1x = 0 i c10 = 0. What mut the form of the function c1x for 0 6 x 6 L Figure P40.70 be in order to atify both the U(x) Schrödinger equation and thi ` boundary condition? (b) The wave function mut remain finite a x S q. What mut the form of U 0 the function c1x for x 7 L be in order to atify both the x Schrödinger equation and thi 0 L boundary condition at infinity? (c) Impoe the boundary condition that c and dc>dx are continuou at x = L. Show that the energie of the allowed level are obtained from olution of the equation k cot kl = -k, where k = me>u and k = m1u 0 - E>U Section 40. conidered a box with wall at x = 0 and Figure P40.71 x = L. Now conider a box with width L but centered at x = 0, o that it extend from x = -L> to ` U(x) ` x = +L> (Fig. P40.71). Note that thi box i ymmetric about x x = 0. (a) Conider poible wave L/ 0 L/ function of the form c1x = A in kx. Apply the boundary condition at the wall to obtain the allowed energy level. (b) Another et of poible wave function are function of the form c1x = Aco kx. Apply the boundary condition at the wall to obtain the allowed energy level. (c) Compare the energie obtained in part (a) and (b) to the et of energie given in Eq. (40.31). (d) An odd function ƒ atifie the condition ƒ1x = -ƒ1-x, and an even function g atifie g1x = g1-x. Of the wave function from part (a) and (b), which are even and which are odd? CHALLENGE PROBLEMS CALC The WKB Approximation. It can be a challenge to olve the Schrödinger equation for the bound-tate energy level of an arbitrary potential well. An alternative approach that can yield good approximate reult for the energy level i the WKB approximation (named for the phyicit Gregor Wentzel, Hendrik Kramer, and Léon Brillouin, who pioneered it application to quantum mechanic). The WKB approximation begin from three phyical tatement: (i) According to de Broglie, the magnitude of momentum p of a quantum-mechanical particle i p = h>l. (ii) The magnitude of momentum i related to the kinetic energy K by the relationhip K = p >m. (iii) If there are no nonconervative force, then in Newtonian mechanic the energy E for a particle i contant and equal at each point to the um of the kinetic and potential energie at that point: E = K + U1x, where x i the coordinate. (a) Combine thee three relationhip to how that the wavelength of the particle at a coordinate x can be written a l1x = h m3e - U1x4 Thu we enviion a quantum-mechanical particle in a potential well U1x a being like a free particle, but with a wavelength l1x that i a function of poition. (b) When the particle move into a region of increaing potential energy, what happen to it wavelength? (c) At a point where E = U1x, Newtonian mechanic ay that the particle ha zero kinetic energy and mut be intantaneouly at ret. Such a point i called a claical turning point, ince thi i where a Newtonian particle mut top it motion and

42 136 CHAPTER 40 Quantum Mechanic revere direction. A an example, an object ocillating in imple harmonic motion with amplitude A move back and forth between the point x = -A and x = +A; each of thee i a claical turning point, ince there the potential energy 1 k x equal the total 1 energy k A. In the WKB expreion for l1x, what i the wavelength at a claical turning point? (d) For a particle in a box with length L, the wall of the box are claical turning point (ee Fig. 40.8). Furthermore, the number of wavelength that fit within the box mut be a half-integer (ee Fig ), o that L = 1n>l and hence L>l = n>, where n = 1,, 3, Á. [Note that thi i a retatement of Eq. (40.9).] The WKB cheme for finding the allowed bound-tate energy level of an arbitrary potential well i an extenion of thee obervation. It demand that for an allowed energy E, there mut be a half-integer number of wavelength between the claical turning point for that energy. Since the wavelength in the WKB approximation i not a contant but depend on x, the number of wavelength between the claical turning point a and b for a given value of the energy i the integral of 1>l1x between thoe point: Uing the expreion for l1x you found in part (a), how that the WKB condition for an allowed bound-tate energy can be written a La b (e) A a check on the expreion in part (d), apply it to a particle in a box with wall at x = 0 and x = L. Evaluate the integral and how that the allowed energy level according to the WKB approximation are the ame a thoe given by Eq. (40.31). (Hint: Since the wall of the box are infinitely high, the point x = 0 and x = L are claical turning point for any energy E. Inide the box, the potential energy i zero.) (f ) For the finite quare well hown in Fig , how that the WKB expreion given in part (d) predict the ame bound-tate energie a for an infinite quare well of the ame width. (Hint: Aume E 6 U 0. Then the claical turning point are at x = 0 and x = L. ) Thi how that the WKB approximation doe a poor job when the potential-energy function change dicontinuouly, a for a finite potential well. In the next two problem we conider ituation in which the potential-energy function change gradually and the WKB approximation i much more ueful CALC The WKB approximation (ee Challenge Problem 40.7) can be ued to calculate the energy level for a harmonic ocillator. In thi approximation, the energy level are the olution to the equation La m3e - U1x4 dx = nh b La b dx l1x = n m3e - U1x4 dx = nh 1n = 1,, 3, Á 1n = 1,, 3, Á n = 1,, 3, Á Here E i the energy, U1x i the potential-energy function, and x = a and x = b are the claical turning point (the point at which E i equal to the potential energy, o the Newtonian kinetic energy would be zero). (a) Determine the claical turning point for a harmonic ocillator with energy E and force contant k. (b) Carry out the integral in the WKB approximation and how that the energy level in thi approximation are E n =Uv, where v = k >m and n = 1,, 3, Á. (Hint: Recall that U=h>p. A ueful tandard integral i L A - x dx = 1 cxa - x + A arcina x bd ƒaƒ where arcin denote the invere ine function. Note that the integrand i even, o the integral from -x to x i equal to twice the integral from 0 to x. (c) How do the approximate energy level found in part (b) compare with the true energy level given by Eq. (40.46)? Doe the WKB approximation give an underetimate or an overetimate of the energy level? CALC Proton, neutron, and many other particle are made of more fundamental particle called quark and antiquark (the antimatter equivalent of quark). A quark and an antiquark can form a bound tate with a variety of different energy level, each of which correpond to a different particle oberved in the laboratory. A an example, the c particle i a low-energy bound tate of a o-called charm quark and it antiquark, with a ret energy of 3097 MeV; the c1s particle i an excited tate of thi ame quark antiquark combination, with a ret energy of 3686 MeV. A implified repreentation of the potential energy of interaction between a quark and an antiquark i U1x = Aƒxƒ, where A i a poitive contant and x repreent the ditance between the quark and the antiquark. You can ue the WKB approximation (ee Challenge Problem 40.7) to determine the bound-tate energy level for thi potential-energy function. In the WKB approximation, the energy level are the olution to the equation La b m3e - U1x4dx = nh 1n = 1,, 3, Á Here E i the energy, U1x i the potential-energy function, and x = a and x = b are the claical turning point (the point at which E i equal to the potential energy, o the Newtonian kinetic energy would be zero). (a) Determine the claical turning point for the potential U1x = Aƒxƒ and for an energy E. (b) Carry out the above integral and how that the allowed energy level in the WKB approximation are given by E n = 1 m a 3mAh >3 b n >3 1n = 1,, 3, Á 4 (Hint: The integrand i even, o the integral from -x to x i equal to twice the integral from 0 to x. ) (c) Doe the difference in energy between ucceive level increae, decreae, or remain the ame a n increae? How doe thi compare to the behavior of the energy level for the harmonic ocillator? For the particle in a box? Can you ugget a imple rule that relate the difference in energy between ucceive level to the hape of the potential-energy function?

43 Anwer 1363 Anwer Chapter Opening Quetion? When an electron in one of thee particle called quantum dot make a tranition from an excited level to a lower level, it emit a photon whoe energy i equal to the difference in energy between the level. The maller the quantum dot, the larger the energy pacing between level and hence the horter (bluer) the wavelength of the emitted photon. See Example 40.6 (Section 40.3) for more detail. Tet Your Undertanding Quetion 40.1 Anwer: no Equation (40.19) repreent a uperpoition of wave function with different value of wave number k and hence different value of energy E =U k >m. The tate that thi combined wave function repreent i not a tate of definite energy, and therefore not a tationary tate. Another way to ee thi i to note that there i a factor e -iet>u inide the integral in Eq. (40.19), with a different value of E for each value of k. Thi wave function therefore ha a very complicated time dependence, and the probability ditribution function ƒ 1x, tƒ doe depend on time. 40. Anwer: (v) Our derivation of the tationary-tate wave function for a particle in a box how that they are uperpoition of wave propagating in oppoite direction, jut like a tanding wave on a tring. One wave ha momentum in the poitive x-direction, while the other wave ha an equal magnitude of momentum in the negative x-direction. The total x-component of momentum i zero Anwer: (i) The energy level are arranged a hown in Fig b if U where E 1-IDW = p U >ml 0 = 6E 1-IDW, i the ground-level energy of an infinite well. If the well width L i reduced to one-half of it initial value, E 1-IDW increae by a factor of four and o U 0 mut alo increae by a factor of four. The energie E 1, E, and E 3 hown in Fig b are all pecific fraction of U 0, o they will alo increae by a factor of four Anwer: ye Figure 40.0 how a poible wave function c1x for tunneling. Since c1x i not zero within the barrier 10 x L, there i ome probability that the particle can be found there Anwer: (ii) If the econd photon ha a longer wavelength and hence lower energy than the firt photon, the difference in energy between the firt and econd excited level mut be le than the difference between the ground level and the firt excited level. Thi i the cae for the hydrogen atom, for which the energy difference between level decreae a the energy increae (ee Fig. 39.4). By contrat, the energy difference between ucceive level increae for a particle in a box (ee Fig b) and i contant for a harmonic ocillator (ee Fig. 40.5). Bridging Problem Anwer: (a) ƒ 1x, tƒ = 1 px c in L L + in px L + px in L in px L 3p U (b) no (c) ye (d) (e) 0.905p U ml ml co a 1E - E 1 t bd U

44 41 ATOMIC STRUCTURE LEARNING GOALS By tudying thi chapter, you will learn: How to extend quantum-mechanical calculation to three-dimenional problem. How to olve the Schrödinger equation for a particle trapped in a cubical box. How to decribe the tate of a hydrogen atom in term of quantum number. How magnetic field affect the orbital motion of atomic electron. How we know that electron have their own intrinic angular momentum. How to analyze the tructure of many-electron atom. How x ray emitted by atom reveal their inner tructure.? Lithium (with three electron per atom) i a metal that burn pontaneouly in water, while helium (with two electron per atom) i a ga that undergoe almot no chemical reaction. How can one extra electron make thee two element o dramatically different? Some phyicit claim that all of chemitry i contained in the Schrödinger equation. Thi i omewhat of an exaggeration, but thi equation can teach u a great deal about the chemical behavior of element and the nature of chemical bond. It provide inight into the periodic table of the element and the microcopic bai of magnetim. In order to learn about the quantum-mechanical tructure of atom, we ll firt contruct a three-dimenional verion of the Schrödinger equation. We ll try thi equation out by looking at a three-dimenional verion of a particle in a box: a particle confined to a cubical volume. We ll then ee that we can learn a great deal about the tructure and propertie of all atom from the olution to the Schrödinger equation for the hydrogen atom. Thee olution have quantized value of angular momentum; we don t need to make a eparate tatement about quantization a we did with the Bohr model. We label the tate with a et of quantum number, which we ll ue later with many-electron atom a well. We ll find that the electron alo ha an intrinic pin angular momentum in addition to the orbital angular momentum aociated with it motion. We ll alo encounter the excluion principle, a kind of microcopic zoning ordinance that i the key to undertanding many-electron atom. Thi principle ay that no two electron in an atom can have the ame quantum-mechanical tate. Finally, we ll ue the principle of thi chapter to explain the characteritic x-ray pectra of atom. 1364

45 41.1 The Schrödinger Equation in Three Dimenion The Schrödinger Equation in Three Dimenion We have dicued the Schrödinger equation and it application only for onedimenional problem, the analog of a Newtonian particle moving along a traight line. The traight-line model i adequate for ome application, but to undertand atomic tructure, we need a three-dimenional generalization. It not difficult to gue what the three-dimenional Schrödinger equation hould look like. Firt, the wave function i a function of time and all three pace coordinate 1x, y, z. In general, the potential-energy function alo depend on all three coordinate and can be written a U1x, y, z. Next, recall from Section 40.1 that the term -1U >m0 >0x in the one-dimenional Schrödinger equation, Eq. (40.0), i related to the kinetic energy of the particle in the tate decribed by the wave function. For example, if we inert into thi term the wave function 1x, t = Ae ikx e -ivt for a free particle with magnitude of momentum p =Ukand kinetic energy K = p >m, we obtain -1U >m1ik Ae ikx e -ivt = 1U k >mae ikx e -ivt = 1p >m 1x, t = K 1x, t. If the particle can move in three dimenion, it momentum ha three component 1p x, p y, p z and it kinetic energy i K = p x m + p y m + p z m (41.1) Thee obervation, taken together, ugget that the correct generalization of the Schrödinger equation to three dimenion i - U m a 0 1x, y, z, t 0x + 0 1x, y, z, t 0y + 0 1x, y, z, t 0z b 0 1x, y, z, t + U1x, y, z 1x, y, z, t = iu 0t (general three-dimenional Schrödinger equation) (41.) The three-dimenional wave function 1x, y, z, t ha a imilar interpretation a in one dimenion. The wave function itelf i a complex quantity with both a real part and an imaginary part, but ƒ 1x, y, z, tƒ the quare of it abolute value, equal to the product of 1x, y, z, t and it complex conjugate * 1x, y, z, t i real and either poitive or zero at every point in pace. We interpret ƒ 1x, y, z, tƒ dv a the probability of finding the particle within a mall volume dv centered on the point 1x, y, z at time t, o ƒ 1x, y, z, tƒ i the probability ditribution function in three dimenion. The normalization condition on the wave function i that the probability that the particle i omewhere in pace i exactly 1. Hence the integral of ƒ 1x, y, z, tƒ over all pace mut equal 1: L ƒ 1x, y, z, tƒ dv = 1 (normalization condition in three dimenion) (41.3) If the wave function 1x, y, z, t repreent a tate of a definite energy E that i, a tationary tate we can write it a the product of a patial wave function c1x, y, z and a function of time e -iet>u : 1x, y, z, t = c1x, y, ze -iet>u (time-dependent wave function for a tate of definite energy) (41.4) (Compare thi to Eq. (40.1) for a one-dimenional tate of definite energy.) If we ubtitute Eq. (41.4) into Eq. (41.), the right-hand ide of the equation become iuc1x, y, z1-ie>ue -iet>u = Ec1x, y, ze -iet>u. We can then divide

46 1366 CHAPTER 41 Atomic Structure both ide by the factor, leaving the time-independent Schrödinger equation in three dimenion for a tationary tate: e -iet>u - U m a 0 c1x, y, z 0x + 0 c1x, y, z 0y + 0 c1x, y, z 0z b + U1x, y, zc1x, y, z = Ec1x, y, z (three-dimenional time-independent (41.5) Schrödinger equation) The probability ditribution function for a tationary tate i jut the quare of the abolute value of the patial wave function: ƒc1x, y, ze -iet>u ƒ = c * 1x, y, ze +iet>u c1x, y, ze -iet>u = ƒc1x, y, zƒ. Note that thi doen t depend on time. (A we dicued in Section 40.1, that why we call thee tate tationary.) Hence for a tationary tate the wave function normalization condition, Eq. (41.3), become L ƒc1x, y, zƒ dv = 1 (normalization condition for a tationary tate in three dimenion) (41.6) We won t pretend that we have derived Eq. (41.) and (41.5). Like their onedimenional verion, thee equation have to be teted by comparion of their prediction with experimental reult. Happily, Eq. (41.) and (41.5) both pa thi tet with flying color, o we are confident that they are the correct equation. An important topic that we will addre in thi chapter i the olution for Eq. (41.5) for the tationary tate of the hydrogen atom. The potential-energy function for an electron in a hydrogen atom i pherically ymmetric; it depend only on the ditance r = 1x + y + z 1> from the origin of coordinate. To take advantage of thi ymmetry, it bet to ue pherical coordinate rather than the Carteian coordinate 1x, y, z to olve the Schrödinger equation for the hydrogen atom. Before introducing thee new coordinate and invetigating the hydrogen atom, it ueful to look at the three-dimenional verion of the particle in a box that we conidered in Section 40.. Solving thi impler problem will give u inight into the more complicated tationary tate found in atomic phyic. Tet Your Undertanding of Section 41.1 In a certain region of pace the potential-energy function for a quantum-mechanical particle i zero. In thi region the wave function c1x, y, z for a certain tationary tate atifie 0 c>0x 7 0, 0 c>0y 7 0, and 0 c>0z 7 0. The particle ha a definite energy E that i poitive. What can you conclude about c1x, y, z in thi region? (i) It mut be poitive; (ii) it mut be negative; (iii) it mut be zero; (iv) not enough information given to decide A particle i confined in a cubical box with wall at x = 0, x = L, y = 0, y = L, z = 0, and z = L. z z 5 L x 5 L x O y 5 L y 41. Particle in a Three-Dimenional Box Conider a particle encloed within a cubical box of ide L. Thi could repreent an electron that free to move anywhere within the interior of a olid metal cube but cannot ecape the cube. We ll chooe the origin to be at one corner of the box, with the x-, y-, and z-axe along edge of the box. Then the particle i confined to the region 0 x L, 0 y L, 0 z L (Fig. 41.1). What are the tationary tate of thi ytem? A for the particle in a one-dimenional box that we conidered in Section 40., we ll ay that the potential energy i zero inide the box but infinite outide. Hence the patial wave function c1x, y, z mut be zero outide the box in order that the term U1x, y, zc1x, y, z in the time-independent Schrödinger equation, Eq. (41.5), not be infinite. Hence the probability ditribution function ƒc1x, y, zƒ i zero outide the box, and there i zero probability that the particle will be found

47 41. Particle in a Three-Dimenional Box 1367 there. Inide the box, the patial wave function for a tationary tate obey the time-independent Schrödinger equation, Eq. (41.5), with U1x, y, z = 0: - U m a 0 c1x, y, z 0x + 0 c1x, y, z 0y + 0 c1x, y, z 0z b = Ec1x, y, z (particle in a three-dimenional box) (41.7) In order for the wave function to be continuou from the inide to the outide of the box, c1x, y, z mut equal zero on the wall. Hence our boundary condition are that c1x, y, z = 0 at x = 0, x = L, y = 0, y = L, z = 0, and z = L. Gueing a olution to a complicated partial differential equation like Eq. (41.7) eem like quite a challenge. To make progre, recall that we wrote the timedependent wave function for a tationary tate a the product of one function that depend only on the patial coordinate x, y, and z and a econd function that depend only on the time t: 1x, y, z, t = c1x, y, ze -iet>u. In the ame way, let try a technique called eparation of variable: We ll write the patial wave function c1x, y, z a a product of one function X that depend only on x, a econd function Y that depend only on y, and a third function Z that depend only on z: c1x, y, z = X1xY1yZ1z If we ubtitute Eq. (41.8) into Eq. (41.7), we get (41.8) - U m ay1yz1zd X1x dx = EX1xY1yZ1z + X1xZ1z d Y1y dy + X1xY1y d Z1z dz b (41.9) The partial derivative in Eq. (41.7) have become ordinary derivative ince they act on function of a ingle variable. Now we divide both ide of Eq. (41.9) by the product X1xY1yZ1z: a - U 1 m X1x d X1x dx b + a - U m 1 Y1y d Y1y dy b + a - U 1 d Z1z m Z1z dz b = E (41.10) The right-hand ide of Eq. (41.10) i the energy of the tationary tate, which doe not and cannot depend on the value of x, y, or z. For thi to be true, the lefthand ide of the equation mut alo be independent of the value of x, y, and z. Hence the firt term in parenthee on the left-hand ide of Eq. (41.10) mut equal a contant that doen t depend on x, the econd term in parenthee mut equal another contant that doen t depend on y, and the third term in parenthee mut equal a third contant that doen t depend on z. Let call thee contant E X, E Y, and E Z, repectively. We then have a eparate equation for each of the three function X1x, Y1y, and Z1z: - U d X1x m dx - U d Y1y m dy - U d Z1z m dz = E X X1x = E Y Y1y = E Z Z1z (41.11a) (41.11b) (41.11c) To atify the boundary condition that c1x, y, z = X1xY1yZ1z be equal to zero on the wall of the box, we demand that X1x = 0 at x = 0 and x = L, Y1y = 0 at y = 0 and y = L, and Z1z = 0 at z = 0 and z = L.

48 1368 CHAPTER 41 Atomic Structure How can we interpret the three contant E X, E Y, and E Z in Eq. (41.11)? From Eq. (41.10), they are related to the energy E by (41.1) Equation (41.1) hould remind you of Eq. (41.1) in Section 41.1, which tate that the kinetic energy of a particle i the um of contribution coming from it x-, y-, and z-component of momentum. Hence the contant E X, E Y, and E Z tell u how much of the particle energy i due to motion along each of the three coordinate axe. (Inide the box the potential energy i zero, o the particle energy i purely kinetic.) Equation (41.11) repreent an enormou implification; we ve reduced the problem of olving a fairly complex partial differential equation with three independent variable to the much impler problem of olving three eparate ordinary differential equation with one independent variable each. What more, each of thee ordinary differential equation i jut the ame a the time-independent Schrödinger equation for a particle in a one-dimenional box, Eq. (40.5), and with exactly the ame boundary condition at 0 and L. (The only difference are that ome of the quantitie are labeled by different ymbol.) By comparing with our work in Section 40., you can ee that the olution to Eq. (41.11) are (41.13a) (41.13b) (41.13c) where C X, C Y, and C Z are contant. The correponding value of E X, E Y, and are E Z X nx 1x = C X in n Xpx L 1n X = 1,, 3, Á Y ny (y = C Y in n Y py L 1n Y = 1,, 3, Á Z nz 1z = C Z in n Zpz L 1n Z = 1,, 3, Á E X = n X p U ml 1n X = 1,, 3, Á E Y = n Y p U ml 1n Y = 1,, 3, Á E Z = n Z p U ml 1n Z = 1,, 3, Á (41.14a) (41.14b) (41.14c) There i only one quantum number n for the one-dimenional particle in a box, but three quantum number n X, n Y, and n Z for the three-dimenional box. If we ubtitute Eq. (41.13) back into Eq. (41.8) for the total patial wave function, c1x, y, z = X1xY1yZ1z, we get the following tationary-tate wave function for a particle in a three-dimenional cubical box: c nx,n Y,n Z 1x, y, z = Cin n X px L E X + E Y + E Z = E in n Y py in n Z pz L L 1n X = 1,, 3, Á ; n Y = 1,, 3, Á ; n Z = 1,, 3, Á (41.15) where C = C X C Y C Z. The value of the contant C i determined by the normalization condition, Eq. (41.6). In Section 40. we aw that the tationary-tate wave function for a particle in a one-dimenional box were analogou to tanding wave on a tring. In a imilar way, the three-dimenional wave function given by Eq. (41.15) are analogou to tanding electromagnetic wave in a cubical cavity like the interior of a microwave oven (ee Section 3.5). In a microwave oven there are dead pot where the wave intenity i zero, correponding to the node of the tanding wave. (The rotating platform in a microwave oven enure even cooking by making ure that no part of the food it at any dead pot. ) In a imilar fahion, the probability ditribution function correponding to Eq. (41.15) can have dead

49 41. Particle in a Three-Dimenional Box Probability ditribution function ƒ c nx,n for equal to (a) (, 1, 1), (b) (1,, 1), and (c) (1, 1, ). The value Y,n Z 1x, y, z ƒ 1n X, n Y, n Z of ƒ c ƒ i proportional to the denity of dot. The wave function i zero on the wall of the box and on the midplane of the box, o ƒ c ƒ = 0 at thee location. (a) c, 1, 1 z (b) c 1,, 1 z (c) c 1, 1, z x x x y y y pot where there i zero probability of finding the particle. A an example, conider the cae 1n X, n Y, n Z = 1, 1, 1. From Eq. (41.15), the probability ditribution function for thi cae i ƒc,1,1 1x, y, zƒ = ƒcƒ in px py pz in in L L L A Fig. 41.a how, the probability ditribution function i zero on the plane x = L>, where in 1px>L = in p = 0. The particle i mot likely to be found near where all three of the ine-quared function are greatet, at 1x, y, z = 1L>4, L>, L> or 1x, y, z = 13L>4, L>, L>. Figure 41.b and 41.c how the imilar cae 1n X, n Y, n Z = 11,, 1 and 1n X, n Y, n Z = 11, 1,. For higher value of the quantum number n X, n Y, and n Z there are additional plane on which the probability ditribution function equal zero, jut a the probability ditribution function ƒc1xƒ for a one-dimenional box ha more zero for higher value of n (ee Fig. 40.1). Example 41.1 Probability in a three-dimenional box (a) Find the value of the contant C that normalize the wave function of Eq. (41.15). (b) Find the probability that the particle will be found omewhere in the region 0 x L>4 (Fig. 41.3) for the cae (i) (n X, n Y, n Z = 11,, 1, (ii) (n X, n Y, n Z = 1, 1, 1, and (iii) 1n X, n Y, n Z ) = 13, 1, 1. SOLUTION IDENTIFY and SET UP: Equation (41.6) tell u that to normalize the wave function, we have to chooe the value of C o that the 41.3 What i the probability that the particle i in the darkcolored quarter of the box? x 5 L x z O z 5 L x 5 L/4 y 5 L y integral of the probability ditribution function ƒc nx, n Y, n Z 1x, y, zƒ over the volume within the box equal 1. (The integral i actually over all pace, but the particle-in-a-box wave function are zero outide the box.) The probability of finding the particle within a certain volume within the box equal the integral of the probability ditribution function over that volume. Hence in part (b) we ll integrate ƒc nx, n Y, n Z 1x, y, zƒ for the given value of 1n X, n Y, n Z over the volume 0 x L>4, 0 y L, 0 z L. EXECUTE: (a) From Eq. (41.15), ƒc nx,n Y,n Z 1x, y, zƒ = ƒcƒ in n Xpx in n Ypy n Zpz L L in L Hence the normalization condition i L ƒc n X,n Y,n Z 1x, y, zƒ dv = ƒcƒ L x=l x=0 Ly=0 Lz=0 x=l = ƒcƒ a in n y=l Xpx dxba in n Ypy dyb L x=0 L L y=0 L z=l * a in n Zpz dzb = 1 L L z=0 y=l z=l in n Xpx L n Ypy n Zpz in L in dxdydz L Continued

50 1370 CHAPTER 41 Atomic Structure We can ue the identity in u = cou and the variable ubtitution u = n X px>l to how that n Xpx L in L If we evaluate thi integral between x = 0 and x = L, the reult i L> (recall that in 0 = 0 and in n X p = 0 for any integer n X ). The y- and z-integral each yield the ame reult, o the normalization condition i or ƒcƒ = 1>L 3. If we chooe C to be real and poitive, then C = 1>L 3>. (b) We have the ame y- and z-integral a in part (a), but now the limit of integration on the x-integral are x = 0 and x = L>4: x=l>4 P = ƒc nx,n Y,n Z ƒ dv = ƒcƒ a in n Xpx dxb L0 x L>4 L L The x-integral i L x=l>4 x=0 dx = ƒcƒ a L bal bal b = ƒcƒ a L 3 b = 1 y=l * a in n z=l Ypy dyba in n Zpz dzb L L L L y=0 = x - L 4n X p ina n Xpx b L in n Xpx L L n X p c n Xpx L - 1 ina n Xpx bd L z=0 x=0 dx = a x - L 4n X p ina n Xpx L = L 8 - L 4n X p ina n Xp b x=l>4 bd x=0 Hence the probability of finding the particle omewhere in the region 0 x L>4 i Thi depend only on the value of, not on or. Hence for the three cae we have (i) : P = p ina p b = X = 1 p 11 (ii) n X = : (iii) n X = 3: P = a L b 3 a L 8 - = n X p ina n Xp b = p = P = p ina p b = p in p = = 0.50 P = p ina 3p b = p 1-1 = p = L 4n X p ina n Xp bbal bal b EVALUATE: You can ee why the probabilitie in part (b) are different by looking at part (b) of Fig. 40.1, which how in n X px>l for n X = 1,, and 3. For n X = the area under the curve between x = 0 and x = L>4 (equal to the integral between thee two 1 point) i exactly 4 of the total area between x = 0 and x = L. For 1 n X = 1 the area between x = 0 and x = L>4 i le than 4 of the 1 total area, and for n X = 3 it i greater than of the total area. n X 4 n Y n Z Energy Level, Degeneracy, and Symmetry From Eq. (41.1) and (41.14), the allowed energie for a particle of ma m in a cubical box of ide L are E nx,n Y,n Z = 1n X + n Y + n Z p U ml 1n X = 1,, 3, Á ; n Y = 1,, 3, Á ; n Z = 1,, 3, Á (41.16) (energy level, particle in a three-dimenional cubical box) Figure 41.4 how the ix lowet energy level given by Eq. (41.16). Note that mot energy level correpond to more than one et of quantum number 1n X, n Y, n Z and hence to more than one quantum tate. Having two or more ditinct quantum tate with the ame energy i called degeneracy, and tate with the ame energy are aid to be degenerate. For example, Fig how that the tate 1n X, n Y, n Z = 1, 1, 1, 11,, 1, and 11, 1, are degenerate. By comparion, for a particle in a one-dimenional box there i jut one tate for each energy level (ee Fig a) and no degeneracy. The reaon the cubical box exhibit degeneracy i that it i ymmetric: All ide of the box have the ame dimenion. A an illutration, Fig. 41. how the probability ditribution function for the three tate 1n X, n Y, n Z = 1, 1, 1, 11,, 1, and 11, 1,. You can tranform any one of thee three tate into a different one by imply rotating the cubical box by 90. Thi rotation doen t change the energy, o the three tate are degenerate.

51 41. Particle in a Three-Dimenional Box 1371 (3,, 1), (3, 1, ), (1, 3, ), (, 3, 1), (1,, 3), (, 1, 3) (,, ) (3, 1, 1), (1, 3, 1), (1, 1, 3) (,, 1), (, 1, ), (1,, ) (, 1, 1), (1,, 1), (1, 1, ) E 14 E 3 1, 1, 1 4E 1, 1, 1 11 E 3 1, 1, 1 3E 1, 1, 1 E 1, 1, Energy-level diagram for a particle in a three-dimenional cubical box. We label each level with the quantum number of the tate 1n X, n Y, n Z with that energy. Several of the level are degenerate (more than one tate ha the ame energy). The lowet (ground) level, 1n X, n Y, n Z = 11, 1, 1, ha energy E 1,1,1 = p U >ml = 3p U >ml ; we how the energie of the other level a multiple of E 1,1,1. (1, 1, 1) E 1, 1, 1 Since degeneracy i a conequence of ymmetry, we can remove the degeneracy by making the box aymmetric. We do thi by giving the three ide of the box different length L X, L Y, and L Z. If we repeat the tep that we followed to olve the time-independent Schrödinger equation, we find that the energy level are given by E nx,n Y,n Z = a n X L + n Y X L + n Z Y L b p U Z m 1n X = 1,, 3, Á ; n Y = 1,, 3, Á ; n Z = 1,, 3, Á (41.17) (energy level, particle in a three-dimenional box with ide of length,, and L Z ) L X L Y E 5 0 If L X, L Y, and L Z are all different, the tate 1n X, n Y, n Z = 1, 1, 1, 11,, 1, and 11, 1, have different energie and hence are no longer degenerate. Note that Eq. (41.17) reduce to Eq. (41.16) if the length are all the ame ( L X = L Y = L Z = L). Let ummarize the key difference between the three-dimenional particle in a box and the one-dimenional cae that we examined in Section 40.: We can write the wave function for a three-dimenional tationary tate a a product of three function, one for each patial coordinate. Only a ingle function of the coordinate x i needed in one dimenion. In the three-dimenional cae, three quantum number are needed to decribe each tationary tate. Only one quantum number i needed in the one-dimenional cae. Mot of the energy level for the three-dimenional cae are degenerate: More than one tationary tate ha thi energy. There i no degeneracy in the one-dimenional cae. For a tationary tate of the three-dimenional cae, there are urface on which the probability ditribution function ƒcƒ i zero. In the onedimenional cae there are poition on the x-axi where ƒcƒ i zero. We ll ee thee ame feature in the following ection a we examine a threedimenional ituation that more realitic than a particle in a box: a hydrogen atom in which a negatively charged electron orbit a poitively charged nucleu. Tet Your Undertanding of Section 41. Rank the following tate of a particle in a cubical box of ide L in order from highet to lowet energy: (i) 1n X, n Y, n Z = 1, 3, ; (ii) 1n X, n Y, n Z = 14, 1, 1; (iii) 1n X, n Y, n Z = 1,, 3; (iv) 1n X, n Y, n Z = 11, 3, 3.

52 137 CHAPTER 41 Atomic Structure 41.3 The Hydrogen Atom Let continue the dicuion of the hydrogen atom that we began in Chapter 39. In the Bohr model, electron move in circular orbit like Newtonian particle, but with quantized value of angular momentum. While thi model gave the correct energy level of the hydrogen atom, a deduced from pectra, it had many conceptual difficultie. It mixed claical phyic with new and eemingly contradictory concept. It provided no inight into the proce by which photon are emitted and aborbed. It could not be generalized to atom with more than one electron. It predicted the wrong magnetic propertie for the hydrogen atom. And perhap mot important, it picture of the electron a a localized point particle wa inconitent with the more general view we developed in Chapter 39 and 40. To go beyond the Bohr model, let apply the Schrödinger equation to find the wave function for tationary tate (tate of definite energy) of the hydrogen atom. A in Section 39.3, we include the motion of the nucleu by imply replacing the electron ma m with the reduced ma m r The Schrödinger equation for the hydrogen atom can be olved mot readily uing pherical coordinate. Nucleu, charge 1e, at the origin x z f u Electron, charge e, at coordinate (r, u, f) r y The Schrödinger Equation for the Hydrogen Atom We dicued the three-dimenional verion of the Schrödinger equation in Section The potential-energy function i pherically ymmetric: It depend only on the ditance r = 1x + y + z 1> from the origin of coordinate: (41.18) The hydrogen-atom problem i bet formulated in pherical coordinate 1r, u, f, hown in Fig. 41.5; the pherically ymmetric potential-energy function depend only on r, not on u or f. The Schrödinger equation with thi potential-energy function can be olved exactly; the olution are combination of familiar function. Without going into a lot of detail, we can decribe the mot important feature of the procedure and the reult. Firt, we find the olution uing the ame method of eparation of variable that we employed for a particle in a cubical box in Section 41.. We expre the wave function c1r, u, f a a product of three function, each one a function of only one of the three coordinate: (41.19) That i, the function R1r depend only on r, 1u depend only on u, and 1f depend only on f. Jut a for a particle in a three-dimenional box, when we ubtitute Eq. (41.19) into the Schrödinger equation, we get three eparate ordinary differential equation. One equation involve only r and R1r, a econd involve only u and 1u, and a third involve only f and 1f: - U m r r d dr 1 in u d du U1r = - 1 e 4pP 0 r c1r, u, f = R1r 1u 1f dr1r ar b + a U l1l + 1 dr m r r + U1rbR1r = ER1r d 1u) ain u b + al1l m l du in u b 1u = 0 d 1f df + m l 1f = 0 (41.0a) (41.0b) (41.0c) In Eq. (41.0) E i the energy of the tationary tate and l and m l are contant that we ll dicu later. (Be careful! Don t confue the contant m l with the reduced ma m r.) We won t attempt to olve thi et of three equation, but we can decribe how it done. A for the particle in a cubical box, the phyically acceptable olution of thee three equation are determined by boundary condition. The radial function R1r in Eq. (41.0a) mut approach zero at large r, becaue we are decribing

53 41.3 The Hydrogen Atom 1373 bound tate of the electron that are localized near the nucleu. Thi i analogou to the requirement that the harmonic-ocillator wave function (ee Section 40.5) mut approach zero at large x. The angular function 1u and 1f in Eq. (41.0b) and (41.0c) mut be finite for all relevant value of the angle. For example, there are olution of the equation that become infinite at u = 0 and u = p; thee are unacceptable, ince c1r, u, f mut be normalizable. Furthermore, the angular function 1f in Eq. (41.0c) mut be periodic. For example, 1r, u, f and 1r, u, f + p decribe the ame point, o 1f + p mut equal 1f. The allowed radial function R1r turn out to be an exponential function e -ar (where a i poitive) multiplied by a polynomial in r. The function 1u are polynomial containing variou power of in u and co u, and the function 1f are imply proportional to e imlf, where i = 1-1 and m l i an integer that may be poitive, zero, or negative. In the proce of finding olution that atify the boundary condition, we alo find the correponding energy level. We denote the energie of thee level [E in Eq. (41.0a)] by E n 1n = 1,, 3, Á. Thee turn out to be identical to thoe from the Bohr model, a given by Eq. (39.15), with the electron ret ma m replaced by the reduced ma m r. Rewriting that equation uing U=h>p, we have 1 m r e ev E n = - 14pP 0 n = - U n (energy level of hydrogen) (41.1) A in Section 39.3, we call n the principal quantum number for the level of energy E n. Equation (41.1) i an important validation of our Schrödinger-equation analyi of the hydrogen atom. The Schrödinger analyi i quite different from the Bohr model, both mathematically and conceptually, yet both yield the ame energy-level cheme a cheme that agree with the energie determined from pectra. A we will ee, the Schrödinger analyi can explain many more apect of the hydrogen atom than can the Bohr model. Quantization of Orbital Angular Momentum The olution to Eq. (41.0) that atify the boundary condition mentioned above alo have quantized value of orbital angular momentum. That i, only certain dicrete value of the magnitude and component of orbital angular momentum are permitted. In dicuing the Bohr model in Section 39.3, we mentioned that quantization of angular momentum wa a reult with little fundamental jutification. With the Schrödinger equation it appear automatically. The poible value of the magnitude L of orbital angular momentum L S are determined by the requirement that the 1u function in Eq. (41.0b) mut be finite at u = 0 and u = p. In a level with energy E n and principal quantum number n, the poible value of L are L = l1l + 1U 1l = 0, 1,, Á, n - 1 (magnitude of orbital angular momentum) (41.) The orbital angular-momentum quantum number l, which i the ame l that appear in Eq. (41.0a) and (41.0b), i called the orbital quantum number for hort. In the Bohr model, each energy level correponded to a ingle value of angular momentum. Equation (41.) how that in fact there are n different poible value of L for the nth energy level. An intereting feature of Eq. (41.) i that the orbital angular momentum i zero for l = 0 tate. Thi reult diagree with the Bohr model, in which the electron alway moved in a circle of definite radiu and L wa never zero. The l = 0 wave function c depend only on r; for thee tate, the function 1u and 1f are contant. Thu the wave function for l = 0 tate are pherically

54 1374 CHAPTER 41 Atomic Structure ymmetric. There i nothing in their probability ditribution ƒcƒ to favor one direction over any other, and there i no orbital angular momentum. The permitted value of the component of L S in a given direction, ay the z-component L z, are determined by the requirement that the 1f function mut equal 1f + p. The poible value of are L z L z = m l U 1m l = 0, 1,, Á, l (component of orbital angular momentum) (41.3) The quantum number m l i the ame a that in Eq. (41.0b) and (41.0c). We ee that m l can be zero or a poitive or negative integer up to, but no larger in magnitude than, l. That i, ƒm l ƒ l. For example, if l = 1, m l can equal 1, 0, or -1. For reaon that will emerge later, we call m l the orbital magnetic quantum number, or magnetic quantum number for hort. The component L z can never be quite a large a L (unle both are zero). For example, when l =, the larget poible value of m l i alo ; then Eq. (41.) and (41.3) give L = 1 + 1U = 6U =.45U 41.6 (a) When l =, the magnitude of the angular monentum vector L S i 16U =.45U, but L S doe not have a definite direction. In thi emiclaical vector picture, L S make an angle of 35.3 with the z-axi when the z-component ha it maximum value of U. (b) Thee cone how the poible direction of L S for different value of. L z (a) z S L z h L 6h L z h L z 0 L z h L z h l (b) z L z h L z L S L z h u L L z 0 L z h L z h L z = U Figure 41.6 how the ituation. The minimum value of the angle vector L S and the z-axi i given by u L = arcco L z L = arcco.45 = 35.3 between the That ƒl z ƒ i alway le than L i alo required by the uncertainty principle. Suppoe we could know the precie direction of the orbital angular momentum vector. Then we could let that be the direction of the z-axi, and L z would equal L. Thi correpond to a particle moving in the xy-plane only, in which cae the S z-component of the linear momentum p would be zero with no uncertainty p z. Then the uncertainty principle z p z ÚUrequire infinite uncertainty z in the coordinate z. Thi i impoible for a localized tate; we conclude that we can t know the direction of L S preciely. Thu, a we ve already tated, the component of L S in a given direction can never be quite a large a it magnitude L. Alo, if we can t know the direction of L S preciely, we can t determine the component L x and L preciely. Thu we how cone of poible direction for L S y in Fig. 41.6b. You may wonder why we have ingled out the z-axi for pecial attention. There no fundamental reaon for thi; the atom certainly doen t care what coordinate ytem we ue. The point i that we can t determine all three component of orbital angular momentum with certainty, o we arbitrarily pick one a the component we want to meaure. When we dicu interaction of the atom with a magnetic field, we will conitently chooe the poitive z-axi to be in the direction of B S. Quantum Number Notation The wave function for the hydrogen atom are determined by the value of three quantum number n, l, and m l. (Compare thi to the particle in a three-dimenional box that we conidered in Section 41.. There, too, three quantum number were needed to decribe each tationary tate.) The energy E n i determined by the principal quantum number n according to Eq. (41.1). The magnitude of orbital angular momentum i determined by the orbital quantum number l, a in Eq. (41.). The component of orbital angular momentum in a pecified axi direction (cutomarily the z-axi) i determined by the magnetic quantum number m l, a in Eq. (41.3). The energy doe not depend on the value of l or m l (Fig. 41.7), o for each energy level E n given by Eq. (41.1), there i more than one ditinct tate having the ame energy but different quantum number. That i, thee tate are degenerate, jut like mot of the tate of a particle in a u L

55 41.3 The Hydrogen Atom 1375 three-dimenional box. A for the three-dimenional box, degeneracy arie becaue the hydrogen atom i ymmetric: If you rotate the atom through any angle, the potential-energy function at a ditance r from the nucleu ha the ame value. State with variou value of the orbital quantum number l are often labeled with letter, according to the following cheme: l = 0: tate l = 1: p tate l = : d tate l = 3: ƒ tate l = 4: g tate l = 5: h tate and o on alphabetically. Thi eemingly irrational choice of the letter, p, d, and ƒ originated in the early day of pectrocopy and ha no fundamental ignificance. In an important form of pectrocopic notation that we ll ue often, a tate with n = and l = 1 i called a p tate; a tate with n = 4 and l = 0 i a 4 tate; and o on. Only tate 1l = 0 are pherically ymmetric. Here another bit of notation. The radial extent of the wave function increae with the principal quantum number n, and we can peak of a region of pace aociated with a particular value of n a a hell. Epecially in dicuion of many-electron atom, thee hell are denoted by capital letter: n = 1: K hell n = : L hell n = 3: M hell n = 4: N hell and o on alphabetically. For each n, different value of l correpond to different ubhell. For example, the L hell 1n = contain the and p ubhell. Table 41.1 how ome of the poible combination of the quantum number n, l, and m l for hydrogen-atom wave function. The pectrocopic notation and the hell notation for each are alo hown The energy for an orbiting atellite uch a the Hubble Space Telecope depend on the average ditance between the atellite and the center of the earth. It doe not depend on whether the orbit i circular (with a large orbital angular momentum L) or elliptical (in which cae L i maller). In the ame way, the energy of a hydrogen atom doe not depend on the orbital angular momentum. Table 41.1 Quantum State of the Hydrogen Atom n l m l Spectrocopic Notation Shell K , 0, 1 p V L , 0, 1 3p v M 3 -,-1, 0, 1, 3d N and o on Problem-Solving Strategy 41.1 Atomic Structure IDENTIFY the relevant concept: Many problem in atomic tructure can be olved imply by reference to the quantum number n, l, and m l that decribe the total energy E, the magnitude of the orbital angular momentum L S, the z-component of L S, and other propertie of an atom. SET UP the problem: Identify the target variable and chooe the appropriate equation, which may include Eq. (41.1), (41.), and (41.3). EXECUTE the olution a follow: 1. Be ure you undertand the poible value of the quantum number n, l, and m l for the hydrogen atom. They are all integer; n i alway greater than zero, l can be zero or poitive up to n - 1, and m l can range from -l to l. You hould know how to count the number of (n, l, m l ) tate in each hell ( K, L, M, and o on) and ubhell (3, 3p, 3d, and o on). Be able to contruct Table 41.1, not jut to write it from memory.. Solve for the target variable. EVALUATE your anwer: It help to be familiar with typical magnitude in atomic phyic. For example, the electric potential energy of a proton and electron 0.10 nm apart (typical of atomic dimenion) i about -15 ev. Viible light ha wavelength around 500 nm and frequencie around 5 * Hz. Problem-Solving Strategy 39.1 (Section 39.1) give other typical magnitude.

56 1376 CHAPTER 41 Atomic Structure Example 41. Counting hydrogen tate How many ditinct 1n, l, m l tate of the hydrogen atom with 1n, l, m l tate with n = 3 i therefore = 9. (In n = 3 are there? What are their energie? Section 41.5 we ll find that the total number of n = 3 tate i in fact twice thi, or 18, becaue of electron pin.) SOLUTION IDENTIFY and SET UP: Thi problem ue the relationhip among The energy of a hydrogen-atom tate depend only on n, o all 9 of thee tate have the ame energy. From Eq. (41.1), the principal quantum number n, orbital quantum number l, magnetic quantum number m l, and energy of a tate for the hydrogen E 3 = ev atom. We ue the rule that l can have n integer value, from 0 to 3 = ev n - 1, and that m l can have l + 1 value, from -l to l. Equation EVALUATE: For a given value of n, the total number of 1n, l, m (41.1) give the energy of any particular tate. l tate turn out to be n. In thi cae n = 3 and there are 3 = 9 EXECUTE: When n = 3, l can be 0, 1, or. When l = 0, m l can be only 0 (1 tate). When l = 1, m l can be -1, 0, or 1 (3 tate). When tate. Remember that the ground level of hydrogen ha n = 1 and E 1 = ev; the n = 3 excited tate have a higher (le negative) l =, can be -, -1, 0, 1, or (5 tate). The total number of energy. m l Example 41.3 Angular momentum in an excited level of hydrogen Conider the n = 4 tate of hydrogen. (a) What i the maximum magnitude L of the orbital angular momentum? (b) What i the maximum value of L (c) What i the minimum angle between L S z? and the z-axi? Give your anwer to part (a) and (b) in term of U. SOLUTION IDENTIFY and SET UP: We again need to relate the principal quantum number n and the orbital quantum number l for a hydrogen atom. We alo need to relate the value of l and the magnitude and poible direction of the orbital angular momentum vector. We ll ue Eq. (41.) in part (a) to determine the maximum value of L; then we ll ue Eq. (41.3) in part (b) to determine the maximum value of L The angle between L S z. and the z-axi i minimum when L i maximum (o that L S z i mot nearly aligned with the poitive z-axi). EXECUTE: (a) When n = 4, the maximum value of the orbital angular-momentum quantum number l i 1n - 1 = 14-1 = 3; from Eq. (41.), L max = U =1U =3.464U (b) For l = 3 the maximum value of i 3. From Eq. (41.3), 1L z max (c) The minimum allowed angle between and the z-axi correpond to the maximum allowed value of L z and m l (Fig. 41.6b how an l = example). For the tate with l = 3 and m l = 3, u min = arcco 1L z max L = 3U 3U = arcco 3.464U = 30.0 EVALUATE: A a check, you can verify that u i greater than 30.0 for all tate with maller value of l. m l L S Electron Probability Ditribution Rather than picturing the electron a a point particle moving in a precie circle, the Schrödinger equation give an electron probability ditribution urrounding the nucleu. The hydrogen-atom probability ditribution are three-dimenional, o they are harder to viualize than the two-dimenional circular orbit of the Bohr model. It helpful to look at the radial probability ditribution function P1r that i, the probability per radial length for the electron to be found at variou ditance from the proton. From Section 41.1 the probability for finding the electron in a mall volume element dv i ƒcƒ dv. (We aume that c i normalized in accordance with Eq. (41.6) that i, that the integral of ƒcƒ dv over all pace equal unity o that there i 100% probability of finding the electron omewhere in the univere.) Let take a our volume element a thin pherical hell with inner radiu r and outer radiu r + dr. The volume dv of thi hell i approximately it area 4pr multiplied by it thickne dr: dv = 4pr dr (41.4)

57 41.3 The Hydrogen Atom 1377 We denote by P1r dr the probability of finding the particle within the radial range dr; then, uing Eq. (41.4), P1r) dr = ƒcƒ dv = ƒcƒ 4pr dr (probability that the electron i between r and r + dr (41.5) For wave function that depend on u and f a well a r, we ue the value of ƒcƒ averaged over all angle in Eq. (41.5). Figure 41.8 how graph of P1r for everal hydrogen-atom wave function. The r cale are labeled in multiple of a, the mallet ditance between the electron and the nucleu in the Bohr model: a = P 0h pm r e = 4pP 0U m r e = 5.9 * m (mallet r, Bohr model) (41.6) Jut a for a particle in a cubical box (ee Section 41.), there are ome location where the probability i zero. Thee urface are plane for a particle in a box; for a hydrogen atom thee are pherical urface (that i, urface of contant r). But again, the uncertainty principle tell u not to worry; we can t localize the electron exactly anyway. Note that for the tate having the larget poible l for each n (uch a 1, p, 3d, and 4ƒ tate), P1r ha a ingle maximum at n a. For thee tate, the electron i mot likely to be found at the ditance from the nucleu that i predicted by the Bohr model, r = n a. Figure 41.8 how radial probability ditribution function P1r = 4pr ƒcƒ, which indicate the relative probability of finding the electron within a thin pherical hell of radiu r. By contrat, Fig and how the three-dimenional probability ditribution function ƒcƒ, which indicate the relative probability of finding the electron within a mall box at a given poition. The darker the blue cloud, the greater the value of ƒcƒ. (Thee are imilar to the cloud hown in Fig. 41..) Figure 41.9 how cro ection of the pherically ymmetric probability cloud for the lowet three ubhell, for which ƒcƒ depend only on the radial coordinate r. Figure how cro ection of the cloud for other electron tate for which ƒcƒ depend on both r and u. For thee tate the probability ditribution function i zero for certain value of u a well a for certain value of r. In any tationary tate of the hydrogen atom, ƒcƒ i independent of f Radial probability ditribution function P1r for everal hydrogen-atom wave function, plotted a function of the ratio r>a [ee Eq. (41.6)]. For each function, the number of maxima i 1n - l. The curve for which l = n , p, 3d, Á have only one maximum, located at r = n a. 0.6 P(r) State with l r/a P(r) 0.0 p State with l p 4p 0.04 r/a P(r) State with l 5 or l d f 4d 0.04 r/a Three-dimenional probability ditribution function ƒ c ƒ for the pherically ymmetric 1,, and 3 hydrogen-atom wave function. c c c 0 1 r/a 0 r/a 0 3 r/a 1 3 r r r

58 1378 CHAPTER 41 Atomic Structure Cro ection of three-dimenional probability ditribution for a few quantum tate of the hydrogen atom. They are not to the ame cale. Mentally rotate each drawing about the z-axi to obtain the three-dimenional repreentation of ƒ c ƒ. For example, the p, m l = ;1 probability ditribution look like a fuzzy donut. z z z z 1, m l 0, m l 0 p, m l 1 p, m l 0 z z z z z 3p, m l 0 3p, m l 1 3d, m l 0 3d, m l 1 3d, m l Example 41.4 A hydrogen wave function The ground-tate wave function for hydrogen (a 1 tate) i c 1 1r = 1 pa 3 e-r>a (a) Verify that thi function i normalized. (b) What i the probability that the electron will be found at a ditance le than a from the nucleu? SOLUTION IDENTIFY and SET UP: Thi example i imilar to Example 41.1 in Section 41.. We need to how that thi wave function atifie the condition that the probability of finding the electron omewhere i 1. We then need to find the probability that it will be found in the region r 6 a. In part (a) we ll carry out the integral 1 ƒcƒ dv over all pace; if it i equal to 1, the wave function i normalized. In part (b) we ll carry out the ame integral over a pherical volume that extend from the origin (the nucleu) out to a ditance a from the nucleu. EXECUTE: (a) Since the wave function depend only on the radial coordinate r, we can chooe our volume element to be pherical hell of radiu r, thickne dr, and volume dv given by Eq. (41.4). We then have q ƒc 1 ƒ 1 dv = Lall pace L 0 pa 3 e-r>a 14pr dr = 4 a 3 L0 q r e -r>a dr You can find the following indefinite integral in a table of integral or by integrating by part: Evaluating thi between the limit r = 0 and r = q i imple; it i zero at r = q becaue of the exponential factor, and at r = 0 only the lat term in the parenthee urvive. Thu the value of the definite integral i a 3 >4. Putting it all together, we find L0 L r e -r>a dr = a - ar - a r - a3 4 be-r>a q ƒc 1 ƒ dv = 4 a 3 L0 q r e -r>a dr = 4 a 3 a 3 4 = 1 The wave function i normalized. (b) To find the probability P that the electron i found within r 6 a, we carry out the ame integration but with the limit 0 and a. We ll leave the detail to you (Exercie 41.15). From the upper limit we get -5e - a 3 >4; the final reult i a P = ƒc 1 ƒ 4pr dr = 4 L 0 a 3 a - 5a3 e - + a3 4 4 b = (-5e - + 1) = 1-5e - = 0.33 EVALUATE: Our reult tell u that in a ground tate we expect to 1 find the electron at a ditance from the nucleu le than a about 3 of the time and at a greater ditance about 3 of the time. It hard to tell, but in Fig. 41.8, about 3 of the area under the 1 curve i at ditance greater than a (that i, r>a 7 1). Hydrogenlike Atom Two generalization that we dicued with the Bohr model in Section 39.3 are equally valid in the Schrödinger analyi. Firt, if the atom i not compoed of a ingle proton and a ingle electron, uing the reduced ma m r of the ytem in Eq. (41.1) and (41.6) will lead to change that are ubtantial for ome exotic

59 41.4 The Zeeman Effect 1379 ytem. One example i poitronium, in which a poitron and an electron orbit each other; another i a muonic atom, in which the electron i replaced by an untable particle called a muon that ha the ame charge a an electron but i 07 time more maive. Second, our analyi i applicable to ingle-electron ion, uch a He +, Li +, and o on. For uch ion we replace e by Ze in Eq. (41.1) and (41.6), where Z i the number of proton (the atomic number). Tet Your Undertanding of Section 41.3 Rank the following tate of the hydrogen atom in order from highet to lowet probability of finding the electron in the vicinity of r = 5a: (i) n = 1, l = 0, m l = 0; (ii) n =, l = 1, m l = +1; (iii) n =, l = 1, m l = The Zeeman Effect The Zeeman effect i the plitting of atomic energy level and the aociated pectral line when the atom are placed in a magnetic field (Fig ). Thi effect confirm experimentally the quantization of angular momentum. The dicuion in thi ection, which aume that the only angular momentum i the orbital angular momentum of a ingle electron, alo how why we call m l the magnetic quantum number. Atom contain charge in motion, o it hould not be urpriing that magnetic force caue change in that motion and in the energy level. A early a the middle of the 19th century, phyicit peculated that the ource of viible light might be vibrating electric charge on an atomic cale. In 1896 the Dutch phyicit Pieter Zeeman wa the firt to how that in the preence of a magnetic field, ome pectral line were plit into group of cloely paced line (Fig. 41.1). Thi effect now bear hi name. Magnetic Moment of an Orbiting Electron Let begin our analyi of the Zeeman effect by reviewing the concept of magnetic dipole moment or magnetic moment, introduced in Section 7.7. A plane current loop with vector area A S carrying current I ha a magnetic moment M S given by M S IA S (41.7) When a magnetic dipole of moment i placed in a magnetic field B S, the field exert a torque S T M S : B S on the dipole. The potential energy U aociated with thi interaction i given by Eq. (7.7): U = -M # S S B (41.8) Now let ue Eq. (41.7) and (41.8) and the Bohr model to look at the interaction of a hydrogen atom with a magnetic field. The orbiting electron with peed v i equivalent to a current loop with radiu r and area pr. The average current I i the average charge per unit time that pae a given point of the orbit. Thi i equal to the charge magnitude e divided by the time T for one revolution, given by T = pr>v. Thu I = ev>pr, and from Eq. (41.7) the magnitude m of the magnetic moment i m = IA = (41.9) We can alo expre thi in term of the magnitude L of the orbital angular momentum. From Eq. (10.8) the angular momentum of a particle in a circular orbit i L = mvr, o Eq. (41.9) become m = M S ev pr pr = evr e m L (41.30) Magnetic effect on the pectrum of unlight. (a) The lit of a pectrograph i poitioned along the black line croing a portion of a unpot. (b) The 0.4-T magnetic field in the unpot (a thouand time greater than the earth field) plit the middle pectral line into three line. (a) Slit poitioned along thi line (b) No S B field With S B field Spectral line plit into three Sunpot 41.1 The normal Zeeman effect. Compare thi to the magnetic plitting in the olar pectrum hown in Fig b. When an excited ga i placed in a magnetic field, the interaction of orbital magnetic moment with the field plit individual pectral line of the ga into et of three line.

60 1380 CHAPTER 41 Atomic Structure The ratio of the magnitude of to the magnitude of i m>l = e>m and i called the gyromagnetic ratio. In the Bohr model, L = nh>p = nu, where n = 1,, Á. For an n = 1 tate (a ground tate), Eq. (41.30) become m = 1e>mU. Thi quantity i a natural unit for magnetic moment; it i called one Bohr magneton, denoted by m B : m B = eu m Evaluating Eq. (41.31) give (41.31) Note that the unit and A # m J>T are equivalent. We defined thi quantity previouly in Section 8.8. While the Bohr model ugget that the orbital motion of an atomic electron give rie to a magnetic moment, thi model doe not give correct prediction about magnetic interaction. A an example, the Bohr model predict that an electron in a hydrogen-atom ground tate ha an orbital magnetic moment of magnitude m B. But the Schrödinger picture tell u that uch a ground-tate electron i in an tate with zero angular momentum, o the orbital magnetic moment mut be zero! To get the correct reult, we mut decribe the tate by uing Schrödinger wave function. It turn out that in the Schrödinger formulation, electron have the ame ratio of m to L (gyromagnetic ratio) a in the Bohr model namely, e>m. Suppoe the magnetic field B S i directed along the +z-axi. From Eq. (41.8) the interaction energy U of the atom magnetic moment with the field i (41.3) where m i the of the vector M S z z-component. Now we ue Eq. (41.30) to find m z, recalling that e i the magnitude of the electron charge and that the actual charge i -e. Becaue the electron charge i negative, the orbital angular momentum and magnetic moment vector are oppoite. We find (41.33) For the Schrödinger wave function, L z = m l U, with m l = 0, 1,, Á, l, o Eq. (41.33) become m z = - e m L z = -m l eu m (41.34) CAUTION Two ue of the ymbol m Be careful not to confue the electron ma m with the magnetic quantum number. m l M S (definition of the Bohr magneton) m B = * 10-5 ev>t = 9.74 * 10-4 J>T or A # m U = -m z B m z = - e m L z Finally, we can expre the interaction energy, Eq. (41.3), a U = -m z B = m l eu m B (m l = 0, 1,, Á, l (orbital magnetic interaction energy) (41.35) In term of the Bohr magneton m B = eu>m, we can write Eq. (41.35) a L S U = m l m B B (orbital magnetic interaction energy) (41.36) The magnetic field hift the energy of each orbital tate by an amount U. The interaction energy U depend on the value of becaue determine the m l m l

61 41.4 The Zeeman Effect 1381 orientation of the orbital magnetic moment relative to the magnetic field. Thi dependence i the reaon m l i called the magnetic quantum number. The value of m l range from -l to +l in tep of one, o an energy level with a particular value of the orbital quantum number l contain 1l + 1 different orbital tate. Without a magnetic field thee tate all have the ame energy; that i, they are degenerate. The magnetic field remove thi degeneracy. In the preence of a magnetic field they are plit into l + 1 ditinct energy level; adjacent level differ in energy by 1eU>mB = m B B. We can undertand thi in term of the connection between degeneracy and ymmetry. With a magnetic field applied along the z-axi, the atom i no longer completely ymmetric under rotation: There i a preferred direction in pace. By removing the ymmetry, we remove the degeneracy of tate. Figure how the effect on the energy level of hydrogen. Spectral line correponding to tranition from one et of level to another et are correpondingly plit and appear a a erie of three cloely paced pectral line replacing a ingle line. A the following example how, the plitting of pectral line i quite mall becaue the value of m B B i mall even for ubtantial magnetic field Thi energy-level diagram for hydrogen how how the level are plit when the electron orbital magnetic moment interact with an external magnetic field. The value of m l are hown adjacent to the variou level. The relative magnitude of the level plitting are exaggerated for clarity. The n = 4 plitting are not hown; can you draw them in? n 5 4 n 5 3 n 5 m l m l m l l l 5 E ev 1.51 ev 3.40 ev n l ev Example 41.5 An atom in a magnetic field An atom in a tate with l = 1 emit a photon with wavelength nm a it decay to a tate with l = 0. If the atom i placed in a magnetic field with magnitude B =.00 T, what are the hift in the energy level and in the wavelength that reult from the interaction between the atom orbital magnetic moment and the magnetic field? SOLUTION IDENTIFY and SET UP: Thi problem concern the plitting of atomic energy level by a magnetic field (the Zeeman effect). We ue Eq. (41.35) or (41.36) to determine the energy-level hift. The relationhip E = hc>l between the energy and wavelength of a photon then let u calculate the wavelength emitted during tranition from the l = 1 tate to the l = 0 tate. EXECUTE: The energy of a 600-nm photon i E = hc * ev # * 10 8 m> = l 600 * 10-9 m =.07 ev If there i no external magnetic field, that i the difference in energy between the l = 0 and l = 1 level. With a.00-t field preent, Eq. (41.36) how that there i no hift of the l = 0 tate (which ha m l = 0). For the l = 1 tate, the plitting of level i given by U = m l m B B = m l * 10-5 ev>t1.00 T = m l * 10-4 ev = m l * 10-3 J The poible value of m l for l = 1 are -1, 0, and +1, and the three correponding level are eparated by equal interval of Continued

62 138 CHAPTER 41 Atomic Structure 1.16 * 10-4 ev. Thi i a mall fraction of the.07-ev photon The correponding wavelength hift are approximately * energy: nm = nm. The original nm line i plit E E = 1.16 * into a triplet with wavelength , , and nm ev = 5.60 * ev The poible value of m l for l 1 are -1, 0, and +1, and the three correponding level are eparated by equal interval of 1.16 * 10-4 ev. Thi i a mall fraction of the.07-ev photon energy: 1.16 * 10-4 ev.07 ev = 5.60 * 10-5 EVALUATE: Even though.00 T would be a trong field in mot laboratorie, the wavelength plitting are extremely mall. Nonethele, modern pectrograph have more than enough chromatic reolving power to meaure thee plitting (ee Section 36.5) Thi figure how how the plitting of the energy level of a d tate (l = ) depend on the magnitude B of an external magnetic field, auming only an orbital magnetic moment. E E d B 5 0 B increaing DE 1m B B 1m B B 0 m B B m B B Selection Rule Figure how what happen to a et of d tate 1l = a the magnetic field increae. With zero field the five tate m l = -, -1, 0, 1, and are degenerate (have the ame energy), but the applied field pread the tate out. Figure how the plitting of both the 3d and p tate. Equal energy difference 1eU>mB = m B B eparate adjacent level. In the abence of a magnetic field, a tranition from a 3d to a p tate would yield a ingle pectral line with photon energy E i - E f. With the level plit a hown, it might eem that there are five poible photon energie. In fact, there are only three poibilitie. Not all combination of initial and final level are poible becaue of a retriction aociated with conervation of angular momentum. The photon ordinarily carrie off one unit 1U of angular momentum, which lead to the requirement that in a tranition l mut change by 1 and m l mut change by 0 or 1. Thee requirement are called election rule. Tranition that obey thee rule are called allowed tranition; thoe that don t are forbidden tranition. In Fig we how the allowed tranition by olid arrow. You hould count the poible tranition energie to convince yourelf that the nine olid arrow give only three poible energie; the zero-field value E i - E f, and that value plu or minu E = 1eU>mB = m B B. Figure 41.1 how the correponding pectral line. What we have decribed i called the normal Zeeman effect. It i baed entirely on the orbital angular momentum of the electron. However, it leave out a very important conideration: the electron pin angular momentum, the ubject of the next ection The caue of the normal Zeeman effect. The magnetic field plit the level, but election rule allow tranition with only three different energy change, giving three different photon frequencie and wavelength. 3d Solid line: allowed tranition Dahed line: forbidden tranition m l DE 1 DE 0 DE 1 DE l 5 E i E f p l 5 1 DE DE m l Tet Your Undertanding of Section 41.4 In thi ection we aumed that the magnetic field point in the poitive z-direction. Would the reult be different if the magnetic field pointed in the poitive x-direction?

63 41.5 Electron Spin Electron Spin Depite the ucce of the Schrödinger equation in predicting the energy level of the hydrogen atom, experimental obervation indicate that it doen t tell the whole tory of the behavior of electron in atom. Firt, pectrocopit have found magnetic-field plitting into other than the three line we ve explained, ometime unequally paced. Before thi effect wa undertood, it wa called the anomalou Zeeman effect to ditinguih it from the normal effect dicued in the preceding ection. Figure how both kind of plitting. Second, ome energy level how plitting that reemble the Zeeman effect even when there i no external magnetic field. For example, when the line in the hydrogen pectrum are examined with a high-reolution pectrograph, ome line are found to conit of et of cloely paced line called multiplet. Similarly, the orange-yellow line of odium, correponding to the tranition 4p S 3 of the outer electron, i found to be a doublet 1l = 589.0, nm, uggeting that the 4p level might in fact be two cloely paced level. The Schrödinger equation in it original form didn t predict any of thi Illutration of the normal and anomalou Zeeman effect for two element, zinc and odium. The bracket under each illutration how the normal plitting predicted by neglecting the effect of electron pin. Zinc inglet Sodium principal doublet Zinc harp triplet (only one of three pattern hown) No S B field No S B field D D 1 With S B field With S B field Predicted normal plitting Normal pattern: Experiment agree with predicted normal plitting. Predicted normal plitting Predicted normal plitting Predicted normal plitting Anomalou pattern: Experiment doe not agree with predicted normal plitting. The Stern Gerlach Experiment Similar anomalie appeared in 19 in atomic-beam experiment performed in Germany by Otto Stern and Walter Gerlach. When they paed a beam of neutral atom through a nonuniform magnetic field (Fig ), atom were deflected PhET: Stern Gerlach Experiment 1 A beam of atom i directed parallel to the y-axi The Stern Gerlach experiment. N z Oven Slit Specially haped magnet pole produce a trongly nonuniform magnetic field that exert a net force on the magnetic moment of the atom. Magnet S Gla plate detector 3 Each atom i deflected upward or downward according to the orientation of it magnetic moment. x y

64 1384 CHAPTER 41 Atomic Structure according to the orientation of their magnetic moment with repect to the field. Thee experiment demontrated the quantization of angular momentum in a very direct way. If there were only orbital angular momentum, the deflection would plit the beam into an odd number 1l + 1 of different component. However, ome atomic beam were plit into an even number of component. If we ue a different ymbol j for an angular momentum quantum number, etting j + 1 equal to an even number give j = 1 3 5,,, Á, uggeting a half-integer angular momentum. Thi can t be undertood on the bai of the Bohr model and imilar picture of atomic tructure. In 195 two graduate tudent in the Netherland, Samuel Goudmidt and George Uhlenbeck, propoed that the electron might have ome additional motion. Uing a emiclaical model, they uggeted that the electron might behave like a pinning phere of charge intead of a particle. If o, it would have an additional pin angular momentum and magnetic moment. If thee were quantized in much the ame way a orbital angular momentum and magnetic moment, they might help to explain the oberved energy-level anomalie. An Analogy for Electron Spin To introduce the concept of electron pin, let tart with an analogy. The earth travel in a nearly circular orbit around the un, and at the ame time it rotate on it axi. Each motion ha it aociated angular momentum, which we call the orbital and pin angular momentum, repectively. The total angular momentum of the earth i the vector um of the two. If we were to model the earth a a ingle point, it would have no moment of inertia about it pin axi and thu no pin angular momentum. But when our model include the finite ize of the earth, pin angular momentum become poible. In the Bohr model, uppoe the electron i not jut a point charge but a mall pinning phere moving in orbit. Then the electron ha not only orbital angular momentum but alo pin angular momentum aociated with the rotation of it ma about it axi. The phere carrie an electric charge, o the pinning motion lead to current loop and to a magnetic moment, a we dicued in Section 7.7. In a magnetic field, the pin magnetic moment ha an interaction energy in addition to that of the orbital magnetic moment (the normal Zeeman-effect interaction that we dicued in Section 41.4). We hould ee additional Zeeman hift due to the pin magnetic moment. A we mentioned, uch hift are indeed oberved in precie pectrocopic analyi. Thi and a variety of other experimental evidence have hown concluively that the electron doe have a pin angular momentum and a pin magnetic moment that do not depend on it orbital motion but are intrinic to the electron itelf. The origin of thi pin angular momentum i fundamentally quantummechanical, o it not correct to model the electron a a pinning charged phere. But jut a the Bohr model can be a ueful conceptual picture for the motion of an electron in an atom, the pinning-phere analogy can help you viualize the intrinic pin angular momentum of an electron. Spin Quantum Number Like orbital angular momentum, the pin angular momentum of an electron S (denoted by ) i found to be quantized. Suppoe we have an apparatu that S meaure a particular component of, ay the z-component S z. We find that the only poible value are S z = 1 U (component of pin angular momentum) (41.37) Thi relationhip i reminicent of the expreion L z = m l U for the z-component of orbital angular momentum, except that ƒs z ƒ i one-half of U intead of an

65 41.5 Electron Spin 1385 integer multiple. Equation (41.37) alo ugget that the magnitude S of the pin angular momentum i given by an expreion analogou to Eq. (41.) with the orbital quantum number l replaced by the pin quantum number = 1 : S = 1 A 1 + 1BU =3 4 U (magnitude of pin angular momentum) (41.38) The electron i often called a pin - 1 particle. To viualize the quantized pin of an electron in a hydrogen atom, think of the electron probability ditribution function ƒcƒ a a cloud urrounding the nucleu like thoe hown in Fig and Then imagine many tiny pin arrow ditributed throughout the cloud, either all with component in the +z-direction or all with component in the -z-direction. But don t take thi picture too eriouly. To label completely the tate of the electron in a hydrogen atom, we now need a fourth quantum number m to pecify the electron pin orientation. In analogy to the orbital magnetic quantum number m l, we call m the pin magnetic quantum number. For an electron we give m the value + 1 or - 1 to agree with Eq. (41.37): S z = m U Am = 1 B (allowed value of m and S z for an electron) S (41.39) The pin angular momentum vector can have only two orientation in pace relative to the z-axi: pin up with a z-component of + 1 and pin down with a z-component of - 1 U. U The z-component of the aociated pin magnetic moment 1m z turn out to be related to S z by m z = e (41.40) m S z where -e and m are (a uual) the charge and ma of the electron. When the atom i placed in a magnetic field, the interaction energy -M S S # B of the pin magnetic dipole moment with the field caue further plitting in energy level and in the correponding pectral line. Equation (41.40) how that the gyromagnetic ratio for electron pin i approximately twice a great a the value e>m for orbital angular momentum and magnetic dipole moment. Thi reult ha no claical analog. But in 198 Paul Dirac developed a relativitic generalization of the Schrödinger equation for electron. Hi equation gave a pin gyromagnetic ratio of exactly 1e>m. It took another two decade to develop the area of phyic called quantum electrodynamic, abbreviated QED, which predict the value we ve given to only ix ignificant figure a.003. QED now predict a value that agree with a recent (006) meaurement of (15), making QED the mot precie theory in all cience. Example 41.6 Energy of electron pin in a magnetic field Calculate the interaction energy for an electron in an l = 0 tate in a magnetic field with magnitude.00 T. SOLUTION IDENTIFY and SET UP: For l 0 the electron ha zero orbital angular momentum and zero orbital magnetic moment. Hence the only magnetic interaction i that between the B S field and the pin magnetic moment M S. From Eq. (41.8), the interaction energy i U = -M S S # B. A in Section 41.4, we take B S to be in the poitive z-direction o that U = -m z B [Eq. (41.3)]. Equation (41.40) give in term of, and Eq. (41.37) give. m z S z S z EXECUTE: Combining Eq. (41.37) and (41.40), we have m z = a e m b A 1 UB = 1 eu 1.003a m b = m B = * 10-4 J>T = 9.85 * 10-4 J>T = * 10-5 ev>t Continued

66 1386 CHAPTER 41 Atomic Structure Then from Eq. (41.3), U = -m z B = * 10-4 J>T1.00 T = 1.86 * 10-3 J = 1.16 * 10-4 ev The poitive value of U and the negative value of m z correpond to S z = + 1 (pin up); the negative value of and the poitive value of m U U z correpond to S z = - 1 U (pin down). EVALUATE: Let check the ign of our reult. If the electron i pin S down, point generally oppoite to B S. Then the magnetic S moment M S (which i oppoite to becaue the electron charge i negative) point generally parallel to B S, and m i poitive. From Eq. (41.8), U = -M S S z # B, the interaction energy i negative if M S and B S are parallel. Our reult how that U i indeed negative in thi cae. We can imilarly confirm that U mut be poitive and m z negative for a pin-up electron. The red line in Fig how how the interaction energie for the two pin tate vary with the magnetic field magnitude B. The graph are traight line becaue, from Eq. (41.3), U i proportional to B An l = 0 level of a ingle electron i plit by interaction of the pin magnetic moment with an external magnetic field. The greater the magnitude B of the magnetic field, the greater the plitting. The quantity * 10-5 ev>t i jut m B. E E Spin up m 51 1 B 5 0 Spin down m 5 1 B increaing E 1 ( ev/t)b E ( ev/t)b Spin-Orbit Coupling We mentioned earlier that the pin magnetic dipole moment alo give plitting of energy level even when there i no external field. One caue involve the orbital motion of the electron. In the Bohr model, oberver moving with the electron would ee the poitively charged nucleu revolving around them (jut a to earthbound oberver the un eem to be orbiting the earth). Thi apparent motion of charge caue a magnetic field at the location of the electron, a meaured in the electron moving frame of reference. The reulting interaction with the pin magnetic moment caue a twofold plitting of thi level, correponding to the two poible orientation of electron pin. Dicuion baed on the Bohr model can t be taken too eriouly, but a imilar reult can be derived from the Schrödinger equation. The interaction energy U can be expreed in term of the calar product of the angular momentum vector S S L and. Thi effect i called pin-orbit coupling; it i reponible for the mall energy difference between the two cloely paced, lowet excited level of odium hown in Fig a and for the correponding doublet (589.0, nm) in the pectrum of odium. Example 41.7 An effective magnetic field To ix ignificant figure, the wavelength of the two pectral line that make up the odium doublet are l 1 = nm and l = nm. Calculate the effective magnetic field experienced by the electron in the 3p level of the odium atom. SOLUTION IDENTIFY and SET UP: The two line in the odium doublet reult from tranition from the two 3p level, which are plit by pinorbit coupling, to the 3 level, which i not plit becaue it ha L = 0. We picture the pin-orbit coupling a an interaction between the electron pin magnetic moment and an effective magnetic field due to the nucleu. Thi example i like Example 41.6 in revere: There we were given B and found the difference between the energie of the two pin tate, while here we ue the energy difference to find the target variable B. The difference in energy between the two 3p level i equal to the difference in energy between the two photon of the odium doublet. We ue thi relationhip and the reult of Example 41.6 to determine B. EXECUTE: The energie of the two photon are E 1 = hc>l 1 and E = hc>l. Here E 1 7 E becaue l 1 6 l, o the difference in their energie i E = hc - hc = hca l - l 1 b l 1 l l l 1 = * ev # * 10 8 m> * * 10-9 m * 10-9 m * 10-9 m * 10-9 m = ev = 3.41 * 10 - J Thi equal the energy difference between the two 3p level. The pin-orbit interaction raie one level by 1.70 * 10 - J (one-half

67 41.6 Many-Electron Atom and the Excluion Principle 1387 of 3.41 * 10 - J) and lower the other by 1.70 * 10 - J. From EVALUATE: The electron experience a very trong effective magnetic Example 41.6, the amount each tate i raied or lowered i field. To produce a teady, macrocopic field of thi magni- ƒuƒ = m B B, o tude in the laboratory require tate-of-the-art electromagnet. U B = ` ` = 1.70 * 10- J m B 9.8 * 10-4 J>T = 18.0 T Combining Orbital and Spin Angular Momenta The orbital and pin angular momenta ( L S and S, repectively) can combine in S S variou way. The vector um of L S and i the total angular momentum J: S S S J L (41.41) The poible value of the magnitude J are given in term of a quantum number j by J = j1j + 1U (41.4) We can then have tate in which j = ƒl 1 The l + 1 tate correpond to the S cae in which the vector L S ƒ. and have parallel z-component; for the l - 1 tate, S 1 L S and have antiparallel z-component. For example, when l = 1, j can be or 3 In another pectrocopic notation thee tate are labeled and. p P 1> P 3>, repectively. The upercript i the number of poible pin orientation, the letter P (now capitalized) indicate tate with l = 1, and the ubcript i the value of j. We ued thi cheme to label the energy level of the odium atom in Fig a. The variou line plitting reulting from magnetic interaction are collectively called fine tructure. There are alo additional, much maller plitting aociated with the fact that the nucleu of the atom ha a magnetic dipole moment that interact with the orbital and> or pin magnetic dipole moment of the electron. Thee effect are called hyperfine tructure. For example, the ground level of hydrogen i plit into two tate, eparated by only 5.9 * 10-6 ev. The photon that i emitted in the tranition between thee tate ha a wavelength of 1 cm. Radio atronomer ue thi wavelength to map cloud of intertellar hydrogen ga that are too cold to emit viible light (Fig ) In a viible-light image (top), thee three ditant galaxie appear to be unrelated. But in fact thee galaxie are connected by immene treamer of hydrogen ga. Thi i revealed by by the falecolor image (bottom) made with a radio telecope tuned to the 1-cm wavelength emitted by hydrogen atom. Galaxie in viible light (negative image; galaxie appear dark) Radio image at wavelength 1 cm Tet Your Undertanding of Section 41.5 In which of the following ituation i the magnetic moment of an electron perfectly aligned with a magnetic field that point in the poitive z-direction? (i) m = + 1 (ii) m = - 1 ; ; (iii) both (i) and (ii); (iv) neither (i) nor (ii) Many-Electron Atom and the Excluion Principle So far our analyi of atomic tructure ha concentrated on the hydrogen atom. That natural; neutral hydrogen, with only one electron, i the implet atom. If we can t undertand hydrogen, we certainly can t undertand anything more complex. But now let move on to many-electron atom. In general, an atom in it normal (electrically neutral) tate ha Z electron and Z proton. Recall from Section 41.3 that we call Z the atomic number. The total electric charge of uch an atom i exactly zero becaue the neutron ha no charge while the proton and electron charge have the ame magnitude but oppoite ign. We can apply the Schrödinger equation to thi general atom. However, the complexity of the analyi increae very rapidly with increaing Z. Each of the Z electron interact not only with the nucleu but alo with every other electron.

68 1388 CHAPTER 41 Atomic Structure The wave function and the potential energy are function of 3Z coordinate, and the equation contain econd derivative with repect to all of them. The mathematical problem of finding olution of uch equation i o complex that it ha not been olved exactly even for the neutral helium atom, which ha only two electron. Fortunately, variou approximation cheme are available. The implet approximation i to ignore all interaction between electron and conider each electron a moving under the action only of the nucleu (conidered to be a point charge). In thi approximation the wave function for each individual electron i a function like that for the hydrogen atom, pecified by four quantum number 1n, l, m l, m. The nuclear charge i Ze intead of e, o we replace every factor of e in the wave function and the energy level by Ze. In particular, the energy level are given by Eq. (41.1) with e 4 replaced by Z e 4 : E n = - 1 m r Z e 4 14pP 0 n U = - Z n ev (41.43) Thi approximation i fairly dratic; when there are many electron, their interaction with each other are a important a the interaction of each with the nucleu. So thi model in t very ueful for quantitative prediction. The Central-Field Approximation A le dratic and more ueful approximation i to think of all the electron together a making up a charge cloud that i, on average, pherically ymmetric. We can then think of each individual electron a moving in the total electric field due to the nucleu and thi averaged-out cloud of all the other electron. There i a correponding pherically ymmetric potential-energy function U1r. Thi picture i called the central-field approximation; it provide a ueful tarting point for undertanding atomic tructure. In the central-field approximation we can again deal with one-electron wave function. The Schrödinger equation differ from the equation for hydrogen only in that the 1>r potential-energy function i replaced by a different function U1r. But it turn out that U1r doe not enter the differential equation for 1u and 1f, o thoe angular function are exactly the ame a for hydrogen, and the orbital angular-momentum tate are alo the ame a before. The quantum number l, m l, and m have the ame meaning a before, and Eq. (41.) and (41.3) again give the magnitude and z-component of the orbital angular momentum. The radial wave function and probabilitie are different than for hydrogen becaue of the change in U1r, o the energy level are no longer given by Eq. (41.1). We can till label a tate uing the four quantum number 1n, l, m l, m. In general, the energy of a tate now depend on both n and l, rather than jut on n a with hydrogen. The retriction on value of the quantum number are the ame a before: n Ú 1 0 l n - 1 ƒm l ƒ l m = 1 (allowed value of quantum number) (41.44) The Excluion Principle To undertand the tructure of many-electron atom, we need an additional principle, the excluion principle. To ee why thi principle i needed, let conider the lowet-energy tate, or ground tate, of a many-electron atom. In the oneelectron tate of the central-field model, there i a lowet-energy tate (correponding to an n = 1 tate of hydrogen). We might expect that in the ground tate of a complex atom, all the electron hould be in thi lowet tate. If o, then we hould ee only gradual change in phyical and chemical propertie when we look at the behavior of atom with increaing number of electron 1Z.

69 41.6 Many-Electron Atom and the Excluion Principle 1389 Such gradual change are not what i oberved. Intead, propertie of element vary widely from one to the next, with each element having it own ditinct peronality. For example, the element fluorine, neon, and odium have 9, 10, and 11 electron, repectively, per atom. Fluorine 1Z = 9 i a halogen; it tend trongly to form compound in which each fluorine atom acquire an extra electron. Sodium 1Z = 11 i an alkali metal; it form compound in which each odium atom loe an electron. Neon 1Z = 10 i a noble ga, forming no compound at all. Such obervation how that in the ground tate of a complex atom the electron cannot all be in the lowet-energy tate. But why not? The key to thi puzzle, dicovered by the Autrian phyicit Wolfgang Pauli (Fig. 41.0) in 195, i called the excluion principle. Thi principle tate that no two electron can occupy the ame quantum-mechanical tate in a given ytem. That i, no two electron in an atom can have the ame value of all four quantum number 1n, l, m l, m. Each quantum tate correpond to a certain ditribution of the electron cloud in pace. Therefore the principle alo ay, in effect, that no more than two electron with oppoite value of the quantum number m can occupy the ame region of pace. We houldn t take thi lat tatement too eriouly becaue the electron probability function don t have harp, definite boundarie. But the excluion principle limit the amount by which electron wave function can overlap. Think of it a the quantummechanical analog of a univerity rule that allow only one tudent per dek The key to undertanding the periodic table of the element wa the dicovery by Wolfgang Pauli ( ) of the excluion principle. Pauli received the 1945 Nobel Prize in phyic for hi accomplihment. Thi photo how Pauli (on the left) and Niel Bohr watching the phyic of a toy top pinning on the floor a macrocopic analog of a microcopic electron with pin. CAUTION The meaning of the excluion principle Don t confue the excluion principle with the electric repulion between electron. While both effect tend to keep electron within an atom eparated from each other, they are very different in character. Two electron can alway be puhed cloer together by adding energy to combat electric repulion; in contrat, nothing can overcome the excluion principle and force two electron into the ame quantum-mechanical tate. Table 41. lit ome of the et of quantum number for electron tate in an atom. It imilar to Table 41.1 (Section 41.3), but we ve added the number of tate in each ubhell and hell. Becaue of the excluion principle, the number of tate i the ame a the maximum number of electron that can be found in thoe tate. For each tate, m can be either + 1 or - 1. A with the hydrogen wave function, different tate correpond to different patial ditribution; electron with larger value of n are concentrated at larger ditance from the nucleu. Figure 41.8 (Section 41.3) how thi effect. When an atom ha more than two electron, they can t all huddle down in the low-energy n = 1 tate nearet to the nucleu becaue there are only two of thee tate; the excluion principle forbid multiple occupancy of a tate. Some electron are forced into tate farther away, with higher energie. Each value of n correpond roughly to a region of pace around the nucleu in the form of a pherical hell. Hence we peak of the K hell a the region that i occupied by the electron in the n = 1 tate, the L hell a the region of the n = tate, and o on. State with the ame n but different l form ubhell, uch a the 3p ubhell. Table 41. Quantum State of Electron in the Firt Four Shell n l m l Spectrocopic Notation Number of State Shell K 0 0 V 1-1, 0, 1 p 6 8 L , 0, 1 3p 6 18 M 3 -, -1, 0, 1, 3d , 0, 1 4p 6 4 -, -1, 0, 1, 4d 10 3 N 4 3-3, -, -1, 0, 1,, 3 4ƒ 14

70 1390 CHAPTER 41 Atomic Structure 41.1 Schematic repreentation of the charge ditribution in a lithium atom. The nucleu ha a charge of +3e. On average, the electron i coniderably farther from the nucleu than the 1 electron. Therefore, it experience a net nuclear charge of approximately 13e e 51e (rather than 13e). Nucleu 1 ubhell 13e e e ubhell 41. Salt (odium chloride, NaCl) diolve readily in water, making eawater alty. Thi i due to the electron configuration of odium and chlorine: Sodium can eaily loe an electron to form an Na + ion, and chlorine can eaily gain an electron to form a Cl - ion. Thee ion are held in olution becaue they are attracted to the polar end of water molecule (ee Fig. 1.30a). The Periodic Table We can ue the excluion principle to derive the mot important feature of the tructure and chemical behavior of multielectron atom, including the periodic table of the element. Let imagine contructing a neutral atom by tarting with a bare nucleu with Z proton and then adding Z electron, one by one. To obtain the ground tate of the atom a a whole, we fill the lowet-energy electron tate (thoe cloet to the nucleu, with the mallet value of n and l) firt, and we ue ucceively higher tate until all the electron are in place. The chemical propertie of an atom are determined principally by interaction involving the outermot, or valence, electron, o we particularly want to learn how thee electron are arranged. Let look at the ground-tate electron configuration for the firt few atom (in order of increaing Z). For hydrogen the ground tate i 1; the ingle electron i in a tate n = 1, l = 0, m l = 0, and m = 1. In the helium atom 1Z =, both electron are in 1 tate, with oppoite pin; one ha m = - 1 and the other ha m We denote the helium ground tate a 1 = (The upercript i not an exponent; the notation 1 tell u that there are two electron in the 1 ubhell. Alo, the upercript 1 i undertood, a in.) For helium the K hell i completely filled, and all other are empty. Helium i a noble ga; it ha no tendency to gain or loe an electron, and it form no compound. Lithium 1Z = 3 ha three electron. In it ground tate, two are in 1 tate and one i in a tate, o we denote the lithium ground tate a 1.? On average, the electron i coniderably farther from the nucleu than are the 1 electron (Fig. 41.1). According to Gau law, the net charge Q encl attracting the electron i nearer to +e than to the value +3e it would have without the two 1 electron preent. A a reult, the electron i looely bound; only 5.4 ev i required to remove it, compared with the 30.6 ev given by Eq. (41.43) with Z = 3 and n =. In chemical behavior, lithium i an alkali metal. It form ionic compound in which each lithium atom loe an electron and ha a valence of +1. Next i beryllium 1Z = 4; it ground-tate configuration i 1, with it two valence electron filling the ubhell of the L hell. Beryllium i the firt of the alkaline earth element, forming ionic compound in which the valence of the atom i +. Table 41.3 how the ground-tate electron configuration of the firt 30 element. The L hell can hold eight electron. At Z = 10, both the K and L hell are filled, and there are no electron in the M hell. We expect thi to be a particularly table configuration, with little tendency to gain or loe electron. Thi element i neon, a noble ga with no known compound. The next element after neon i odium 1Z = 11, with filled K and L hell and one electron in the M hell. It noble-ga-plu-one-electron tructure reemble that of lithium; both are alkali metal. The element before neon i fluorine, with Z = 9. It ha a vacancy in the L hell and ha an affinity for an extra electron to fill the hell. Fluorine form ionic compound in which it ha a valence of -1. Thi behavior i characteritic of the halogen (fluorine, chlorine, bromine, iodine, and atatine), all of which have noble-ga-minu-one configuration (Fig. 41.). Proceeding down the lit, we can undertand the regularitie in chemical behavior diplayed by the periodic table of the element (Appendix D) on the bai of electron configuration. The imilarity of element in each group (vertical column) of the periodic table i the reult of imilarity in outer-electron configuration. All the noble gae (helium, neon, argon, krypton, xenon, and radon) have filled-hell or filled-hell plu filled p ubhell configuration. All the alkali metal (lithium, odium, potaium, rubidium, ceium, and francium) have noble-ga-plu-one configuration. All the alkaline earth metal (beryllium, magneium, calcium, trontium, barium, and radium) have noble-ga-plu-two configuration, and, a we jut mentioned, all the halogen (fluorine, chlorine, bromine, iodine, and atatine) have noble-ga-minu-one tructure.

71 41.6 Many-Electron Atom and the Excluion Principle 1391 Table 41.3 Ground-State Electron Configuration Atomic Element Symbol Number (Z) Electron Configuration Hydrogen H 1 1 Helium He 1 Lithium Li 3 1 Beryllium Be 4 1 Boron B 5 1 p Carbon C 6 1 p Nitrogen N 7 1 p 3 Oxygen O 8 1 p 4 Fluorine F 9 1 p 5 Neon Ne 10 1 p 6 Sodium Na 11 1 p 6 3 Magneium Mg 1 1 p 6 3 Aluminum Al 13 1 p 6 3 3p Silicon Si 14 1 p 6 3 3p Phophoru P 15 1 p 6 3 3p 3 Sulfur S 16 1 p 6 3 3p 4 Chlorine Cl 17 1 p 6 3 3p 5 Argon Ar 18 1 p 6 3 3p 6 Potaium K 19 1 p 6 3 3p 6 4 Calcium Ca 0 1 p 6 3 3p 6 4 Scandium Sc 1 1 p 6 3 3p 6 4 3d Titanium Ti 1 p 6 3 3p 6 4 3d Vanadium V 3 1 p 6 3 3p 6 4 3d 3 Chromium Cr 4 1 p 6 3 3p 6 43d 5 Manganee Mn 5 1 p 6 3 3p 6 4 3d 5 Iron Fe 6 1 p 6 3 3p 6 4 3d 6 Cobalt Co 7 1 p 6 3 3p 6 4 3d 7 Nickel Ni 8 1 p 6 3 3p 6 4 3d 8 Copper Cu 9 1 p 6 3 3p 6 43d 10 Zinc Zn 30 1 p 6 3 3p 6 4 3d 10 A light complication occur with the M and N hell becaue the 3d and 4 ubhell level ( n = 3, l =, and n = 4, l = 0, repectively) have imilar energie. (We ll dicu in the next ubection why thi happen.) Argon 1Z = 18 ha all the 1,, p, 3, and 3p ubhell filled, but in potaium 1Z = 19 the additional electron goe into a 4 energy tate rather than a 3d tate (becaue the 4 tate ha lightly lower energy). The next everal element have one or two electron in the 4 ubhell and increaing number in the 3d ubhell. Thee element are all metal with rather imilar chemical and phyical propertie; they form the firt tranition erie, tarting with candium 1Z = 1 and ending with zinc 1Z = 30, for which all the 3d and 4 ubhell are filled. Something imilar happen with Z = 57 through Z = 71, which have one or two electron in the 6 ubhell but only partially filled 4ƒ and 5d ubhell. Thee are the rare earth element; they all have very imilar phyical and chemical propertie. Yet another uch erie, called the actinide erie, tart with Z = 91. Application Electron Configuration and Bone Cancer Radiotherapy The orange pot in thi colored x-ray image are bone cancer tumor. One method of treating bone cancer i to inject a radioactive iotope of trontium 1 89 Sr into a patient vein. Strontium i chemically imilar to calcium becaue in both atom the two outer electron are in an tate (the tructure are 1 p 6 3 3p 6 4 3d 10 4p 6 5 for trontium and 1 p 6 3 3p 6 4 for calcium). Hence the trontium i readily taken up by the tumor, where calcium turnover i more rapid than in healthy bone. Radiation from the trontium help to detroy the tumor. Screening We have mentioned that in the central-field picture, the energy level depend on l a well a n. Let take odium 1Z = 11 a an example. If 10 of it electron fill it K and L hell, the energie of ome of the tate for the remaining electron are found experimentally to be 3 tate: ev 3p tate: ev 3d tate: ev 4 tate: ev

72 139 CHAPTER 41 Atomic Structure The 3 tate are the lowet (mot negative); one i the ground tate for the 11th electron in odium. The energy of the 3d tate i quite cloe to the energy of the n = 3 tate in hydrogen. The urprie i that the 4 tate energy i 0.46 ev below the 3d tate, even though the 4 tate ha larger n. We can undertand thee reult uing Gau law and the radial probability ditribution. For any pherically ymmetric charge ditribution, the electric-field magnitude at a ditance r from the center i Q encl >4pP 0 r, where Q encl i the total charge encloed within a phere with radiu r. Mentally remove the outer (valence) electron atom from a odium atom. What you have left i a pherically ymmetric collection of 10 electron (filling the K and L hell) and 11 proton, o Q encl = -10e + 11e = +e. If the 11th electron i completely outide thi collection of charge, it i attracted by an effective charge of +e, not +11e. Thi i a more extreme example of the effect depicted in Fig Thi effect i called creening; the 10 electron creen 10 of the 11 proton, leaving an effective net charge of +e. In general, an electron that pend all it time completely outide a poitive charge Z eff e ha energy level given by the hydrogen expreion with e replaced by Z eff e. From Eq. (41.43) thi i E n = - Z eff ev (energy level with creening) n (41.45) If the 11th electron in the odium atom i completely outide the remaining charge ditribution, then Z eff = 1. CAUTION Different equation for different atom Equation (41.1), (41.43), and (41.45) all give value of E in term of ev>n n, but they don t apply in general to the ame atom. Equation (41.1) i only for hydrogen. Equation (41.43) i only for the cae in which there i no interaction with any other electron (and i thu accurate only when the atom ha jut one electron). Equation (41.45) i ueful when one electron i creened from the nucleu by other electron. Now let ue the radial probability function hown in Fig to explain why the energy of a odium 3d tate i approximately the ame a the n = 3 value of hydrogen, ev. The ditribution for the 3d tate (for which l ha the maximum value n - 1) ha one peak, and it mot probable radiu i outide the poition of the electron with n = 1 or. (Thoe electron alo are pulled cloer to the nucleu than in hydrogen becaue they are le effectively creened from the poitive charge 11e of the nucleu.) Thu in odium a 3d electron pend mot of it time well outide the n = 1 and n = tate (the K and L hell). The 10 electron in thee hell creen about ten-eleventh of the charge of the 11 proton, leaving a net charge of about Z eff e = 11e. Then, from Eq. (41.45), the correponding energy i approximately ev>3 = ev. Thi approximation i very cloe to the experimental value of ev. Looking again at Fig. 41.8, we ee that the radial probability denity for the 3p tate (for which l = n - ha two peak and that for the 3 tate 1l = n - 3 ha three peak. For odium the firt mall peak in the 3p ditribution give a 3p electron a higher probability (compared to the 3d tate) of being inide the charge ditribution for the electron in the n = tate. That i, a 3p electron i le completely creened from the nucleu than i a 3d electron becaue it pend ome of it time within the filled K and L hell. Thu for the 3p electron, Z eff i greater than unity. From Eq. (41.45) the 3p energy i lower (more negative) than the 3d energy of ev. The actual value i ev. A 3 electron pend even more time within the inner electron hell than a 3p electron doe, giving an even larger and an even more negative energy. Z eff

73 41.7 X-Ray Spectra 1393 Example 41.8 Determining Z eff experimentally The meaured energy of a 3 tate of odium i ev. Calculate the value of Z eff. Sodium 11 proton are creened by an average of 11 - EVALUATE: The effective charge attracting a 3 electron i 1.84e = 9.16 electron intead of 10 electron becaue the 3 electron pend ome time within the inner 1K and L) hell. SOLUTION Each alkali metal (lithium, odium, potaium, rubidium, and IDENTIFY and SET UP: Sodium ha a ingle electron in the M hell ceium) ha one more electron than the correponding noble ga outide filled K and L hell. The ten K and L electron partially (helium, neon, argon, krypton, and xenon). Thi extra electron i creen the ingle M electron from the 11e charge of the nucleu; motly outide the other electron in the filled hell and ubhell. our goal i to determine the extent of thi creening. We are given Therefore all the alkali metal behave imilarly to odium. n = 3 and E n = ev, o we can ue Eq. (41.45) to determine Z eff. EXECUTE: Solving Eq. (41.45) for Z eff, we have Zeff = - n E n 13.6 ev = ev = ev Z eff = 1.84 Example 41.9 Energie for a valence electron The valence electron in potaium ha a 4 ground tate. Calculate the approximate energy of the n 4 tate having the mallet Z eff, and dicu the relative energie of the 4, 4p, 4d, and 4ƒ tate. SOLUTION IDENTIFY and SET UP: The tate with the mallet i the one in which the valence electron pend the mot time outide the inner filled hell and ubhell, o that it i mot effectively creened from the charge of the nucleu. Once we have determined which tate ha the mallet Z eff, we can ue Eq. (41.45) to determine the energy of thi tate. EXECUTE: A 4ƒ tate ha n = 4 and l = 3 = 4-1. Thu it i the tate of greatet orbital angular momentum for n = 4, and thu the tate in which the electron pend the mot time outide the electron charge cloud of the inner filled hell and ubhell. Thi make for a 4ƒ tate cloe to unity. Equation (41.45) then give Z eff Z eff E 4 = - Z eff n ev = ev = ev 4 Thi approximation agree with the meaured energy of the odium 4f tate to the preciion given. An electron in a 4d tate pend a bit more time within the inner hell, and it energy i therefore a bit more negative (meaured to be ev). For the ame reaon, a 4p tate ha an even lower energy (meaured to be -.73 ev) and a 4 tate ha the lowet energy (meaured to be ev). EVALUATE: We can extend thi analyi to the ingly ionized alkaline earth element: Be +, Mg +, Ca +, Sr +, and Ba +. For any allowed value of n, the highet-l tate 1l = n - 1 of the one remaining outer electron ee an effective charge of almot +e, o for thee tate, Z A 3d tate for Mg + eff =., for example, ha an energy of about ev>3 = -6.0 ev. Tet Your Undertanding of Section 41.6 If electron did not obey the excluion principle, would it be eaier or more difficult to remove the firt electron from odium? 41.7 X-Ray Spectra X-ray pectra provide yet another example of the richne and power of the Schrödinger equation and of the model of atomic tructure that we derived from it in the preceding ection. In Section 38. we dicued x-ray production on the bai of the photon concept. With the development of x-ray diffraction technique (ee Section 36.6) by von Laue, Bragg, and other, beginning in 191, it became poible to meaure x-ray wavelength quite preciely (to within 0.1% or le). Detailed tudie of x-ray pectra howed a continuou pectrum of wavelength (ee Fig in Section 38.), with minimum wavelength (correponding to maximum frequency and photon energy) determined by the accelerating

74 1394 CHAPTER 41 Atomic Structure voltage V AC in the x-ray tube, according to the relationhip derived in Section 38. for bremtrahlung procee: l min = (41.46) Thi continuou-pectrum radiation i nearly independent of the target material in the x-ray tube. hc ev AC 41.3 Graph of intenity per unit wavelength a a function of wavelength for x ray produced with an accelerating voltage of 35 kv and a molybdenum target. The curve i a mooth function imilar to the bremtrahlung pectra in Fig (Section 38.), but with two harp pike correponding to part of the characteritic x-ray pectrum for molybdenum. I(l) Sharp peak K a Moeley Law and Atomic Energy Level Depending on the accelerating voltage and the target element, we may find harp peak uperimpoed on thi continuou pectrum, a in Fig Thee peak are at different wavelength for different element; they form what i called a characteritic x-ray pectrum for each target element. In 1913 the Britih cientit Henry G. J. Moeley tudied thee pectra in detail uing x-ray diffraction technique. He found that the mot intene hort-wavelength line in the characteritic x-ray pectrum from a particular target element, called the K a line, varied moothly with that element atomic number Z (Fig. 41.4). Thi i in harp contrat to optical pectra, in which element with adjacent Z-value have pectra that often bear no reemblance to each other. Moeley found that the relationhip could be expreed in term of x-ray frequencie ƒ by a imple formula called Moeley law: 0 Continuou pectrum l min K b l (pm) ƒ = 1.48 * Hz1Z - 1 (Moeley law (41.47) Moeley went far beyond thi empirical relationhip; he howed how characteritic x-ray pectra could be undertood on the bai of energy level of atom in the target. Hi analyi wa baed on the Bohr model, publihed in the ame year. We will recat it omewhat, uing the idea of atomic tructure that we dicued in Section Firt recall that the outer electron of an atom are reponible for optical pectra. Their excited tate are uually only a few electron volt above their ground tate. In tranition from excited tate to the ground tate, they uually emit photon in or near the viible region. Characteritic x ray, by contrat, are emitted in tranition involving the inner hell of a complex atom. In an x-ray tube the electron may trike the target with enough energy to knock electron out of the inner hell of the target atom. Thee inner electron are much cloer to the nucleu than are the electron in the outer hell; they are much more tightly bound, and hundred or thouand of electron volt may be required to remove them. Suppoe one electron i knocked out of the K hell. Thi proce leave a vacancy, which we ll call a hole. (One electron remain in the K hell.) The hole can then be filled by an electron falling in from one of the outer hell, uch a the L, M, N, Á hell. Thi tranition i accompanied by a decreae in the energy of the 41.4 The quare root of Moeley meaured frequencie of the K a line for 14 element. f (10 8 Hz 1/ ) 4 The graph of f veru Z i a traight line... Zr Y with an intercept at Z 5 1, confirming Moeley law, Ti Eq. (41.47) Cl K Al Si Cu Co Cr V Fe Ni Zn Z

75 41.7 X-Ray Spectra 1395 atom (becaue le energy would be needed to remove an electron from an L, M, N, Á hell), and an x-ray photon i emitted with energy equal to thi decreae. Each tate ha definite energy, o the emitted x ray have definite wavelength; the emitted pectrum i a line pectrum. We can etimate the energy and frequency of K a x-ray photon uing the concept of creening from Section A K a x-ray photon i emitted when an electron in the L hell 1n = drop down to fill a hole in the K hell 1n = 1. A the electron drop down, it i attracted by the Z proton in the nucleu creened by the one remaining electron in the K hell. We therefore approximate the energy by Eq. (41.45), with Z eff = Z - 1, n i =, and n f. The energy before the tranition i Application X Ray in Forenic Science When a handgun i fired, a cloud of gunhot reidue (GSR) i ejected from the barrel. The x-ray emiion pectrum of GSR include characteritic peak from lead (Pb), antimony (Sb), and barium (Ba). If a ample taken from a upect kin or clothing ha an x-ray emiion pectrum with thee characteritic, it indicate that the upect recently fired a gun. 1Z - 1 E i L ev = -1Z ev and the energy after the tranition i 1Z - 1 E f L ev = -1Z ev The energy of the 13.6 ev. That i, K a x-ray photon i E Ka = E i - E f L 1Z ev + Pb Ba E Ka L 1Z ev The frequency of the photon i it energy divided by Planck contant: (41.48) Sb ƒ = E h L 1Z ev * ev # = 1.47 * Hz1Z - 1 Thi relationhip agree almot exactly with Moeley experimental law, Eq. (41.47). Indeed, conidering the approximation we have made, the agreement i better than we have a right to expect. But our calculation doe how how Moeley law can be undertood on the bae of creening and tranition between energy level. The hole in the K hell may alo be filled by an electron falling from the M or N hell, auming that thee are occupied. If o, the x-ray pectrum of a large group of atom of a ingle element how a erie, named the K erie, of three line, called the K a, K b, and K g line. Thee three line reult from tranition in which the K-hell hole i filled by an L, M, or N electron, repectively. Figure 41.5 how the K erie for tungten 1Z = 74, molybdenum 1Z = 4, and copper 1Z = 9. There are other erie of x-ray line, called the L, M, and N erie, that are produced after the ejection of electron from the L, M, and N hell rather than the K hell. Electron in thee outer hell are farther away from the nucleu and are not held a tightly a are thoe in the K hell, o removing thee outer electron require le energy. Hence the x-ray photon that are emitted when thee vacancie are filled have lower energy than thoe in the K erie. W Mo Cu gb a gb a gb a 41.5 Wavelength of the K a, K b, and K g line of tungten (W), molybdenum (Mo), and copper (Cu) l (pm) The three line in each erie are called the K a, K b, and K g line. The K a line i produced by the tranition of an L electron to the vacancy in the K hell, the K b line by an M electron, and the K g line by an N electron.

76 1396 CHAPTER 41 Atomic Structure Example Chemical analyi by x-ray emiion You meaure the K a wavelength for an unknown element, obtaining the value nm. What i the element? correponding to the element molybdenum. We know that Z ha to be an integer; we conclude that Z = 4, EVALUATE: If you re worried that our calculation did not give an SOLUTION integer for Z, remember that Moeley law i an empirical relationhip. There are light variation from one atom to another due IDENTIFY and SET UP: To determine which element thi i, we need to know it atomic number Z. We can find thi uing Moeley to difference in the tructure of the electron hell. Nonethele, law, which relate the frequency of an element K a x-ray emiion thi example ugget the power of Moeley law. line to that element atomic number Z. We ll ue the relationhip Niel Bohr commented that it wa Moeley obervation, not ƒ = c>l to calculate the frequency for the K a line, and then ue the alpha-particle cattering experiment of Rutherford, Geiger, Eq. (41.47) to find the correponding value of the atomic number and Marden (ee Section 39.), that truly convinced phyicit Z. We ll then conult the periodic table (Appendix D) to determine that the atom conit of a poitive nucleu urrounded by electron in motion. Unlike Bohr or Rutherford, Moeley did not which element ha thi atomic number. receive a Nobel Prize for hi important work; thee award are EXECUTE: The frequency i given only to living cientit, and Moeley wa killed in combat ƒ = c l = 3.00 * during the Firt World War. 108 m> * 10-9 m = 4.3 * 1018 Hz Solving Moeley law for Z, we get ƒ Z = 1 + B.48 * Hz = * Hz B.48 * Hz = When a beam of x ray i paed through a lab of molybdenum, the extent to which the beam i aborbed depend on the energy E of the x-ray photon. A harp increae in aborption occur at the K aborption edge at 0 kev. The increae occur becaue photon with energie above thi value can excite an electron from the K hell of a molybdenum atom into an empty tate. Aborption K aborption edge X-Ray Aborption Spectra We can alo oberve x-ray aborption pectra. Unlike optical pectra, the aborption wavelength are uually not the ame a thoe for emiion, epecially in many-electron atom, and do not give imple line pectra. For example, the K a emiion line reult from a tranition from the L hell to a hole in the K hell. The revere tranition doen t occur in atom with Z Ú 10 becaue in the atom ground tate, there i no vacancy in the L hell. To be aborbed, a photon mut have enough energy to move an electron to an empty tate. Since empty tate are only a few electron volt in energy below the free-electron continuum, the minimum aborption energie in many-electron atom are about the ame a the minimum energie that are needed to remove an electron from it hell. Experimentally, if we gradually increae the accelerating voltage and hence the maximum photon energy, we oberve udden increae in aborption when we reach thee minimum energie. Thee udden jump of aborption are called aborption edge (Fig. 41.6). Characteritic x-ray pectra provide a very ueful analytical tool. Satellite-borne x-ray pectrometer are ued to tudy x-ray emiion line from highly excited atom in ditant atronomical ource. X-ray pectra are alo ued in air-pollution monitoring and in tudie of the abundance of variou element in rock. 0 E (kev) Tet Your Undertanding of Section 41.7 A beam of photon i paed through a ample of high-temperature atomic hydrogen. At what photon energy would you expect there to be an aborption edge like that hown in Fig. 41.6? (i) ev; (ii) 3.40 ev; (iii) 1.51 ev; (iv) all of thee; (v) none of thee.

77 CHAPTER 41 SUMMARY Three-dimenional problem: The time-independent Schrödinger equation for three-dimenional problem i given by Eq. (41.5). - U m a 0 c1x, y, z 0x + 0 c1x, y, z 0y + 0 c1x, y, z 0z b + U1x, y, zc1x, y, z = Ec1x, y, z (three-dimenional time-independent Schrödinger equation) (41.5) Particle in a three-dimenional box: The wave function for a particle in a cubical box i the product of a function of x only, a function of y only, and a function of z only. Each tationary tate i decribed by three quantum number 1n X, n Y, n Z. Mot of the energy level given by Eq. (41.16) exhibit degeneracy: More than one quantum tate ha the ame energy. (See Example 41.1.) E nx,n Y,n Z = 1n X + n Y + n Z p U ml 1n X = 1,, 3, Á ; n Y = 1,, 3, Á ; n Z = 1,, 3, Á (energy level, particle in a threedimenional cubical box) (41.16) x 5 L x z z 5 L O y 5 L y The hydrogen atom: The Schrödinger equation for the hydrogen atom give the ame energy level a the Bohr model. If the nucleu ha charge Ze, there i an additional factor of Z in the numerator of Eq. (41.1). The poible magnitude L of orbital angular momentum are given by Eq. (41.), and the poible value of the z-component of orbital angular momentum are given by Eq. (41.3). (See Example 41. and 41.3.) The probability that an atomic electron i between r and r + dr from the nucleu i P1r dr, given by Eq. (41.5). Atomic ditance are often meaured in unit of a, the mallet ditance between the electron and the nucleu in the Bohr model. (See Example 41.4.) 1 m r e ev E n = - 14pP 0 n = - U n (energy level of hydrogen) (41.1) L = l1l + 1U 1l = 0, 1,, Á, n - 1 L z = m l U 1m l = 0, 1,, Á, l P1r dr = ƒcƒ dv = ƒcƒ 4pr dr a = P 0h pm r e = 4pP 0 U m r e 4 (41.) (41.3) (41.5) z L z h L z h L z 0 L z h L z h S L 6h l = 5.9 * m (41.6) The Zeeman effect: The interaction energy of an electron (ma m) with magnetic quantum number m l in a magnetic field B S along the +z-direction i given by Eq. (41.35) or (41.36), where m B = eu>m i called the Bohr magneton. (See Example 41.5.) U = -m z B = m l eu m B = m lm B B 1m l = 0, 1,, Á, l (41.35), (41.36) E E d B 5 0 B increaing DE 1m B B 1m B B 0 m B B m B B 1397

78 1398 CHAPTER 41 Atomic Structure Electron pin: An electron ha an intrinic pin angular momentum of magnitude S, given by Eq. (41.38). The poible value of the z-component of the pin angular momentum are S z = m where m = 1 U,. (See Example 41.6 and 41.7.) S = 1 A 1 + 1BU =3 4 U S z = 1 U (41.38) (41.37) E Spin up E B 0 Spin down B increaing E ( ev/t)b E ( ev/t)b Many-electron atom: In a hydrogen atom, the quantum number n, l, m l, and m of the electron have certain allowed value given by Eq. (41.44). In a many-electron atom, the allowed quantum number for each electron are the ame a in hydrogen, but the energy level depend on both n and l becaue of creening, the partial cancellation of the field of the nucleu by the inner electron. If the effective (creened) charge attracting an electron i Z eff e, the energie of the level are given approximately by Eq. (41.45). (See Example 41.8 and 41.9.) n Ú 1 0 l n - 1 ƒm l ƒ l m = 1 E n = - Z eff n ev (41.44) (41.45) Nucleu 13e e e 1 ubhell ubhell X-ray pectra: Moeley law tate that the frequency of a K a x ray from a target with atomic number Z i given by Eq. (41.47). Characteritic x-ray pectra reult from tranition to a hole in an inner energy level of an atom. (See Example ) ƒ = 1.48 * Hz1Z - 1 (41.47) f (10 8 Hz 1/ ) 4 Cu Cr Zr Co Y 16 Cl Ti 8 Al K Si V Fe Ni Zn Z BRIDGING PROBLEM A Many-Electron Atom in a Box An atom of titanium (Ti) ha electron and ha a radiu of EXECUTE 1.47 * m. A a imple model of thi atom, imagine putting 3. Ue your knowledge of geometry to find the length of each ide electron into a cubical box that ha the ame volume a a titanium atom. (a) What i the length of each ide of the box? (b) 4. Each electron tate i decribed by four quantum number: of the box. What will be the configuration of the electron? (c) Find the n X, n Y, and n Z a decribed in Section 41. and the pin magnetic quantum number m decribed in Section Ue the energie of each of the level occupied by the electron. (Ignore the electric force that the electron exert on each other.) (d) You excluion principle to determine the quantum number of each remove one of the electron from the lowet level. A a reult, one of the electron in the atom. (Hint: Figure 41.4 in Section of the electron from the highet occupied level drop into the lowet level to fill the hole, emitting a photon in the proce. What i tive to the ground level E 1,1,1.) 41. how the firt everal energy level of a cubical box rela- the energy of thi photon? How doe thi compare to the energy of 5. Ue your reult from tep 3 and 4 to find the energie of each the K a photon for titanium a predicted by Moeley law? of the occupied level. 6. Ue your reult from tep 5 to find the energy of the photon SOLUTION GUIDE emitted when an electron make a tranition from the highet occupied level to the ground level. Compare thi to the energy See MateringPhyic tudy area for a Video Tutor olution. calculated for titanium uing Moeley law. IDENTIFY and SET UP 1. In thi problem you ll ue idea from Section 41. about a particle in a cubical box. You ll alo apply the excluion principle 7. I thi cubical atom a ueful model for titanium? Why or why EVALUATE from Section 41.6 to find the electron configuration of thi not? cubical atom. The idea about x-ray pectra from Section 8. In thi problem you ignored the electric interaction between electron. To etimate how large thee are, find the electrotatic poten are alo important.. The target variable are (a) the dimenion of the box, (b) the tial energy of two electron eparated by half the length of the electron configuration (like thoe given in Table 41.3 for real box. How doe thi compare to the energy level you calculated in atom), (c) the occupied energy level of the cubical box, and tep 5? I it a good approximation to ignore thee interaction? (d) the energy of the emitted photon.

79 Exercie 1399 Problem For intructor-aigned homework, go to : Problem of increaing difficulty. CP: Cumulative problem incorporating material from earlier chapter. CALC: Problem requiring calculu. BIO: Biocience problem. DISCUSSION QUESTIONS Q41.1 Particle A i decribed by the wave function c1x, y, z. Particle B i decribed by the wave function c1x, y, ze if, where f i a real contant. How doe the probability of finding particle A within a volume dv around a certain point in pace compare with the probability of finding particle B within thi ame volume? Q41. What are the mot ignificant difference between the Bohr model of the hydrogen atom and the Schrödinger analyi? What are the imilaritie? Q41.3 For a body orbiting the un, uch a a planet, comet, or ateroid, i there any retriction on the z-component of it orbital angular momentum uch a there i with the z-component of the electron orbital angular momentum in hydrogen? Explain. Q41.4 Why i the analyi of the helium atom much more complex than that of the hydrogen atom, either in a Bohr type of model or uing the Schrödinger equation? Q41.5 The Stern Gerlach experiment i alway performed with beam of neutral atom. Wouldn t it be eaier to form beam uing ionized atom? Why won t thi work? Q41.6 (a) If two electron in hydrogen atom have the ame principal quantum number, can they have different orbital angular momentum? How? (b) If two electron in hydrogen atom have the ame orbital angular-momentum quantum number, can they have different principal quantum number? How? Q41.7 In the Stern Gerlach experiment, why i it eential for the magnetic field to be inhomogeneou (that i, nonuniform)? Q41.8 In the ground tate of the helium atom one electron mut have pin down and the other pin up. Why? Q41.9 An electron in a hydrogen atom i in an level, and the atom i in a magnetic field B S Bk N. Explain why the pin up tate Am = + 1 ha a higher energy than the pin down tate Am = - 1 B B. Q41.10 The central-field approximation i more accurate for alkali metal than for tranition metal uch a iron, nickel, or copper. Why? Q41.11 Table 41.3 how that for the ground tate of the potaium atom, the outermot electron i in a 4 tate. What doe thi tell you about the relative energie of the 3d and 4 level for thi atom? Explain. Q41.1 Do gravitational force play a ignificant role in atomic tructure? Explain. Q41.13 Why do the tranition element 1Z = 1 to 30 all have imilar chemical propertie? Q41.14 Ue Table 41.3 to help determine the ground-tate electron configuration of the neutral gallium atom (Ga) a well a the ion Ga + and Ga -. Gallium ha an atomic number of 31. Q41.15 On the bai of the Pauli excluion principle, the tructure of the periodic table of the element how that there mut be a fourth quantum number in addition to n, l, and m l. Explain. Q41.16 A mall amount of magnetic-field plitting of pectral line occur even when the atom are not in a magnetic field. What caue thi? Q41.17 The ionization energie of the alkali metal (that i, the lowet energy required to remove one outer electron when the atom i in it ground tate) are about 4 or 5 ev, while thoe of the noble gae are in the range from 11 to 5 ev. Why i there a difference? Q41.18 The energy required to remove the 3 electron from a odium atom in it ground tate i about 5 ev. Would you expect the energy required to remove an additional electron to be about the ame, or more, or le? Why? Q41.19 What i the central-field approximation and why i it only an approximation? Q41.0 The nucleu of a gold atom contain 79 proton. How doe the energy required to remove a 1 electron completely from a gold atom compare with the energy required to remove the electron from the ground level in a hydrogen atom? In what region of the electromagnetic pectrum would a photon with thi energy for each of thee two atom lie? Q41.1 (a) Can you how that the orbital angular momentum of an electron in any given direction (e.g., along the z-axi) i alway le than or equal to it total orbital angular momentum? In which cae would the two be equal to each other? (b) I the reult in part (a) true for a claical object, uch a a pinning top or planet? Q41. An atom in it ground level aborb a photon with energy equal to the K aborption edge. Doe aborbing thi photon ionize thi atom? Explain. Q41.3 Can a hydrogen atom emit x ray? If o, how? If not, why not? EXERCISES Section 41. Particle in a Three-Dimenional Box For a particle in a three-dimenional box, what i the degeneracy (number of different quantum tate with the ame energy) of the following energy level: (a) 3p U >ml and (b) 9p U >ml? 41.. CP Model a hydrogen atom a an electron in a cubical box with ide length L. Set the value of L o that the volume of the box equal the volume of a phere of radiu a = 5.9 * m, the Bohr radiu. Calculate the energy eparation between the ground and firt excited level, and compare the reult to thi energy eparation calculated from the Bohr model CP A photon i emitted when an electron in a threedimenional box of ide length 8.00 * m make a tranition from the n X =, n Y =, n Z = 1 tate to the n X = 1, n Y = 1, n Z = 1 tate. What i the wavelength of thi photon? For each of the following tate of a particle in a threedimenional box, at what point i the probability ditribution function a maximum: (a) n X = 1, n Y = 1, n Z = 1 and (b) n X =, n Y =, n Z = 1? A particle i in the three-dimenional box of Section For the tate n X =, n Y =, n Z = 1, for what plane (in addition to the wall of the box) i the probability ditribution function zero? Compare thi number of plane to the correponding number of plane where ƒcƒ i zero for the lower-energy tate n X =, n Y = 1, n Z = 1 and for the ground tate n X = 1, n Y = 1, n Z = 1.

80 1400 CHAPTER 41 Atomic Structure What i the energy difference between the two lowet energy level for a proton in a cubical box with ide length 1.00 * m, the approximate diameter of a nucleu? Section 41.3 The Hydrogen Atom Conider an electron in the N hell. (a) What i the mallet orbital angular momentum it could have? (b) What i the larget orbital angular momentum it could have? Expre your anwer in term of U and in SI unit. (c) What i the larget orbital angular momentum thi electron could have in any choen direction? Expre your anwer in term of U and in SI unit. (d) What i the larget pin angular momentum thi electron could have in any choen direction? Expre your anwer in term of U and in SI unit. (e) For the electron in part (c), what i the ratio of it pin angular momentum in the z-direction to it orbital angular momentum in the z-direction? An electron i in the hydrogen atom with n = 5. (a) Find the poible value of L and L z for thi electron, in unit of U. (b) For each value of L, find all the poible angle between L S and the z-axi. (c) What are the maximum and minimum value of the magnitude of the angle between L S and the z-axi? The orbital angular momentum of an electron ha a magnitude of * kg # m >. What i the angular-momentum quantum number l for thi electron? Conider tate with angular-momentum quantum number l =. (a) In unit of U, what i the larget poible value of L z? (b) In unit of U, what i the value of L? Which i larger: L or the maximum poible L z? (c) For each allowed value of L z, what angle doe the vector L S make with the +z-axi? How doe the minimum angle for l = compare to the minimum angle for l = 3 calculated in Example 41.3? Calculate, in unit of U, the magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom for tate with a principal quantum number of, 0, and 00. Compare each with the value of nu potulated in the Bohr model. What trend do you ee? (a) Make a chart howing all the poible et of quantum number l and m l for the tate of the electron in the hydrogen atom when n = 5. How many combination are there? (b) What are the energie of thee tate? (a) How many different 5g tate doe hydrogen have? (b) Which of the tate in part (a) ha the larget angle between L S and the z-axi, and what i that angle? (c) Which of the tate in part (a) ha the mallet angle between L S and the z-axi, and what i that angle? CALC (a) What i the probability that an electron in the 1 tate of a hydrogen atom will be found at a ditance le than a> from the nucleu? (b) Ue the reult of part (a) and of Example 41.4 to calculate the probability that the electron will be found at ditance between a> and a from the nucleu CALC In Example 41.4 fill in the miing detail that how that P = 1-5e Show that 1f = e imlf = 1f + p (that i, how that 1f i periodic with period p) if and only if m l i retricted to the value 0, 1,, Á. (Hint: Euler formula tate that e if = co f + iin f. ) Section 41.4 The Zeeman Effect A hydrogen atom in a 3p tate i placed in a uniform external magnetic field B S. Conider the interaction of the magnetic field with the atom orbital magnetic dipole moment. (a) What field magnitude B i required to plit the 3p tate into multiple lev- el with an energy difference of.71 * 10-5 ev between adjacent level? (b) How many level will there be? A hydrogen atom i in a d tate. In the abence of an external magnetic field the tate with different m l value have (approximately) the ame energy. Conider the interaction of the magnetic field with the atom orbital magnetic dipole moment. (a) Calculate the plitting (in electron volt) of the m l level when the atom i put in a T magnetic field that i in the +z-direction. (b) Which m l level will have the lowet energy? (c) Draw an energy-level diagram that how the d level with and without the external magnetic field A hydrogen atom in the 5g tate i placed in a magnetic field of T that i in the z-direction. (a) Into how many level i thi tate plit by the interaction of the atom orbital magnetic dipole moment with the magnetic field? (b) What i the energy eparation between adjacent level? (c) What i the energy eparation between the level of lowet energy and the level of highet energy? CP A hydrogen atom undergoe a tranition from a p tate to the 1 ground tate. In the abence of a magnetic field, the energy of the photon emitted i 1 nm. The atom i then placed in a trong magnetic field in the z-direction. Ignore pin effect; conider only the interaction of the magnetic field with the atom orbital magnetic moment. (a) How many different photon wavelength are oberved for the p S 1 tranition? What are the m l value for the initial and final tate for the tranition that lead to each photon wavelength? (b) One oberved wavelength i exactly the ame with the magnetic field a without. What are the initial and final m l value for the tranition that produce a photon of thi wavelength? (c) One oberved wavelength with the field i longer than the wavelength without the field. What are the initial and final m l value for the tranition that produce a photon of thi wavelength? (d) Repeat part (c) for the wavelength that i horter than the wavelength in the abence of the field. Section 41.5 Electron Spin CP Claical Electron Spin. (a) If you treat an electron a a claical pherical object with a radiu of 1.0 * m, what angular peed i neceary to produce a pin angular momentum of magnitude 3 4 U? (b) Ue v = rv and the reult of part (a) to calculate the peed v of a point at the electron equator. What doe your reult ugget about the validity of thi model? A hydrogen atom in the n = 1, m = - 1 tate i placed in a magnetic field with a magnitude of T in the +z-direction. (a) Find the magnetic interaction energy (in electron volt) of the electron with the field. (b) I there any orbital magnetic dipole moment interaction for thi tate? Explain. Can there be an orbital magnetic dipole moment interaction for n Z 1? Calculate the energy difference between the m = 1 ( pin up ) and m = - 1 ( pin down ) level of a hydrogen atom in the 1 tate when it i placed in a 1.45-T magnetic field in the negative z-direction. Which level, m = 1 or m = - 1, ha the lower energy? CP The hyperfine interaction in a hydrogen atom between the magnetic dipole moment of the proton and the pin magnetic dipole moment of the electron plit the ground level into two level eparated by 5.9 * 10-6 ev. (a) Calculate the wavelength and frequency of the photon emitted when the atom make a tranition between thee tate, and compare your anwer to the value given at the end of Section In what part of the electromagnetic pectrum doe thi lie? Such photon are emitted by cold hydrogen cloud in intertellar pace; by detecting thee photon,

81 Problem 1401 atronomer can learn about the number and denity of uch cloud. (b) Calculate the effective magnetic field experienced by the electron in thee tate (ee Fig ). Compare your reult to the effective magnetic field due to the pin-orbit coupling calculated in Example A hydrogen atom in a particular orbital angular momentum tate i found to have j quantum number 7 9 and. What i the letter that label the value of l for the tate? Section 41.6 Many-Electron Atom and the Excluion Principle For germanium 1Ge, Z = 3, make a lit of the number of electron in each ubhell 11,, p, Á. Ue the allowed value of the quantum number along with the excluion principle; do not refer to Table Make a lit of the four quantum number n, l, m l, and m for each of the 10 electron in the ground tate of the neon atom. Do not refer to Table 41. or (a) Write out the ground-tate electron configuration (1,, Á ) for the carbon atom. (b) What element of next-larger Z ha chemical propertie imilar to thoe of carbon? Give the ground-tate electron configuration for thi element (a) Write out the ground-tate electron configuration (1,, Á ) for the beryllium atom. (b) What element of next-larger Z ha chemical propertie imilar to thoe of beryllium? Give the ground-tate electron configuration of thi element. (c) Ue the procedure of part (b) to predict what element of next-larger Z than in (b) will have chemical propertie imilar to thoe of the element you found in part (b), and give it ground-tate electron configuration For magneium, the firt ionization potential i 7.6 ev. The econd ionization potential (additional energy required to remove a econd electron) i almot twice thi, 15 ev, and the third ionization potential i much larger, about 80 ev. How can thee number be undertood? The 5 electron in rubidium (Rb) ee an effective charge of.771e. Calculate the ionization energy of thi electron The energie of the 4, 4p, and 4d tate of potaium are given in Example Calculate Z eff for each tate. What trend do your reult how? How can you explain thi trend? (a) The doubly charged ion N + i formed by removing two electron from a nitrogen atom. What i the ground-tate electron configuration for the N + ion? (b) Etimate the energy of the leat trongly bound level in the L hell of N +. (c) The doubly charged ion P + i formed by removing two electron from a phophoru atom. What i the ground-tate electron configuration for the P + ion? (d) Etimate the energy of the leat trongly bound level in the M hell of P (a) The energy of the tate of lithium i ev. Calculate the value of Z eff for thi tate. (b) The energy of the 4 tate of potaium i ev. Calculate the value of Z eff for thi tate. (c) Compare Z eff for the tate of lithium, the 3 tate of odium (ee Example 41.8), and the 4 tate of potaium. What trend do you ee? How can you explain thi trend? Etimate the energy of the highet-l tate for (a) the L hell of Be + and (b) the N hell of Ca +. Section 41.7 X-Ray Spectra A K a x ray emitted from a ample ha an energy of 7.46 kev. Of which element i the ample made? Calculate the frequency, energy (in kev), and wavelength of the K a x ray for the element (a) calcium 1Ca, Z = 0; (b) cobalt 1Co, Z = 7; (c) cadmium 1Cd, Z = The energie for an electron in the K, L, and M hell of the tungten atom are -69,500 ev, -1,000 ev, and -00 ev, repectively. Calculate the wavelength of the K a and K b x ray of tungten. PROBLEMS E 1,1, In term of the ground-tate energy, what i the energy of the highet level occupied by an electron when 10 electron are placed into a cubical box? CALC A particle in the three-dimenional box of Section 41. i in the ground tate, where n X = n Y = n Z = 1. (a) Calculate the probability that the particle will be found omewhere between x = 0 and x = L>. (b) Calculate the probability that the particle will be found omewhere between x = L>4 and x = L>. Compare your reult to the reult of Example 41.1 for the probability of finding the particle in the region x = 0 to x = L> CALC A particle i in the three-dimenional box of Section 41.. (a) Conider the cubical volume defined by 0 x L>4, 0 y L>4, and 0 z L>4. What fraction of the total volume of the box i thi cubical volume? (b) If the particle i in the ground tate ( n X = 1, n Y = 1, n Z = 1) calculate the probability that the particle will be found in the cubical volume defined in part (a). (c) Repeat the calculation of part (b) when the particle i in the tate n X =, n Y = 1, n Z = CALC A particle i decribed by the normalized wave function c1x, y, z) = Axe -ax e -by e -gz, where A, a, b, and g are all real, poitive contant. The probability that the particle will be found in the infiniteimal volume dx dy dz centered at the point 1x i ƒ c1x 0, y 0, z 0 ƒ 0, y 0, z 0 ) dx dy dz. (a) At what value of x 0 i the particle mot likely to be found? (b) Are there value of x 0 for which the probability of the particle being found i zero? If o, at what x 0? CALC A particle i decribed by the normalized wave function c1x, y, z) = Ae -a1x +y +z, where A and a are real, poitive contant. (a) Determine the probability of finding the particle at a ditance between r and r + dr from the origin. (Hint: See Problem Conider a pherical hell centered on the origin with inner radiu r and thickne dr.) (b) For what value of r doe the probability in part (a) have it maximum value? I thi the ame value of r for which ƒ c1x, y, zƒ i a maximum? Explain any difference CP CALC A Three-Dimenional Iotropic Harmonic Ocillator. An iotropic harmonic ocillator ha the potentialenergy function U1x, y, z = 1 k 1x + y + z. (Iotropic mean that the force contant k i the ame in all three coordinate direction.) (a) Show that for thi potential, a olution to Eq. (41.5) i given by c = c nx 1xc ny 1yc nz 1z. In thi expreion, c nx 1x i a olution to the one-dimenional harmonic ocillator Schrödinger equation, Eq. (40.44), with energy E nx = An x + 1 BUv. The function c ny 1y and c nz 1z are analogou one-dimenional wave function for ocillation in the y- and z-direction. Find the energy aociated with thi c. (b) From your reult in part (a) what are the ground-level and firt-excited-level energie of the threedimenional iotropic ocillator? (c) Show that there i only one tate (one et of quantum number n x, n y, and n z ) for the ground level but three tate for the firt excited level CP CALC Three-Dimenional Aniotropic Harmonic Ocillator. An ocillator ha the potential-energy function U1x, y, z = 1 where k 1 œ 7 k. œ k 11x œ + y + 1 k z œ, Thi ocillator i called aniotropic becaue the force contant i not the ame in all three coordinate direction. (a) Find a general expreion

82 140 CHAPTER 41 Atomic Structure for the energy level of the ocillator (ee Problem 41.44). (b) From your reult in part (a), what are the ground-level and firt-excited-level energie of thi ocillator? (c) How many tate (different et of quantum number n x, n y, and n z ) are there for the ground level and for the firt excited level? Compare to part (c) of Problem An electron in hydrogen i in the 5ƒ tate. (a) Find the larget poible value of the z-component of it angular momentum. (b) Show that for the electron in part (a), the correponding x- and y-component of it angular momentum atify the equation L x + L y =U (a) Show that the total number of atomic tate (including different pin tate) in a hell of principal quantum number n i n. [Hint: The um of the firt N integer Á + N i equal to N1N + 1>. ] (b) Which hell ha 50 tate? (a) What i the lowet poible energy (in electron volt) of an electron in hydrogen if it orbital angular momentum i 1U? (b) What are the larget and mallet value of the z-component of the orbital angular momentum (in term of U) for the electron in part (a)? (c) What are the larget and mallet value of the pin angular momentum (in term of U) for the electron in part (a)? (d) What are the larget and mallet value of the orbital angular momentum (in term of U) for an electron in the M hell of hydrogen? Conider an electron in hydrogen having total energy ev. (a) What are the poible value of it orbital angular momentum (in term of U? (b) What wavelength of light would it take to excite thi electron to the next higher hell? I thi photon viible to human? (a) Show all the ditinct tate for an electron in the N hell of hydrogen. Include all four quantum number. (b) For an ƒ electron in the N hell, what i the larget poible orbital angular momentum and the greatet poitive value for the component of thi angular momentum along any choen direction (the z-axi)? What i the magnitude of it pin angular momentum? Expre thee quantitie in unit of U. (c) For an electron in the d tate of the N hell, what are the maximum and minimum angle between it angular momentum vector and any choen direction (the z-axi)? (d) What i the larget value of the orbital angular momentum for an ƒ electron in the M hell? (a) The energy of an electron in the 4 tate of odium i ev. What i the effective net charge of the nucleu een by thi electron? On the average, how many electron creen the nucleu? (b) For an outer electron in the 4p tate of potaium, on the average 17. inner electron creen the nucleu. (i) What i the effective net charge of the nucleu een by thi outer electron? (ii) What i the energy of thi outer electron? CALC For a hydrogen atom, the probability P1r of finding the electron within a pherical hell with inner radiu r and outer radiu r + dr i given by Eq. (41.5). For a hydrogen atom in the 1 ground tate, at what value of r doe P1r have it maximum value? How doe your reult compare to the ditance between the electron and the nucleu for the n = 1 tate in the Bohr model, Eq. (41.6)? CALC Conider a hydrogen atom in the 1 tate. (a) For what value of r i the potential energy U1r equal to the total energy E? Expre your anwer in term of a. Thi value of r i called the claical turning point, ince thi i where a Newtonian particle would top it motion and revere direction. (b) For r greater than the claical turning point, U1r 7 E. Claically, the particle cannot be in thi region, ince the kinetic energy cannot be negative. Calculate the probability of the electron being found in thi claically forbidden region CP Rydberg Atom. Rydberg atom are atom whoe outermot electron i in an excited tate with a very large principal quantum number. Rydberg atom have been produced in the laboratory and detected in intertellar pace. (a) Why do all neutral Rydberg atom with the ame n value have eentially the ame ionization energy, independent of the total number of electron in the atom? (b) What i the ionization energy for a Rydberg atom with a principal quantum number of 350? What i the radiu in the Bohr model of the Rydberg electron orbit? (c) Repeat part (b) for n = CALC The wave function for a hydrogen atom in the tate i c 1r = 1 3pa a - r 3 a be-r>a (a) Verify that thi function i normalized. (b) In the Bohr model, the ditance between the electron and the nucleu in the n = tate i exactly 4a. Calculate the probability that an electron in the tate will be found at a ditance le than 4a from the nucleu CALC The normalized wave function for a hydrogen atom in the tate i given in Problem (a) For a hydrogen atom in the tate, at what value of r i P(r) maximum? How doe your reult compare to 4a, the ditance between the electron and the nucleu in the n = tate of the Bohr model? (b) At what value of r (other than r = 0 or r = q) i P(r) equal to zero, o that the probability of finding the electron at that eparation from the nucleu i zero? Compare your reult to Fig (a) For an excited tate of hydrogen, how that the mallet angle that the orbital angular momentum vector have with the z-axi i n - 1 1u L min = arcco n1n - 1 can (b) What i the correponding expreion for 1u L ) max, the larget poible angle between L S and the z-axi? (a) If the value of L z i known, we cannot know either L x or L y preciely. But we can know the value of the quantity Lx + Ly. Write an expreion for thi quantity in term of l, m and (b) What i the meaning of Lx + L l, U. y? (c) For a tate of nonzero orbital angular momentum, find the maximum and minimum value of Lx + Ly. Explain your reult CALC The normalized radial wave function for the p tate of the hydrogen atom i R p = A1> 4a 5 Bre -r>a. After we average over the angular variable, the radial probability function become P1r dr = 1R p r dr. At what value of r i P(r) for the p tate a maximum? Compare your reult to the radiu of the n = tate in the Bohr model CP Stern Gerlach Experiment. In a Stern Gerlach experiment, the deflecting force on the atom i F z = -m z 1dB z >dz, where m z i given by Eq. (41.40) and db z >dz i the magnetic-field gradient. In a particular experiment the magnetic-field region i 50.0 cm long; aume the magnetic-field gradient i contant in thi region. A beam of ilver atom enter the magnetic field with a peed of 55 m>. What value of db z >dz i required to give a eparation of 1.0 mm between the two pin component a they exit the field? (Note: The magnetic dipole moment of ilver i the ame a that for hydrogen, ince it valence electron i in an l = 0 tate.) Conider the tranition from a 3d to a p tate of hydrogen in an external magnetic field. Aume that the effect of electron L S

83 Challenge Problem 1403 pin can be ignored (which i not actually the cae) o that the magnetic field interact only with the orbital angular momentum. Identify each allowed tranition by the m l value of the initial and final tate. For each of thee allowed tranition, determine the hift of the tranition energy from the zero-field value and how that there are three different tranition energie An atom in a 3d tate emit a photon of wavelength nm when it decay to a p tate. (a) What i the energy (in electron volt) of the photon emitted in thi tranition? (b) Ue the election rule decribed in Section 41.4 to find the allowed tranition if the atom i now in an external magnetic field of T. Ignore the effect of the electron pin. (c) For the cae in part (b), if the energy of the 3d tate wa originally ev with no magnetic field preent, what will be the energie of the tate into which it plit in the magnetic field? (d) What are the allowed wavelength of the light emitted during tranition in part (b)? CALC Spectral Analyi. While tudying the pectrum of a ga cloud in pace, an atronomer magnifie a pectral line that reult from a tranition from a p tate to an tate. She find that the line at nm ha actually plit into three line, with adjacent line nm apart, indicating that the ga i in an external magnetic field. (Ignore effect due to electron pin.) What i the trength of the external magnetic field? A hydrogen atom make a tranition from an n = 3 tate to an n = tate (the Balmer H a line) while in a magnetic field in the +z-direction and with magnitude 1.40 T. (a) If the magnetic quantum number i m l = in the initial 1n = 3 tate and m l = 1 in the final 1n = tate, by how much i each energy level hifted from the zero-field value? (b) By how much i the wavelength of the H a line hifted from the zero-field value? I the wavelength increaed or decreaed? Diregard the effect of electron pin. [Hint: Ue the reult of Problem 39.86(c).] CP A large number of hydrogen atom in 1 tate are placed in an external magnetic field that i in the +z-direction. Aume that the atom are in thermal equilibrium at room temperature, T = 300 K. According to the Maxwell Boltzmann ditribution (ee Section 39.4), what i the ratio of the number of atom in the m tate to the number in the m = - 1 = 1 tate when the magnetic-field magnitude i (a) 5.00 * 10-5 T (approximately the earth field); (b) T; (c) 5.00 T? Effective Magnetic Field. An electron in a hydrogen atom i in the p tate. In a imple model of the atom, aume that the electron circle the proton in an orbit with radiu r equal to the Bohr-model radiu for n =. Aume that the peed v of the orbiting electron can be calculated by etting L = mvr and taking L to have the quantum-mechanical value for a p tate. In the frame of the electron, the proton orbit with radiu r and peed v. Model the orbiting proton a a circular current loop, and calculate the magnetic field it produce at the location of the electron Weird Univere. In another univere, the electron i a pin- 3 rather than a pin- 1 particle, but all other phyic are the ame a in our univere. In thi univere, (a) what are the atomic number of the lightet two inert gae? (b) What i the groundtate electron configuration of odium? For an ion with nuclear charge Z and a ingle electron, the electric potential energy i -Ze >4pP 0 r and the expreion for the energie of the tate and for the normalized wave function are obtained from thoe for hydrogen by replacing e by Ze. Conider the N 6+ ion, with even proton and one electron. (a) What i the ground-tate energy in electron volt? (b) What i the ionization energy, the energy required to remove the electron from the N 6+ ion if it i initially in the ground tate? (c) What i the ditance a [given for hydrogen by Eq. (41.6)] for thi ion? (d) What i the wavelength of the photon emitted when the N 6+ ion make a tranition from the n = tate to the n = 1 ground tate? A hydrogen atom in an n =, l = 1, m l = -1 tate emit a photon when it decay to an n = 1, l = 0, m l = 0 ground tate. (a) In the abence of an external magnetic field, what i the wavelength of thi photon? (b) If the atom i in a magnetic field in the +z-direction and with a magnitude of.0 T, what i the hift in the wavelength of the photon from the zero-field value? Doe the magnetic field increae or decreae the wavelength? Diregard the effect of electron pin. [Hint: Ue the reult of Problem 39.86(c).] A lithium atom ha three electron, and the S 1> groundtate electron configuration i 1. The 1 p excited tate i plit into two cloely paced level, P and 3> P 1>, by the pinorbit interaction (ee Example 41.7 in Section 41.5). A photon with wavelength mm i emitted in the P 3> S S 1> tranition, and a photon with wavelength mm i emitted in the P 1> S S 1> tranition. Calculate the effective magnetic field een by the electron in the 1 p tate of the lithium atom. How doe your reult compare to that for the 3p level of odium found in Example 41.7? Etimate the minimum and maximum wavelength of the characteritic x ray emitted by (a) vanadium 1Z = 3 and (b) rhenium 1Z = 45. Dicu any approximation that you make CP Electron Spin Reonance. Electron in the lower of two pin tate in a magnetic field can aborb a photon of the right frequency and move to the higher tate. (a) Find the magnetic-field magnitude B required for thi tranition in a hydrogen atom with n = 1 and l = 0 to be induced by microwave with wavelength l. (b) Calculate the value of B for a wavelength of 3.50 cm. CHALLENGE PROBLEMS Each of N electron (ma m) i free to move along the x-axi. The potential-energy function for each electron i U1x = 1 k x, where k i a poitive contant. The electric and magnetic interaction between electron can be ignored. Ue the excluion principle to how that the minimum energy of the ytem of N electron i UN k >m. (Hint: See Section 40.5 and the hint given in Problem ) CP Conider a imple model of the helium atom in which two electron, each with ma m, move around the nucleu (charge +e) in the ame circular orbit. Each electron ha orbital angular momentum U (that i, the orbit i the mallet-radiu Bohr orbit), and the two electron are alway on oppoite ide of the nucleu. Ignore the effect of pin. (a) Determine the radiu of the orbit and the orbital peed of each electron. [Hint: Follow the procedure ued in Section 39.3 to derive Eq. (39.8) and (39.9). Each electron experience an attractive force from the nucleu and a repulive force from the other electron.] (b) What i the total kinetic energy of the electron? (c) What i the potential energy of the ytem (the nucleu and the two electron)? (d) In thi model, how much energy i required to remove both electron to infinity? How doe thi compare to the experimental value of 79.0 ev? CALC Repeat the calculation of Problem for a one-electron ion with nuclear charge Z. (See Problem ) How doe the probability of the electron being found in the claically forbidden region depend on Z?

84 1404 CHAPTER 41 Atomic Structure Anwer Chapter Opening Quetion? Helium i inert becaue it ha a filled K hell, while odium i very reactive becaue it third electron i looely bound in an L hell. See Section 41.6 for more detail. Tet Your Undertanding Quetion 41.1 Anwer: (iv) If U1x, y, z = 0 in a certain region of pace, we can rewrite the time-independent Schrödinger equation [Eq. (41.5)] for that region a 0 c>0x +0 c>0y + 0 c>0z = 1-mE>U c. We are told that all of the econd derivative of c1x, y, z are poitive in thi region, o the left-hand ide of thi equation i poitive. Hence the right-hand ide 1-mE>U c mut alo be poitive. Since E 7 0, the quantity -me>u i negative, and o c1x, y, z mut be negative. 41. Anwer: (iv), (ii), (i) and (iii) (tie) Equation (41.16) how that the energy level for a cubical box are proportional to the quantity n X + n Y + n Z. Hence ranking in order of thi quantity i the ame a ranking in order of energy. For the four cae we are given, we have (i) n X + n Y + n Z = = 17; (ii) n ; (iii) n X + n X + n Y + n Z = = 18 Y + n ; and (iv) n X + n Y + n Z = = 17 Z = = 19. The tate 1n X, n Y, n Z = 1, 3, and 1n X, n Y, n Z = 1,, 3 have the ame energy (they are degenerate) Anwer: (ii) and (iii) (tie), (i) An electron in a tate with principal quantum number n i mot likely to be found at r = n a. Thi reult i independent of the value of the quantum number l and m l. Hence an electron with n = (mot likely to be found at r = 4a) i more likely to be found near r = 5a than an electron with n = 1 (mot likely to be found at r = a) Anwer: no All that matter i the component of the electron orbital magnetic moment along the direction of B S. We called thi quantity m z in Eq. (41.3) becaue we defined the poitive z-axi to be in the direction of B S. In reality, the name of the axe are entirely arbitrary Anwer: (iv) For the magnetic moment to be perfectly aligned with the z-direction, the z-component of the pin vector S S would have to have the ame abolute value a. However, the poible value of S z are 1 U [Eq. (41.37)], while the magnitude S of the pin vector i S = 3 4 U [Eq. (41.38)]. Hence can never be perfectly aligned with any one direction in pace Anwer: more difficult If there were no excluion principle, all 11 electron in the odium atom would be in the level of lowet energy (the 1 level) and the configuration would be Conequently, it would be more difficult to remove the firt electron. (In a real odium atom the valence electron i in a creened 3 tate, which ha a comparatively high energy.) 41.7 Anwer: (iv) An aborption edge appear if the photon energy i jut high enough to remove an electron in a given energy level from the atom. In a ample of high-temperature hydrogen we expect to find atom whoe electron i in the ground level 1n = 1, the firt excited level 1n =, and the econd excited level 1n = 3. From Eq. (41.1) thee level have energie E n = ev)>n = ev, ev, and ev (ee Fig. 38.9b). Bridging Problem 11, 1, 1, + 1, Anwer: (a).37 * m (b) Value of X, n Y, n Z, m for the electron: 11, 1, 1, -1, 1, 1, 1, +1, 1, 1, 1, -1, 11,, 1, + 1, 11,, 1, -1, 11, 1,, +1, 11, 1,, -1, 1,, 1, + 1, 1,, 1, -1, 1, 1,, +1, 1, 1,, -1, 11,,, + 1, 11,,, -1, 13, 1, 1, +1, 13, 1, 1, -1, 11, 3, 1, + 1, 11, 3, 1, -1, 11, 1, 3, +1, 11, 1, 3, -1, 1,,, + 1, 1,,, - 1 (c) 0.1 ev, 40. ev, 60.3 ev, 73.7 ev, and 80.4 ev (d) 60.3 ev veru 4.5 * 10 3 ev

85 MOLECULES AND CONDENSED MATTER 4 LEARNING GOALS By tudying thi chapter, you will learn:? Thi fale-color image of Venu how it thick, cloud-hrouded atmophere, which i 96.5% carbon dioxide (CO ). The atmophere raie the urface temperature of Venu to 735 K 146 C = 863 F hotter even than Mercury, the cloet planet to the un. What property of CO molecule make them a potent agent for raiing Venu temperature? In Chapter 41 we dicued the tructure and propertie of iolated atom. But uch atom are the exception; uually we find atom combined to form molecule or more extended tructure we call condened matter (liquid or olid). It the attractive force between atom, called molecular bond, that caue them to combine. In thi chapter we ll tudy everal kind of bond a well a the energy level and pectra aociated with diatomic molecule. We will ee that jut a atom have quantized energie determined by the quantum-mechanical tate of their electron, o molecule have quantized energie determined by their rotational and vibrational tate. The ame phyical principle behind molecular bond alo apply to the tudy of condened matter, in which variou type of bonding occur. We ll explore the concept of energy band and ee how it help u undertand the propertie of olid. Then we ll look more cloely at the propertie of a pecial cla of olid called emiconductor. Device uing emiconductor are found in every radio, TV, pocket calculator, and computer ued today; they have revolutionized the entire field of electronic during the pat half-century. The variou type of bond that hold atom together. How the rotational and vibrational dynamic of molecule are revealed by molecular pectra. How and why atom form into crytalline tructure. How to ue the energy-band concept to explain the electrical propertie of olid. A imple model for metal that explain many of their phyical propertie. How the character of a emiconductor can be radically tranformed by adding mall amount of an impurity. Some of the technological application of emiconductor device. Why certain material become uperconductor at low temperature. 4.1 Type of Molecular Bond We can ue our dicuion of atomic tructure in Chapter 41 a a bai for exploring the nature of molecular bond, the interaction that hold atom together to form table tructure uch a molecule and olid. 1405

86 1406 CHAPTER 4 Molecule and Condened Matter 4.1 When the eparation r between two oppoitely charged ion i large, the potential energy U(r) i proportional to 1/r a for point charge and the force i attractive. A r decreae, the charge cloud of the two atom overlap and the force become le attractive. If r i le than the equilibrium eparation r 0, the force i repulive. U(r) 0 r 0 U 0 r, r 0 : U decreae with increaing eparation r; force i repulive. r. r 0 : U decreae with decreaing eparation r; force i attractive. r Ionic Bond The ionic bond i an interaction between oppoitely charged ionized atom. The mot familiar example i odium chloride (NaCl), in which the odium atom give it one 3 electron to the chlorine atom, filling the vacancy in the 3p ubhell of chlorine. Let look at the energy balance in thi tranaction. Removing the 3 electron from a neutral odium atom require ev of energy; thi i called the ionization energy of odium. The neutral chlorine atom can attract an extra electron into the vacancy in the 3p ubhell, where it i incompletely creened by the other electron and therefore i attracted to the nucleu. Thi tate ha ev lower energy than a neutral chlorine atom and a ditant free electron; ev i the magnitude of the electron affinity of chlorine. Thu creating the well-eparated Na + and Cl - ion require a net invetment of only ev ev = 1.55 ev. When the two oppoitely charged ion are brought together by their mutual attraction, the magnitude of their negative potential energy i determined by how cloely they can approach each other. Thi in turn i limited by the excluion principle, which forbid extenive overlap of the electron cloud of the two ion. A the ditance decreae, the excluion principle ditort the charge cloud, o the ion no longer interact like point charge and the interaction eventually become repulive (Fig. 4.1). The minimum electric potential energy for NaCl turn out to be -5.7 ev at a eparation of 0.4 nm. The net energy releaed in creating the ion and letting them come together to the equilibrium eparation of 0.4 nm i 5.7 ev ev = 4. ev. Thu, if the kinetic energy of the ion i neglected, 4. ev i the binding energy of the NaCl molecule, the energy that i needed to diociate the molecule into eparate neutral atom. Ionic bond can involve more than one electron per atom. For intance, alkaline earth element form ionic compound in which an atom loe two electron; an example i magneium chloride, or Mg + 1Cl -. Ionic bond that involve a lo of more than two electron are relatively rare. Intead, a different kind of bond, the covalent bond, come into operation. We ll dicu thi type of bond below. Example 4.1 Electric potential energy of the NaCl molecule Find the electric potential energy of an Na + ion and a Cl - ion eparated by 0.4 nm. Conider the ion a point charge. SOLUTION IDENTIFY and SET UP: Equation (3.9) in Section 3.1 tell u that the electric potential energy of two point charge q and q 0 eparated by a ditance r i U = qq 0 >4pP 0 r. EXECUTE: We have q = +e (for Na + ), q (for Cl - 0 = -e ), and r = 0.4 nm = 0.4 * 10-9 m. From Eq. (3.9), U = - 1 e = * 10 9 N # m >C 11.6 * C 4pP 0 r * 10-9 m = -9.6 * J = -6.0 ev EVALUATE: Thi reult agree fairly well with the oberved value of 5.7 ev. The reaon for the difference i that when the two ion are at their equilibrium eparation of 0.4 nm, the outer region of their electron cloud overlap. Hence the two ion don t behave exactly like point charge. PhET: Double Well and Covalent Bond Covalent Bond The covalent bond i characterized by a more egalitarian participation of the two atom than occur with the ionic bond. The implet covalent bond i found in the hydrogen molecule, a tructure containing two proton and two electron. A the eparate atom (Fig. 4.a) come together, the electron wave function are ditorted and become more concentrated in the region between the two proton (Fig. 4.b). The net attraction of the electron for each proton more than balance the repulion of the two proton and of the two electron. The attractive interaction i then upplied by a pair of electron, one contributed by each atom, with charge cloud that are concentrated primarily in the region between the two atom. The energy of the covalent bond in the hydrogen molecule i ev. H

87 4.1 Type of Molecular Bond 1407 A we aw in Chapter 41, the excluion principle permit two electron to occupy the ame region of pace (that i, to be in the ame patial quantum tate) only when they have oppoite pin. When the pin are parallel, the excluion principle forbid the molecular tate that would be mot favorable from energy conideration (with both electron in the region between atom). Oppoite pin are an eential requirement for a covalent bond, and no more than two electron can participate in uch a bond. However, an atom with everal electron in it outermot hell can form everal covalent bond. The bonding of carbon and hydrogen atom, of central importance in organic chemitry, i an example. In the methane molecule (CH 4 ) the carbon atom i at the center of a regular tetrahedron, with a hydrogen atom at each corner. The carbon atom ha four electron in it L hell, and each of thee four electron form a covalent bond with one of the four hydrogen atom (Fig. 4.3). Similar pattern occur in more complex organic molecule. Becaue of the role played by the excluion principle, covalent bond are highly directional. In the methane molecule the wave function for each of carbon four valence electron i a combination of the and p wave function called a hybrid wave function. The probability ditribution for each one ha a lobe protruding toward a corner of a tetrahedron. Thi ymmetric arrangement minimize the overlap of wave function for the electron pair, minimizing their repulive potential energy. Ionic and covalent bond repreent two extreme in molecular bonding, but there i no harp diviion between the two type. Often there i a partial tranfer of one or more electron from one atom to another. A a reult, many molecule that have diimilar atom have electric dipole moment that i, a preponderance of poitive charge at one end and of negative charge at the other. Such molecule are called polar molecule. Water molecule have large electric dipole moment; thee are reponible for the exceptionally large dielectric contant of liquid water (ee Section 4.4 and 4.5). van der Waal Bond Ionic and covalent bond, with typical bond energie of 1 to 5 ev, are called trong bond. There are alo two type of weaker bond. One of thee, the van der Waal bond, i an interaction between the electric dipole moment of atom or molecule; typical energie are 0.1 ev or le. The bonding of water molecule in the liquid and olid tate reult partly from dipole dipole interaction. No atom ha a permanent electric dipole moment, nor do many molecule. However, fluctuating charge ditribution can lead to fluctuating dipole moment; thee in turn can induce dipole moment in neighboring tructure. Overall, the reulting dipole dipole interaction i attractive, giving a weak bonding of atom or molecule. The interaction potential energy drop off very quickly with ditance r between molecule, uually a 1/r 6. The liquefaction and olidification of the inert gae and of molecule uch a H, O, and N are due to induceddipole van der Waal interaction. Not much thermal-agitation energy i needed to break thee weak bond, o uch ubtance uually exit in the liquid and olid tate only at very low temperature. 4. Covalent bond in a hydrogen molecule. (a) Separate hydrogen atom H Individual H atom are uually widely eparated and do not interact. (b) H molecule Nucleu (proton) Covalent bond: the charge cloud for the two electron with oppoite pin are concentrated in the region between the nuclei. H H H C H H 4.3 Schematic diagram of the methane 1CH 4 molecule. The carbon atom i at the center of a regular tetrahedron and form four covalent bond with the hydrogen atom at the corner. Each covalent bond include two electron with oppoite pin, forming a charge cloud that i concentrated between the carbon atom and a hydrogen atom. H Hydrogen Bond In the other type of weak bond, the hydrogen bond, a proton (H + ion) get between two atom, polarizing them and attracting them by mean of the induced dipole. Thi bond i unique to hydrogen-containing compound becaue only hydrogen ha a ingly ionized tate with no remaining electron cloud; the hydrogen ion i a bare proton, much maller than any other ingly ionized atom. The bond energy i uually le than 0.5 ev. The hydrogen bond i reponible for the cro-linking of long-chain organic molecule uch a polyethylene (ued in platic bag). Hydrogen bonding alo play a role in the tructure of ice.

88 1408 CHAPTER 4 Molecule and Condened Matter Application Molecular Zipper A DNA molecule function like a twited zipper. Each of the two trand of the zipper conit of an outer backbone and inward-facing nucleotide teeth ; hydrogen bond between facing teeth zip the trand together. The covalent bond that hold together the atom of each trand are trong, wherea the hydrogen bond are relatively weak, o that the cell biochemical machinery can eaily eparate the trand for reading or copying. All thee bond type hold the atom together in olid a well a in molecule. Indeed, a olid i in many repect a giant molecule. Still another type of bonding, the metallic bond, come into play in the tructure of metallic olid. We ll return to thi ubject in Section 4.3. Tet Your Undertanding of Section 4.1 If electron obeyed the excluion principle but did not have pin, how many electron could participate in a covalent bond? (i) one; (ii) two; (iii) three; (iv) more than three. 4. Molecular Spectra Molecule have energy level that are aociated with rotational motion of a molecule a a whole and with vibrational motion of the atom relative to each other. Jut a tranition between energy level in atom lead to atomic pectra, tranition between rotational and vibrational level in molecule lead to molecular pectra. 4.4 A diatomic molecule modeled a two point mae m 1 and m eparated by a ditance r 0. The ditance of the mae from the center of ma are r 1 and r, where r 1 + r = r 0. r r r m 1 cm m 4.5 The ground level and firt four excited rotational energy level for a diatomic molecule. The level are not equally paced. E 10h /I 6h /I 3h /I h /I 0 l 5 4 l 5 3 l 5 l 5 1 l 5 0 Rotational Energy Level In thi dicuion we ll concentrate motly on diatomic molecule, to keep thing a imple a poible. In Fig. 4.4 we picture a diatomic molecule a a rigid dumbbell (two point mae m 1 and m eparated by a contant ditance r 0 ) that can rotate about axe through it center of ma, perpendicular to the line joining them. What are the energy level aociated with thi motion? We howed in Section 10.5 that when a rigid body rotate with angular peed v about a perpendicular axi through it center of ma, the magnitude L of it angular momentum i given by Eq. (10.8), L = Iv, where I i it moment of inertia about that ymmetry axi. It kinetic energy i given by Eq. (9.17), K = 1 Combining thee two equation, we find K = L Iv. >I. There i no potential energy U, o the kinetic energy K i equal to the total mechanical energy E: (4.1) Zero potential energy mean that U doe not depend on the angular coordinate of the molecule. But the potential-energy function U for the hydrogen atom (ee Section 41.3) alo ha no dependence on angular coordinate. Thu the angular olution to the Schrödinger equation for rigid-body rotation are the ame a for the hydrogen atom, and the angular momentum i quantized in the ame way. A in Eq. (41.1), Combining Eq. (4.1) and (4.), we obtain the rotational energy level: E l = l1l + 1 U I 1l = 0, 1,, Á (rotational energy level, diatomic molecule) (4.) (4.3) Figure 4.5 i an energy-level diagram howing thee rotational level. The ground level ha zero quantum number l, correponding to zero angular momentum (no rotation and zero rotational energy E). The pacing of adjacent level increae with increaing l. We can expre the moment of inertia I in Eq. (4.1) and (4.3) in term of the reduced ma of the molecule: m r L = l1l + 1U 1l = 0, 1,, Á m r = E = L I m 1 m m 1 + m (4.4) We introduced the reduced ma in Section 39.3 to accommodate the finite nuclear ma of the hydrogen atom. In Fig. 4.4 the ditance and are the r 1 r

89 4. Molecular Spectra 1409 ditance from the center of ma to the nuclei of the atom. By definition of the center of ma, m 1 r 1 = m r, and the figure alo how that r 0 = r 1 + r. Solving thee equation for r 1 and r, we find PhET: The Greenhoue Effect r 1 = m m 1 + m r 0 r = m 1 m 1 + m r 0 (4.5) The moment of inertia i I = m 1 r 1 + m r ; ubtituting Eq. (4.5), we find m I = m 1 1m 1 + m r 0 m 1 + m 1m 1 + m r 0 m 1 m = r m 1 + m 0 or I = m r r 0 (moment of inertia of a diatomic molecule) (4.6) The reduced ma enable u to reduce thi two-body problem to an equivalent one-body problem (a particle of ma m r moving around a circle with radiu r 0 ), jut a we did with the hydrogen atom. Indeed, the only difference between thi problem and the hydrogen atom i the difference in the radial force. To conerve angular momentum and account for the angular momentum of the emitted or aborbed photon, the allowed tranition between rotational tate mut atify the ame election rule that we dicued in Section 41.4 for allowed tranition between the tate of an atom: l mut change by exactly one unit, that i, l = 1. Example 4. Rotational pectrum of carbon monoxide The two nuclei in the carbon monoxide (CO) molecule are nm apart. The ma of the mot common carbon atom i * 10-6 kg; that of the mot common oxygen atom i.656 * 10-6 kg. (a) Find the energie of the lowet three rotational energy level of CO. Expre your reult in mev11 mev = 10-3 ev. (b) Find the wavelength of the photon emitted in the tranition from the l = to the l = 1 level. SOLUTION IDENTIFY and SET UP: Thi problem ue the idea developed in thi ection about the rotational energy level of molecule. We are given the ditance r 0 between the atom and their mae m 1 and m. We find the reduced ma m r uing Eq. (4.4), the moment of inertia I uing Eq. (4.6), and the energie E l uing Eq. (4.3). The energy E of the emitted photon i equal to the difference in energy between the l = and l = 1 level. (Thi tranition obey the l = 1 election rule, ince l = 1 - = -1.) We determine the photon wavelength uing E = hc>l. EXECUTE: (a) From Eq. (4.4) and (4.6), the reduced ma and moment of inertia of the CO molecule are: m 1 m m r = m 1 + m * 10-6 kg1.656 * 10-6 kg = * 10-6 kg * 10-6 kg = * 10-6 kg I = m r r0 = * 10-6 kg * 10-9 m = * kg # m The rotational level are given by Eq. (4.3): E l = l1l + 1 U I = l1l * 10-3 J = l1l mev 11 mev = 10-3 ev. Subtituting l = 0, 1,, we find E 0 = 0 E 1 = mev E = mev (b) The photon energy and wavelength are: E = E - E 1 = mev l = hc E * J # = l1l * kg # m * ev # * 10 8 m> = * 10-3 ev = 1.9 * 10-3 m = 1.9 mm EVALUATE: The difference between the firt few rotational energy level of CO are very mall (about 1 mev = 10-3 ev) compared to the difference between atomic energy level (typically a few ev). Hence a photon emitted by a CO molecule in a tranition from the l = to the l = 1 level ha very low energy and a very long wavelength compared to the viible light emitted by excited atom. Photon wavelength for rotational tranition in molecule are typically in the microwave and far infrared region of the pectrum. In thi example we were given the equilibrium eparation between the atom, alo called the bond length, and we ued it to calculate one of the wavelength emitted by excited CO molecule. In actual experiment, cientit work thi problem backward: By meauring the long-wavelength emiion of a ample of diatomic molecule, they determine the moment of inertia of the molecule and hence the bond length.

90 1410 CHAPTER 4 Molecule and Condened Matter 4.6 A diatomic molecule modeled a two point mae m 1 and m connected by a pring with force contant k. m 1 E 3 cm k m 4.7 The ground level and firt three excited vibrational level for a diatomic molecule, auming mall diplacement from equilibrium o we can treat the ocillation a imple harmonic. (Compare Fig ) E hv hv hv 1 hv 0 r 0 n 5 3 n 5 n 5 1 n Energy-level diagram for vibrational and rotational energy level of a diatomic molecule. For each vibrational level 1n there i a erie of more cloely paced rotational level 1l. Several tranition correponding to a ingle band in a band pectrum are hown. Thee tranition obey the election rule l = 1. n = l = 0 Vibrational Energy Level Molecule are never completely rigid. In a more realitic model of a diatomic molecule we repreent the connection between atom not a a rigid rod but a a pring (Fig. 4.6). Then, in addition to rotating, the atom of the molecule can vibrate about their equilibrium poition along the line joining them. For mall ocillation the retoring force can be taken a proportional to the diplacement from the equilibrium eparation r 0 (like a pring that obey Hooke law with a force contant k ), and the ytem i a harmonic ocillator. We dicued the quantum-mechanical harmonic ocillator in Section The energy level are given by Eq. (40.46) with the ma m replaced by the reduced ma m r : E n = 1n + 1 Uv = 1n + 1 U k 1n = 0, 1,, Á A m r (vibrational energy level of a diatomic molecule) (4.7) Thi repreent a erie of level equally paced in energy, with an energy eparation of k E =Uv =U (4.8) A m r Figure 4.7 i an energy-level diagram howing thee vibrational level. A an example, for the carbon monoxide molecule of Example 4. the pacing Uv between vibrational energy level i ev. From Eq. (4.8) thi correpond to a force contant of 1.90 * 10 3 N>m, which i a fairly looe pring. (To tretch a macrocopic pring with thi value of k by 1.0 cm would require a force of only 19 N, or about 4 lb.) Force contant for diatomic molecule are typically about 100 to 000 N> m. CAUTION Watch out for k, k, and K A in Section 40.5 we re again uing k for the force contant, thi time to minimize confuion with Boltzmann contant k, the ga contant per molecule (introduced in Section 18.3). Beide the quantitie k and k, we alo ue the abolute temperature unit 1K = 1 kelvin. Rotation and Vibration Combined Viible-light photon have energie between 1.65 ev and 3.6 ev. The eV energy difference between vibrational level for carbon monoxide correpond to a photon of wavelength μm, in the infrared region of the pectrum. Thi i much cloer to the viible region than i the photon in the rotational tranition in Example 4.. In general the energy difference for molecular vibration are much maller than thoe that produce atomic pectra, but much larger than the energy difference for molecular rotation. When we include both rotational and vibrational energie, the energy level for our diatomic molecule are n = 1 n = 0 Ground level l = l = 0 E nl = l1l + 1 U I + 1n + 1 U k A (4.9) Figure 4.8 how the energy-level diagram. For each value of n there are many value of l, forming a erie of cloely paced level. The red arrow in Fig. 4.8 how everal poible tranition in which a molecule goe from a level with n = to a level with n = 1 by emitting a photon. A we mentioned above, thee tranition mut obey the election rule l = 1 in order to conerve angular momentum. An additional election rule tate that if the vibrational level change, the vibrational quantum number n in Eq. (4.9) mut increae by 1 1 n = 1 if a photon i aborbed or decreae by 1 1 n = -1 if a photon i emitted. m r

91 4. Molecular Spectra A typical molecular band pectrum. A an illutration of thee election rule, Fig. 4.8 how that a molecule in the n =, l = 4 level can emit a photon and drop into the n = 1, l = 5 level 1 n = -1, l = +1 or the n = 1, l = 3 level 1 n = -1, l = -1, but i forbidden from making a n = -1, l = 0 tranition into the n = 1, l = 4 level. Tranition between tate with variou pair of n-value give different erie of pectrum line, and the reulting pectrum ha a erie of band. Each band correpond to a particular vibrational tranition, and each individual line in a band repreent a particular rotational tranition, with the election rule l = 1. Figure 4.9 how a typical band pectrum. All molecule can have excited tate of the electron in addition to the rotational and vibrational tate that we have decribed. In general, thee lie at higher energie than the rotational and vibrational tate, and there i no imple rule relating them. When there i a tranition between electronic tate, the n = 1 election rule for the vibrational level no longer hold. Example 4.3 Vibration-rotation pectrum of carbon monoxide Conider again the CO molecule of Example 4.. Find the wavelength of the photon emitted by a CO molecule when it vibrational energy change and it rotational energy i (a) initially zero and (b) finally zero. SOLUTION IDENTIFY and SET UP: We need to ue the election rule for the vibrational and rotational tranition of a diatomic molecule. Since a photon i emitted a the vibrational energy change, the election rule n = -1 tell u that the vibrational quantum number n decreae by 1 in both part (a) and (b). In part (a) the initial value of l i zero; the election rule l = 1 tell u that the final value of l i 1, o the rotational energy increae in thi cae. In part (b) the final value of l i zero; l = 1 then tell u that the initial value of l i 1, and the rotational energy decreae. In each cae the energy E of the emitted photon i the difference between the initial and final energie of the molecule, accounting for the change in both vibrational and rotational energy. In part (a) E equal the difference Uv between adjacent vibrational energy level minu the rotational energy that the molecule gain; in part (b) E equal Uv plu the rotational energy that the molecule loe. Example 4. tell u that the difference between the l = 0 and l = 1 rotational energy level i mev = ev, and we learned above that the vibrational energy-level eparation for CO i Uv = ev. We ue E = hc>l to determine the correponding wavelength (our target variable). EXECUTE: (a) In thi tranition the CO molecule loe Uv = ev of vibrational energy and gain ev of rotational energy. Hence the energy E that goe into the emitted photon equal ev le ev, or ev. The photon wavelength i l = hc E = * 10-6 m = mm (b) Now the CO molecule loe Uv = ev of vibrational energy and alo loe ev of rotational energy, o the energy that goe into the photon i E = ev ev = ev. The wavelength i l = hc E * ev # * 10 8 m> = ev * ev # * 10 8 m> = ev = * 10-6 m = mm EVALUATE: In part (b) the molecule loe more energy than it doe in part (a), o the emitted photon mut have greater energy and a horter wavelength. That jut what our reult how. Complex Molecule We can apply thee ame principle to more complex molecule. A molecule with three or more atom ha everal different kind or mode of vibratory motion. Each mode ha it own et of energy level, related to it frequency by Eq. (4.7).

92 141 CHAPTER 4 Molecule and Condened Matter 4.10 The carbon dioxide molecule can vibrate in three different mode. For clarity, the atom are not hown to cale: The eparation between atom i actually comparable to their diameter. (a) Bending mode O O C (b) Symmetric tretching mode C (c) Aymmetric tretching mode O O In nearly all cae the aociated radiation lie in the infrared region of the electromagnetic pectrum. Infrared pectrocopy ha proved to be an extremely valuable analytical tool. It provide information about the trength, rigidity, and length of molecular bond and the tructure of complex molecule. Alo, becaue every molecule (like every atom) ha it characteritic pectrum, infrared pectrocopy can be ued to identify unknown compound. One molecule that can readily aborb and emit infrared radiation i carbon dioxide (CO ). Figure 4.10 how the three poible mode of vibration of a? CO molecule. A number of tranition are poible between excited level of the ame vibrational mode a well a between level of different vibrational mode. The energy difference are le than 1 ev in all of thee tranition, and o involve infrared photon of wavelength longer than 1 μm. Hence a ga of CO can readily aborb light at a number of different infrared wavelength. Thi make CO a very effective greenhoue ga (ee Section 17.7) even on earth, where carbon dioxide i jut 0.04% of the atmophere by volume. On Venu, however, the atmophere ha more than 90 time the total ma of our atmophere and i almot entirely CO. The reulting greenhoue effect i tremendou: The urface temperature on Venu i more than 400 kelvin greater than what it would be if the planet had no atmophere at all. O C O Tet Your Undertanding of Section 4. A rotating diatomic molecule emit a photon when it make a tranition from level l to level l - 1. If the value of l increae, doe the wavelength of the emitted photon (i) increae, (ii) decreae, or (iii) remain unchanged? 4.3 Structure of Solid The term condened matter include both olid and liquid. In both tate, the interaction between atom or molecule are trong enough to give the material a definite volume that change relatively little with applied tre. In condened matter, adjacent atom attract one another until their outer electron charge cloud begin to overlap ignificantly. Thu the ditance between adjacent atom in condened matter are about the ame a the diameter of the atom themelve, typically 0.1 to 0.5 nm. Alo, when we peak of the ditance between atom, we mean the center-to-center (nucleu-to-nucleu) ditance. Ordinarily, we think of a liquid a a material that can flow and of a olid a a material with a definite hape. However, if you heat a horizontal gla rod in the flame of a burner, you ll find that the rod begin to ag (flow) more and more eaily a it temperature rie. Gla ha no definite tranition from olid to liquid, and no definite melting point. On thi bai, we can conider gla at room temperature a being an extremely vicou liquid. Tar and butter how imilar behavior. What i the microcopic difference between material like gla or butter and olid like ice or copper, which do have definite melting point? Ice and copper are example of crytalline olid in which the atom have long-range order, a recurring pattern of atomic poition that extend over many atom. Thi pattern i called the crytal tructure. In contrat, gla at room temperature i an example of an amorphou olid, one that ha no long-range order, but only hort-range order (correlation between neighboring atom or molecule). Liquid alo have only hort-range order. The boundarie between crytalline olid, amorphou olid, and liquid may be ometime blurred. Some olid, crytalline when perfect, can form with o many imperfection in their tructure that they have almot no long-range order. Converely, ome liquid crytal (organic compound compoed of cylindrical molecule that tend to line up parallel to each other) have a fairly high degree of long-range order. Nearly everything we know about crytal tructure ha been learned from diffraction experiment, initially with x ray and later with electron and neutron.

93 4.3 Structure of Solid Portion of ome common type of crytal lattice. (a) Simple cubic (c) (b) Face-centered cubic (fcc) (c) Body-centered cubic (bcc) (d) Hexagonal cloe packed (hcp) (e) Top view, hexagonal cloe packed A typical ditance between atom i of the order of 0.1 nm. You can how that 1.4-keV x ray, 150-eV electron, and eV neutron all have wavelength l = 0.1 nm. Crytal Lattice and Structure A crytal lattice i a repeating pattern of mathematical point that extend throughout pace. There are 14 general type of uch pattern; Fig how mall portion of ome common example. The imple cubic lattice (c) ha a lattice point at each corner of a cubic array (Fig. 4.11a). The face-centered cubic lattice (fcc) i like the imple cubic but with an additional lattice point at the center of each cube face (Fig. 4.11b). The body-centered cubic lattice (bcc) i like the imple cubic but with an additional point at the center of each cube (Fig. 4.11c). The hexagonal cloe-packed lattice ha layer of lattice point in hexagonal pattern, each hexagon made up of ix equilateral triangle (Fig. 4.11d and 4.11e). CAUTION A perfect crytal lattice i infinitely large Figure 4.11 how jut enough lattice point o you can eaily viualize the pattern; the lattice, a mathematical abtraction, extend throughout pace. Thu the lattice point hown repeat endlely in all direction. In a crytal tructure, a ingle atom or a group of atom i aociated with each lattice point. The group may contain the ame or different kind of atom. Thi atom or group of atom i called a bai. Thu a complete decription of a crytal tructure include both the lattice and the bai. We initially conider perfect crytal, or ideal ingle crytal, in which the crytal tructure extend uninterrupted throughout pace. The bcc and fcc tructure are two common imple crytal tructure. The alkali metal have a bcc tructure that i, a bcc lattice with a bai of one atom at each lattice point. Each atom in a bcc tructure ha eight nearet neighbor (Fig. 4.1a). The element Al, Ca, Cu, Ag, and Au have an fcc tructure that i, an fcc lattice with a bai of one atom at each lattice point. Each atom in an fcc tructure ha 1 nearet neighbor (Fig. 4.1b). Figure 4.13 how a repreentation of the tructure of odium chloride (NaCl, ordinary alt). It may look like a imple cubic tructure, but it in t. The odium and chloride ion each form an fcc tructure, o we can think roughly of the odium chloride tructure a being compoed of two interpenetrating fcc tructure. More correctly, the odium chloride crytal tructure of Fig ha an fcc lattice with one chloride ion at each lattice point and one odium ion half a cube length above it. That i, it bai conit of one chloride and one odium ion. Another example i the diamond tructure; it called that becaue it i the crytal tructure of carbon in the diamond form. It alo the crytal tructure of ilicon, germanium, and gray tin. The diamond lattice i fcc; the bai conit of one atom at each lattice point and a econd identical atom diplaced a quarter of a cube length in each of the three cube-edge direction. Figure 4.14 will help you 4.1 (a) The bcc tructure i compoed of a bcc lattice with a bai of one atom for each lattice point. (b) The fcc tructure i compoed of an fcc lattice with a bai of one atom for each lattice point. Thee tructure repeat preciely to make up perfect crytal. (a) The bcc tructure (b) The fcc tructure 4.13 Repreentation of part of the odium chloride crytal tructure. The ditance between ion are exaggerated. Na + Cl Face-centered cubic tructure of odium ion Face-centered cubic tructure of chloride ion

94 1414 CHAPTER 4 Molecule and Condened Matter 4.14 The diamond tructure, hown a two interpenetrating face-centered cubic tructure with ditance between atom exaggerated. Relative to the correponding green atom, each purple atom i hifted up, back, and to the left by a ditance a/4. a/4 a viualize thi. The haded volume in Fig how the bottom right front eighth of the baic cube; the four atom at alternate corner of thi cube are at the corner of a regular tetrahedron, and there i an additional atom at the center. Thu each atom in the diamond tructure i at the center of a regular tetrahedron with four nearet-neighbor atom at the corner. In the diamond tructure, both the purple and green phere in Fig repreent identical atom for example, both carbon or both ilicon. In the cubic zinc ulfide tructure, the purple phere repreent one type of atom and the green phere repreent a different type. For example, in zinc ulfide (ZnS) each zinc atom (purple in Fig. 4.14) i at the center of a regular tetrahedron with four ulfur atom (green in Fig. 4.14) at it corner, and vice vera. Gallium arenide (GaA) and imilar compound have thi ame tructure. Bonding in Solid The force that are reponible for the regular arrangement of atom in a crytal are the ame a thoe involved in molecular bond, plu one additional type. Not urpriingly, ionic and covalent molecular bond are found in ionic and covalent crytal, repectively. The mot familiar ionic crytal are the alkali halide, uch a ordinary alt (NaCl). The poitive odium ion and the negative chloride ion occupy alternate poition in a cubic arrangement (ee Fig. 4.13). The attractive force are the familiar Coulomb -law force between charged particle. Thee force have no preferred direction, and the arrangement in which the material crytallize i partly determined by the relative ize of the two ion. Such a tructure i table in the ene that it ha lower total energy than the eparated ion (ee the following example). The negative potential energie of pair of oppoite charge are greater in abolute value than the poitive energie of pair of like charge becaue the pair of unlike charge are cloer together, on average. Example 4.4 Potential energy of an ionic crytal Imagine a one-dimenional ionic crytal coniting of a very large number of alternating poitive and negative ion with charge e and -e, with equal pacing a along a line. Show that the total interaction potential energy i negative, which mean that uch a crytal i table. SOLUTION IDENTIFY and SET UP: We treat each ion a a point charge and ue our reult from Section 3.1 for the electric potential energy of a collection of point charge. Equation (3.10) and (3.11) tell u to conider the electric potential energy U of each pair of charge. The total potential energy of the ytem i the um of the value of U for every poible pair; we take the number of pair to be infinite. EXECUTE: Let pick an ion omewhere in the middle of the line and add the potential energie of it interaction with all the ion to one ide of it. From Eq. (3.11), that um i a U = - e 1 4pP 0 a + e 1 4pP 0 a - = - e 4pP 0 a Á e 1 4pP 0 3a + Á You may notice that the erie in parenthee reemble the Taylor erie for the function ln11 + x: ln11 + x = x - x + x x Á When x = 1 thi become the erie in parenthee above, o a U = - 4pP 0 a ln Thi i certainly a negative quantity. The atom on the other ide of the ion we re conidering make an equal contribution to the potential energy. And if we include the potential energie of all pair of atom, the um i certainly negative. EVALUATE: We conclude that thi one-dimenional ionic crytal i table: It ha lower energy than the zero electric potential energy that i obtained when all the ion are infinitely far apart from each other. e Type of Crytal Carbon, ilicon, germanium, and tin in the diamond tructure are imple example of covalent crytal. Thee element are in Group IV of the periodic table,

95 4.3 Structure of Solid 1415 meaning that each atom ha four electron in it outermot hell. Each atom form a covalent bond with each of four adjacent atom at the corner of a tetrahedron (Fig. 4.14). Thee bond are trongly directional becaue of the aymmetric electron ditribution dictated by the excluion principle, and the reult i the tetrahedral diamond tructure. A third crytal type, le directly related to the chemical bond than are ionic or covalent crytal, i the metallic crytal. In thi tructure, one or more of the outermot electron in each atom become detached from the parent atom (leaving a poitive ion) and are free to move through the crytal. Thee electron are not localized near the individual ion. The correponding electron wave function extend over many atom. Thu we can picture a metallic crytal a an array of poitive ion immered in a ea of freed electron whoe attraction for the poitive ion hold the crytal together (Fig. 4.15). Thee electron alo give metal their high electrical and thermal conductivitie. Thi ea of electron ha many of the propertie of a ga, and indeed we peak of the electron-ga model of metallic olid. The implet verion of thi model i the free-electron model, which ignore interaction with the ion completely (except at the urface). We ll return to thi model in Section 4.5. In a metallic crytal the freed electron are not localized but are hared among many atom. Thi give a bonding that i neither localized nor trongly directional. The crytal tructure i determined primarily by conideration of cloe packing that i, the maximum number of atom that can fit into a given volume. The two mot common metallic crytal lattice are the face-centered cubic and hexagonal cloe-packed (ee Fig. 4.11b, 4.11d, and 4.11e). In tructure compoed of thee lattice with a bai of one atom, each atom ha 1 nearet neighbor. A we mentioned in Section 4.1, van der Waal interaction and hydrogen bonding alo play a role in the tructure of ome olid. In polyethylene and imilar polymer, covalent bonding of atom form long-chain molecule, and hydrogen bonding form cro-link between adjacent chain. In olid water, both van der Waal force and hydrogen bond are ignificant in determining the crytal tructure of ice. Our dicuion ha centered on perfect crytal, or ideal ingle crytal. Real crytal how a variety of departure from thi idealized tructure. Material are often polycrytalline, compoed of many mall ingle crytal bonded together at grain boundarie. There may be point defect within a ingle crytal: Intertitial atom may occur in place where they do not belong, and there may be vacancie, poition that hould be occupied by an atom but are not. A point defect of particular interet in emiconductor, which we will dicu in Section 4.6, i the ubtitutional impurity, a foreign atom replacing a regular atom (for example, arenic in a ilicon crytal). There are everal baic type of extended defect called dilocation. One type i the edge dilocation, hown chematically in Fig. 4.16, in which one plane of atom lip relative to another. The mechanical propertie of metallic crytal are influenced trongly by the preence of dilocation. The ductility and malleability of ome metal depend on the preence of dilocation that can move through the crytal during platic deformation. Solid-tate phyicit often point out that the bigget extended defect of all, preent in all real crytal, i the urface of the material with it dangling bond and abrupt change in potential energy In a metallic olid, one or more electron are detached from each atom and are free to wander around the crytal, forming an electron ga. The wave function for thee electron extend over many atom. The poitive ion vibrate around fixed location in the crytal. Poitive ion 4.16 An edge dilocation in two dimenion. In three dimenion an edge dilocation would look like an extra plane of atom lipped partway into the crytal. Tet Your Undertanding of Section 4.3 If a i the ditance in an NaCl crytal from an Na + ion to one of it nearet-neighbor Cl - ion, what i the ditance from an Na + ion to one of it next-to-nearet-neighbor Cl - ion? (i) a1; (ii) a13; (iii) a; (iv) none of thee. The irregularity i een mot eaily by viewing the figure from variou direction at a grazing angle with the page.

96 1416 CHAPTER 4 Molecule and Condened Matter 4.17 The concept of energy band wa firt developed by the Swi-American phyicit Felix Bloch ( ) in hi doctoral thei. Our modern undertanding of electrical conductivity tem from that landmark work. Bloch work in nuclear phyic brought him (along with Edward Purcell) the 195 Nobel Prize in phyic Origin of energy band in a olid. (a) A the ditance r between atom decreae, the energy level pread into band. The vertical line at r 0 how the actual atomic pacing in the crytal. (b) Symbolic repreentation of energy band. E O r 0 (a) Actual eparation of atom in the crytal r (b) 4.4 Energy Band The energy-band concept, introduced in 198 (Fig. 4.17), i a great help in undertanding everal propertie of olid. To introduce the idea, uppoe we have a large number N of identical atom, far enough apart that their interaction are negligible. Every atom ha the ame energy-level diagram. We can draw an energy-level diagram for the entire ytem. It look jut like the diagram for a ingle atom, but the excluion principle, applied to the entire ytem, permit each tate to be occupied by N electron intead of jut one. Now we begin to puh the atom uniformly cloer together. Becaue of the electrical interaction and the excluion principle, the wave function begin to ditort, epecially thoe of the outer, or valence, electron. The correponding energie alo hift, ome upward and ome downward, by varying amount, a the valence electron wave function become le localized and extend over more and more atom. Thu the valence tate that formerly gave the ytem a tate with a harp energy level that could accommodate N electron now give a band containing N cloely paced level (Fig. 4.18). Ordinarily, N i very large, omewhere near the order of Avogadro number 110 4, o we can accurately treat the level a forming a continuou ditribution of energie within a band. Between adjacent energy band are gap or forbidden region where there are no allowed energy level. The inner electron in an atom are affected much le by nearby atom than are the valence electron, and their energy level remain relatively harp. Inulator, Semiconductor, and Conductor The nature of the energy band determine whether the material i an electrical inulator, a emiconductor, or a conductor. In particular, what matter are the extent to which the tate in each band are occupied and the pacing, or energy gap, between adjacent band. A crucial factor i the excluion principle (ee Section 41.6), which tate that only one electron can occupy a given quantummechanical tate. In an inulator at abolute zero temperature, the highet band that i completely filled, called the valence band, i alo the highet band that ha any electron in it. The next higher band, called the conduction band, i completely empty; there are no electron in it tate (Fig. 4.19a). Imagine what happen if an electric field i applied to a material of thi kind. To move in repone to the field, an electron would have to go into a different quantum tate with a lightly different energy. It can t do that, however, becaue all the neighboring tate are already occupied. The only way uch an electron can move i to jump acro the energy gap into the conduction band, where there are plenty of nearby unoccupied tate. At any temperature above abolute zero 4.19 Three type of energy-band tructure. (a) In an inulator at abolute zero, there are no electron in the conduction band. E (b) A emiconductor ha the ame band tructure a an inulator but a maller gap between the valence and conduction band. E (c) A conductor ha a partially filled conduction band. E Empty Small Partially filled conduction band energy conduction band gap Empty E g Large energy gap conduction band Energy gap E g E g Filled valence band Filled valence band Filled valence band

97 4.4 Energy Band 1417 there i ome probability thi jump can happen, becaue an electron can gain energy from thermal motion. In an inulator, however, the energy gap between the valence and conduction band can be 5 ev or more, and that much thermal energy i not ordinarily available. Hence little or no current flow in repone to an applied electric field, and the electric conductivity (Section 5.) i low. The thermal conductivity (Section 17.7), which alo depend on mobile electron, i likewie low. We aw in Section 4.4 that an inulator become a conductor if it i ubjected to a large enough electric field; thi i called dielectric breakdown. If the electric field i of order V/m, there i a potential difference of a few volt over a ditance comparable to atomic ize. In thi cae the field can do enough work on a valence electron to boot it acro the energy gap and into the conduction band. (In practice dielectric breakdown occur for field much le than V/m, becaue imperfection in the tructure of an inulator provide ome more acceible energy tate within the energy gap.) A in an inulator, a emiconductor at abolute zero ha an empty conduction band above the full valence band. The difference i that in a emiconductor the energy gap between thee band i relatively mall and electron can more readily jump into the conduction band (Fig. 4.19b). A the temperature of a emiconductor increae, the population in the conduction band increae very rapidly, a doe the electric conductivity. For example, in a emiconductor near room temperature with an energy gap of 1 ev, the number of conduction electron double when the temperature rie by jut 10 C. We will ue the concept of energy band to explore emiconductor in more depth in Section 4.6. In a conductor uch a a metal, there are electron in the conduction band even at abolute zero (Fig. 4.19c). The metal odium i an example. An analyi of the atomic energy-level diagram for odium (ee Fig a) how that for an iolated odium atom, the ix lowet excited tate (all 3p tate) are about.1 ev above the two 3 ground tate. In olid odium, however, the atom are o cloe together that the 3 and 3p band pread out and overlap into a ingle band. Each odium atom contribute one electron to the band, leaving an Na + ion behind. Each atom alo contribute eight tate to that band (two 3, ix 3p), o the band i only one-eighth occupied. We call thi tructure a conduction band becaue it i only partially occupied. Electron near the top of the filled portion of the band have many adjacent unoccupied tate available, and they can eaily gain or loe mall amount of energy in repone to an applied electric field. Therefore thee electron are mobile, giving olid odium it high electrical and thermal conductivity. A imilar decription applie to other conducting material. PhET: Band Structure PhET: Conductivity Example 4.5 Photoconductivity in germanium At room temperature, pure germanium ha an almot completely filled valence band eparated by a 0.67-eV gap from an almot completely empty conduction band. It i a poor electrical conductor, but it conductivity increae greatly when it i irradiated with electromagnetic wave of a certain maximum wavelength. What i that wavelength? SOLUTION IDENTIFY and SET UP: The conductivity of a emiconductor increae greatly when electron are excited from the valence band into the conduction band. In germanium, the excitation occur when an electron aborb a photon with an energy of at leat E min = 0.67 ev. From the relationhip E = hc>l, the maximum wavelength l max (our target variable) correpond to thi minimum photon energy. EXECUTE: The wavelength of a photon with energy E min = 0.67 ev i l max = hc * ev # * 10 8 m> = E min 0.67 ev = 1.9 * 10-6 m = 1.9 mm = 1900 nm EVALUATE: Thi wavelength i in the infrared part of the pectrum, o viible-light photon (which have horter wavelength and higher energy) will alo induce conductivity in germanium. A we ll ee in Section 4.7, emiconductor crytal are widely ued a photocell and for many other application.

98 1418 CHAPTER 4 Molecule and Condened Matter Tet Your Undertanding of Section 4.4 One type of thermometer work by meauring the temperature-dependent electrical reitivity of a ample. Which of the following type of material diplay the greatet change in reitivity for a given temperature change? (i) inulator; (ii) emiconductor; (iii) reitor. 4.5 Free-Electron Model of Metal 4.0 A cubical box with ide length L. We tudied thi three-dimenional verion of the infinite quare well in Section 41.. The energy level for a particle in thi box are given by Eq. (4.10). z z 5 L x 5 L x n Z n Y n r O n X y 5 L 4.1 The allowed value of n X, n Y, and n Z are poitive integer for the electron tate in the free-electron ga model. Including pin, there are two tate for each unit volume in n pace. y Studying the energy tate of electron in metal can give u a lot of inight into their electrical and magnetic propertie, the electron contribution to heat capacitie, and other behavior. A we dicued in Section 4.3, one of the ditinguihing feature of a metal i that one or more valence electron are detached from their home atom and can move freely within the metal, with wave function that extend over many atom. The free-electron model aume that thee electron are completely free inide the material, that they don t interact at all with the ion or with each other, but that there are infinite potential-energy barrier at the urface. The idea i that a typical electron move o rapidly within the metal that it ee the effect of the ion and other electron a a uniform potential-energy function, whoe value we can chooe to be zero. We can repreent the urface of the metal by the ame cubical box that we analyzed in Section 41. (the three-dimenional verion of the particle in a box tudied in Section 40.). If the box ha ide of length L (Fig. 4.0), the energie of the tationary tate (quantum tate of definite energy) are 1n X = 1,, 3, Á ; n Y = 1, E nx,n Y,n Z = 1n X + n Y + n Z p U (4.10) ml, 3, Á ; n Z = 1,, 3, Á Each tate i labeled by the three poitive-integer quantum number 1n X, n Y, n Z. Denity of State Later we ll need to know the number dn of quantum tate that have energie in a given range de. The number of tate per unit energy range dn>de i called the denity of tate, denoted by g1e. We ll begin by working out an expreion for g1e. Think of a three-dimenional pace with coordinate 1n X, n Y, n Z (Fig. 4.1). The radiu n r of a phere centered at the origin in that pace i n r = 1n X + n Y + n Z 1>. Each point with integer coordinate in that pace repreent one patial quantum tate. Thu each point correpond to one unit of volume in the pace, and the total number of point with integer coordinate 4 inide a phere equal the volume of the phere, 3 pn r 3. Becaue all our n are 1 poitive, we mut take only one octant of the phere, with 8 the total volume, or pn r 3 = 1 6 pn r 3. The particle are electron, o each point correpond to two tate with oppoite pin component 1m = 1, and the total number n of 1 electron tate correponding to point inide the octant i twice 6 pn r 3, or n = pn r 3 (4.11) 3 The energy E of tate at the urface of the phere can be expreed in term of n r. Equation (4.10) become E = n r p U (4.1) ml We can combine Eq. (4.11) and (4.1) to get a relationhip between E and n that doen t contain n r. We ll leave the detail a an exercie (Exercie 4.4); the reult i n = 1m3/ VE 3/ (4.13) 3p U 3

99 4.5 Free-Electron Model of Metal 1419 where V = L 3 i the volume of the box. Equation (4.13) give the total number of tate with energie of E or le. To get the number of tate dn in an energy interval de, we treat n and E a continuou variable and take differential of both ide of Eq. (4.13). We get dn = 1m3> VE 1> p U 3 de (4.14) The denity of tate g1e i equal to dn/de, o from Eq. (4.14) we get g1e = 1m3> V p U 3 E 1> (denity of tate, free-electron model) (4.15) Fermi Dirac Ditribution Now we need to know how the electron are ditributed among the variou quantum tate at any given temperature. The Maxwell Boltzmann ditribution tate that the average number of particle in a tate of energy E i proportional to e -E>kT (ee Section 18.5 and 39.4). However, there are two very important reaon why it wouldn t be right to ue the Maxwell Boltzmann ditribution. The firt reaon i the excluion principle. At abolute zero the Maxwell Boltzmann function predict that all the electron would go into the two ground tate of the ytem, with n X = n and m = 1 Y = n Z = 1. But the excluion principle allow only one electron in each tate. At abolute zero the electron can fill up the lowet available tate, but there not enough room for all of them to go into the lowet tate. Thu a reaonable gue a to the hape of the ditribution would be Fig. 4.. At abolute zero temperature the tate are filled up to ome value E F0, and all tate above thi value are empty. The econd reaon we can t ue the Maxwell Boltzmann ditribution i more ubtle. That ditribution aume that we are dealing with ditinguihable particle. It might eem that we could put a tag on each electron and know which i which. But overlapping electron in a ytem uch a a metal are inditinguihable. Suppoe we have two electron; a tate in which the firt i in energy level E 1 and the econd i in level E i not ditinguihable from a tate in which the two electron are revered, becaue we can t tell which electron i which. The tatitical ditribution function that emerge from the excluion principle and the inditinguihability requirement i called (after it inventor) the Fermi Dirac ditribution. Becaue of the excluion principle, the probability that a particular tate with energy E i occupied by an electron i the ame a ƒ1e, the fraction of tate with that energy that are occupied: ƒ1e = 1 e 1E-EF>kT + 1 (Fermi Dirac ditribution) (4.16) The energy E F i called the Fermi energy or the Fermi level; we ll dicu it ignificance below. We ue E F0 for it value at abolute zero 1T = 0 and E F for other temperature. We can accurately let E F = E F0 for metal becaue the Fermi energy doe not change much with temperature for olid conductor. However, it i not afe to aume that E F = E F0 for emiconductor, in which the Fermi energy uually doe change with temperature. Figure 4.3 how graph of Eq. (4.16) for three temperature. The trend of thi function a kt approache zero confirm our gue. When E = E F, the exponent i zero and ƒ1e F = 1. That i, the probability i 1 that a tate at the Fermi energy contain an electron. Alternatively, at E = E F, half the tate are filled (and half are empty). 4. The probability ditribution for occupation of free-electron energy tate at abolute zero. At abolute zero, all tate are occupied f(e) (occupation probability 1) at energie up to E F E 0 E F0... and all tate are empty (occupation probability zero) at energie above E F Graph of the Fermi Dirac ditribution function for variou value of kt, auming that the Fermi energy E F i independent of the temperature T. f(e) E F 1 kt = E 40 F 1 kt = E 10 F kt = 1 E 4 F A T increae, more and more of the electron are excited to tate with energy E. E F. E

100 140 CHAPTER 4 Molecule and Condened Matter For E 6 E F the exponent i negative, and ƒ1e 7 1 For the exponent i poitive, and ƒ1e E 7 E F The hape depend on the ratio E F >kt. At T V E F >k thi ratio i very large. Then for E 6 E F the curve very quickly approache 1, and for E 7 E F it quickly approache zero. When T i larger, the change are more gradual. When T i zero, all the tate up to the Fermi level E F0 are filled, and all tate above that level are empty (Fig. 4.). Example 4.6 Probabilitie in the free-electron model For free electron in a olid, at what energy i the probability that a particular tate i occupied equal to (a) 0.01 and (b) 0.99? SOLUTION IDENTIFY and SET UP: Thi problem ak u to explore the Fermi Dirac ditribution. Equation (4.16) give the occupation probability ƒ1e for a given energy E. If we olve thi equation for E, we get an expreion for the energy that correpond to a given occupation probability which i jut what we need to olve thi problem. EXECUTE: Uing Eq. (4.16), you can how that E = E F + kt ln a 1 ƒ1e - 1b (a) When ƒ1e = 0.01, E = E F + kt ln a b = E F + 4.6kT The probability that a tate 4.6kT above the Fermi level i occupied i only 0.01, or 1%. (b) When ƒ1e = 0.99, E = E F + kt ln a b = E F - 4.6kT The probability that a tate 4.6kT below the Fermi level i occupied i 0.99, or 99%. EVALUATE: At very low temperature, 4.6kT i much le than E F. Then the occupation probability of level even lightly below E F i nearly %, and that for level even lightly above E F i nearly zero (ee Fig. 4.3). In general, if the probability i P that a tate with an energy E above E F i occupied, then the probability i 1 - P that a tate E below E F i occupied. We leave the proof to you (Problem 4.50). Electron Concentration and Fermi Energy Equation (4.16) give the probability that any pecific tate with energy E i occupied at a temperature T. To get the actual number of electron in any energy range de, we have to multiply thi probability by the number dn of tate in that range g1e de. Thu the number dn of electron with energie in the range de i dn = g1eƒ1e de = 1m3> VE 1> p U 3 1 e 1E-E F>kT + 1 de (4.17) The Fermi energy E F i determined by the total number N of electron; at any temperature the electron tate are filled up to a point at which all electron are accommodated. At abolute zero there i a imple relationhip between E F0 and N. All tate below E F0 are filled; in Eq. (4.13) we et n equal to the total number of electron N and E to the Fermi energy at abolute zero E F0 : Solving Eq. (4.18) for E F0, we get N = 1m3> VE F0 3> 3p U 3 (4.18) E F0 = 3>3 p 4>3 U m a N V b >3 (4.19) The quantity N>V i the number of free electron per unit volume. It i called the electron concentration and i uually denoted by n. If we replace N>V with n, Eq. (4.19) become E F0 = 3>3 p 4>3 U n >3 m (4.0)

101 4.5 Free-Electron Model of Metal 141 CAUTION Electron concentration and number of electron Don t confue the electron concentration n with any quantum number n. Furthermore, the number of tate i not in general the ame a the total number of electron N. Example 4.7 The Fermi energy in copper At low temperature, copper ha a free-electron concentration n = 8.45 * 10 8 m -3. Uing the free-electron model, find the Fermi energy for olid copper, and find the peed of an electron with a kinetic energy equal to the Fermi energy. SOLUTION IDENTIFY and SET UP: Thi problem ue the relationhip between Fermi energy and free-electron concentration. Becaue copper i a olid conductor, it Fermi energy change very little with temperature and we can ue the expreion for the Fermi energy at abolute zero, Eq. (4.0). We ll find the Fermi peed v F that correpond 1 to kinetic energy E uing the nonrelativitic formula E F EXECUTE: Uing the given value of n, we olve for E F and =v mv F F. F : E F = 3>3 p 4> * J # * 10 8 m -3 > * kg = 1.16 * J = 7.03 ev v F = B E F m = * J = 1.57 * 10 6 m> B 9.11 * kg E F EVALUATE: Our value of and are within the range of typical value for metal, ev and 0.8. * 10 6 m>, repectively. Note that the calculated Fermi peed i far le than the peed of light c = 3.00 * 10 8 m>, which jutifie our ue of the 1 nonrelativitic formula mv F = E F. Our calculated Fermi energy i much larger than kt at ordinary temperature. (At room temperature T = 0 C = 93 K, the quantity kt equal * 10-3 J>K193 K = 4.04 * 10-1 J = ev.) So it i a good approximation to take almot all the tate below E F a completely full and almot all thoe above E F a completely empty (ee Fig. 4.). We can alo ue Eq. (4.15) to find g1e if E and V are known. You can how that if E = 7.03 ev and V = 1 cm 3, g1e i about * 10 tate>ev. Thi huge number how why we were jutified in treating n and E a continuou variable in our denityof-tate derivation. v F Average Free-Electron Energy We can calculate the average free-electron energy in a metal at abolute zero by uing the ame idea that we ued to find E F0. From Eq. (4.17) the number dn of electron with energie in the range de i g1eƒ1e de. The energy of thee electron i E dn = Eg1Eƒ1E de. At abolute zero we ubtitute ƒ1e = 1 from E = 0 to E = E F0 and ƒ1e = 0 for all other energie. Therefore the total energy E tot of all the N electron i E F0 q E tot = Eg1E11 de + L L 0 Eg1E10 de = E F0 L 0 The implet way to evaluate thi expreion i to compare Eq. (4.15) and (4.19), noting that E F0 Eg1E de g1e = 3NE 1> E F0 3> Subtituting thi expreion into the integral and uing E av = E tot >N, we get E av = 3 E 3> F0 L0 (4.1) 3 At abolute zero the average free-electron energy equal 5 of the Fermi energy. E F0 E 3> de = 3 5 E F0 Example 4.8 Free-electron ga veru ideal ga (a) Find the average energy of the free electron in copper at abolute zero (ee Example 4.7). (b) What would be the average kinetic energy of electron if they behaved like an ideal ga at room temperature, 0 C (ee Section 18.3)? What would be the peed of an electron with thi kinetic energy? Compare thee idealga value with the (correct) free-electron value. Continued

102 14 CHAPTER 4 Molecule and Condened Matter SOLUTION IDENTIFY and SET UP: Free electron in a metal behave like a kind of ga. In part (a) we ue Eq. (4.1) to determine the average kinetic energy of free electron in term of the Fermi energy at abolute zero, which we know for copper from Example 4.7. In part (b) we treat electron a an ideal ga at room temperature: Eq. (18.16) then give the average kinetic energy per electron a E and E av = 1 av = 3 kt, mv give the correponding electron peed v. EXECUTE: (a) From Example 4.7, the Fermi energy in copper at abolute zero i 1.16 * J = 7.03 ev. According to Eq. 3 (4.1), the average energy i of thi, or 6.76 * J = 4. ev. (b) In Example 4.7 we found that kt = 4.04 * 10-1 J = ev at room temperature T = 0 C = 93 K. If electron behaved like an ideal ga at thi temperature, the average kinetic 3 energy per electron would be of thi, or 6.07 * 10-1 J = ev. The peed of an electron with thi kinetic energy would be v = B E av m = * 10-1 J B 9.11 * kg = 1.15 * 105 m> EVALUATE: The ideal-ga model predict an average energy that i about 1% of the value given by the free-electron model, and a peed that i about 7% of the free-electron Fermi peed v F = 1.57 * 10 6 m> that we found in Example 4.7. Thu temperature play a very mall role in determining the propertie of electron in metal; their average energie are determined almot entirely by the excluion principle. A imilar analyi allow u to determine the contribution of electron to the heat capacitie of a olid metal. If there i one conduction electron per atom, the principle of equipartition of energy (ee Section 18.4) would predict that the kinetic energie of thee electron contribute 3R> to the molar heat capacity at contant volume C V. But when kt i much maller than E F, which i uually the ituation in metal, only thoe few electron near the Fermi level can find empty tate and change energy appreciably when the temperature change. The number of uch electron i proportional to kt>e F, o we expect that the electron molar heat capacity at contant volume i proportional to 1kT>E F 13R> = 13kT>E F R. A more detailed analyi how that the actual electron contribution to C for a olid metal i 1p V kt>e F R, not far from our prediction. You can verify that if T = 93 K and E F = 7.03 ev, the electron contribution to C V i 0.018R, which i only 1.% of the (incorrect) 3R> prediction of the equipartition principle. Becaue the electron contribution i o mall, the overall heat capacity of mot olid metal i due primarily to vibration of the atom in the crytal tructure (ee Fig in Section 18.4). Tet Your Undertanding of Section 4.5 An ideal ga obey the relationhip pv = nrt (ee Section 18.1). That i, for a given volume V and a number of mole n, a the temperature T decreae, the preure p decreae proportionately and tend to zero a T approache abolute zero. I thi alo true of the free-electron ga in a olid metal? 4.6 Semiconductor PhET: Semiconductor PhET: Conductivity A emiconductor ha an electrical reitivity that i intermediate between thoe of good conductor and of good inulator. The tremendou importance of emiconductor in preent-day electronic tem in part from the fact that their electrical propertie are very enitive to very mall concentration of impuritie. We ll dicu the baic concept uing the emiconductor element ilicon (Si) and germanium (Ge) a example. Silicon and germanium are in Group IV of the periodic table. Both have four electron in the outermot atomic ubhell (3 3p for ilicon, 4 4p for germanium), and both crytallize in the covalently bonded diamond tructure dicued in Section 4.3 (ee Fig. 4.14). Becaue all four of the outer electron are involved in the bonding, at abolute zero the band tructure (ee Section 4.4) ha a completely empty conduction band (ee Fig. 4.19b). A we dicued in Section 4.4, at very low temperature electron cannot jump from the filled valence band into the conduction band. Thi property make thee material inulator at very low temperature; their electron have no nearby tate available into which they can move in repone to an applied electric field. However, in emiconductor the energy gap E g between the valence and conduction band i mall in comparion to the gap of 5 ev or more for many inulator; room-temperature value are 1.1 ev for ilicon and only 0.67 ev for germanium. Thu even at room temperature a ubtantial number of electron can gain enough energy to jump the gap to the conduction band, where they are diociated from their parent atom and are free to move about the crytal. The number of thee electron increae rapidly with temperature.

103 4.6 Semiconductor 143 Example 4.9 Jumping a band gap Conider a material with the band tructure decribed above, with it Fermi energy in the middle of the gap (Fig. 4.4). Find the probability that a tate at the bottom of the conduction band i occupied at T = 300 K, and compare that with the probability at T = 310 K, for band gap of (a) 0.00 ev; (b) 1.00 ev; (c) 5.00 ev. SOLUTION IDENTIFY and SET UP: The Fermi Dirac ditribution function give the probability that a tate of energy E i occupied at temperature T. Figure 4.4 how that the tate of interet at the bottom of the conduction band ha an energy E = E F + E g > that i greater than the Fermi energy E F, with E - E F = E g >. 4.4 Band tructure of a emiconductor. At abolute zero a completely filled valence band i eparated by a narrow energy gap E g of 1 ev or o from a completely empty conduction band. At ordinary temperature, a number of electron are excited to the conduction band. E E F Semiconductor Conduction band Energy gap E g Valence band Figure 4.3 how that the higher the temperature, the larger the fraction of electron with energie greater than the Fermi energy. EXECUTE: (a) When E g = 0.00 ev, E - E F kt = E g ev * 10-5 ev>k1300 K = ƒ1e = e = For T = 310 K, the exponent i 3.74 and ƒ1e = 0.031, a 13% increae in probability for a temperature rie of 10 K. (b) For E g = 1.00 ev, both exponent are five time a large a in part (a), namely 19.3 and 18.7; the value of ƒ1e are 4.0 * 10-9 and 7.4 * In thi cae the (low) probability nearly double with a temperature rie of 10 K. (c) For E g = 5.0 ev, the exponent are 96.7 and 93.6; the value of ƒ1e are 1.0 * 10-4 and.3 * The (extremely low) probability increae by a factor of 3 for a 10 K temperature rie. EVALUATE: Thi example illutrate two important point. Firt, the probability of finding an electron in a tate at the bottom of the conduction band i extremely enitive to the width of the band gap. At room temperature, the probability i about % for a 0.00-eV gap, a few in a thouand million for a 1.00-eV gap, and eentially zero for a 5.00-eV gap. (Pure diamond, with a 5.47-eV band gap, ha eentially no electron in the conduction band and i an excellent inulator.) Second, for any given band gap the probability depend trongly on temperature, and even more trongly for large gap than for mall one. In principle, we could continue the calculation in Example 4.9 to find the actual denity n = N>V of electron in the conduction band at any temperature. To do thi, we would have to evaluate the integral 1g1Eƒ1E de from the bottom of the conduction band to it top. Firt we would need to know the denity of tate function g1e. It wouldn t be correct to ue Eq. (4.15) becaue the energy-level tructure and the denity of tate for real olid are more complex than thoe for the imple free-electron model. However, there are theoretical method for predicting what g1e hould be near the bottom of the conduction band, and uch calculation have been carried out. Once we know n, we can begin to determine the reitivity of the material (and it temperature dependence) uing the analyi of Section 5., which you may want to review. But next we ll ee that the electron in the conduction band don t tell the whole tory about conduction in emiconductor. Hole When an electron i removed from a covalent bond, it leave a vacancy behind. An electron from a neighboring atom can move into thi vacancy, leaving the neighbor with the vacancy. In thi way the vacancy, called a hole, can travel through the material and erve a an additional current carrier. It like decribing the motion of a bubble in a liquid. In a pure, or intrinic, emiconductor, valenceband hole and conduction-band electron are alway preent in equal number. When an electric field i applied, they move in oppoite direction (Fig. 4.5). Thu a hole in the valence band behave like a poitively charged particle, even though the moving charge in that band are electron. The conductivity that we 4.5 Motion of electron in the conduction band and of hole in the valence band of a emiconductor under the action of an applied electric field E S. E E g + + Hole Electric field E Conduction electron S Conduction band Band gap Valence band

104 144 CHAPTER 4 Molecule and Condened Matter jut decribed for a pure emiconductor i called intrinic conductivity. Another kind of conductivity, to be dicued in the next ubection, i due to impuritie. An analogy help to picture conduction in an intrinic emiconductor. The valence band at abolute zero i like a floor of a parking garage that filled bumper to bumper with car (which repreent electron). No car can move becaue there i nowhere for them to go. But if one car i moved to the vacant floor above, it can move freely, jut a electron can move freely in the conduction band. Alo, the empty pace that it leave permit car to move on the nearly filled floor, thereby moving the empty pace jut a hole move in the normally filled valence band. 4.6 An n-type emiconductor. (a) A donor (n-type) impurity atom ha a fifth valence electron that doe not participate in the covalent bonding and i very looely bound. Ge Ge Ge Ge Donor electron Ge Ge Ge Ge Valence electron (b) Energy-band diagram for an n-type emiconductor at a low temperature. One donor electron ha been excited from the donor level into the conduction band. E g 1 ev A Conduction band Donor level Valence band E d 0.01 ev Impuritie Suppoe we mix into melted germanium 1Z = 3 a mall amount of arenic 1Z = 33, the next element after germanium in the periodic table. Thi deliberate addition of impurity element i called doping. Arenic i in Group V; it ha five valence electron. When one of thee electron i removed, the remaining electron tructure i eentially identical to that of germanium. The only difference i that it i maller; the arenic nucleu ha a charge of +33e rather than +3e, and it pull the electron in a little more. An arenic atom can comfortably take the place of a germanium atom a a ubtitutional impurity. Four of it five valence electron form the neceary nearet-neighbor covalent bond. The fifth valence electron i very looely bound (Fig. 4.6a); it doen t participate in the covalent bond, and it i creened from the nuclear charge of +33e by the 3 electron, leaving a net effective charge of about +e. We might gue that the binding energy would be of the ame order of magnitude a the energy of the n = 4 level in hydrogen that i, ev = 0.85 ev. In fact, it i much maller than thi, only about 0.01 ev, becaue the electron probability ditribution actually extend over many atomic diameter and the polarization of intervening atom provide additional creening. The energy level of thi fifth electron correpond in the band picture to an iolated energy level lying in the gap, about 0.01 ev below the bottom of the conduction band (Fig. 4.6b). Thi level i called a donor level, and the impurity atom that i reponible for it i imply called a donor. All Group V element, including N, P, A, Sb, and Bi, can erve a donor. At room temperature, kt i about 0.05 ev. Thi i ubtantially greater than 0.01 ev, o at ordinary temperature, mot electron can gain enough energy to jump from donor level into the conduction band, where they are free to wander through the material. The remaining ionized donor tay at it ite in the tructure and doe not participate in conduction. Example 4.9 how that at ordinary temperature and with a band gap of 1.0 ev, only a very mall fraction (of the order of 10-9 ) of the tate at the bottom of the conduction band in a pure emiconductor contain electron to participate in intrinic conductivity. Thu we expect the conductivity of uch a emiconductor to be about 10-9 a great a that of good metallic conductor, and meaurement bear out thi prediction. However, a concentration of donor a mall a one part in 10 8 can increae the conductivity o dratically that conduction due to impuritie become by far the dominant mechanim. In thi cae the conductivity i due almot entirely to negative charge (electron) motion. We call the material an n-type emiconductor, with n-type impuritie. Adding atom of an element in Group III (B, Al, Ga, In, Tl), with only three valence electron, ha an analogou effect. An example i gallium 1Z = 31; a a ubtitutional impurity in germanium, the gallium atom would like to form four covalent bond, but it ha only three outer electron. It can, however, teal an electron from a neighboring germanium atom to complete the required four covalent bond (Fig. 4.7a). The reulting atom ha the ame electron configuration a Ge but i omewhat larger becaue gallium nuclear charge i maller, +31e intead of +3e.

105 4.7 Semiconductor Device 145 Thi theft leave the neighboring atom with a hole, or miing electron. The hole act a a poitive charge that can move through the crytal jut a with intrinic conductivity. The tolen electron i bound to the gallium atom in a level called an acceptor level about 0.01 ev above the top of the valence band (Fig. 4.7b). The gallium atom, called an acceptor, thu accept an electron to complete it deire for four covalent bond. Thi extra electron give the previouly neutral gallium atom a net charge of -e. The reulting gallium ion i not free to move. In a emiconductor that i doped with acceptor, we conider the conductivity to be almot entirely due to poitive charge (hole) motion. We call the material a p-type emiconductor, with p-type impuritie. Some emiconductor are doped with both n- and p-type impuritie. Such material are called compenated emiconductor. CAUTION The meaning of p-type and n-type Saying that a material i a p-type emiconductor doe not mean that the material ha a poitive charge; ordinarily, it would be neutral. Rather, it mean that it majority carrier of current are poitive hole (and therefore it minority carrier are negative electron). The ame idea hold for an n-type emiconductor; ordinarily, it will not have a negative charge, but it majority carrier are negative electron. We can verify the aertion that the current in n- and p-type emiconductor really i carried by electron and hole, repectively, by uing the Hall effect (ee Section 7.9). The ign of the Hall emf i oppoite in the two cae. Hall-effect device contructed from emiconductor material are ued in probe to meaure magnetic field and the current that caue thoe field. 4.7 A p-type emiconductor. (a) An acceptor (p-type) impurity atom ha only three valence electron, o it can borrow an electron from a neighboring atom. The reulting hole i free to move about the crytal. Ge Ge Ge Ge Acceptor hole Ga Valence electron Ge Ge Ge Ge (b) Energy-band diagram for a p-type emiconductor at a low temperature. One acceptor level ha accepted an electron from the valence band, leaving a hole behind. Conduction band Tet Your Undertanding of Section 4.6 Would there be any advantage to adding n-type or p-type impuritie to copper? E g 1 ev Acceptor level E a 0.01 ev 4.7 Semiconductor Device Valence band Semiconductor device play an indipenable role in contemporary electronic. In the early day of radio and televiion, tranmitting and receiving equipment relied on vacuum tube, but thee have been almot completely replaced in the lat ix decade by olid-tate device, including tranitor, diode, integrated circuit, and other emiconductor device. The only urviving vacuum tube in conumer electronic are the picture tube in older TV receiver and computer monitor; thee are rapidly being replaced by flat-creen diplay. One imple emiconductor device i the photocell (Fig. 4.8). When a thin lab of emiconductor i irradiated with an electromagnetic wave whoe photon have at leat a much energy a the band gap between the valence and conduction band, an electron in the valence band can aborb a photon and jump to the conduction band, where it and the hole it left behind contribute to the conductivity (ee Example 4.5 in Section 4.4). The conductivity therefore increae with wave intenity, thu increaing the current I in the photocell circuit of Fig Hence the ammeter reading indicate the intenity of the light. Detector for charged particle operate on the ame principle. An external circuit applie a voltage acro a emiconductor. An energetic charged particle paing through the emiconductor collide inelatically with valence electron, exciting them from the valence to the conduction band and creating pair of hole and conduction electron. The conductivity increae momentarily, cauing a pule of current in the external circuit. Solid-tate detector are widely ued in nuclear and high-energy phyic reearch. 4.8 A emiconductor photocell in a circuit. The more intene the light falling on the photocell, the greater the conductivity of the photocell and the greater the current meaured by the ammeter (A). Light I Photocell + A The p-n Junction In many emiconductor device the eential principle i the fact that the conductivity of the material i controlled by impurity concentration, which can be varied

106 146 CHAPTER 4 Molecule and Condened Matter 4.9 (a) A emiconductor p-n junction in a circuit. (b) Graph howing the aymmetric current voltage relationhip. The curve i decribed by Eq. (4.). (a) p V n (b) I I Forward bia + Variable emf A Revere bia O I S Forward bia V within wide limit from one region of a device to another. An example i the p-n junction at the boundary between one region of a emiconductor with p-type impuritie and another region containing n-type impuritie. One way of fabricating a p-n junction i to depoit ome n-type material on the very clean urface of ome p-type material. (We can t jut tick p- and n-type piece together and expect the junction to work properly becaue of the impoibility of matching their urface at the atomic level.) When a p-n junction i connected to an external circuit, a in Fig. 4.9a, and the potential difference V p - V n = V acro the junction i varied, the current I varie a hown in Fig. 4.9b. In triking contrat to the ymmetric behavior of reitor that obey Ohm law and give a traight line on an I V graph, a p-n junction conduct much more readily in the direction from p to n than the revere. Such a (motly) one-way device i called a diode rectifier. Later we ll dicu a imple model of p-n junction behavior that predict a current voltage relationhip in the form I = I S 1e ev>kt - 1 (current through a p-n junction) (4.) In the exponent, e = 1.60 * C i the quantum of charge, k i Boltzmann contant, and T i abolute temperature. CAUTION Two different ue of e In e ev>kt the bae of the exponent alo ue the ymbol e, tanding for the bae of the natural logarithm,.7188 Á. Thi e i quite different from e = 1.60 * C in the exponent. Equation (4.) i valid for both poitive and negative value of V; note that V and I alway have the ame ign. A V become very negative, I approache the value -I S. The magnitude (alway poitive) i called the aturation current. I S Current Through a p-n Junction We can undertand the behavior of a p-n junction diode qualitatively on the bai of the mechanim for conductivity in the two region. Suppoe, a in Fig. 4.9a, you connect the poitive terminal of the battery to the p region and the negative terminal to the n region. Then the p region i at higher potential than the n region, correponding to poitive V in Eq. (4.), and the reulting electric field i in the direction p to n. Thi i called the forward direction, and the poitive potential difference i called forward bia. Hole, plentiful in the p region, flow eaily acro the junction into the n region, and free electron, plentiful in the n region, eaily flow into the p region; thee movement of charge contitute a forward current. Connecting the battery with the oppoite polarity give revere bia, and the field tend to puh electron from p to n and hole from n to p. But there are very few free electron in the p region and very few hole in the n region. A a reult, the current in the revere direction i much maller than that with the ame potential difference in the forward direction.

107 4.7 Semiconductor Device A p-n junction in equilibrium, with no externally applied field or potential difference. The generation and recombination current exactly balance. The Fermi energy E i the ame on both ide of the junction. The exce poitive and negative charge on the n and p ide produce an electric field E S F in the direction hown. E p ide Junction n ide The p ide ha an exce of negative charge and i at a lower electric potential, o negatively charged electron have higher energy band here. E F i ng Electron current + S E i nr i pr i pg + Hole current Conduction band The n ide ha an exce of poitive charge and i at a higher electric potential, o negatively charged electron have lower energy band here. Valence band Suppoe you have a box with a barrier eparating the left and right ide: You fill the left ide with oxygen ga and the right ide with nitrogen ga. What happen if the barrier leak? Oxygen diffue to the right, and nitrogen diffue to the left. A imilar diffuion occur acro a p-n junction. Firt conider the equilibrium ituation with no applied voltage (Fig. 4.30). The many hole in the p region act like a hole ga that diffue acro the junction into the n region. Once there, the hole recombine with ome of the many free electron. Similarly, electron diffue from the n region to the p region and fall into ome of the many hole there. The hole and electron diffuion current lead to a net poitive charge in the n region and a net negative charge in the p region, cauing an electric field in the direction from n to p at the junction. The potential energy aociated with thi field raie the electron energy level in the p region relative to the ame level in the n region. There are four current acro the junction, a hown. The diffuion procee lead to recombination current of hole and electron, labeled i pr and i nr in Fig At the ame time, electron hole pair are generated in the junction region by thermal excitation. The electric field decribed above weep thee electron and hole out of the junction; electron are wept oppoite the field to the n ide, and hole are wept in the ame direction a the field to the p ide. The correponding current, called generation current, are labeled i pg and i ng. At equilibrium the magnitude of the generation and recombination current are equal: ƒi pg ƒ = ƒi pr ƒ and ƒi ng ƒ = ƒi nr ƒ (4.3) In thermal equilibrium the Fermi energy i the ame at each point acro the junction. Now we apply a forward bia that i, a poitive potential difference V acro the junction. A forward bia decreae the electric field in the junction region. It alo decreae the difference between the energy level on the p and n ide (Fig. 4.31) E p ide Junction n ide i ng E S i pr i nr i pg Conduction band V 5 0 V. 0 (forward bia) 4.31 A p-n junction under forwardbia condition. The potential difference between p and n region i reduced, a i the electric field within the junction. The recombination current increae but the generation current are nearly contant, cauing a net current from left to right. (Compare Fig ) Valence band

108 148 CHAPTER 4 Molecule and Condened Matter 4.3 Under revere-bia condition the potential-energy difference between the p and n ide of a junction i greater than at equilibrium. If thi difference i great enough, the bottom of the conduction band on the n ide may actually be below the top of the valence band on the p ide. E p ide Junction n ide Conduction band Valence band If a p-n junction under revere bia i thin enough, electron can tunnel from the valence band to the conduction band (a proce called Zener breakdown). Application Swallow Thi Semiconductor Device Thi tiny capule deigned to be wallowed by a patient contain a miniature camera with a CCD light detector, plu ix LED to illuminate the ubject. The capule radio high-reolution image to an external recording unit a it pae painlely through the patient tomach and intetine. Thi technique make it poible to examine the mall intetine, which i not readily acceible with conventional endocopy. by an amount E = -ev. It become eaier for the electron in the n region to climb the potential-energy hill and diffue into the p region and for the hole in the p region to diffue into the n region. Thi effect increae both recombination current by the Maxwell Boltzmann factor e - E>kT = e ev>kt. (We don t have to ue the Fermi Dirac ditribution becaue mot of the available tate for the diffuing electron and hole are empty, o the excluion principle ha little effect.) The generation current don t change appreciably, o the net hole current i The net electron current I = i ptot + i ntot i i ntot i ptot = i pr - ƒi pg ƒ = ƒi pg ƒe ev>kt - ƒi pg ƒ = ƒi pg ƒ1e ev>kt - 1 (4.4) i given by a imilar expreion, o the total current I = I S 1e ev>kt - 1 (4.5) in agreement with Eq. (4.). We can repeat thi entire dicuion for revere bia (negative V and I) with the ame reult. Therefore Eq. (4.) i valid for both poitive and negative value. Several effect make the behavior of practical p-n junction diode more complex than thi imple analyi predict. One effect, avalanche breakdown, occur under large revere bia. The electric field in the junction i o great that the carrier can gain enough energy between colliion to create electron hole pair during inelatic colliion. The electron and hole then gain energy and collide to form more pair, and o on. (A imilar effect occur in dielectric breakdown in inulator, dicued in Section 4.4.) A econd type of breakdown begin when the revere bia become large enough that the top of the valence band in the p region i jut higher in energy than the bottom of the conduction band in the n region (Fig. 4.3). If the junction region i thin enough, the probability become large that electron can tunnel from the valence band of the p region to the conduction band of the n region. Thi proce i called Zener breakdown. It occur in Zener diode, which are ued for voltage regulation and protection againt voltage urge. Semiconductor Device and Light A light-emitting diode (LED) i a p-n junction diode that emit light. When the junction i forward biaed, many hole are puhed from their p region to the junction region, and many electron are puhed from their n region to the junction region. In the junction region the electron fall into hole (recombine). In recombining, the electron can emit a photon with energy approximately equal to the band gap. Thi energy (and therefore the photon wavelength and the color of the light) can be varied by uing material with different band gap. Light-emitting diode are very energy-efficient light ource and have many application, including automobile lamp, traffic ignal, and large tadium diplay. The revere proce i called the photovoltaic effect. Here the material aborb photon, and electron hole pair are created. Pair that are created in the p-n junction, or cloe enough to migrate to it without recombining, are eparated by the electric field we decribed above that weep the electron to the n ide and the hole to the p ide. We can connect thi device to an external circuit, where it become a ource of emf and power. Such a device i often called a olar cell, although unlight in t required. Any light with photon energie greater than the band gap will do. You might have a calculator powered by uch cell. Production of low-cot photovoltaic cell for large-cale olar energy converion i a very active field of reearch. The ame baic phyic i ued in charge-coupled device (CCD) image detector, digital camera, and video camera.

109 4.7 Semiconductor Device 149 Tranitor A bipolar junction tranitor include two p-n junction in a andwich configuration, which may be either p-n-p or n-p-n. Figure 4.33 how uch a p-n-p tranitor. The three region are called the emitter, bae, and collector, a hown. When there i no current in the left loop of the circuit, there i only a very mall current through the reitor R becaue the voltage acro the bae collector junction i in the revere direction. But when a forward bia i applied between emitter and bae, a hown, mot of the hole traveling from emitter to bae travel through the bae (which i typically both narrow and lightly doped) to the econd junction, where they come under the influence of the collector-to-bae potential difference and flow on through the collector to give an increaed current to the reitor. In thi way the current in the collector circuit i controlled by the current in the emitter circuit. Furthermore, V c may be coniderably larger than V e, o the power diipated in R may be much larger than the power upplied to the emitter circuit by the battery V e. Thu the device function a a power amplifier. If the potential drop acro R i greater than V e, it may alo be a voltage amplifier. In thi configuration the bae i the common element between the input and output ide of the circuit. Another widely ued arrangement i the commonemitter circuit, hown in Fig In thi circuit the current in the collector ide of the circuit i much larger than that in the bae ide, and the reult i current amplification. The field-effect tranitor (Fig. 4.35) i an important type. In one variation a lab of p-type ilicon i made with two n-type region on the top, called the ource and the drain; a metallic conductor i fatened to each. A third electrode called the gate i eparated from the lab, ource, and drain by an inulating layer of SiO. When there i no charge on the gate and a potential difference of either polarity i applied between the ource and the drain, there i very little current becaue one of the p-n junction i revere biaed. Now we place a poitive charge on the gate. With dimenion of the order of 10-6 m, it take little charge to provide a ubtantial electric field. Thu there i very little current into or out of the gate. There aren t many free electron in the p-type material, but there are ome, and the effect of the field i to attract them toward the poitive gate. The reulting greatly enhanced concentration of electron near the gate (and between the two junction) permit current to flow between the ource and the drain. The current i very enitive to the gate charge and potential, and the device function a an amplifier. The device jut decribed i called an enhancement-type MOSFET (metal-oxide-emiconductor field-effect tranitor). Integrated Circuit A further refinement in emiconductor technology i the integrated circuit. By ucceively depoiting layer of material and etching pattern to define current path, we can combine the function of everal MOSFET, capacitor, and reitor on a ingle quare of emiconductor material that may be only a few millimeter on a ide. An elaboration of thi idea lead to large-cale integrated circuit. The reulting integrated circuit chip are the heart of all pocket calculator and preent-day computer, large and mall (Fig. 4.36) Schematic diagram of a p-n-p tranitor and circuit. I e Emitter + V e p Bae n Hole flow When V e 5 0, the current i very mall. When a potential V e i applied between emitter and bae, hole travel from the emitter to the bae. When V c i ufficiently large, mot of the hole continue into the collector. + Collector V b V c When V b 5 0, I c i very mall, and mot of the voltage V c appear acro the bae collector junction. A V b increae, the bae collector potential decreae, and more hole can diffue into the collector; thu, I c increae. Ordinarily, I c i much larger than I b. p V c 4.34 A common-emitter circuit. I b + p n p I c Collector Bae Emitter + R R I c i n-type ource Gate + + Inulating layer of SiO i n-type drain 4.35 A field-effect tranitor. The current from ource to drain i controlled by the potential difference between the ource and the drain and by the charge on the gate; no current flow through the gate. p-type ilicon

110 1430 CHAPTER 4 Molecule and Condened Matter 4.36 An integrated circuit chip the ize of your thumb can contain million of tranitor. The firt emiconductor device were invented in Since then, they have completely revolutionized the electronic indutry through miniaturization, reliability, peed, energy uage, and cot. They have found application in communication, computer ytem, control ytem, and many other area. In tranforming thee area, they have changed, and continue to change, human civilization itelf. Tet Your Undertanding of Section 4.7 Suppoe a negative charge i placed on the gate of the MOSFET hown in Fig Will a ubtantial current flow between the ource and the drain? 4.8 Superconductivity Superconductivity i the complete diappearance of all electrical reitance at low temperature. We decribed thi property at the end of Section 5. and the magnetic propertie of type-i and type-ii uperconductor in Section 9.8. In thi ection we ll relate uperconductivity to the tructure and energy-band model of a olid. Although uperconductivity wa dicovered in 1911, it wa not well undertood on a theoretical bai until In that year, the American phyicit John Bardeen, Leon Cooper, and Robert Schrieffer publihed the theory of uperconductivity, now called the BCS theory, that wa to earn them the Nobel Prize in phyic in 197. (It wa Bardeen econd Nobel Prize; he hared hi firt for hi work on the development of the tranitor.) The key to the BCS theory i an interaction between pair of conduction electron, called Cooper pair, caued by an interaction with the poitive ion of the crytal. Here a rough qualitative picture of what happen. A free electron exert attractive force on nearby poitive ion, pulling them lightly cloer together. The reulting light concentration of poitive charge then exert an attractive force on another free electron with momentum oppoite to the firt. At ordinary temperature thi electron-pair interaction i very mall in comparion to energie of thermal motion, but at very low temperature it become ignificant. Bound together thi way, the pair of electron cannot individually gain or loe very mall amount of energy, a they would ordinarily be able to do in a partly filled conduction band. Their pairing give an energy gap in the allowed electron quantum level, and at low temperature there i not enough colliion energy to jump thi gap. Therefore the electron can move freely through the crytal without any energy exchange through colliion that i, with zero reitance. Reearcher have not yet reached a conenu on whether ome modification of the BCS theory can explain the propertie of the high-t C uperconductor that have been dicovered ince There i evidence for pairing, but of a different ort than for conventional uperconductor. Furthermore, the original pairing mechanim of the BCS theory eem too weak to explain the high tranition temperature and critical field of thee new uperconductor.

111 CHAPTER 4 SUMMARY Molecular bond and molecular pectra: The principal type of molecular bond are ionic, covalent, van der Waal, and hydrogen bond. In a diatomic molecule the rotational energy level are given by Eq. (4.3), where I i the moment of inertia of the molecule, m r i it reduced ma, and r 0 i the ditance between the two atom. The vibrational energy level are given by Eq. (4.7), where k i the effective force contant of the interatomic force. (See Example ) E l = l1l + 1 U I I = m r r 0 m 1 m m r = m 1 + m 1l = 0, 1,, Á (4.3) (4.6) (4.4) E n = An + 1 B Uv = An + 1 k B U A m r 1n = 0, 1,, Á (4.7) m 1 k 3 cm r 0 m Solid and energy band: Interatomic bond in olid are of the ame type a in molecule plu one additional type, the metallic bond. Aociating the bai with each lattice point give the crytal tructure. (See Example 4.4.) When atom are bound together in condened matter, their outer energy level pread out into band. At abolute zero, inulator and conductor have a completely filled valence band eparated by an energy gap from an empty conduction band. Conductor, including metal, have partially filled conduction band. (See Example 4.5.) E O r 0 r Free-electron model of metal: In the free-electron model of the behavior of conductor, the electron are treated a completely free particle within the conductor. In thi model the denity of tate i given by Eq. (4.15). The probability that an energy tate of energy E i occupied i given by the Fermi Dirac ditribution, Eq. (4.16), which i a conequence of the excluion principle. In Eq. (4.16), E F i the Fermi energy. (See Example ) g1e = 1m3> V p U 3 E 1> 1 ƒ1e = e 1E-EF>kT + 1 (4.15) (4.16) f(e) 1 kt 5 1 E 40 F kt 5 1 E 1 10 F kt 5 1 E 4 F E 0 E F Semiconductor: A emiconductor ha an energy gap of about 1 ev between it valence and conduction band. It electrical propertie may be dratically changed by the addition of mall concentration of donor impuritie, giving an n-type emiconductor, or acceptor impuritie, giving a p-type emiconductor. (See Example 4.9.) E Conduction electron E g + + Hole S Electric field E Conduction band Band gap Valence band Semiconductor device: Many emiconductor device, including diode, tranitor, and integrated circuit, ue one or more p-n junction. The current voltage relationhip for an ideal p-n junction diode i given by Eq. (4.). I = I S 1e ev>kt - 1 (4.) Forward bia I p n V + p-n diode A Variable emf 1431

112 143 CHAPTER 4 Molecule and Condened Matter BRIDGING PROBLEM Detecting Infrared Photon At 80 K, the band gap in the emiconductor indium antimonide (InSb) i 0.30 ev. A photon emitted by a hydrogen fluoride (HF) molecule undergoing a vibration-rotation tranition from 1n = 1, l = 0 to 1n = 0, l = 1 i aborbed by an electron at the top of the valence band of InSb. (a) How far above the top of the band gap (in ev) i the final tate of the electron? (b) What i the probability that the final tate wa already occupied? The vibration frequency for HF i 1.4 * Hz, the ma of a hydrogen atom i 1.67 * 10-7 kg, the ma of a fluorine atom i 3.15 * 10-6 kg, and the equilibrium ditance between the two nuclei i 0.09 nm. Aume that the Fermi energy for InSb i in the middle of the gap. SOLUTION GUIDE See MateringPhyic tudy area for a Video Tutor olution. IDENTIFY and SET UP 1. Thi problem involve what you learned about molecular tranition in Section 4., about the Fermi Dirac ditribution in Section 4.5, and about emiconductor in Section Equation (4.9) give the combined vibrational-rotational energy in the initial and final molecular tate. The difference between the initial and final molecular energie equal the energy E of the emitted photon, which i in turn equal to the energy gained by the InSb valence electron when it aborb that photon. The probability that the final tate i occupied i given by the Fermi Dirac ditribution, Eq. (4.16). EXECUTE 3. Before you can ue Eq. (4.9), you ll firt need to ue the data given to calculate the moment of inertia I and the quantity Uv for the HF molecule. (Hint: Be careful not to confue frequency ƒ and angular frequency v.) 4. Ue your reult from tep 3 to calculate the initial and final energie of the HF molecule. (Hint: Doe the vibrational energy increae or decreae? What about the rotational energy?) 5. Ue your reult from tep 4 to find the energy imparted to the InSb electron. Determine the final energy of thi electron relative to the bottom of the conduction band. 6. Ue your reult from tep 5 to determine the probability that the InSb final tate i already occupied. EVALUATE 7. I the molecular tranition of the HF molecule allowed? Which i larger: the vibrational energy change or the rotational energy change? 8. I it likely that the excited InSb electron will be blocked from entering a tate in the conduction band? Problem For intructor-aigned homework, go to : Problem of increaing difficulty. CP: Cumulative problem incorporating material from earlier chapter. CALC: Problem requiring calculu. BIO: Biocience problem. DISCUSSION QUESTIONS Q4.1 Ionic bond reult from the electrical attraction of oppoitely charged particle. Are other type of molecular bond alo electrical in nature, or i ome other interaction involved? Explain. Q4. In ionic bond, an electron i tranferred from one atom to another and thu no longer belong to the atom from which it came. Are there imilar tranfer of ownerhip of electron with other type of molecular bond? Explain. Q4.3 Van der Waal bond occur in many molecule, but hydrogen bond occur only with material that contain hydrogen. Why i thi type of bond unique to hydrogen? Q4.4 The bonding of gallium arenide (GaA) i aid to be 31% ionic and 69% covalent. Explain. Q4.5 The H + molecule conit of two hydrogen nuclei and a ingle electron. What kind of molecular bond do you think hold thi molecule together? Explain. Q4.6 The moment of inertia for an axi through the center of ma of a diatomic molecule calculated from the wavelength emitted in an l = 19 S l = 18 tranition i different from the moment of inertia calculated from the wavelength of the photon emitted in an l = 1 S l = 0 tranition. Explain thi difference. Which tranition correpond to the larger moment of inertia? Q4.7 Analyi of the photon aborption pectrum of a diatomic molecule how that the vibrational energy level for mall value of n are very nearly equally paced but the level for large n are not equally paced. Dicu the reaon for thi obervation. Do you expect the adjacent level to move cloer together or farther apart a n increae? Explain. Q4.8 Dicu the difference between the rotational and vibrational energy level of the deuterium ( heavy hydrogen ) molecule D and thoe of the ordinary hydrogen molecule H. A deuterium atom ha twice the ma of an ordinary hydrogen atom. Q4.9 Variou organic molecule have been dicovered in intertellar pace. Why were thee dicoverie made with radio telecope rather than optical telecope? Q4.10 The air you are breathing contain primarily nitrogen 1N and oxygen 1O. Many of thee molecule are in excited rotational energy level 1l = 1,, 3, Á, but almot all of them are in the vibrational ground level 1n = 0. Explain thi difference between the rotational and vibrational behavior of the molecule. Q4.11 In what way do atom in a diatomic molecule behave a though they were held together by a pring? In what way i thi a poor decription of the interaction between the atom? Q4.1 Individual atom have dicrete energy level, but certain olid (which are made up of only individual atom) how energy band and gap. What caue the olid to behave o differently from the atom of which they are compoed?

113 Exercie 1433 Q4.13 What factor determine whether a material i a conductor of electricity or an inulator? Explain. Q4.14 Ionic crytal are often tranparent, wherea metallic crytal are alway opaque. Why? Q4.15 Speed of molecule in a ga vary with temperature, wherea peed of electron in the conduction band of a metal are nearly independent of temperature. Why are thee behavior o different? Q4.16 Ue the band model to explain how it i poible for ome material to undergo a emiconductor-to-metal tranition a the temperature or preure varie. Q4.17 An iolated zinc atom ha a ground-tate electron configuration of filled 1,, p, 3, 3p, and 4 ubhell. How can zinc be a conductor if it valence ubhell i full? Q4.18 The aumption of the free-electron model of metal may eem contrary to reaon, ince electron exert powerful electrical force on each other. Give ome reaon why thee aumption actually make phyical ene. Q4.19 Why are material that are good thermal conductor alo good electrical conductor? What kind of problem doe thi poe for the deign of appliance uch a clothe iron and electric heater? Are there material that do not follow thi general rule? Q4.0 What i the eential characteritic for an element to erve a a donor impurity in a emiconductor uch a Si or Ge? For it to erve a an acceptor impurity? Explain. Q4.1 There are everal method for removing electron from the urface of a emiconductor. Can hole be removed from the urface? Explain. Q4. A tudent aert that ilicon and germanium become good inulator at very low temperature and good conductor at very high temperature. Do you agree? Explain your reaoning. Q4.3 The electrical conductivitie of mot metal decreae gradually with increaing temperature, but the intrinic conductivity of emiconductor alway increae rapidly with increaing temperature. What caue the difference? Q4.4 How could you make compenated ilicon that ha twice a many acceptor a donor? Q4.5 For electronic device uch a amplifier, what are ome advantage of tranitor compared to vacuum tube? What are ome diadvantage? Are there any ituation in which vacuum tube cannot be replaced by olid-tate device? Explain your reaoning. Q4.6 Why doe tunneling limit the miniaturization of MOSFET? Q4.7 The aturation current I S for a p-n junction, Eq. (4.), depend trongly on temperature. Explain why. EXERCISES Section 4.1 Type of Molecular Bond 4.1. If the energy of the H covalent bond i ev, what wavelength of light i needed to break that molecule apart? In what part of the electromagnetic pectrum doe thi light lie? 4.. An Ionic Bond. (a) Calculate the electric potential energy for a K + ion and a Br - ion eparated by a ditance of 0.9 nm, the equilibrium eparation in the KBr molecule. Treat the ion a point charge. (b) The ionization energy of the potaium atom i 4.3 ev. Atomic bromine ha an electron affinity of 3.5 ev. Ue thee data and the reult of part (a) to etimate the binding energy of the KBr molecule. Do you expect the actual binding energy to be higher or lower than your etimate? Explain your reaoning We know from Chapter 18 that the average kinetic energy 3 of an ideal-ga atom or molecule at Kelvin temperature T i kt. For what value of T doe thi energy correpond to (a) the bond energy of the van der Waal bond in He 17.9 * 10-4 ev and (b) the bond energy of the covalent bond in H (4.48 ev)? (c) The kinetic energy in a colliion between molecule can go into diociating one or both molecule, provided the kinetic energy i higher than the bond energy. At room temperature (300 K), i it likely that He molecule will remain intact after a colliion? What about H molecule? Explain Light of wavelength 3.10 mm trike and i aborbed by a molecule. I thi proce mot likely to alter the rotational, vibrational, or atomic energy level of the molecule? Explain your reaoning. (b) If the light in part (a) had a wavelength of 07 nm, which energy level would it mot likely affect? Explain For the H molecule the equilibrium pacing of the two proton i nm. The ma of a hydrogen atom i 1.67 * 10-7 kg. Calculate the wavelength of the photon emitted in the rotational tranition l = to l = (a) A molecule decreae it vibrational energy by 0.50 ev by giving up a photon of light. What wavelength of light doe it give up during thi proce, and in what part of the electromagnetic pectrum doe that wavelength of light lie? (b) An atom decreae it energy by 8.50 ev by giving up a photon of light. What wavelength of light doe it give up during thi proce, and in what part of the electromagnetic pectrum doe that wavelength of light lie? (c) A molecule decreae it rotational energy by 3.0 * 10-3 ev by giving up a photon of light. What wavelength of light doe it give up during thi proce, and in what part of the electromagnetic pectrum doe that wavelength of light lie? Section 4. Molecular Spectra 4.7. A hypothetical NH molecule make a rotational-level tranition from l = 3 to l = 1 and give off a photon of wavelength nm in doing o. What i the eparation between the two atom in thi molecule if we model them a point mae? The ma of hydrogen i 1.67 * 10-7 kg, and the ma of nitrogen i.33 * 10-6 kg The water molecule ha an l = 1 rotational level 1.01 * 10-5 ev above the l = 0 ground level. Calculate the wavelength and frequency of the photon aborbed by water when it undergoe a rotational-level tranition from l = 0 to l = 1. The magnetron ocillator in a microwave oven generate microwave with a frequency of 450 MHz. Doe thi make ene, in view of the frequency you calculated in thi problem? Explain In Example 4. the moment of inertia for CO wa calculated uing Eq. (4.6). (a) In CO, how far i each atom from the center of ma of the molecule? (b) Ue I = m 1 r 1 + m r to calculate the moment of inertia of CO about an axi through the center of ma and perpendicular to the line joining the center of the two atom. Doe your reult agree with the value obtained in Example 4.? Two atom of ceium (C) can form a C molecule. The equilibrium ditance between the nuclei in a C molecule i nm. Calculate the moment of inertia about an axi through the center of ma of the two nuclei and perpendicular to the line joining them. The ma of a ceium atom i.1 * 10-5 kg CP The rotational energy level of CO are calculated in Example 4.. If the energy of the rotating molecule i decribed by the claical expreion K = 1 Iv, for the l = 1 level what are (a) the angular peed of the rotating molecule; (b) the linear peed

114 1434 CHAPTER 4 Molecule and Condened Matter of each atom (ue the reult of Exercie 4.9); (c) the rotational period (the time for one rotation)? 4.1. If a odium chloride (NaCl) molecule could undergo an n S n - 1 vibrational tranition with no change in rotational quantum number, a photon with wavelength 0.0 mm would be emitted. The ma of a odium atom i 3.8 * 10-6 kg, and the ma of a chlorine atom i 5.81 * 10-6 kg. Calculate the force contant k for the interatomic force in NaCl A lithium atom ha ma 1.17 * 10-6 kg, and a hydrogen atom ha ma 1.67 * 10-7 kg. The equilibrium eparation between the two nuclei in the LiH molecule i nm. (a) What i the difference in energy between the l = 3 and l = 4 rotational level? (b) What i the wavelength of the photon emitted in a tranition from the l = 4 to the l = 3 level? When a hypothetical diatomic molecule having atom nm apart undergoe a rotational tranition from the l = tate to the next lower tate, it give up a photon having energy * 10-4 ev. When the molecule undergoe a vibrational tranition from one energy tate to the next lower energy tate, it give up ev. Find the force contant of thi molecule (a) Show that the energy difference between rotational level with angular-momentum quantum number l and l - 1 i lu >I. (b) In term of l, U, and I, what i the frequency of the photon emitted in the pure rotation tranition l S l - 1? The vibrational and rotational energie of the CO molecule are given by Eq. (4.9). Calculate the wavelength of the photon aborbed by CO in each of the following vibration rotation tranition: (a) n = 0, l = 1 S n = 1, l = ; (b) n = 0, l = S n = 1, l = 1; (c) n = 0, l = 3 S n = 1, l =. Section 4.3 Structure of Solid Denity of NaCl. The pacing of adjacent atom in a crytal of odium chloride i 0.8 nm. The ma of a odium atom i 3.8 * 10-6 kg, and the ma of a chlorine atom i 5.89 * 10-6 kg. Calculate the denity of odium chloride Potaium bromide (KBr) ha a denity of.75 * 10 3 kg>m 3 and the ame crytal tructure a NaCl. The ma of a potaium atom i 6.49 * 10-6 kg, and the ma of a bromine atom i 1.33 * 10-5 kg. (a) Calculate the average pacing between adjacent atom in a KBr crytal. (b) How doe the value calculated in part (a) compare with the pacing in NaCl (ee Exercie 4.17)? I the relationhip between the two value qualitatively what you would expect? Explain. Section 4.4 Energy Band The maximum wavelength of light that a certain ilicon photocell can detect i 1.11 mm. (a) What i the energy gap (in electron volt) between the valence and conduction band for thi photocell? (b) Explain why pure ilicon i opaque The gap between valence and conduction band in diamond i 5.47 ev. (a) What i the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic pectrum doe thi photon lie? (b) Explain why pure diamond i tranparent and colorle. (c) Mot gem diamond have a yellow color. Explain how impuritie in the diamond can caue thi color The gap between valence and conduction band in ilicon i 1.1 ev. A nickel nucleu in an excited tate emit a gamma-ray photon with wavelength 9.31 * 10-4 nm. How many electron can be excited from the top of the valence band to the bottom of the conduction band by the aborption of thi gamma ray? Section 4.5 Free-Electron Model of Metal 4.. Calculate v rm for free electron with average kinetic 3 energy kt at a temperature of 300 K. How doe your reult compare to the peed of an electron with a kinetic energy equal to the Fermi energy of copper, calculated in Example 4.7? Why i there uch a difference between thee peed? 4.3. Calculate the denity of tate g1e for the free-electron model of a metal if E = 7.0 ev and V = 1.0 cm 3. Expre your anwer in unit of tate per electron volt Supply the detail in the derivation of Eq. (4.13) from Eq. (4.11) and (4.1) CP Silver ha a Fermi energy of 5.48 ev. Calculate the electron contribution to the molar heat capacity at contant volume of ilver, C V, at 300 K. Expre your reult (a) a a multiple of R and (b) a a fraction of the actual value for ilver, C V = 5.3 J>mol # K. (c) I the value of C V due principally to the electron? If not, to what i it due? (Hint: See Section 18.4.) 4.6. The Fermi energy of odium i 3.3 ev. (a) Find the average energy E av of the electron at abolute zero. (b) What i the peed of an electron that ha energy E av? (c) At what Kelvin temperature T i kt equal to E F? (Thi i called the Fermi temperature for the metal. It i approximately the temperature at which molecule in a claical ideal ga would have the ame kinetic energy a the fatet-moving electron in the metal.) For a olid metal having a Fermi energy of ev, what i the probability, at room temperature, that a tate having an energy of 8.50 ev i occupied by an electron? Section 4.6 Semiconductor 4.8. Pure germanium ha a band gap of 0.67 ev. The Fermi energy i in the middle of the gap. (a) For temperature of 50 K, 300 K, and 350 K, calculate the probability ƒ1e that a tate at the bottom of the conduction band i occupied. (b) For each temperature in part (a), calculate the probability that a tate at the top of the valence band i empty Germanium ha a band gap of 0.67 ev. Doping with arenic add donor level in the gap 0.01 ev below the bottom of the conduction band. At a temperature of 300 K, the probability i 4.4 * 10-4 that an electron tate i occupied at the bottom of the conduction band. Where i the Fermi level relative to the conduction band in thi cae? Section 4.7 Semiconductor Device (a) Suppoe a piece of very pure germanium i to be ued a a light detector by oberving, through the aborption of photon, the increae in conductivity reulting from generation of electron hole pair. If each pair require 0.67 ev of energy, what i the maximum wavelength that can be detected? In what portion of the pectrum doe it lie? (b) What are the anwer to part (a) if the material i ilicon, with an energy requirement of 1.14 ev per pair, correponding to the gap between valence and conduction band in that element? CP At a temperature of 90 K, a certain p-n junction ha a aturation current I S = ma. (a) Find the current at thi temperature when the voltage i (i) 1.00 mv, (ii) mv, (iii) 100 mv, and (iv) -100 mv. (b) I there a region of applied voltage where the diode obey Ohm law? 4.3. For a certain p-n junction diode, the aturation current at room temperature (0 C) i ma. What i the reitance of thi diode when the voltage acro it i (a) 85.0 mv and (b) mv?

115 Problem (a) A forward-bia voltage of 15.0 mv produce a poitive current of 9.5 ma through a p-n junction at 300 K. What doe the poitive current become if the forward-bia voltage i reduced to 10.0 mv? (b) For revere-bia voltage of mv and mv, what i the revere-bia negative current? A p-n junction ha a aturation current of 3.60 ma. (a) At a temperature of 300 K, what voltage i needed to produce a poitive current of 40.0 ma? (b) For a voltage equal to the negative of the value calculated in part (a), what i the negative current? PROBLEMS A hypothetical diatomic molecule of oxygen 1ma =.656 * 10-6 kg and hydrogen 1ma = 1.67 * 10-7 kg emit a photon of wavelength.39 mm when it make a tranition from one vibrational tate to the next lower tate. If we model thi molecule a two point mae at oppoite end of a male pring, (a) what i the force contant of thi pring, and (b) how many vibration per econd i the molecule making? When a diatomic molecule undergoe a tranition from the l = to the l = 1 rotational tate, a photon with wavelength 63.8 mm i emitted. What i the moment of inertia of the molecule for an axi through it center of ma and perpendicular to the line connecting the nuclei? CP (a) The equilibrium eparation of the two nuclei in an NaCl molecule i 0.4 nm. If the molecule i modeled a charge +e and -e eparated by 0.4 nm, what i the electric dipole moment of the molecule (ee Section 1.7)? (b) The meaured electric dipole moment of an NaCl molecule i 3.0 * 10-9 C # m. If thi dipole moment arie from point charge +q and -q eparated by 0.4 nm, what i q? (c) A definition of the fractional ionic character of the bond i q>e. If the odium atom ha charge +e and the chlorine atom ha charge -e, the fractional ionic character would be equal to 1. What i the actual fractional ionic character for the bond in NaCl? (d) The equilibrium ditance between nuclei in the hydrogen iodide (HI) molecule i 0.16 nm, and the meaured electric dipole moment of the molecule i 1.5 * C # m. What i the fractional ionic character for the bond in HI? How doe your anwer compare to that for NaCl calculated in part (c)? Dicu reaon for the difference in thee reult The binding energy of a potaium chloride molecule (KCl) i 4.43 ev. The ionization energy of a potaium atom i 4.3 ev, and the electron affinity of chlorine i 3.6 ev. Ue thee data to etimate the equilibrium eparation between the two atom in the KCl molecule. Explain why your reult i only an etimate and not a precie value (a) For the odium chloride molecule (NaCl) dicued at the beginning of Section 4.1, what i the maximum eparation of the ion for tability if they may be regarded a point charge? That i, what i the larget eparation for which the energy of an Na + ion and a Cl - ion, calculated in thi model, i lower than the energy of the two eparate atom Na and Cl? (b) Calculate thi ditance for the potaium bromide molecule, decribed in Exercie The rotational pectrum of HCl contain the following wavelength (among other): 60.4 mm, 69.0 mm, 80.4 mm, 96.4 mm, and 10.4 mm. Ue thi pectrum to find the moment of inertia of the HCl molecule about an axi through the center of ma and perpendicular to the line joining the two nuclei (a) Ue the reult of Problem 4.40 to calculate the equilibrium eparation of the atom in an HCl molecule. The ma of a chlorine atom i 5.81 * 10-6 kg, and the ma of a hydrogen atom i 1.67 * 10-7 kg. (b) The value of l change by 1 in rotational tranition. What i the value of l for the upper level of the tranition that give rie to each of the wavelength lited in Problem 4.40? (c) What i the longet-wavelength line in the rotational pectrum of HCl? (d) Calculate the wavelength of the emitted light for the correponding tranition in the deuterium chloride (DCl) molecule. In thi molecule the hydrogen atom in HCl i replaced by an atom of deuterium, an iotope of hydrogen with a ma of 3.34 * 10-7 kg. Aume that the equilibrium eparation between the atom i the ame a for HCl When a NaF molecule make a tranition from the l = 3 to the l = rotational level with no change in vibrational quantum number or electronic tate, a photon with wavelength 3.83 mm i emitted. A odium atom ha ma 3.8 * 10-6 kg, and a fluorine atom ha ma 3.15 * 10-6 kg. Calculate the equilibrium eparation between the nuclei in a NaF molecule. How doe your anwer compare with the value for NaCl given in Section 4.1? I thi reult reaonable? Explain CP Conider a ga of diatomic molecule (moment of inertia I) at an abolute temperature T. If E g i a ground-tate energy and E ex i the energy of an excited tate, then the Maxwell Boltzmann ditribution (ee Section 39.4) predict that the ratio of the number of molecule in the two tate i n ex n g = e -1E ex-e g >kt (a) Explain why the ratio of the number of molecule in the lth rotational energy level to the number of molecule in the ground (l = 0) rotational level i n l n 0 = 1l + 1e -3l1l+1U 4>lkT (Hint: For each value of l, how many tate are there with different value of m l?) (b) Determine the ratio n l >n 0 for a ga of CO molecule at 300 K for the cae (i) l = 1; (ii) l = ; (iii) l = 10; (iv) l = 0; (v) l = 50. The moment of inertia of the CO molecule i given in Example 4. (Section 4.). (c) Your reult in part (b) how that a l i increaed, the ratio n l >n 0 firt increae and then decreae. Explain why Our galaxy contain numerou molecular cloud, region many light-year in extent in Figure P4.44 which the denity i high enough and the temperature low enough for atom to form into molecule. Mot of the molecule are H, but a mall fraction of the molecule are carbon monoxide (CO). Such a molecular cloud in the contellation Orion i hown in Fig. P4.44. The left-hand image wa made with an ordinary viible-light telecope; the right-hand image how the molecular cloud in Orion a imaged with a radio telecope tuned to a wavelength emitted by CO in a rotational tranition. The different color in the radio image indicate region of the cloud that are moving either toward u (blue) or away from u (red) relative to the motion of the cloud a a whole, a determined by the Doppler hift of the radiation. (Since a

116 1436 CHAPTER 4 Molecule and Condened Matter molecular cloud ha about 10,000 hydrogen molecule for each CO molecule, it might eem more reaonable to tune a radio telecope to emiion from H than to emiion from CO. Unfortunately, it turn out that the H molecule in molecular cloud do not radiate in either the radio or viible portion of the electromagnetic pectrum.) (a) Uing the data in Example 4. (Section 4.), calculate the energy and wavelength of the photon emitted by a CO molecule in an l = 1 S l = 0 rotational tranition. (b) A a rule, molecule in a ga at temperature T will be found in a certain excited rotational energy level provided the energy of that level i no higher than kt (ee Problem 4.43). Ue thi rule to explain why atronomer can detect radiation from CO in molecular cloud even though the typical temperature of a molecular cloud i a very low 0 K Spectral Line from Iotope. The equilibrium eparation for NaCl i nm. The ma of a odium atom i * 10-6 kg. Chlorine ha two table iotope, 35 Cl and 37 Cl, that have different mae but identical chemical propertie. The atomic ma of 35 Cl i * 10-6 kg, and the atomic ma of 37 Cl i * 10-6 kg. (a) Calculate the wavelength of the photon emitted in the l = S l = 1 and l = 1 S l = 0 tranition for Na 35 Cl. (b) Repeat part (a) for Na 37 Cl. What are the difference in the wavelength for the two iotope? When an OH molecule undergoe a tranition from the n = 0 to the n = 1 vibrational level, it internal vibrational energy increae by ev. Calculate the frequency of vibration and the force contant for the interatomic force. (The ma of an oxygen atom i.66 * 10-6 kg, and the ma of a hydrogen atom i 1.67 * 10-7 kg.) The force contant for the internuclear force in a hydrogen molecule 1H i k =576 N>m. A hydrogen atom ha ma 1.67 * 10-7 kg. Calculate the zero-point vibrational energy for H (that i, the vibrational energy the molecule ha in the n = 0 ground vibrational level). How doe thi energy compare in magnitude with the H bond energy of ev? Suppoe the hydrogen atom in HF (ee the Bridging Problem for thi chapter) i replaced by an atom of deuterium, an iotope of hydrogen with a ma of 3.34 * 10-7 kg. The force contant i determined by the electron configuration, o it i the ame a for the normal HF molecule. (a) What i the vibrational frequency of thi molecule? (b) What wavelength of light correpond to the energy difference between the n = 1 and n = 0 level? In what region of the pectrum doe thi wavelength lie? The hydrogen iodide (HI) molecule ha equilibrium eparation nm and vibrational frequency 6.93 * Hz. The ma of a hydrogen atom i 1.67 * 10-7 kg, and the ma of an iodine atom i.11 * 10-5 kg. (a) Calculate the moment of inertia of HI about a perpendicular axi through it center of ma. (b) Calculate the wavelength of the photon emitted in each of the following vibration rotation tranition: (i) n = 1, l = 1 S n = 0, l = 0; (ii) n = 1, l = S n = 0, l = 1; (iii) n =, l = S n = 1, l = Prove thi tatement: For free electron in a olid, if a tate that i at an energy E above E F ha probability P of being occupied, then the probability i 1 - P that a tate at an energy E below E F i occupied Compute the Fermi energy of potaium by making the imple approximation that each atom contribute one free electron. The denity of potaium i 851 kg>m 3, and the ma of a ingle potaium atom i 6.49 * 10-6 kg Hydrogen i found in two naturally occurring iotope; normal hydrogen (containing a ingle proton in it nucleu) and deu- terium (having a proton and a neutron). Auming that both molecule are the ame ize and that the proton and neutron have the ame ma (which i almot the cae), find the ratio of (a) the energy of any given rotational tate in a diatomic hydrogen molecule to the energy of the ame tate in a diatomic deuterium molecule and (b) the energy of any given vibrational tate in hydrogen to the ame tate in deuterium (auming that the force contant i the ame for both molecule). Why i it phyically reaonable that the force contant would be the ame for hydrogen and deuterium molecule? Metallic lithium ha a bcc crytal tructure. Each unit cell i a cube of ide length a = 0.35 nm. (a) For a bcc lattice, what i the number of atom per unit volume? Give your anwer in term of a. (Hint: How many atom are there per unit cell?) (b) Ue the reult of part (a) to calculate the zero-temperature Fermi energy E F0 for metallic lithium. Aume there i one free electron per atom CALC The one-dimenional calculation of Example 4.4 (Section 4.3) can be extended to three dimenion. For the threedimenional fcc NaCl lattice, the reult for the potential energy of a pair of Na + and Cl - ion due to the electrotatic interaction with all of the ion in the crytal i U = -ae >4pP 0 r, where a = 1.75 i the Madelung contant. Another contribution to the potential energy i a repulive interaction at mall ionic eparation r due to overlap of the electron cloud. Thi contribution can be repreented by A>r 8, where A i a poitive contant, o the expreion for the total potential energy i U tot = - ae 4pP 0 r + A r 8 (a) Let r 0 be the value of the ionic eparation r for which U tot i a minimum. Ue thi definition to find an equation that relate r 0 and A, and ue thi to write U tot in term of r 0. For NaCl, r 0 = 0.81 nm. Obtain a numerical value (in electron volt) of U tot for NaCl. (b) The quantity -U tot i the energy required to remove a Na + ion and a Cl - ion from the crytal. Forming a pair of neutral atom from thi pair of ion involve the releae of 5.14 ev (the ionization energy of Na) and the expenditure of 3.61 ev (the electron affinity of Cl). Ue the reult of part (a) to calculate the energy required to remove a pair of neutral Na and Cl atom from the crytal. The experimental value for thi quantity i 6.39 ev; how well doe your calculation agree? CALC Conider a ytem of N free electron within a volume V. Even at abolute zero, uch a ytem exert a preure p on it urrounding due to the motion of the electron. To calculate thi preure, imagine that the volume increae by a mall amount dv. The electron will do an amount of work pdv on their urrounding, which mean that the total energy E tot of the electron will change by an amount de tot = -pdv. Hence p = -de tot >dv. (a) Show that the preure of the electron at abolute zero i p = 3>3 p 4>3 U a N 5>3 5m V b (b) Evaluate thi preure for copper, which ha a free-electron concentration of 8.45 * 10 8 m -3. Expre your reult in pacal and in atmophere. (c) The preure you found in part (b) i extremely high. Why, then, don t the electron in a piece of copper imply explode out of the metal? CALC When the preure p on a material increae by an amount p, the volume of the material will change from V to V + V, where V i negative. The bulk modulu B of the mate-

117 Anwer 1437 ƒ ƒ rial i defined to be the ratio of the preure change p to the 1.99 * 10-6 kg; and (ii) all ix of the electron from each carbon abolute value V>V of the fractional volume change. The atom are able to move freely throughout the tar. (d) I it a good greater the bulk modulu, the greater the preure increae required approximation to ignore relativitic effect in the tructure of a for a given fractional volume change, and the more incompreible white dwarf tar? Explain. the material (ee Section 11.4). Since V 6 0, the bulk modulu CP A variable DC battery i connected in erie with a can be written a B = - p>1 V>V 0. In the limit that the preure and volume change are very mall, thi become 15-Æ reitor and a p-n junction diode that ha a aturation current of 0.65 ma at room temperature 10 C. When a voltmeter acro the 15-Æ reitor read 35.0 V, what are (a) the voltage B = -V dp acro the diode and (b) the reitance of the diode? dv (a) Ue the reult of Problem 4.55 to how that the bulk modulu for a ytem of N free electron in a volume V at low temperature i B = 5 3 p. (Hint: The quantity p in the expreion B = -V1dp>dV i the external preure on the ytem. Can you explain why thi i equal to the internal preure of the ytem itelf, a found in Problem 4.55?) (b) Evaluate the bulk modulu for the electron in copper, which ha a free-electron concentration of 8.45 * 10 8 m -3. Expre your reult in pacal. (c) The actual bulk modulu of copper i 1.4 * Pa. Baed on your reult in part (b), what fraction of thi i due to the free electron in copper? (Thi reult how that the free electron in a metal play a major role in making the metal reitant to compreion.) What do you think i reponible for the remaining fraction of the bulk modulu? In the dicuion of free electron in Section 4.5, we aumed that we could ignore the effect of relativity. Thi i not a afe aumption if the Fermi energy i greater than about mc (that i, more than about 1% of the ret energy of an electron). (a) Aume that the Fermi energy at abolute zero, a given 1 by Eq. (4.19), i equal to 100 mc. Show that the electron concentration i N V = 3> m 3 c p U 3 and determine the numerical value of N>V. (b) I it a good approximation to ignore relativitic effect for electron in a metal uch a copper, for which the electron concentration i 8.45 * 10 8 m -3? Explain. (c) A white dwarf tar i what i left behind by a tar like the un after it ha ceaed to produce energy by nuclear reaction. (Our own un will become a white dwarf tar in another 6 * 10 9 year or o.) A typical white dwarf ha ma * kg (comparable to the un) and radiu 6000 km (comparable to that of the earth). The gravitational attraction of different part of the white dwarf for each other tend to compre the tar; what prevent it from compreing i the preure of free electron within the tar (ee Problem 4.55). Etimate the electron concentration within a typical white dwarf tar uing the following aumption: (i) the white dwarf tar i made of carbon, which ha a ma per atom of CHALLENGE PROBLEMS CP Van der Waal bond arie from the interaction between two permanent or induced electric dipole moment in a pair of atom or molecule. (a) Conider two identical dipole, each coniting of charge +q and -q eparated by a ditance d and oriented a hown in Fig. P4.59a. Calculate the electric potential energy, expreed in term of the electric dipole moment p = qd, for the ituation where r W d. I the interaction attractive or repulive, and how doe thi potential energy vary with r, the eparation between the center of the two dipole? (b) Repeat part (a) for the orientation of the dipole hown in Fig. P4.59b. The dipole interaction i more complicated when we have to average over the relative orientation of the two dipole due to thermal motion or when the dipole are induced rather than permanent. Figure P4.59 (a) (b) + 1q q + + 1q q CP CALC (a) Conider the hydrogen molecule (H ) to be a imple harmonic ocillator with an equilibrium pacing of nm, and etimate the vibrational energy-level pacing for H The ma of a hydrogen atom i 1.67 * kg. (Hint: Etimate the force contant by equating the change in Coulomb repulion of the proton, when the atom move lightly cloer together than r 0, to the pring force. That i, aume that the chemical binding force remain approximately contant a r i decreaed lightly from r 0.) (b) Ue the reult of part (a) to calculate the vibrational energy-level pacing for the deuterium molecule, D. Aume that the pring contant i the ame for D a for H. The ma of a deuterium atom i 3.34 * 10-7 kg. r r + 1q q q 1q Anwer Chapter Opening Quetion? Venu mut radiate energy into pace at the ame rate that it receive energy in the form of unlight. However, carbon dioxide 1CO molecule in the atmophere aborb infrared radiation emitted by the urface of Venu and re-emit it toward the ground. To compenate for thi and to maintain the balance between emitted and received energy, the urface temperature of Venu and hence the rate of blackbody radiation from the urface both increae.

118 1438 CHAPTER 4 Molecule and Condened Matter Tet Your Undertanding Quetion 4.1 Anwer: (i) The excluion principle tate that only one electron can be in a given tate. Real electron have pin, o two electron (one pin up, one pin down) can be in a given patial tate and hence two can participate in a given covalent bond between two atom. If electron obeyed the excluion principle but did not have pin, that tate of an electron would be completely decribed by it patial ditribution and only one electron could participate in a covalent bond. (We will learn in Chapter 44 that thi ituation i wholly imaginary: There are ubatomic particle without pin, but they do not obey the excluion principle.) 4. Anwer: (ii) Figure 4.5 how that the difference in energy between adjacent rotational level increae with increaing l. Hence, a l increae, the energy E of the emitted photon increae and the wavelength l = hc>e decreae. 4.3 Anwer: (ii) In Fig let a be the ditance between adjacent Na + and Cl - ion. Thi figure how that the Cl - ion that i the next nearet neighbor to a Na + ion i on the oppoite corner of a cube of ide a. The ditance between thee two ion i a + a + a = 3a = a Anwer: (ii) A mall temperature change caue a ubtantial increae in the population of electron in a emiconductor conduction band and a comparably ubtantial increae in conductivity. The conductivity of conductor and inulator varie more gradually with temperature. 4.5 Anwer: no The kinetic-molecular model of an ideal ga (ee Section 18.3) how that the ga preure i proportional to the average tranlational kinetic energy E av of the particle that make up the ga. In a claical ideal ga, E av i directly proportional to the average temperature T, o the preure decreae a T decreae. In a free-electron ga, the average kinetic energy per electron i not related imply to T; a Example 4.8 how, for the free-electron ga in a metal, E av i almot completely a conequence of the excluion principle at room temperature and colder. Hence the preure of a free-electron ga in a olid metal doe not change appreciably between room temperature and abolute zero. 4.6 Anwer: no Pure copper i already an excellent conductor ince it ha a partially filled conduction band (Fig. 4.19c). Furthermore, copper form a metallic crytal (Fig. 4.15) a oppoed to the covalent crytal of ilicon or germanium, o the cheme of uing an impurity to donate or accept an electron doe not work for copper. In fact, adding impuritie to copper decreae the conductivity becaue an impurity tend to catter electron, impeding the flow of current. 4.7 Anwer: no A negative charge on the gate will repel, not attract, electron in the p-type ilicon. Hence the electron concentration in the region between the two p-n junction will be made even maller. With o few charge carrier preent in thi region, very little current will flow between the ource and the drain. Bridging Problem Anwer: (a) 0.78 ev (b) 1.74 * 10-5

119 NUCLEAR PHYSICS 43? Thi culpture of a wooly mammoth, jut 3.7 cm (1.5 in.) in length, wa carved from a mammoth ivory tuk by an artit who lived in outhwetern Germany 35,000 year ago. What phyical principle make it poible to date biological pecimen uch a thee? LEARNING GOALS By tudying thi chapter, you will learn: Some key propertie of atomic nuclei, including radii, denitie, pin, and magnetic moment. How the binding energy of a nucleu depend on the number of proton and neutron that it contain. The mot important way in which untable nuclei undergo radioactive decay. How the decay rate of a radioactive ubtance depend on time. Some of the biological hazard and medical ue of radiation. During the pat century, application of nuclear phyic have had enormou effect on humankind, ome beneficial, ome catatrophic. Many people have trong opinion about application uch a bomb and reactor. Ideally, thoe opinion hould be baed on undertanding, not on prejudice or emotion, and we hope thi chapter will help you to reach that ideal. Every atom contain at it center an extremely dene, poitively charged nucleu, which i much maller than the overall ize of the atom but contain mot of it total ma. We will look at everal important general propertie of nuclei and of the nuclear force that hold them together. The tability or intability of a particular nucleu i determined by the competition between the attractive nuclear force among the proton and neutron and the repulive electrical interaction among the proton. Untable nuclei decay, tranforming themelve pontaneouly into other nuclei by a variety of procee. Nuclear reaction can alo be induced by impact on a nucleu of a particle or another nucleu. Two clae of reaction of pecial interet are fiion and fuion. We could not urvive without the energy releaed by one nearby fuion reactor, our un. How to analyze ome important type of nuclear reaction. What happen in a nuclear fiion chain reaction, and how it can be controlled. The equence of nuclear reaction that allow the un and tar to hine Propertie of Nuclei A we decribed in Section 39., Rutherford found that the nucleu i ten of thouand of time maller in radiu than the atom itelf. Since Rutherford initial experiment, many additional cattering experiment have been performed, uing high-energy proton, electron, and neutron a well a alpha particle (helium-4 nuclei). Thee experiment how that we can model a nucleu a a phere with a radiu R that depend on the total number of nucleon (neutron 1439

120 1440 CHAPTER 43 Nuclear Phyic and proton) in the nucleu. Thi number i called the nucleon number A. The radii of mot nuclei are repreented quite well by the equation R = R 0 A 1>3 1radiu of a nucleu (43.1) where R 0 i an experimentally determined contant: R 0 = 1. * m = 1. fm The nucleon number A in Eq. (43.1) i alo called the ma number becaue it i the nearet whole number to the ma of the nucleu meaured in unified atomic ma unit (u). (The proton ma and the neutron ma are both approximately 1 u.) The bet current converion factor i 1 u = * 10-7 kg In Section 43. we ll dicu the mae of nuclei in more detail. Note that when we peak of the mae of nuclei and particle, we mean their ret mae. Nuclear Denity The volume V of a phere i equal to 4pR 3 >3, o Eq. (43.1) how that the volume of a nucleu i proportional to A. Dividing A (the approximate ma in u) by the volume give u the approximate denity and cancel out A. Thu all nuclei have approximately the ame denity. Thi fact i of crucial importance in undertanding nuclear tructure. Example 43.1 Calculating nuclear propertie The mot common kind of iron nucleu ha ma number A = 56. The volume V of the nucleu (which we treat a a phere of radiu Find the radiu, approximate ma, and approximate denity of the R) and it denity r are nucleu. V = 4 3 pr3 = 4 3 pr 0 3 A = 4 3 p14.6 * m 3 SOLUTION = 4.1 * m 3 IDENTIFY and SET UP: Equation (43.1) tell u how the nuclear radiu R depend on the ma number A. The ma of the nucleu in atomic ma unit i approximately equal to the value of A, and the denity r i ma divided by volume. EXECUTE: The radiu and approximate ma are R = R 0 A 1>3 = 11. * m156 1>3 = 4.6 * m = 4.6 fm m L 156 u11.66 * 10-7 kg>u = 9.3 * 10-6 kg r = m V L 9.3 * 10-6 kg 4.1 * m 3 =.3 * 1017 kg>m 3 EVALUATE: A we mentioned above, all nuclei have approximately thi ame denity. The denity of olid iron i about 7000 kg>m 3 ; the iron nucleu i more than time a dene a iron in bulk. Such denitie are alo found in neutron tar, which are imilar to gigantic nuclei made almot entirely of neutron. A 1-cm cube of material with thi denity would have a ma of.3 * kg, or 30 million metric ton! Nuclide and Iotope The building block of the nucleu are the proton and the neutron. In a neutral atom, the nucleu i urrounded by one electron for every proton in the nucleu. We introduced thee particle in Section 1.1; we ll recount the dicovery of the neutron and proton in Chapter 44. The mae of thee particle are Proton: Neutron: Electron: m p = u = * 10-7 kg m n = u = * 10-7 kg m e = u = * kg

121 43.1 Propertie of Nuclei 1441 The number of proton in a nucleu i the atomic number Z. The number of neutron i the neutron number N. The nucleon number or ma number A i the um of the number of proton Z and the number of neutron N: A = Z + N (43.) A ingle nuclear pecie having pecific value of both Z and N i called a nuclide. Table 43.1 lit value of A, Z, and N for ome nuclide. The electron tructure of an atom, which i reponible for it chemical propertie, i determined by the charge Ze of the nucleu. The table how ome nuclide that have the ame Z but different N. Thee nuclide are called iotope of that element; they have different mae becaue they have different number of neutron in their nuclei. A familiar example i chlorine (Cl, Z = 17). About 76% of chlorine nuclei have N = 18; the other 4% have N = 0. Different iotope of an element uually have lightly different phyical propertie uch a melting and boiling temperature and diffuion rate. The two common iotope of uranium with A = 35 and 38 are uually eparated indutrially by taking advantage of the different diffuion rate of gaeou uranium hexafluoride 1UF 6 containing the two iotope. Table 43.1 alo how the uual notation for individual nuclide: the ymbol of the element, with a pre-ubcript equal to Z and a pre-upercript equal to the ma number A. The general format for an element El i The iotope of chlorine mentioned above, with and 37, are written Z A El. A = 35 Cl and Cl and pronounced chlorine-35 and chlorine-37, repectively. Thi name of the element determine the atomic number Z, o the pre-ubcript Z i ometime omitted, a in 35 Cl. Table 43. give the mae of ome common atom, including their electron. Note that thi table give mae of neutral atom (with Z electron) rather than mae of bare nuclei, becaue it i much more difficult to meaure mae of bare nuclei with high preciion. The ma of a neutral carbon-1 atom i exactly 1 u; that how the unified atomic ma unit i defined. The mae of other atom are approximately equal to A atomic ma unit, a we tated earlier. In fact, the atomic mae are le than the um of the mae of their part (the Z proton, the Z electron, and the N neutron). We ll explain thi very important ma difference in the next ection. Table 43.1 Compoition of Some Common Nuclide Z = atomic number (number of proton) N = neutron number A = Z + N = ma number (total number of nucleon) Nucleu 1 1H 1H 4 He 6 3Li 7 3Li 9 4Be 10 5 B 11 5 B 1 6 C 13 6 C 14 7 N 16 8 O 3 11Na 65 9Cu 00 80Hg 35 9U 38 9U Z N A Z N

122 144 CHAPTER 43 Nuclear Phyic Table 43. Neutral Atomic Mae for Some Light Nuclide Element and Atomic Neutron Atomic Ma Iotope Number, Z Number, N Ma (u) Number, A Hydrogen H Deuterium 1 1 H Tritium H Helium 1 3 He Helium 1 4 He Lithium Li Lithium Li Beryllium Be Boron 1 5 B Boron 1 5 B Carbon 1 6 C Carbon 1 6 C Nitrogen 1 7 N Nitrogen 1 7 N Oxygen 1 8 O Oxygen 1 8 O Oxygen 1 8 O Source: A. H. Waptra and G. Audi, Nuclear Phyic A595, 4 (1995). Nuclear Spin and Magnetic Moment Like electron, proton and neutron are alo pin- 1 particle with pin angular momenta given by the ame equation a in Section The magnitude of the S pin angular momentum of a nucleon i S = 1 A 1 + 1B U=3 4 U (43.3) and the z-component i S z = 1 U (43.4) In addition to the pin angular momentum of the nucleon, there may be orbital angular momentum aociated with their motion within the nucleu. The orbital angular momentum of the nucleon i quantized in the ame way a that of electron in atom. S The total angular momentum J of the nucleu i the vector um of the individual pin and orbital angular momenta of all the nucleon. It ha magnitude J = j1 j + 1U (43.5) and z-component J z = m j U 1m j = -j, -j + 1, Á, j - 1, j (43.6) When the total number of nucleon A i even, j i an integer; when it i odd, j i a half-integer. All nuclide for which both Z and N are even have J = 0, which ugget that pairing of particle with oppoite pin component may be an important conideration in nuclear tructure. The total nuclear angular momentum quantum number j i uually called the nuclear pin, even though in general it refer to a combination of the orbital and pin angular momenta of the nucleon that make up the nucleu. Aociated with nuclear angular momentum i a magnetic moment. When we dicued electron magnetic moment in Section 41.4, we introduced the Bohr magneton m B = eu>m e a a natural unit of magnetic moment. We found that the

123 43.1 Propertie of Nuclei 1443 magnitude of the z-component of the electron-pin magnetic moment i almot exactly equal to m B ; that i, ƒm z ƒ electron L m B. In dicuing nuclear magnetic moment, we can define an analogou quantity, the nuclear magneton m n : m n = eu = * 10-7 J>T = * 10-8 ev>t m p (nuclear magneton) (43.7) where m p i the proton ma. Becaue the proton ma m p i 1836 time larger than the electron ma m e, the nuclear magneton m n i 1836 time maller than the Bohr magneton m B. We might expect the magnitude of the z-component of the pin magnetic moment of the proton to be approximately m n. Intead, it turn out to be ƒm z ƒ proton =.798m n (43.8) Even more urpriing, the neutron, which ha zero charge, ha a pin magnetic moment; it z-component ha magnitude ƒm z ƒ neutron = m n (43.9) The proton ha a poitive charge; a expected, it pin magnetic moment i S S parallel to it pin angular momentum. However, M S and are oppoite for a neutron, a would be expected for a negative charge ditribution. Thee anomalou magnetic moment arie becaue the proton and neutron aren t really fundamental particle but are made of impler particle called quark. We ll dicu quark in ome detail in Chapter 44. The magnetic moment of an entire nucleu i typically a few nuclear magneton. When a nucleu i placed in an external magnetic field B there i an interaction energy U = M # S, S S B = -mz B jut a with atomic magnetic moment. The component of the magnetic moment in the direction of the field m z are quantized, o a erie of energy level reult from thi interaction. M S Example 43. Proton pin flip Proton are placed in a.30-t magnetic field that point in the poitive z-direction. (a) What i the energy difference between tate with the z-component of proton pin angular momentum parallel and antiparallel to the field? (b) A proton can make a tranition from one of thee tate to the other by emitting or aborbing a photon with the appropriate energy. Find the frequency and wavelength of uch a photon. SOLUTION IDENTIFY and SET UP: The proton i a pin- 1 particle with a magnetic moment M S in the ame direction a it pin, o it energy S depend on the orientation of it pin relative to an applied magnetic field B S. If the z-component of i aligned with B S, then m z i S equal to the poitive value given in Eq. (43.8). If the z-component S of i oppoite B S, then m z i the negative of thi value. The interaction energy in either cae i U = -m z B; the difference between thee energie i our target variable in part (a). We find the photon frequency and wavelength uing E = hƒ = hc>l. S EXECUTE: (a) When the z-component of and M S are parallel to B S, the interaction energy i U = - ƒ m z ƒ B = * 10-8 ev>t1.30 T = -.05 * 10-7 ev S When the z-component of S and are antiparallel to the field, the energy i +.05 * 10-7 ev. Hence the energy difference between the tate i E = (.05 * 10-7 ev) = 4.05 * 10-7 ev (b) The correponding photon frequency and wavelength are ƒ = E = 9.79 * 10 7 Hz = 97.9 MHz h = 4.05 * 10-7 ev * ev # l = c = 3.00 * 108 m> f 9.79 * 10 7 = 3.06 m -1 EVALUATE: Thi frequency i in the middle of the FM radio band. When a hydrogen pecimen i placed in a.30-t magnetic field and irradiated with radio wave of thi frequency, proton pin flip can be detected by the aborption of energy from the radiation. M S

124 1444 CHAPTER 43 Nuclear Phyic 43.1 Magnetic reonance imaging (MRI). (a) Random pin of hydrogen proton B S B S B S Proton, the nuclei of hydrogen atom in the tiue under tudy, normally have random pin orientation. In the preence of a trong magnetic field, the pin become aligned with a S component parallel to B. A brief radio ignal caue the pin to flip orientation. A the proton realign with the S B field, they emit radio wave that are picked up by enitive detector. S (b) Since B ha a different value at different location in the tiue, the radio wave from different location have different frequencie. Thi make it poible to contruct an image. (c) An electromagnet ued for MRI Main coil upplie uniform S B field. x coil varie S B field from left to right. z coil varie S B field from head to toe. y coil varie S B field from top to bottom. Tranceiver end and receive ignal that create image. PhET: Simplified MRI Nuclear Magnetic Reonance and MRI Spin-flip experiment of the ort referred to in Example 43. are called nuclear magnetic reonance (NMR). They have been carried out with many different nuclide. Frequencie and magnetic field can be meaured very preciely, o thi technique permit precie meaurement of nuclear magnetic moment. An elaboration of thi baic idea lead to magnetic reonance imaging (MRI), a noninvaive imaging technique that dicriminate among variou body tiue on the bai of the differing environment of proton in the tiue (Fig. 43.1). The magnetic moment of a nucleu i alo the ource of a magnetic field. In an atom the interaction of an electron magnetic moment with the field of the nucleu magnetic moment caue additional plitting in atomic energy level and pectra. We called thi effect hyperfine tructure in Section Meaurement of the hyperfine tructure may be ued to directly determine the nuclear pin. Tet Your Undertanding of Section 43.1 (a) By what factor mut 3 the ma number of a nucleu increae to double it volume? (i) ; (ii) ; (iii) ; (iv) 4; (v) 8. (b) By what factor mut the ma number increae to double 3 the radiu of the nucleu? (i) ; (ii) ; (iii) ; (iv) 4; (v) Nuclear Binding and Nuclear Structure Becaue energy mut be added to a nucleu to eparate it into it individual proton and neutron, the total ret energy E 0 of the eparated nucleon i greater than the ret energy of the nucleu. The energy that mut be added to eparate the

125 43. Nuclear Binding and Nuclear Structure 1445 nucleon i called the binding energy E B ; it i the magnitude of the energy by which the nucleon are bound together. Thu the ret energy of the nucleu i E 0 - E B. Uing the equivalence of ret ma and energy (ee Section 37.8), we ee that the total ma of the nucleon i alway greater than the ma of the nucleu by an amount E B >c called the ma defect. The binding energy for a nucleu containing Z proton and N neutron i defined a Z A M E B = 1ZM H + Nm n - A ZMc (nuclear binding energy) (43.10) where i the ma of the neutral atom containing the nucleu, the quantity in the parenthee i the ma defect, and c = MeV>u. Note that Eq. (43.10) doe not include Zm p, the ma of Z proton. Rather, it contain ZM H, the ma of Z proton and Z electron combined a Z neutral atom, to balance the Z electron included in Z A 1 1 H M, the ma of the neutral atom. The implet nucleu i that of hydrogen, a ingle proton. Next come the nucleu of 1 H, the iotope of hydrogen with ma number, uually called deuterium. It nucleu conit of a proton and a neutron bound together to form a particle called the deuteron. By uing value from Table 43. in Eq. (43.10), we find that the binding energy of the deuteron i E B = u u u MeV>u =.4 MeV Thi much energy would be required to pull the deuteron apart into a proton and a neutron. An important meaure of how tightly a nucleu i bound i the binding energy per nucleon, E B >A. At 1.4 MeV>1 nucleon = 1.11 MeV per nucleon, H ha the lowet binding energy per nucleon of all nuclide. 1 Application Deuterium and Heavy Water Toxicity A crucial tep in plant and animal cell diviion i the formation of a pindle, which eparate the two et of daughter chromoome. If a plant i given only heavy water in which one or both of the hydrogen atom in an H O molecule are replaced with a deuterium atom cell diviion top and the plant top growing. The reaon i that deuterium i more maive than ordinary hydrogen, o the O H bond in heavy water ha a lightly different binding energy and heavy water ha lightly different propertie a a olvent. The biochemical reaction that occur during cell diviion are very enitive to thee olvent propertie, o a pindle never form and the cell cannot reproduce. Chromoome Spindle Problem-Solving Strategy 43.1 Nuclear Propertie IDENTIFY the relevant concept: The key propertie of a nucleu are it ma, radiu, binding energy, ma defect, binding energy per nucleon, and angular momentum. SET UP the problem: Once you have identified the target variable, aemble the equation needed to olve the problem. A relatively mall number of equation from thi ection and Section 43.1 are all you need. EXECUTE the olution: Solve for the target variable. Bindingenergy calculation uing Eq. (43.10) often involve ubtracting two nearly equal quantitie. To get enough preciion in the difference, you may need to carry a many a nine ignificant figure, if that many are available. EVALUATE your anwer: It ueful to be familiar with the following benchmark magnitude. Proton and neutron are about 1840 time a maive a electron. Nuclear radii are of the order of m. The electric potential energy of two proton in a nucleu i roughly J or 1 MeV, o nuclear interaction energie are typically a few MeV rather than a few ev a with atom. The binding energy per nucleon i about 1% of the nucleon ret energy. (The ionization energy of the hydrogen atom i only 0.003% of the electron ret energy.) Angular momenta are determined only by the value of U, o they are of the ame order of magnitude in both nuclei and atom. Nuclear magnetic moment, however, are about a factor of 1000 maller than thoe of electron in atom becaue nuclei are o much more maive than electron. Example 43.3 The mot trongly bound nuclide Find the ma defect, the total binding energy, and the binding 6 energy per nucleon of 8Ni, which ha the highet binding energy per nucleon of all nuclide (Fig. 43.). The neutral atomic ma of 6 8Ni i u. SOLUTION IDENTIFY and SET UP: The ma defect M i the difference between the ma of the nucleu and the combined ma of it contituent nucleon. The binding energy i thi quantity E B multiplied by c, and the binding energy per nucleon i E B divided by the ma number A. We ue Eq. (43.10), M = ZM H + Nm n - A ZM, to determine both the ma defect and the binding energy. EXECUTE: With Z = 8, M H = u, N = A - Z = A 6-8 = 34, m n = u, and ZM = u, Eq. (43.10) give M u. The binding energy i then E B = u MeV>u = MeV Continued

126 1446 CHAPTER 43 Nuclear Phyic The binding energy per nucleon i E B >A = MeV>6, or MeV per nucleon. EVALUATE: Our reult mean that it would take a minimum of MeV to pull a 8Ni completely apart into 8 proton and 6 34 neutron. The ma defect of 8Ni i about 1% of the atomic (or the nuclear) ma. The binding energy i therefore about 1% of the ret energy of the nucleu, and the binding energy per nucleon i about 1% of the ret energy of a nucleon. Note that the ma defect i more than half the ma of a nucleon, which ugget how tightly bound nuclei are. Nearly all table nuclide, from the lightet to the mot maive, have binding energie in the range of 7 9 MeV per nucleon. Figure 43. i a graph of binding energy per nucleon a a function of the ma number A. Note the pike at A = 4, howing the unuually large binding energy per nucleon of the 4 He nucleu (alpha particle) relative to it neighbor. To explain thi curve, we mut conider the interaction among the nucleon. The Nuclear Force The force that bind proton and neutron together in the nucleu, depite the electrical repulion of the proton, i an example of the trong interaction that we mentioned in Section 5.5. In the context of nuclear tructure, thi interaction i called the nuclear force. Here are ome of it characteritic. Firt, it doe not depend on charge; neutron a well a proton are bound, and the binding i the ame for both. Second, it ha hort range, of the order of nuclear dimenion that i, m. (Otherwie, the nucleu would grow by pulling in additional proton and neutron.) But within it range, the nuclear force i much tronger than electrical force; otherwie, the nucleu could never be table. It would be nice if we could write a imple equation like Newton law of gravitation or Coulomb law for thi force, but phyicit have yet to fully determine it dependence on the eparation r. Third, the nearly contant denity of nuclear matter and the nearly contant binding energy per nucleon of larger nuclide how that a particular nucleon cannot interact imultaneouly with all the other nucleon in a nucleu, but only with thoe few in it immediate vicinity. Thi i different from electrical force; every proton in the nucleu repel every other one. Thi limited number of interaction i called aturation; it i analogou to covalent bonding in molecule and olid. Finally, the nuclear force favor binding of pair of proton or neutron with oppoite pin and of pair of pair that i, a pair of proton and a pair of neutron, each pair having oppoite pin. Hence the alpha particle (two proton and two neutron) i an exceptionally table nucleu for it ma number. We ll ee other evidence for pairing effect in nuclei in the 43. Approximate binding energy per nucleon a a function of ma number A (the total number of nucleon) for table nuclide. E B/A (MeV/nucleon) C 4 He 6 8 Ni 38 9 U O 1 H The curve reache a peak of about 8.8 MeV/nucleon at A 5 6, correponding to the element nickel. The pike at A 5 4 how the unuual tability of 4 the He tructure. A

127 43. Nuclear Binding and Nuclear Structure 1447 next ubection. (In Section 4.8 we decribed an analogou pairing that bind oppoite-pin electron in Cooper pair in the BCS theory of uperconductivity.) The analyi of nuclear tructure i more complex than the analyi of manyelectron atom. Two different kind of interaction are involved (electrical and nuclear), and the nuclear force i not yet completely undertood. Even o, we can gain ome inight into nuclear tructure by the ue of imple model. We ll dicu briefly two rather different but ucceful model, the liquid-drop model and the hell model. The Liquid-Drop Model The liquid-drop model, firt propoed in 198 by the Ruian phyicit George Gamow and later expanded on by Niel Bohr, i uggeted by the obervation that all nuclei have nearly the ame denity. The individual nucleon are analogou to molecule of a liquid, held together by hort-range interaction and urface-tenion effect. We can ue thi imple picture to derive a formula for the etimated total binding energy of a nucleu. We ll include five contribution: 1. We ve remarked that nuclear force how aturation; an individual nucleon interact only with a few of it nearet neighbor. Thi effect give a bindingenergy term that i proportional to the number of nucleon. We write thi term a C 1 A, where C 1 i an experimentally determined contant.. The nucleon on the urface of the nucleu are le tightly bound than thoe in the interior becaue they have no neighbor outide the urface. Thi decreae in the binding energy give a negative energy term proportional to the urface area 4pR. Becaue R i proportional to A 1>3, thi term i proportional to A >3 ; we write it a -C A >3, where C i another contant. 3. Every one of the Z proton repel every one of the 1Z - 1 other proton. The total repulive electric potential energy i proportional to Z1Z - 1 and inverely proportional to the radiu R and thu to A 1/3. Thi energy term i negative becaue the nucleon are le tightly bound than they would be without the electrical repulion. We write thi correction a -C 3 Z(Z - 1)>A 1/3. 4. To be in a table, low-energy tate, the nucleu mut have a balance between the energie aociated with the neutron and with the proton. Thi mean that N i cloe to Z for mall A and N i greater than Z (but not too much greater) for larger A. We need a negative energy term correponding to the difference ƒn - Zƒ. The bet agreement with oberved binding energie i obtained if thi term i proportional to 1N - Z >A. If we ue N = A - Z to expre thi energy in term of A and Z, thi correction i -C 4 1A - Z >A. 5. Finally, the nuclear force favor pairing of proton and of neutron. Thi energy term i poitive (more binding) if both Z and N are even, negative (le binding) if both Z and N are odd, and zero otherwie. The bet fit to the data occur with the form C 5 A -4>3 for thi term. E B The total etimated binding energy i the um of thee five term: E B = C 1 A - C A >3 Z1Z - 1 1A - Z - C 3 A 1>3 - C 4 C A 5 A -4>3 (43.11) (nuclear binding energy) The contant C 1, C, C 3, C 4, and C 5, choen to make thi formula bet fit the oberved binding energie of nuclide, are C 1 = MeV C = MeV C 3 = MeV C 4 = 3.69 MeV C 5 = 39 MeV

128 1448 CHAPTER 43 Nuclear Phyic The contant C 1 i the binding energy per nucleon due to the aturated nuclear force. Thi energy i almot 16 MeV per nucleon, about double the total binding energy per nucleon in mot nuclide. If we etimate the binding energy E B uing Eq. (43.11), we can olve Eq. (43.10) to ue it to etimate the ma of any neutral atom: Z A M = ZM H + Nm n - E B c (emiempirical ma formula) (43.1) Equation (43.1) i called the emiempirical ma formula. The name i apt; it i empirical in the ene that the C have to be determined empirically (experimentally), yet it doe have a ound theoretical bai. Example 43.4 Etimating binding energy and ma 6 For the nuclide 8Ni of Example 43.3, (a) calculate the five term in the binding energy and the total etimated binding energy, and (b) find the neutral atomic ma uing the emiempirical ma formula. SOLUTION IDENTIFY and SET UP: We ue the liquid-drop model of the nucleu and it five contribution to the binding energy, a given by Eq. (43.11), to calculate the total binding energy E B. We then 6 ue Eq. (43.1) to find the neutral atomic ma 8M. EXECUTE: (a) With Z = 8, A = 6, and N = 34, the five term in Eq. (43.11) are 1. C 1 A = MeV16 = MeV. -C A >3 = MeV16 >3 = MeV Z1Z C 3 A 1>3 = MeV >3 = MeV 4. -C 4 1A - Z A 5. +C 5 A -4>3 = 139 MeV16-4>3 = 0. MeV The pairing correction (term 5) i by far the mallet of all the term; it i poitive becaue both Z and N are even. The um of all five term i the total etimated binding energy, E B MeV. (b) We ue MeV in Eq. (43.1): E B = MeV 6 = MeV 6 8M = u u MeV MeV>u = u EVALUATE: The binding energy of 8Ni calculated in part (a) i only about 0.6% larger than the true value of MeV found in Example 43.3, and the ma calculated in part (b) i only about 0.005% maller than the meaured value of u. The emiempirical ma formula can be quite accurate! 6 The liquid-drop model and the ma formula derived from it are quite ucceful in correlating nuclear mae, and we will ee later that they are a great help in undertanding decay procee of untable nuclide. Some other apect of nuclei, uch a angular momentum and excited tate, are better approached with different model. The Shell Model The hell model of nuclear tructure i analogou to the central-field approximation in atomic phyic (ee Section 41.6). We picture each nucleon a moving in a potential that repreent the averaged-out effect of all the other nucleon. Thi may not eem to be a very promiing approach; the nuclear force i very trong, very hort range, and therefore trongly ditance dependent. However, in ome repect, thi model turn out to work fairly well. The potential-energy function for the nuclear force i the ame for proton a for neutron. Figure 43.3a how a reaonable aumption for the hape of thi function: a pherical verion of the quare-well potential we dicued in Section The corner are omewhat rounded becaue the nucleu doen t have a harply defined urface. For proton there i an additional potential energy aociated with electrical repulion. We conider each proton to interact with a phere of uniform charge denity, with radiu R and total charge 1Z - 1e. Figure 43.3b how the nuclear, electric, and total potential energie for a proton a function of the ditance r from the center of the nucleu.

129 43.3 Nuclear Stability and Radioactivity 1449 In principle, we could olve the Schrödinger equation for a proton or neutron moving in uch a potential. For any pherically ymmetric potential energy, the angular-momentum tate are the ame a for the electron in the central-field approximation in atomic phyic. In particular, we can ue the concept of filled hell and ubhell and their relationhip to tability. In atomic tructure we found that the value Z =, 10, 18, 36, 54, and 86 (the atomic number of the noble gae) correpond to particularly table electron arrangement. A comparable effect occur in nuclear tructure. The number are different becaue the potential-energy function i different and the nuclear pin-orbit interaction i much tronger and of oppoite ign than in atom, o the ubhell fill up in a different order from thoe for electron in an atom. It i found that when the number of neutron or the number of proton i, 8, 0, 8, 50, 8, or 16, the reulting tructure i unuually table that i, ha an unuually high binding energy. (Nuclide with Z = 16 have not been oberved in nature.) Thee number are called magic number. Nuclide in which Z i a magic number tend to have an above-average number of table iotope. There are everal doubly magic nuclide for which both Z and N are magic, including 4 He 16 8O 0 40 Ca 0 48 Ca 08 8Pb All thee nuclide have ubtantially higher binding energy per nucleon than do nuclide with neighboring value of N or Z. They alo all have zero nuclear pin. The magic number correpond to filled-hell or -ubhell configuration of nucleon energy level with a relatively large jump in energy to the next allowed level Approximate potential-energy function for a nucleon in a nucleu. The approximate nuclear radiu i R. (a) The potential energy U nuc due to the nuclear force i the ame for proton and neutron. For neutron, it i the total potential energy U nuc (MeV) 0 (b) For proton, the total potential energy U tot i the um of the nuclear (U nuc ) and electric (U el ) potential energie. U (MeV) 0 U el 0 0 R R r r U tot 5 U el U nuc Tet Your Undertanding of Section 43. Rank the following nuclei in order from larget to mallet value of the binding energy per nucleon. (i) 4 He; (ii) 4 5 Cr; (iii) (iv) (v) 15 6Sm; 00 80Hg; 5 9Cf. 40 U nuc 43.3 Nuclear Stability and Radioactivity Among about 500 known nuclide, fewer than 300 are table. The other are untable tructure that decay to form other nuclide by emitting particle and electromagnetic radiation, a proce called radioactivity. The time cale of thee decay procee range from a mall fraction of a microecond to billion of year. The table nuclide are hown by dot on the graph in Fig. 43.4, where the neutron number N and proton number (or atomic number) Z for each nuclide are plotted. Such a chart i called a Segrè chart, after it inventor, the Italian- American phyicit Emilio Segrè ( ). Each blue line perpendicular to the line N = Z repreent a pecific value of the ma number A = Z + N. Mot line of contant A pa through only one or two table nuclide; that i, there i uually a very narrow range of tability for a given ma number. The line at A = 0, A = 40, A = 60, and A = 80 are example. In four cae thee line pa through three table nuclide namely, at A = 96, 14, 130, and 136. Four table nuclide have both odd Z and odd N: ActivPhyic 19.: Nuclear Binding Energy ActivPhyic 19.4: Radioactivity Thee are called odd-odd nuclide. The abence of other odd-odd nuclide how the influence of pairing. Alo, there i no table nuclide with A = 5 or A = 8. The doubly magic 4 He nucleu, with a pair of proton and a pair of neutron, ha no interet in accepting a fifth particle into it tructure. Collection of eight nucleon decay to maller nuclide, with a nucleu immediately plitting into two Be He nuclei. 1 H 3 6 Li 10 5B 14 7N

130 1450 CHAPTER 43 Nuclear Phyic 43.4 Segrè chart howing neutron number and proton number for table nuclide. N 144 A A A A Neutron number A A A A 5 14 A 5 10 A 5 80 A A N 5 Z A A 5 0 A 5 40 A the ma number A increae, table nuclide have an increaing ratio of neutron to proton Proton number 96 Z The point on the Segrè chart repreenting table nuclide define a rather narrow tability region. For low ma number, the number of proton and neutron are approximately equal, N L Z. The ratio N>Z increae gradually with A, up to about 1.6 at large ma number, becaue of the increaing influence of the electrical repulion of the proton. Point to the right of the tability region repreent nuclide that have too many proton relative to neutron to be table. In thee cae, repulion win, and the nucleu come apart. To the left are nuclide with too many neutron relative to proton. In thee cae the energy aociated with the neutron i out of balance with that aociated with the proton, and the nuclide decay in a proce that convert neutron to proton. The graph alo how that no nuclide with A 7 09 or Z 7 83 i table. A nucleu i untable if it i too big. Note that there i no table nuclide with Z = 43 (technetium) or 61 (promethium). PhET: Alpha Decay Alpha Decay Nearly 90% of the 500 known nuclide are radioactive; they are not table but decay into other nuclide. When untable nuclide decay into different nuclide, they uually emit alpha 1a or beta 1b particle. An alpha particle i a 4 He nucleu, two proton and two neutron bound together, with total pin zero. Alpha emiion occur principally with nuclei that are too large to be table. When a nucleu emit an alpha particle, it N and Z value each decreae by and A decreae by 4, moving it cloer to table territory on the Segrè chart.

131 43.3 Nuclear Stability and Radioactivity Alpha decay of the untable radium nuclide 6 88 Ra. (a) 6 Ra 88 88p 138n 86p 136n Rn 86 The nuclide 6 88 Ra decay by alpha emiion to 86Rn. a (p, n) 4 He (b) Potential-energy curve for an a particle and Rn nucleu 86 U (MeV) R The a particle tunnel through the potential-energy barrier. r (c) Energy-level diagram for the ytem 6 Ra can a-decay directly 88 to the Rn ground level MeV MeV MeV a g a... or it can a-decay to an excited level 86 Rn*, which can then decay to 6 the Rn ground level by 88 Ra 86 emitting a MeV photon (g). 86 Rn* 86 Rn 6 88Ra A familiar example of an alpha emitter i radium, (Fig. 43.5a). The peed of the emitted alpha particle, determined from the curvature of it path in a tranvere magnetic field, i about 1.5 * 10 7 m>. Thi peed, although large, i only 5% of the peed of light, o we can ue the nonrelativitic kinetic-energy expreion K = 1 mv : K = * 10-7 kg11.5 * 10 7 m> = 7.67 * J = 4.79 MeV Alpha particle are alway emitted with definite kinetic energie, determined by conervation of momentum and energy. Becaue of their charge and ma, alpha particle can travel only everal centimeter in air, or a few tenth or hundredth of a millimeter through olid, before they are brought to ret by colliion. Some nuclei can pontaneouly decay by emiion of a particle becaue energy i releaed in their alpha decay. You can ue conervation of ma-energy to how that alpha decay i poible whenever the ma of the original neutral atom i greater than the um of the mae of the final neutral atom and the neutral helium-4 atom. In alpha decay, the a particle tunnel through a potential-energy barrier, a Fig. 43.5b how. You may want to review the dicuion of tunneling in Section Example 43.5 Alpha decay of radium 6 Show that the -emiion proce 88Ra S 86Rn a He i energetically poible, and calculate the kinetic energy of the emitted The difference in ma between the original nucleu and the decay EXECUTE: From Table 43., the ma of the He atom i u. 6 a particle. The neutral atomic mae are u for 88Ra product i and u for 86Rn u u u = u SOLUTION IDENTIFY and SET UP: Alpha emiion i poible if the ma of 6 the 88Ra atom i greater than the um of the atomic mae of 4 86Rn and He. The ma difference between the initial radium atom and the final radon and helium atom correpond (through E = mc ) to the energy E releaed in the decay. Becaue momentum i conerved a well a energy, both the alpha particle and the 86Rn atom are in motion after the decay; we will have to account for thi in determining the kinetic energy of the alpha particle. Since thi i poitive, a decay i energetically poible. The energy equivalent of thi ma difference i E = u MeV>u = MeV Thu we expect the decay product to emerge with total kinetic energy MeV. Momentum i alo conerved; if the parent 6 88Ra nucleu i at ret, the daughter 86Rn nucleu and the a particle have momenta of equal magnitude p but oppoite direction. Kinetic Continued

132 145 CHAPTER 43 Nuclear Phyic energy i K = 1 mv = p >m: Since p i the ame for the two particle, the kinetic energy divide inverely a their mae. Hence the a particle get >1 + 4 of the total, or 4.78 MeV. EVALUATE: Experiment how that 88Ra doe undergo alpha decay, and the oberved a-particle energy i 4.78 MeV. You can check your reult by verifying that the alpha particle and the 6 86Rn nucleu produced in the decay have the ame magnitude of momentum p = mv. You can calculate the peed v of each of the decay product from it repective kinetic energy. You ll find that the alpha particle move at a prightly c = 1.5 * 10 7 m>; if momentum i conerved, you hould find that the 86Rn nucleu 4 move a fat. Doe it? Beta Decay There are three different imple type of beta decay: beta-minu, beta-plu, and electron capture. A beta-minu particle (b - ) i an electron. It not obviou how a nucleu can emit an electron if there aren t any electron in the nucleu. Emiion of a b - involve tranformation of a neutron into a proton, an electron, and a third particle called an antineutrino. In fact, if you freed a neutron from a nucleu, it would decay into a proton, an electron, and an antineutrino in an average time of about 15 minute. Beta particle can be identified and their peed can be meaured with technique that are imilar to the Thomon experiment we decribed in Section 7.5. The peed of beta particle range up to of the peed of light, o their motion i highly relativitic. They are emitted with a continuou pectrum of energie. Thi would not be poible if the only two particle were the b - and the recoiling nucleu, ince energy and momentum conervation would then require a definite peed for the b -. Thu there mut be a third particle involved. From conervation of charge, it mut be neutral, and from conervation of angular momentum, it mut be a pin- 1 particle. Thi third particle i an antineutrino, the antiparticle of a neutrino. The ymbol for a neutrino i n e (the Greek letter nu). Both the neutrino and the antineutrino have zero charge and zero (or very mall) ma and therefore produce very little obervable effect when paing through matter. Both evaded detection until 1953, when Frederick Reine and Clyde Cowan ucceeded in oberving the antineutrino directly. We now know that there are at leat three varietie of neutrino, each with it correponding antineutrino; one i aociated with beta decay and the other two are aociated with the decay of two untable particle, the muon and the tau particle. We ll dicu thee particle in more detail in Chapter 44. The antineutrino - that i emitted in b - decay i denoted a n e. The baic proce of b decay i n S p + b - + n e (43.13) Beta-minu decay uually occur with nuclide for which the neutron-toproton ratio N>Z i too large for tability. In b - decay, N decreae by 1, Z increae by 1, and A doen t change. You can ue conervation of ma-energy to how that beta-minu decay can occur whenever the ma of the original neutral atom i larger than that of the final atom. Example 43.6 Why cobalt-60 i a beta-minu emitter 60 The nuclide 7Co, an odd-odd untable nucleu, i ued in medical We mut firt identify the nuclide that will reult if 7Co undergoe b - decay and then compare it neutral atomic ma to that and indutrial application of radiation. Show that it i untable relative to b - 60 decay. The atomic mae you need are u of 7Co for 7Co and u for 8Ni. 60 EXECUTE: In the preumed b - decay of 7Co, Z increae by 60 1 from 7 to 8 and A remain at 60, o the final nuclide i 8Ni. SOLUTION The neutral atomic ma of 7Co i greater than that of 8Ni by IDENTIFY and SET UP: Beta-minu decay i poible if the ma u, o b - decay can occur. of the original neutral atom i greater than that of the final atom. 60

133 43.3 Nuclear Stability and Radioactivity 1453 EVALUATE: With three decay product in decay the 8Ni nucleu, the electron, and the antineutrino the energy can be hared in many different way that are conitent with conervation of energy and momentum. It impoible to predict preciely how the energy will b - 60 be hared for the decay of a particular 7Co nucleu. By contrat, in alpha decay there are jut two decay product, and their energie and momenta are determined uniquely (ee Example 43.5). 60 We have noted that decay occur with nuclide that have too large a neutron-to-proton ratio N>Z. Nuclide for which N>Z i too mall for tability can emit a poitron, the electron antiparticle, which i identical to the electron but with poitive charge. (We ll dicu the poitron in more detail in Chapter 44.) The baic proce, called beta-plu decay 1b +, i p S n + b + + n e (43.14) where b + i a poitron and i the electron neutrino. b - n e Beta-plu decay can occur whenever the ma of the original neutral atom i at leat two electron mae larger than that of the final atom. You can how thi uing conervation of ma-energy. The third type of beta decay i electron capture. There are a few nuclide for which b + emiion i not energetically poible but in which an orbital electron (uually in the K hell) can combine with a proton in the nucleu to form a neutron and a neutrino. The neutron remain in the nucleu and the neutrino i emitted. The baic proce i p + b - S n + n e You can ue conervation of ma-energy to how that (43.15) electron capture can occur whenever the ma of the original neutral atom i larger than that of the final atom. In all type of beta decay, A remain contant. However, in beta-plu decay and electron capture, N increae by 1 and Z decreae by 1 a the neutron proton ratio increae toward a more table value. The reaction of Eq. (43.15) alo help to explain the formation of a neutron tar, mentioned in Example CAUTION Beta decay inide and outide nuclei The beta-decay reaction given by Eq. (43.13), (43.14), and (43.15) occur within a nucleu. Although the decay of a neutron outide the nucleu proceed through the reaction of Eq. (43.13), the reaction of Eq. (43.14) i forbidden by conervation of ma-energy for a proton outide the nucleu. The reaction of Eq. (43.15) can occur outide the nucleu only with the addition of ome extra energy, a in a colliion. Example 43.7 Why cobalt-57 i not a beta-plu emitter 57 The nuclide 7Co i an odd-even untable nucleu. Show that it EXECUTE: The original nuclide i 7Co. In both the preumed cannot undergo b + decay, but that it can decay by electron capture. decay and electron capture, Z decreae by 1 from 7 to 6, and The atomic mae you need are u for 7Co and A remain at 57, o the final nuclide i 6Fe. It ma i le than u for 6Fe. that of 7Co by u, a value maller than u (two electron mae), o b + decay cannot occur. However, the ma of SOLUTION the original atom i greater than the ma of the final atom, o electron capture can occur. IDENTIFY and SET UP: Beta-plu decay i poible if the ma of the original neutral atom i greater than that of the final atom plu EVALUATE: In electron capture there are jut two decay product, two electron mae ( u). Electron capture i poible if the final nucleu and the emitted neutrino. A in alpha decay the ma of the original atom i greater than that of the final atom. (Example 43.5) but unlike in b - decay (Example 43.6), the decay 57 We mut firt identify the nuclide that will reult if 7Co undergoe product of electron capture have unique energie and momenta. In b + decay or electron capture and then find the correponding ma Section 43.4 we ll ee how to relate the probability that electron difference. capture will occur to the half-life of thi nuclide. 57 b +

134 1454 CHAPTER 43 Nuclear Phyic Gamma Decay The energy of internal motion of a nucleu i quantized. A typical nucleu ha a et of allowed energy level, including a ground tate (tate of lowet energy) and everal excited tate. Becaue of the great trength of nuclear interaction, excitation energie of nuclei are typically of the order of 1 MeV, compared with a few ev for atomic energy level. In ordinary phyical and chemical tranformation the nucleu alway remain in it ground tate. When a nucleu i placed in an excited tate, either by bombardment with high-energy particle or by a radioactive tranformation, it can decay to the ground tate by emiion of one or more photon called gamma ray or gamma-ray photon, with typical energie of 10 kev to 5 MeV. Thi proce i called gamma 1g decay. For example, alpha particle emitted from 6 Ra have two poible kinetic energie, either MeV or 4.60 MeV. Including the recoil energy of the reulting Rn nucleu, thee correpond to a total releaed energy of MeV or MeV, repectively. When an alpha particle with the maller energy i emitted, the Rn nucleu i left in an excited tate. It then decay to it ground tate by emitting a gamma-ray photon with energy MeV = MeV A photon with thi energy i oberved during thi decay (Fig. 43.5c). CAUTION G decay v. A and B decay In both a and b decay, the Z value of a nucleu change and the nucleu of one element become the nucleu of a different element. In g decay, the element doe not change; the nucleu merely goe from an excited tate to a le excited tate Earthquake are caued in part by the radioactive decay of 38 U in the earth interior. Thee decay releae energy that help to produce convection current in the earth interior. Such current drive the motion of the earth crut, including the udden harp motion that we call earthquake (like the one that caued thi damage). Natural Radioactivity Many radioactive element occur in nature. For example, you are very lightly radioactive becaue of untable nuclide uch a carbon-14 and potaium-40 that are preent throughout your body. The tudy of natural radioactivity began in 1896, one year after Röntgen dicovered x ray. Henri Becquerel dicovered a radiation from uranium alt that eemed imilar to x ray. Intenive invetigation in the following two decade by Marie and Pierre Curie, Ernet Rutherford, and many other revealed that the emiion conit of poitively and negatively charged particle and neutral ray; they were given the name alpha, beta, and gamma becaue of their differing penetration characteritic. The decaying nucleu i uually called the parent nucleu; the reulting nucleu i the daughter nucleu. When a radioactive nucleu decay, the daughter nucleu may alo be untable. In thi cae a erie of ucceive decay occur until a table configuration i reached. Several uch erie are found in nature. The mot abundant radioactive nuclide found on earth i the uranium iotope 38 U, which undergoe a erie of 14 decay, including eight a emiion and ix b - emiion, terminating at a table iotope of lead, 06 Pb (Fig. 43.6). Radioactive decay erie can be repreented on a Segrè chart, a in Fig The neutron number N i plotted vertically, and the atomic number Z i plotted horizontally. In alpha emiion, both N and Z decreae by. In b - emiion, N decreae by 1 and Z increae by 1. The decay can alo be repreented in equation form; the firt two decay in the erie are written a 38 U S 34 Th + a 34 Th S 34 Pa + b - + n e or more briefly a 38 U S a 34 Th 34 Th S b- 34 Pa

135 43.3 Nuclear Stability and Radioactivity 1455 N y 38 U 34 Th 4.10 d 34 Pa y 34 U 43.7 Segrè chart howing the uranium 38 U decay erie, terminating with the table nuclide 06 Pb. The time are halflive (dicued in the next ection), given in year (y), day (d), hour (h), minute (m), or econd () y 30 Th y 6 Ra d Rn Pb 7 m 10 Tl 19.9 m 3.10 m 1.30 m 10 Pb.3 y 5.01 d 14 Bi 19.9 m Bi 18 Po 14 Po 10 Po a decay b decay Pb d Hg Tl Pb Bi Po At Rn Fr Ra Ac Th Pa U Np Z In the econd proce, the beta decay leave the daughter nucleu in an excited tate, from which it decay to the ground tate by emitting a gamma-ray photon. An excited tate i denoted by an aterik, o we can repreent the g emiion a 34 Pa * S 34 Pa + g 34 Pa or 34 Pa * S g 34 Pa 38 U An intereting feature of the decay erie i the branching that occur at 14 Bi. Thi nuclide decay to 10 Pb by emiion of an a and a b -, which can occur in either order. We alo note that the erie include untable iotope of everal element that alo have table iotope, including thallium (Tl), lead (Pb), and bimuth (Bi). The untable iotope of thee element that occur in the 38 U erie all have too many neutron to be table. Many other decay erie are known. Two of thee occur in nature, one tarting with the uncommon iotope 35 U and ending with 07 Pb, the other tarting with thorium 1 3 Th and ending with 08 Pb.

136 1456 CHAPTER 43 Nuclear Phyic Tet Your Undertanding of Section 43.3 A nucleu with atomic number Z and neutron number N undergoe two decay procee. The reult i a nucleu with atomic number Z - 3 and neutron number N - 1. Which decay procee may have taken place? (i) two b - decay; (ii) two b + decay; (iii) two a decay; (iv) an a decay and a b - decay; (v) an a decay and a b + decay Activitie and Half-Live Suppoe you need to dipoe of ome radioactive wate that contain a certain number of nuclei of a particular radioactive nuclide. If no more are produced, that number decreae in a imple manner a the nuclei decay. Thi decreae i a tatitical proce; there i no way to predict when any individual nucleu will decay. No change in phyical or chemical environment, uch a chemical reaction or heating or cooling, greatly affect mot decay rate. The rate varie over an extremely wide range for different nuclide. Radioactive Decay Rate Let N1t be the (very large) number of radioactive nuclei in a ample at time t, and let dn1t be the (negative) change in that number during a hort time interval dt. (We ll ue N1t to minimize confuion with the neutron number N.) The number of decay during the interval dt i -dn1t. The rate of change of N1t i the negative quantity dn1t>dt; thu -dn1t>dt i called the decay rate or the activity of the pecimen. The larger the number of nuclei in the pecimen, the more nuclei decay during any time interval. That i, the activity i directly proportional to N1t; it equal a contant l multiplied by N1t: - dn1t dt = ln1t (43.16) The contant l i called the decay contant, and it ha different value for different nuclide. A large value of l correpond to rapid decay; a mall value correpond to lower decay. Solving Eq. (43.16) for l how u that l i the ratio of the number of decay per time to the number of remaining radioactive nuclei; l can then be interpreted a the probability per unit time that any individual nucleu will decay. The ituation i reminicent of a dicharging capacitor, which we tudied in Section 6.4. Equation (43.16) ha the ame form a the negative of Eq. (6.15), with q and 1>RC replaced by N1t and l. Then we can make the ame ubtitution in Eq. (6.16), with the initial number of nuclei N10 = N 0, to find the exponential function: N1t = N 0 e -lt (number of remaining nuclei) (43.17) 43.8 The number of nuclei in a ample of a radioactive element a a function of time. The ample activity ha an exponential decay curve with the ame hape. N(t) N 0 N 0/ Figure 43.8 i a graph of thi function, howing the number of remaining nuclei N1t a a function of time. The half-life T 1> i the time required for the number of radioactive nuclei to decreae to one-half the original number N 0. Then half of the remaining radioactive nuclei decay during a econd interval T 1>, and o on. The number remaining after ucceive half-live are N 0 >, N 0 >4, N 0 >8, Á. To get the relationhip between the half-life T 1> and the decay contant l, we et N1t>N 0 = 1 and t = T 1> in Eq. (43.17), obtaining 1 = e-lt 1> N 0/4 N 0/8 N 0/16 0 T 1 / T 1/ 3T 1/ 4T 1/ t We take logarithm of both ide and olve for T 1> : T 1> = ln l = l (43.18)

137 43.4 Activitie and Half-Live 1457 The mean lifetime T mean, generally called the lifetime, of a nucleu or untable particle i proportional to the half-life T 1> : T mean = 1 l = T 1> ln = T 1> (lifetime T mean, decay contant l, and half-life T 1> ) (43.19) In particle phyic the life of an untable particle i uually decribed by the lifetime, not the half-life. Becaue the activity -dn1t>dt at any time equal ln1t, Eq. (43.17) tell u that the activity alo depend on time a e -lt. Thu the graph of activity veru time ha the ame hape a Fig Alo, after ucceive half-live, the activity i one-half, one-fourth, one-eighth, and o on of the original activity. CAUTION A half-life may not be enough It i ometime implied that any radioactive ample will be afe after a half-life ha paed. That wrong. If your radioactive wate initially ha ten time too much activity for afety, it i not afe after one half-life, when it till ha five time too much. Even after three half-live it till ha 5% more activity than i afe. The number of radioactive nuclei and the activity approach zero only a t approache infinity. A common unit of activity i the curie, abbreviated Ci, which i defined to be 3.70 * decay per econd. Thi i approximately equal to the activity of one gram of radium. The SI unit of activity i the becquerel, abbreviated Bq. One becquerel i one decay per econd, o 1 Ci = 3.70 * Bq = 3.70 * decay> Example 43.8 Activity of 57 Co The iotope 57 Co decay by electron capture to 57 Fe with a halflife of 7 d. The 57 Fe nucleu i produced in an excited tate, and (b) The activity -dn1t>dt i given a.00 mci, o it almot intantaneouly emit gamma ray that we can detect. - dn1t =.00 mci = 1.00 * * (a) Find the mean lifetime and decay contant for 57 dt Co. (b) If the activity of a 57 Co radiation ource i now.00 mci, how many = 7.40 * 10 4 decay> 57 Co nuclei doe the ource contain? (c) What will be the activity From Eq. (43.16) thi i equal to ln(t), o we find after one year? N1t = - dn1t>dt l SOLUTION IDENTIFY and SET UP: Thi problem ue the relationhip among decay contant l, lifetime T mean, and activity -dn1t>dt. In part (a) we ue Eq. (43.19) to find l and T mean from T 1>. In part (b), we ue Eq. (43.16) to calculate the number of nuclei N1t from the activity. Finally, in part (c) we ue Eq. (43.16) and (43.17) to find the activity after one year. EXECUTE: (a) It convenient to convert the half-life to econd: T 1> = 17 d186,400 >d =.35 * 10 7 From Eq. (43.19), the mean lifetime and decay contant are T mean = T 1> ln =.35 * 107 = 3.39 * 10 7 = 39 day l = =.95 * T mean = 7.40 * * =.51 * 101 nuclei If you feel we re being too cavalier about the unit decay and nuclei, you can ue decay>1nucleu # a the unit for l. (c) From Eq. (43.17) the number N1t of nuclei remaining after one year * 10 7 i N1t = N 0 e -lt = N 0 e * *10 7 = 0.394N 0 The number of nuclei ha decreaed to of the original number. Equation (43.16) ay that the activity i proportional to the number of nuclei, o the activity ha decreaed by thi ame factor to mci = mci. EVALUATE: The number of nuclei found in part (b) i equivalent to 4.17 * 10-1 mol, with a ma of.38 * g. Thi i a far maller ma than even the mot enitive balance can meaure. After one 7-day half-life, the number of 57 Co nuclei ha decreaed to N 0 >; after 17 d = 544 d, it ha decreaed to N 0 > = N 0 >4. Thi reult agree with our anwer to part (c), which ay that after 365 d the number of nuclei i between N 0 > and N 0 >4.

138 1458 CHAPTER 43 Nuclear Phyic Radioactive Dating An intereting application of radioactivity i the dating of archaeological and geological pecimen by meauring the concentration of radioactive iotope. The mot familiar example i carbon dating. The untable iotope 14 C, produced during nuclear reaction in the atmophere that reult from comic-ray bombardment, give a mall proportion of 14 C in the CO in the atmophere. Plant that obtain their carbon from thi ource contain the ame proportion of 14 C a the atmophere. When a plant die, it top taking in carbon, and it 14 C b - decay to 14 N with a half-life of 5730 year. By meauring the proportion of 14 C in the remain, we can determine how long ago the organim died. One difficulty with radiocarbon dating i that the 14 C concentration in the atmophere change over long time interval. Correction can be made on the bai of other data uch a meaurement of tree ring that how annual growth cycle. Similar radioactive technique are ued with other iotope for dating geological pecimen. Some rock, for example, contain the untable potaium iotope 40 K, a beta emitter that decay to the table nuclide 40 Ar with a half-life of.4 * 10 8 y. The age of the rock can be determined by comparing the concentration of 40 K and 40 Ar.? Example 43.9 Radiocarbon dating Before 1900 the activity per unit ma of atmopheric carbon due to the preence of 14 C averaged about 0.55 Bq per gram of carbon. (a) What fraction of carbon atom were 14 C? (b) In analyzing an archaeological pecimen containing 500 mg of carbon, you oberve 174 decay in one hour. What i the age of the pecimen, auming that it activity per unit ma of carbon when it died wa that average value of the air? SOLUTION IDENTIFY and SET UP: The key idea i that the preent-day activity of a biological ample containing 14 C i related to both the elaped time ince it topped taking in atmopheric carbon and it activity at that time. We ue Eq. (43.16) and (43.17) to olve for the age t of the pecimen. In part (a) we determine the number of 14 C atom N1t from the activity -dn1t>dt uing Eq. (43.16). We find the total number of carbon atom in 500 mg by uing the molar ma of carbon (1.011 g>mol, given in Appendix D), and we ue the reult to calculate the fraction of carbon atom that are 14 C. The activity decay at the ame rate a the number of 14 C nuclei; we ue thi and Eq. (43.17) to olve for the age t of the pecimen. EXECUTE: (a) To ue Eq. (43.16), we mut firt find the decay contant l from Eq. (43.18): T 1> = 5730 y = y * 10 7 >y = * l = ln T 1> = * = 3.83 * Then, from Eq. (43.16), N1t = -dn>dt l The total number of C atom in 1 gram 11>1.011 mol i 11> * 10 3 = 5.01 * 10. The ratio of 14 C atom to all C atom i Only four carbon atom in every 3 * 10 1 are 14 C. (b) Auming that the activity per gram of carbon in the pecimen when it died 1t = 0 wa 0.55 Bq>g = # g >h = 918 h -1 # g -1, the activity of 500 mg of carbon then wa g1918 h -1 # g -1 = 459 h -1. The oberved activity now, at time t, i 174 h -1. Since the activity i proportional to the number of radioactive nuclei, the activity ratio 174>459 = equal the number ratio N1t>N 0. Now we olve Eq. (43.17) for t and inert value for N1t>N 0 and l: t = ln1n1t>n 0 -l 6.65 * = 1.33 * * 10 = = * = 6.65 * 1010 atom ln * =.53 * 1011 = 800 y 14 C EVALUATE: After 800 y the activity ha decreaed from 459 to 174 decay per hour. The pecimen died and topped taking CO out of the air about 8000 year ago. Radiation in the Home Rn, A eriou health hazard in ome area i the accumulation in houe of an inert, colorle, odorle radioactive ga. Looking at the 38 U decay chain in Fig. 43.7, we ee that the half-life of Rn i 3.8 day. If o, why not jut move out of the houe for a while and let it decay away? The anwer i that Rn i continuouly being produced by the decay of 6 Ra, which i found in minute quantitie in the rock and oil on which ome houe are built. It a dynamic

139 43.5 Biological Effect of Radiation 1459 equilibrium ituation, in which the rate of production equal the rate of decay. The reaon Rn i a bigger hazard than the other element in the 38 U decay erie i that it a ga. During it hort half-life of 3.8 day it can migrate from the oil into your houe. If a Rn nucleu decay in your lung, it emit a damaging a particle and it daughter nucleu 18 Po, which i not chemically inert and i likely to tay in your lung until it decay, emit another damaging a particle and o on down the 38 U decay erie. How much of a hazard i radon? Although report indicate value a high a 3500 pci>l, the average activity per volume in the air inide American home due to Rn i about 1.5 pci>l (over a thouand decay each econd in an average-ized room). If your environment ha thi level of activity, it ha been etimated that a lifetime expoure would reduce your life expectancy by about 40 day. For comparion, moking one pack of cigarette per day reduce life expectancy by 6 year, and it i etimated that the average emiion from all the nuclear power plant in the world reduce life expectancy by anywhere from 0.01 day to 5 day. Thee figure include catatrophe uch a the 1986 nuclear reactor diater at Chernobyl, for which the local effect on life expectancy i much greater. Tet Your Undertanding of Section 43.4 Which ample contain a greater number of nuclei: a 5.00-mCi ample of 40 Pu (half-life 6560 y) or a 4.45-mCi ample of 43 Am (half-life 7370 y)? (i) the 40 Pu ample; (ii) the ample; (iii) both have the ame number of nuclei. 43 Am 43.5 Biological Effect of Radiation The above dicuion of radon introduced the interaction of radiation with living organim, a topic of vital interet and importance. Under radiation we include radioactivity (alpha, beta, gamma, and neutron) and electromagnetic radiation uch a x ray. A thee particle pa through matter, they loe energy, breaking molecular bond and creating ion hence the term ionizing radiation. Charged particle interact directly with the electron in the material. X ray and g ray interact by the photoelectric effect, in which an electron aborb a photon and break looe from it ite, or by Compton cattering (ee Section 38.3). Neutron caue ionization indirectly through colliion with nuclei or aborption by nuclei with ubequent radioactive decay of the reulting nuclei. Thee interaction are extremely complex. It i well known that exceive expoure to radiation, including unlight, x ray, and all the nuclear radiation, can detroy tiue. In mild cae it reult in a burn, a with common unburn. Greater expoure can caue very evere illne or death by a variety of mechanim, including maive detruction of tiue cell, alteration of genetic material, and detruction of the component in bone marrow that produce red blood cell. Calculating Radiation Doe Radiation doimetry i the quantitative decription of the effect of radiation on living tiue. The aborbed doe of radiation i defined a the energy delivered to the tiue per unit ma. The SI unit of aborbed doe, the joule per kilogram, i called the gray (Gy); 1 Gy = 1 J>kg. Another unit i the rad, defined a 0.01 J>kg: 1 rad = 0.01 J>kg = 0.01 Gy Aborbed doe by itelf i not an adequate meaure of biological effect becaue equal energie of different kind of radiation caue different extent of biological effect. Thi variation i decribed by a numerical factor called the relative biological effectivene (RBE), alo called the quality factor (QF), of each pecific radiation. X ray with 00 kev of energy are defined to have an

140 1460 CHAPTER 43 Nuclear Phyic Table 43.3 Relative Biological Effectivene (RBE) for Several Type of Radiation Radiation RBE (Sv/Gy or rem/rad) X ray and g ray 1 Electron Slow neutron 3 5 Proton 10 a particle 0 Heavy ion 0 RBE of unity, and the effect of other radiation can be compared experimentally. Table 43.3 how approximate value of RBE for everal radiation. All thee value depend omewhat on the kind of tiue in which the radiation i aborbed and on the energy of the radiation. The biological effect i decribed by the product of the aborbed doe and the RBE of the radiation; thi quantity i called the biologically equivalent doe, or imply the equivalent doe. The SI unit of equivalent doe for human i the ievert (Sv): Equivalent doe 1Sv = RBE * Aborbed doe 1Gy (43.0) A more common unit, correponding to the rad, i the rem (an abbreviation of röntgen equivalent for man): Equivalent doe 1rem = RBE * Aborbed doe 1rad Thu the unit of the RBE i 1 Sv>Gy or 1 rem/rad, and 1 rem = 0.01 Sv. (43.1) Example Doe from a medical x ray During a diagnotic x-ray examination a 1.-kg portion of a broken leg receive an equivalent doe of 0.40 msv. (a) What i the equivalent doe in mrem? (b) What i the aborbed doe in mrad and in mgy? (c) If the x-ray energy i 50 kev, how many x-ray photon are aborbed? SOLUTION IDENTIFY and SET UP: We are aked to relate the equivalent doe (the biological effect of the radiation, meaured in ievert or rem) to the aborbed doe (the energy aborbed per ma, meaured in gray or rad). In part (a) we ue the converion factor 1 rem = 0.01 Sv for equivalent doe. Table 43.3 give the RBE for x ray; we ue thi value in part (b) to determine the aborbed doe uing Eq. (43.0) and (43.1). Finally, in part (c) we ue the ma and the definition of aborbed doe to find the total energy aborbed and the total number of photon aborbed. EXECUTE: (a) The equivalent doe in mrem i 0.40 msv = 40 mrem 0.01 Sv>rem (b) For x ray, RBE = 1 rem>rad or 1Sv>Gy, o the aborbed doe i 40 mrem = 40 mrad 1 rem>rad 0.40 msv 1Sv>Gy = 0.40 mgy = 4.0 * 10-4 J>kg (c) The total energy aborbed i 14.0 * 10-4 J>kg11. kg = 4.8 * 10-4 J = 3.0 * ev The number of x-ray photon i 3.0 * ev 5.0 * 10 4 ev>photon = 6.0 * 1010 photon EVALUATE: The aborbed doe i relatively large becaue x ray have a low RBE. If the ionizing radiation had been a beam of a particle, for which RBE = 0, the aborbed doe needed for an equivalent doe of 0.40 msv would be only 0.00 mgy, correponding to a maller total aborbed energy of.4 * 10-5 J. Radiation Hazard Here are a few number for perpective. To convert from Sv to rem, imply multiply by 100. An ordinary chet x-ray exam deliver about msv to about 5 kg of tiue. Radiation expoure from comic ray and natural radioactivity in oil, building material, and o on i of the order of 3 msv per year at ea level and twice that at an elevation of 1500 m (5000 ft). A whole-body doe of up to about 0.0 Sv caue no immediately detectable effect. A hort-term wholebody doe of 5 Sv or more uually caue death within a few day or week. A localized doe of 100 Sv caue complete detruction of the expoed tiue. The long-term hazard of radiation expoure in cauing variou cancer and genetic defect have been widely publicized, and the quetion of whether there i any afe level of radiation expoure ha been hotly debated. U.S. government regulation are baed on a maximum yearly expoure, from all except natural reource, of to 5 msv. Worker with occupational expoure to radiation are permitted 50 msv per year. Recent tudie ugget that thee limit are too high and that even extremely mall expoure carry hazard, but it i very difficult to

141 43.5 Biological Effect of Radiation 1461 gather reliable tatitic on the effect of low doe. It ha become clear that any ue of x ray for medical diagnoi hould be preceded by a very careful etimation of the relationhip of rik to poible benefit. Another harply debated quetion i that of radiation hazard from nuclear power plant. The radiation level from thee plant i not negligible. However, to make a meaningful evaluation of hazard, we mut compare thee level with the alternative, uch a coal-powered plant. The health hazard of coal moke are eriou and well documented, and the natural radioactivity in the moke from a coal-fired power plant i believed to be roughly 100 time a great a that from a properly operating nuclear plant with equal capacity. But the comparion i not thi imple; the poibility of a nuclear accident and the very eriou problem of afe dipoal of radioactive wate from nuclear plant mut alo be conidered. It i clearly impoible to eliminate all hazard to health. Our goal hould be to try to take a rational approach to the problem of minimizing the hazard from all ource. Figure 43.9 how one etimate of the variou ource of radiation expoure for the U.S. population. Ionizing radiation i a two-edged word; it poe very eriou health hazard, yet it alo provide many benefit to humanity, including the diagnoi and treatment of dieae and a wide variety of analytical technique. Beneficial Ue of Radiation Radiation i widely ued in medicine for intentional elective detruction of tiue uch a tumor. The hazard are coniderable, but if the dieae would be fatal without treatment, any hazard may be preferable. Artificially produced iotope are often ued a radiation ource. Such iotope have everal advantage over naturally radioactive iotope. They may have horter half-live and correpondingly greater activity. Iotope can be choen that emit the type and energy of radiation deired. Some artificial iotope have been replaced by photon and electron beam from linear accelerator. Nuclear medicine i an expanding field of application. Radioactive iotope have virtually the ame electron configuration and reulting chemical behavior a table iotope of the ame element. But the location and concentration of radioactive iotope can eaily be detected by meaurement of the radiation they emit. A familiar example i the ue of radioactive iodine for thyroid tudie. Nearly all the iodine ingeted i either eliminated or tored in the thyroid, and the body chemical reaction do not dicriminate between the untable iotope 131 I and the table iotope 17 I. A minute quantity of 131 I i fed or injected into the patient, and the peed with which it become concentrated in the thyroid provide a meaure of thyroid function. The half-life i 8.0 day, o there are no longlating radiation hazard. By ue of more ophiticated canning detector, one can alo obtain a picture of the thyroid, which how enlargement and other abnormalitie. Thi procedure, a type of autoradiography, i comparable to photographing the glowing filament of an incandecent light bulb by uing the light emitted by the filament itelf. If thi proce dicover cancerou thyroid nodule, they can be detroyed by much larger quantitie of 131 I. Another ueful nuclide for nuclear medicine i technetium Tc, which i formed in an excited tate by the b - decay of molybdenum 1 99 Mo. The technetium then decay to it ground tate by emitting a g-ray photon with energy 143 kev. The half-life i 6.01 hour, unuually long for g emiion. (The ground tate of 99 Tc i alo untable, with a half-life of.11 * 10 5 y; it decay by b - emiion to the table ruthenium nuclide 99 Ru. ) The chemitry of technetium i uch that it can readily be attached to organic molecule that are taken up by variou organ of the body. A mall quantity of uch technetium-bearing molecule i injected into a patient, and a canning detector or gamma camera i ued to produce an image, or cintigram, that reveal which part of the body take up thee g-emitting molecule. Thi technique, in which 99 Tc act a a radioactive tracer, play an important role in locating cancer, embolim, and other pathologie (Fig ) Contribution of variou ource to the total average radiation expoure in the U.S. population, expreed a percentage of the total. From human activity 18% Other <1% Occupational 0.3% Fallout <0.3% Nuclear fuel cycle 0.1% Micellaneou 0.1% Comic 8% Terretrial 8% Internal 11% Radon 55% Natural 8% Conumer product 3% Nuclear medicine 4% Medical x ray 11% Thi colored cintigram how where a chemical containing radioactive 99 Tc wa taken up by a patient lung. The orange color in the lung on the left indicate trong g-ray emiion by the 99 Tc, which how that the chemical wa able to pa into thi lung through the bloodtream. The lung on the right how weaker emiion, indicating the preence of an embolim (a blood clot or other obtruction in an artery) that i retricting the flow of blood to thi lung. Embolim

142 146 CHAPTER 43 Nuclear Phyic Tracer technique have many other application. Tritium 1 3 H, a radioactive hydrogen iotope, i ued to tag molecule in complex organic reaction; radioactive tag on peticide molecule, for example, can be ued to trace their paage through food chain. In the world of machinery, radioactive iron can be ued to tudy piton-ring wear. Laundry detergent manufacturer have even teted the effectivene of their product uing radioactive dirt. Many direct effect of radiation are alo ueful, uch a trengthening polymer by cro-linking, terilizing urgical tool, dipering of unwanted tatic electricity in the air, and intentionally ionizing the air in moke detector. Gamma ray are alo being ued to terilize and preerve ome food product. Tet Your Undertanding of Section 43.5 Alpha particle have 0 time the relative biological effectivene of 00-keV x ray. Which would be better to ue to radiate tiue deep inide the body? (i) a beam of alpha particle; (ii) a beam of 00-keV x ray; (iii) both are equally effective Nuclear Reaction In the preceding ection we tudied the decay of untable nuclei, epecially pontaneou emiion of an a or b particle, ometime followed by g emiion. Nothing need to be done to initiate thi decay, and nothing can be done to control it. Thi ection examine ome nuclear reaction, rearrangement of nuclear component that reult from a bombardment by a particle rather than a pontaneou natural proce. Rutherford uggeted in 1919 that a maive particle with ufficient kinetic energy might be able to penetrate a nucleu. The reult would be either a new nucleu with greater atomic number and ma number or a decay of the original nucleu. Rutherford bombarded nitrogen 1 14 N with a particle and obtained an oxygen 1 17 O nucleu and a proton: 4 He N S 17 8 O H (43.) Rutherford ued alpha particle from naturally radioactive ource. In Chapter 44 we ll decribe ome of the particle accelerator that are now ued to initiate nuclear reaction. Nuclear reaction are ubject to everal conervation law. The claical conervation principle for charge, momentum, angular momentum, and energy (including ret energie) are obeyed in all nuclear reaction. An additional conervation law, not anticipated by claical phyic, i conervation of the total number of nucleon. The number of proton and neutron need not be conerved eparately; in b decay, neutron and proton change into one another. We ll tudy the bai of the conervation of nucleon number in Chapter 44. When two nuclei interact, charge conervation require that the um of the initial atomic number mut equal the um of the final atomic number. Becaue of conervation of nucleon number, the um of the initial ma number mut alo equal the um of the final ma number. In general, thee are not elatic colliion, and the total initial ma doe not equal the total final ma. Reaction Energy The difference between the mae before and after the reaction correpond to the reaction energy, according to the ma energy relationhip E = mc. If initial particle A and B interact to produce final particle C and D, the reaction energy Q i defined a Q = 1M A + M B - M C - M D c (reaction energy) (43.3) To balance the electron, we ue the neutral atomic mae in Eq. (43.3). That i, we ue the ma of 1 H for a proton, H for a deuteron, 4 He for an a particle, 1 1

143 43.6 Nuclear Reaction 1463 and o on. When Q i poitive, the total ma decreae and the total kinetic energy increae. Such a reaction i called an exoergic reaction. When Q i negative, the ma increae and the kinetic energy decreae, and the reaction i called an endoergic reaction. The term exothermal and endothermal, borrowed from chemitry, are alo ued. In an endoergic reaction the reaction cannot occur at all unle the initial kinetic energy in the center-of-ma reference frame i at leat a great a ƒqƒ. That i, there i a threhold energy, the minimum kinetic energy to make an endoergic reaction go. Example Exoergic and endoergic reaction (a) When a lithium-7 nucleu i bombarded by a proton, two alpha particle 1 4 He are produced. Find the reaction energy. 4 (b) Calculate the reaction energy for the reaction He N S 17 8 O + 1 1H. SOLUTION IDENTIFY and SET UP: The reaction energy Q for any nuclear reaction equal c time the difference between the total initial ma and the total final ma, a in Eq. (43.3). Table 43. give the required mae. 1 EXECUTE: (a) The reaction i 1H + 7 3Li S 4 He + 4 He. The initial and final mae and their repective um are A: 1 1H B: 7 3Li u u u C: 4 He D: 4 He u u u The ma decreae by u. From Eq. (43.3), the reaction energy i Q = u MeV>u = MeV (b) The initial and final mae are A: 4 He B: 14 7N u u u C: 17 8O D: 1 1H u u u The ma increae by u, and the correponding reaction energy i Q = u MeV>u = MeV EVALUATE: The reaction in part (a) i exoergic: The final total kinetic energy of the two eparating alpha particle i MeV greater than the initial total kinetic energy of the proton and the lithium nucleu. The reaction in part (b) i endoergic: In the centerof-ma ytem that i, in a head-on colliion with zero total momentum the minimum total initial kinetic energy required for thi reaction to occur i 1.19 MeV. Ordinarily, the endoergic reaction of part (b) of Example would be produced by bombarding tationary 14 N nuclei with alpha particle from an accelerator. In thi cae an alpha kinetic energy mut be greater than 1.19 MeV. If all the alpha kinetic energy went olely to increaing the ret energy, the final kinetic energy would be zero, and momentum would not be conerved. When a particle with ma m and kinetic energy K collide with a tationary particle with ma M, the total kinetic energy K cm in the center-of-ma coordinate ytem (the energy available to caue reaction) i K cm = M M + m K (43.4) Thi expreion aume that the kinetic energie of the particle and nuclei are much le than their ret energie. We leave the derivation of Eq. (43.4) to you (ee Problem 43.77). In the preent example, K cm = >18.01K, o K mut be at leat > MeV = MeV. For a charged particle uch a a proton or an a particle to penetrate the nucleu of another atom and caue a reaction, it mut uually have enough initial kinetic energy to overcome the potential-energy barrier caued by the repulive electrotatic force. In the reaction of part (a) of Example 43.11, if we treat the proton and the 7 Li nucleu a pherically ymmetric charge with radii given by Eq. (43.1), their center will be 3.5 * m apart when they touch. The repulive potential

144 1464 CHAPTER 43 Nuclear Phyic energy of the proton (charge +e) and the 7 Li nucleu (charge +3e) at thi eparation r i 1 1e13e U = = 19.0 * 10 9 N # m /C * C 4pP 0 r 3.5 * m =.0 * J = 1. MeV Even though the reaction i exoergic, the proton mut have a minimum kinetic energy of about 1. MeV for the reaction to occur, unle the proton tunnel through the barrier (ee Section 40.4). Neutron Aborption Aborption of neutron by nuclei form an important cla of nuclear reaction. Heavy nuclei bombarded by neutron can undergo a erie of neutron aborption alternating with beta decay, in which the ma number A increae by a much a 5. Some of the tranuranic element, element having Z larger than 9, are produced in thi way. Thee element have not been found in nature. Many tranuranic element, having Z poibly a high a 118, have been identified. The analytical technique of neutron activation analyi ue imilar reaction. When bombarded by neutron, many table nuclide aborb a neutron to become untable and then undergo b - decay. The energie of the b - and aociated g emiion depend on the untable nuclide and provide a mean of identifying it and the original table nuclide. Quantitie of element that are far too mall for conventional chemical analyi can be detected in thi way. Tet Your Undertanding of Section 43.6 The reaction decribed in part (a) of Example i exoergic. Can it happen naturally when a ample of olid lithium i placed in a flak of hydrogen ga? 43.7 Nuclear Fiion PhET: Nuclear Fiion Ma ditribution of fiion fragment from the fiion of 36 U * (an excited tate of 36 U), which i produced when 35 U aborb a neutron. The vertical cale i logarithmic. Yield (%) A Nuclear fiion i a decay proce in which an untable nucleu plit into two fragment of comparable ma. Fiion wa dicovered in 1938 through the experiment of Otto Hahn and Fritz Straman in Germany. Puruing earlier work by Fermi, they bombarded uranium 1Z = 9 with neutron. The reulting radiation did not coincide with that of any known radioactive nuclide. Urged on by their colleague Lie Meitner, they ued meticulou chemical analyi to reach the atonihing but inecapable concluion that they had found a radioactive iotope of barium 1Z = 56. Later, radioactive krypton 1Z = 36 wa alo found. Meitner and Otto Frich correctly interpreted thee reult a howing that uranium nuclei were plitting into two maive fragment called fiion fragment. Two or three free neutron uually appear along with the fiion fragment and, very occaionally, a light nuclide uch a 3 H. Both the common iotope (99.3%) 38 U and the uncommon iotope (0.7%) 35 U (a well a everal other nuclide) can be eaily plit by neutron bombardment: 35 U by low neutron (kinetic energy le than 1 ev) but 38 U only by fat neutron with a minimum of about 1 MeV of kinetic energy. Fiion reulting from neutron aborption i called induced fiion. Some nuclide can alo undergo pontaneou fiion without initial neutron aborption, but thi i quite rare. When 35 U aborb a neutron, the reulting nuclide 36 U * i in a highly excited tate and plit into two fragment almot intantaneouly. Strictly peaking, it i 36 U *, not 35 U, that undergoe fiion, but it uual to peak of the fiion of 35 U. Over 100 different nuclide, repreenting more than 0 different element, have been found among the fiion product. Figure how the ditribution of ma number for fiion fragment from the fiion of Mot of the 35 U.

145 43.7 Nuclear Fiion 1465 fragment have ma number from 90 to 100 and from 135 to 145; fiion into two fragment with nearly equal ma i unlikely. Fiion Reaction You hould check the following two typical fiion reaction for conervation of nucleon number and charge: The total kinetic energy of the fiion fragment i enormou, about 00 MeV (compared to typical a and b energie of a few MeV). The reaon for thi i that nuclide at the high end of the ma pectrum (near A = 40) are le tightly bound than thoe nearer the middle (A = 90 to 145). Referring to Fig. 43., we ee that the average binding energy per nucleon i about 7.6 MeV at A = 40 but about 8.5 MeV at A = 10. Therefore a rough etimate of the expected increae in binding energy during fiion i about 8.5 MeV MeV = 0.9 MeV per nucleon, or a total of MeV L 00 MeV. CAUTION Binding energy and ret energy It may eem to be a violation of conervation of energy to have an increae in both the binding energy and the kinetic energy during a fiion reaction. But relative to the total ret energy E 0 of the eparated nucleon, the ret energy of the nucleu i E 0 minu E B. Thu an increae in binding energy correpond to a decreae in ret energy a ret energy i converted to the kinetic energy of the fiion fragment. Fiion fragment alway have too many neutron to be table. We noted in Section 43.3 that the neutron proton ratio 1N>Z for table nuclide i about 1 for light nuclide but almot 1.6 for the heaviet nuclide becaue of the increaing influence of the electrical repulion of the proton. The N>Z value for table nuclide i about 1.3 at A = 100 and 1.4 at A = 150. The fragment have about the ame N>Z a 35 U, about They uually repond to thi urplu of neutron by undergoing a erie of b - decay (each of which increae Z by 1 and decreae N by 1) until a table value of N>Z i reached. A typical example i 140 Ce 35 9U n S 36 9 U * S Ba Kr n 35 9U n S 36 9 U * S Xe Sr n Xe S b C S b Ba S b La S b Ce The nuclide i table. Thi erie of decay produce, on average, about 15 MeV of additional kinetic energy. The neutron exce of fiion fragment alo explain why two or three free neutron are releaed during the fiion. Fiion appear to et an upper limit on the production of tranuranic nuclei, mentioned in Section 43.6, that are relatively table. There are theoretical reaon to expect that nuclei near Z = 114, N = 184 or 196, might be table with repect to pontaneou fiion. In the hell model (ee Section 43.), thee number correpond to filled hell and ubhell in the nuclear energy-level tructure. Such uperheavy nuclei would till be untable with repect to alpha emiion. In 009 it wa confirmed that there are at leat four iotope with Z 114, the longetlived of which ha a half-life due to alpha decay of about.6. Liquid-Drop Model We can undertand fiion qualitatively on the bai of the liquid-drop model of the nucleu (ee Section 43.). The proce i hown in Fig in term of an electrically charged liquid drop. Thee ketche houldn t be taken too literally, but they may help to develop your intuition about fiion. A 35 U nucleu aborb a neutron (Fig. 43.1a), becoming a 36 U * nucleu with exce energy (Fig. 43.1b). Thi exce energy caue violent ocillation, during which a neck between two lobe develop (Fig. 43.1c). The electric repulion of thee two lobe tretche the neck farther (Fig. 43.1d), and finally two maller fragment are formed (Fig. 43.1e) that move rapidly apart. b -

146 1466 CHAPTER 43 Nuclear Phyic 43.1 A liquid-drop model of fiion. (a) A 35 U nucleu aborb a neutron. (b) The reulting 36 U * nucleu i in a highly excited tate and ocillate trongly. (c) A neck develop, and electric repulion puhe the two lobe apart. (d) The two lobe eparate, forming fiion fragment. (e) The fragment emit neutron at the time of fiion (or occaionally a few econd later). Neutron + 35 U nucleu Fiion fragment Neutron Fiion fragment Hypothetical potential-energy function for two fiion fragment in a fiionable nucleu. At ditance r beyond the range of the nuclear force, the potential energy varie approximately a 1>r. Fiion occur if there i an excitation energy greater than U B or an appreciable probability for tunneling through the potentialenergy barrier. U B U 0 r 0 Barrier 1/r electric potential energy Application Making Radioactive Iotope for Medicine The fragment that reult from nuclear fiion are typically untable, neutron-rich iotope. A number of thee are ueful for medical diagnoi and cancer radiotherapy (ee Section 43.5). Thi photograph how a nuclear fiion reactor ued for producing uch iotope. The uranium fuel i kept in a large tank of water for cooling. Some of the neutron-rich fiion fragment undergo beta decay and emit electron that move fater than the peed of light in water (about 0.75c). Like an airplane that produce an intene onic boom when it flie fater than ound (ee Section 16.9), thee ultrafat electron produce a light boom called Cerenkov M radiation that ha a characteritic blue color. r Thi qualitative picture ha been developed into a more quantitative theory to explain why ome nuclei undergo fiion and other don t. Figure how a hypothetical potential-energy function for two poible fiion fragment. If neutron aborption reult in an excitation energy greater than the energy barrier height U B, fiion occur immediately. Even when there in t quite enough energy to urmount the barrier, fiion can take place by quantum-mechanical tunneling, dicued in Section In principle, many table heavy nuclei can fiion by tunneling. But the probability depend very critically on the height and width of the barrier. For mot nuclei thi proce i o unlikely that it i never oberved. Chain Reaction Fiion of a uranium nucleu, triggered by neutron bombardment, releae other neutron that can trigger more fiion, uggeting the poibility of a chain reaction (Fig ). The chain reaction may be made to proceed lowly and in a controlled manner in a nuclear reactor or exploively in a bomb. The energy releae in a nuclear chain reaction i enormou, far greater than that in any chemical reaction. (In a ene, fire i a chemical chain reaction.) For example, when uranium i burned to uranium dioxide in the chemical reaction U + O S UO the heat of combution i about 4500 J>g. Expreed a energy per atom, thi i about 11 ev per atom. By contrat, fiion liberate about 00 MeV per atom, nearly 0 million time a much energy. Nuclear Reactor A nuclear reactor i a ytem in which a controlled nuclear chain reaction i ued to liberate energy. In a nuclear power plant, thi energy i ued to generate team, which operate a turbine and turn an electrical generator. On average, each fiion of a 35 U nucleu produce about.5 free neutron, o 40% of the neutron are needed to utain a chain reaction. A 35 U nucleu i much more likely to aborb a low-energy neutron (le than 1 ev) than one of the higher-energy neutron (1 MeV or o) that are liberated during fiion. In a nuclear reactor the higher-energy neutron are lowed down by colliion with nuclei in the urrounding material, called the moderator, o they are much more likely to caue further fiion. In nuclear power plant, the moderator i often water, occaionally graphite. The rate of the reaction i controlled by inerting or withdrawing control rod made of element (uch a boron or cadmium) whoe nuclei aborb neutron without undergoing any additional reaction. The iotope 38 U can alo aborb neutron, leading to 39 U *, but not with high enough probability for it to utain a chain reaction by itelf. Thu uranium that i ued in reactor i often enriched by increaing the proportion of 35 U above the natural value of 0.7%, typically to 3% or o, by iotope-eparation proceing.

147 43.7 Nuclear Fiion Schematic diagram of a nuclear fiion chain reaction C 35 9 U 35 9U Rb Rb Ba Fiion fragment Firtgeneration 36 Kr 94 neutron 1 0 n 35 U Lot 9 neutron Ba Fiion fragment Secondgeneration neutron 1 0 n 35 9U C 90 Sr U 35 9 U 94 Kr Rb 144 C Xe 143 Xe U Thirdgeneration neutron 1 0 n 90 Fourth- 38 Sr generation neutron 1 0 n The mot familiar application of nuclear reactor i for the generation of electric power. A wa noted above, the fiion energy appear a kinetic energy of the fiion fragment, and it immediate reult i to increae the internal energy of the fuel element and the urrounding moderator. Thi increae in internal energy i tranferred a heat to generate team to drive turbine, which pin the electrical generator. Figure i a chematic diagram of a nuclear power plant. Control rod Reactor core (moderator) Reactor preure veel Water (hot) Pump Water (cool) Primary loop Turbine Steam generator Steam (high preure) Pump Water (high preure) Secondary loop Generator Steam (low preure) Coolant in Steam condener Coolant out Water (low preure) Electric power Schematic diagram of a nuclear power plant.

148 1468 CHAPTER 43 Nuclear Phyic The energetic fiion fragment heat the water urrounding the reactor core. The team generator i a heat exchanger that take heat from thi highly radioactive water and generate nonradioactive team to run the turbine. A typical nuclear plant ha an electric-generating capacity of 1000 MW (or 10 9 W). The turbine are heat engine and are ubject to the efficiency limitation impoed by the econd law of thermodynamic, dicued in Chapter 0. In modern nuclear plant the overall efficiency i about one-third, o 3000 MW of thermal power from the fiion reaction i needed to generate 1000 MW of electrical power. Example 43.1 Uranium conumption in a nuclear reactor What ma of 35 U mut undergo fiion each day to provide Each 35 U atom ha a ma of 135 u11.66 * 10-7 kg>u = 3000 MW of thermal power? 3.9 * 10-5 kg, o the ma of 35 U that undergoe fiion each econd i SOLUTION 19.4 * * 10-5 kg = 3.7 * 10-5 kg = 37 mg IDENTIFY and SET UP: Fiion of 35 U liberate about 00 MeV per atom. We ue thi and the ma of the 35 In one day 186,400, the total conumption of i U atom to determine 35 U the required amount of uranium * 10-5 kg>186,400 = 3. kg EXECUTE: Each econd, we need 3000 MJ or 3000 * 10 6 J. Each EVALUATE: For comparion, a 1000-MW coal-fired power plant fiion provide 00 MeV, or burn 10,600 ton (about 10 million kg) of coal per day! 100 MeV>fiion11.6 * J>MeV = 3. * J>fiion The number of fiion needed each econd i 3000 * 10 6 J 3. * J>fiion = 9.4 * 1019 fiion We mentioned above that about 15 MeV of the energy releaed after fiion of a 35 U nucleu come from the b - decay of the fiion fragment. Thi fact poe a eriou problem with repect to control and afety of reactor. Even after the chain reaction ha been completely topped by inertion of control rod into the core, heat continue to be evolved by the b - decay, which cannot be topped. For a 3000-MW reactor thi heat power i initially very large, about 00 MW. In the event of total lo of cooling water, thi power i more than enough to caue a catatrophic meltdown of the reactor core and poible penetration of the containment veel. The difficulty in achieving a cold hutdown following an accident at the Three Mile Iland nuclear power plant in Pennylvania in March 1979 wa a reult of the continued evolution of heat due to b - decay. The catatrophe of April 6, 1986, at Chernobyl reactor No. 4 in Ukraine reulted from a combination of an inherently untable deign and everal human error committed during a tet of the emergency core cooling ytem. Too many control rod were withdrawn to compenate for a decreae in power caued by a buildup of neutron aborber uch a 135 Xe. The power level roe from 1% of normal to 100 time normal in 4 econd; a team exploion ruptured pipe in the core cooling ytem and blew the heavy concrete cover off the reactor. The graphite moderator caught fire and burned for everal day, and there wa a meltdown of the core. The total activity of the radioactive material releaed into the atmophere ha been etimated a about 10 8 Ci. 35 U Tet Your Undertanding of Section 43.7 The fiion of can be triggered by the aborption of a low neutron by a nucleu. Can a low proton be ued to trigger 35 U fiion?

149 43.8 Nuclear Fuion The proton-proton chain. 1 Two proton combine to 3 A third proton combine with the 4 Two 3 He nuclei fue, forming a form a deuteron ( H)... deuteron, forming a helium nucleu 4 He nucleu and releaing two ( 3 He) and emitting a gamma-ray photon. proton.... a well a a poitron (b ) and an electron neutrino (n e ). p p H p n n e p b p n 3He n p p g p p n 4 He p n n p p n p p p 43.8 Nuclear Fuion In a nuclear fuion reaction, two or more mall light nuclei come together, or fue, to form a larger nucleu. Fuion reaction releae energy for the ame reaon a fiion reaction: The binding energy per nucleon after the reaction i greater than before. Referring to Fig. 43., we ee that the binding energy per nucleon increae with A up to about A = 60, o fuion of nearly any two light nuclei to make a nucleu with A le than 60 i likely to be an exoergic reaction. In comparion to fiion, we are moving toward the peak of thi curve from the oppoite ide. Another way to expre the energy relationhip i that the total ma of the product i le than that of the initial particle. Here are three example of energy-liberating fuion reaction, written in term of the neutral atom: 1 1 H H S 1 H + b + + n e 1 H H S 3 He + g 3 He + 3 He S 4 He H H In the firt reaction, two proton combine to form a deuteron 1 H, with the emiion of a poitron 1b + and an electron neutrino. In the econd, a proton and a deuteron combine to form the nucleu of the light iotope of helium, 3 He, with the emiion of a gamma ray. Now double the firt two reaction to provide the two 3 He nuclei that fue in the third reaction to form an alpha particle 1 4 He and two proton. Together the reaction make up the proce called the proton-proton chain (Fig ). The net effect of the chain i the converion of four proton into one a particle, two poitron, two electron neutrino, and two g. We can calculate the energy releae from thi part of the proce: The ma of an a particle plu two poitron i the ma of neutral 4 He, the neutrino have zero (or negligible) ma, and the gamma have zero ma. Ma of four proton Ma of 4 He Ma difference and energy releae u u u and 4.69 MeV The two poitron that are produced during the firt tep of the proton-proton chain collide with two electron; mutual annihilation of the four particle take place, and their ret energy i converted into MeV =.044 MeV of gamma radiation. Thu the total energy releaed i MeV = 6.73 MeV. The proton-proton chain take place in the interior of the un and other tar (Fig ). Each gram of the un ma contain about 4.5 * 10 3 proton. If all of thee proton were fued into helium, the energy releaed would be about 130,000 kwh. If the un were to continue to radiate at it preent rate, it would take about 75 * 10 9 year to exhaut it upply of proton. A we will ee below, fuion reaction can ActivPhyic 19.3: Fuion The energy releaed a tarlight come from fuion reaction deep within a tar interior. When a tar i firt formed and for mot of it life, it convert the hydrogen in it core into helium. A a tar age, the core temperature can become high enough for additional fuion reaction that convert helium into carbon, oxygen, and other element.

150 1470 CHAPTER 43 Nuclear Phyic occur only at extremely high temperature; in the un, thee temperature are found only deep within the interior. Hence the un cannot fue all of it proton, and can utain fuion for a total of only about 10 * 10 9 year in total. The preent age of the olar ytem (including the un) i 4.54 * 10 9 year, o the un i about halfway through it available tore of proton. Example A fuion reaction Two deuteron fue to form a triton (a nucleu of tritium, or and a proton. How much energy i liberated? 3 H) SOLUTION IDENTIFY and SET UP: Thi i a nuclear reaction of the type dicued in Section We find the energy releaed uing Eq. (43.3). EXECUTE: Adding one electron to each nucleu make each a neutral atom; we find their mae in Table 43.. Subtituting into Eq. (43.3), we find Q = u u u4 * MeV>u = 4.03 MeV EVALUATE: Thu 4.03 MeV i releaed in the reaction; the triton and proton together have 4.03 MeV more kinetic energy than the two deuteron had together Thi target chamber at the National Ignition Facility in California ha aperture for 19 powerful laer beam. The laer deliver 5 * W of power for a few nanoecond to a millimeterized pellet of deuterium and tritium at the center of the chamber, thu triggering thermonuclear fuion. Achieving Fuion For two nuclei to undergo fuion, they mut come together to within the range of the nuclear force, typically of the order of * m. To do thi, they mut overcome the electrical repulion of their poitive charge. For two proton at thi ditance, the correponding potential energy i about 1. * J or 0.7 MeV; thi repreent the total initial kinetic energy that the fuion nuclei mut have for example, 0.6 * J each in a head-on colliion. Atom have thi much energy only at extremely high temperature. The dicuion of Section 18.3 howed that the average tranlational kinetic energy of a ga 3 molecule at temperature T i kt, where k i Boltzmann contant. The temperature at which thi i equal to E = 0.6 * J i determined by the relationhip E = 3 kt T = E 3k = 10.6 * J * 10-3 J>K = 3 * 109 K Fuion reaction are poible at lower temperature becaue the Maxwell Boltzmann ditribution function (ee Section 18.5) give a mall fraction of proton with kinetic energie much higher than the average value. The proton-proton reaction occur at only 1.5 * 10 7 K at the center of the un, making it an extremely low-probability proce; but that why the un i expected to lat o long. At thee temperature the fuion reaction are called thermonuclear reaction. Intenive effort are under way to achieve controlled fuion reaction, which potentially repreent an enormou new reource of energy (ee Fig. 4.11). At the temperature mentioned, light atom are fully ionized, and the reulting tate of matter i called a plama. In one kind of experiment uing magnetic confinement, a plama i heated to extremely high temperature by an electrical dicharge, while being contained by appropriately haped magnetic field. In another, uing inertial confinement, pellet of the material to be fued are heated by a high-intenity laer beam (ee Fig ). Some of the reaction being tudied are 1 H + 1 H S 3 1 H H MeV 1 3 H + 1 H S 4 He n MeV 1 H + 1 H S 3 He n MeV 3 He + 1 H S 4 He H MeV (1) () (3) (4)

151 43.8 Nuclear Fuion 1471 We conidered reaction (1) in Example 43.13; two deuteron fue to form a triton and a proton. In reaction () a triton combine with another deuteron to form an alpha particle and a neutron. The reult of both of thee reaction together i the converion of three deuteron into an alpha particle, a proton, and a neutron, with the liberation of 1.6 MeV of energy. Reaction (3) and (4) together achieve the ame converion. In a plama that contain deuteron, the two pair of reaction occur with roughly equal probability. A yet, no one ha ucceeded in producing thee reaction under controlled condition in uch a way a to yield a net urplu of uable energy. Method of achieving fuion that don t require high temperature are alo being tudied; thee are called cold fuion. One cheme that doe work ue an unuual + hydrogen molecule ion. The uual H ion conit of two proton bound by one hared electron; the nuclear pacing i about 0.1 nm. If the proton are replaced by a deuteron 1 H and a triton 1 3 H and the electron by a muon, which i 08 time a maive a the electron, the pacing i made maller by a factor of 08. The probability then become appreciable for the two nuclei to tunnel through the narrow repulive potential-energy barrier and fue in reaction () above. The propect of making thi proce, called muon-catalyzed fuion, into a practical energy ource i till ditant. Tet Your Undertanding of Section 43.8 exoergic? Are all fuion reaction

152 CHAPTER 43 SUMMARY Nuclear propertie: A nucleu i compoed of A nucleon (Z proton and N neutron). All nuclei have about the ame denity. The radiu of a nucleu with ma number A i given approximately by Eq. (43.1). A ingle nuclear pecie of a given Z and N i called a nuclide. Iotope are nuclide of the ame element (ame Z) that have different number of neutron. Nuclear mae are meaured in atomic ma unit. Nucleon have angular momentum and a magnetic moment. (See Example 43.1 and 43..) R = R 0 A 1>3 1R 0 = 1. * m (43.1) Nuclear binding and tructure: The ma of a nucleu i alway le than the ma of the proton and neutron within it. The ma difference multiplied by c give the binding energy E B. The binding energy for a given nuclide i determined by the nuclear force, which i hort range and favor pair of particle, and by the electric repulion between proton. A nucleu i untable if A or Z i too large or if the ratio N>Z i wrong. Two widely ued model of the nucleu are the liquid-drop model and the hell model; the latter i analogou to the central-field approximation for atomic tructure. (See Example 43.3 and 43.4.) E B = 1ZM H + Nm n - A ZMc (43.10) E B/A (MeV/nucleon) O 1 6 C 6 8 Ni 38 9 U 4 He 1 H A Radioactive decay: Untable nuclide uually emit an alpha particle (a 4 He nucleu) or a beta particle (an electron) in the proce of changing to another nuclide, ometime followed by a gamma-ray photon. The rate of decay of an untable nucleu i decribed by the decay contant l, the half-life T 1>, or the lifetime T mean. If the number of nuclei at time t = 0 i N 0 and no more are produced, the number at time t i given by Eq. (43.17). (See Example ) N1t = N 0 e -lt T mean = 1 l = T 1> ln = T 1> (43.17) (43.19) 6 Ra 88 88p 138n 86 Rn a(p, n) 86p 136n 4 He Biological effect of radiation: The biological effect of any radiation depend on the product of the energy aborbed per unit ma and the relative biological effectivene (RBE), which i different for different radiation. (See Example ) Embolim Nuclear reaction: In a nuclear reaction, two nuclei or particle collide to produce two new nuclei or particle. Reaction can be exoergic or endoergic. Several conervation law, including charge, energy, momentum, angular momentum, and nucleon number, are obeyed. Energy i releaed by the fiion of a heavy nucleu into two lighter, alway untable, nuclei. Energy i alo releaed by the fuion of two light nuclei into a heavier nucleu. (See Example ) p p H p p n b n e p n 3 p n p He n p n p p g 4He p n n p p p 147

153 Dicuion Quetion 1473 BRIDGING PROBLEM Saturation of 18 I Production In an experiment, the iodine iotope I i created by irradiating a both the creation of by the neutron irradiation and the ample of I with a beam of neutron, yielding 1.50 * I decay of any I preent. In the teady tate, how do the rate nuclei per econd. Initially no I nuclei are preent. A I of thee two procee compare? nucleu decay by b - emiion with a half-life of 5.0 min. (a) To what nuclide doe I decay? (b) Could that nuclide decay back to 3. Lit the unknown quantitie for each part of the problem and identify your target variable. 18 I by b + emiion? Why or why not? (c) After the ample ha been irradiated for a long time, what i the maximum number of EXECUTE 18 I atom that can be preent in the ample? What i the maximum 4. Find the value of Z and N of the nuclide produced by the decay activity that can be produced? (Thi teady-tate ituation i called of I. What element i thi? aturation.) (d) Find an expreion for the number of I atom 5. Decide whether thi nuclide can decay back to I. preent in the ample a a function of time. 6. Inpect your equation for dn> dt. What i the value of dn> dt in the teady tate? Ue thi to olve for the teady-tate value of SOLUTION GUIDE N and the activity. 7. Solve your dn> dt equation for the function N(t). (Hint: See See MateringPhyic tudy area for a Video Tutor olution. Section 6.4.) IDENTIFY and SET UP mut be true for decay to be poible? For b + decay to be 8. Your reult from tep 6 tell you the value of N after a long time 1. What happen to the value of Z, N, and A in b - decay? What EVALUATE b poible?. You ll need to write an equation for the rate of change dn> dt of (that i, for large value of t). I thi conitent with your reult from tep 7? What would contitute a long time under thee the number N of atom in the ample, taking account of condition? 18 I 18 I Problem For intructor-aigned homework, go to : Problem of increaing difficulty. CP: Cumulative problem incorporating material from earlier chapter. CALC: Problem requiring calculu. BIO: Biocience problem. DISCUSSION QUESTIONS Q43.1 BIO Neutron have a magnetic dipole moment and can undergo pin flip by aborbing electromagnetic radiation. Why, then, are proton rather than neutron ued in MRI of body tiue? (See Fig ) Q43. In Eq. (43.11), a the total number of nucleon become larger, the importance of the econd term in the equation decreae relative to that of the firt term. Doe thi make phyical ene? Explain. Q43.3 Why aren t the mae of all nuclei integer multiple of the ma of a ingle nucleon? Q43.4 Can you tell from the value of the ma number A whether to ue a plu value, a minu value, or zero for the fifth term of Eq. (43.11)? Explain. Q43.5 What are the ix known element for which Z i a magic number? Dicu what propertie thee element have a a conequence of their pecial value of Z. Q43.6 The binding energy per nucleon for mot nuclide doen t vary much (ee Fig. 43.). I there imilar conitency in the atomic energy of atom, on an energy per electron bai? If o, why? If not, why not? Q43.7 Heavy, untable nuclei uually decay by emitting an a or b particle. Why don t they uually emit a ingle proton or neutron? Q43.8 The only two table nuclide with more proton than neutron are 1H and He. Why i Z 7 N o uncommon? 1 3 Q43.9 Since lead i a table element, why doen t the 38 U decay erie hown in Fig top at lead, 14 Pb? 38 U Q43.10 In the decay erie hown in Fig. 43.7, ome nuclide in the erie are found much more abundantly in nature than other, even though every 38 U nucleu goe through every tep in the erie before finally becoming 06 Pb. Why don t the intermediate nuclide all have the ame abundance? Q43.11 Compared to a particle with the ame energy, b particle can much more eaily penetrate through matter. Why i thi? A Q43.1 If ZEl i repreent the initial nuclide, what i the decay A A-4 proce or procee if the final nuclide i (a) Z+1El f ; (b) Z-El f ; A (c) Z-1El f? Q43.13 In a nuclear decay equation, why can we repreent an electron a -1b -? What are the equivalent repreentation for a 0 poitron, a neutrino, and an antineutrino? Q43.14 Why i the alpha, beta, or gamma decay of an untable nucleu unaffected by the chemical ituation of the atom, uch a the nature of the molecule or olid in which it i bound? The chemical ituation of the atom can, however, have an effect on the halflife in electron capture. Why i thi? Q43.15 In the proce of internal converion, a nucleu decay from an excited tate to a ground tate by giving the excitation energy directly to an atomic electron rather than emitting a gamma-ray photon. Why can thi proce alo produce x-ray photon? Q43.16 In Example 43.9 (Section 43.4), the activity of atmopheric carbon before 1900 wa given. Dicu why thi activity may have changed ince 1900.

154 1474 CHAPTER 43 Nuclear Phyic Q43.17 BIO One problem in radiocarbon dating of biological ample, epecially very old one, i that they can eaily be contaminated with modern biological material during the meaurement proce. What effect would uch contamination have on the etimated age? Why i uch contamination a more eriou problem for ample of older material than for ample of younger material? Q43.18 The mot common radium iotope found on earth, 6 Ra, ha a half-life of about 1600 year. If the earth wa formed well over 10 9 year ago, why i there any radium left now? Q43.19 Fiion reaction occur only for nuclei with large nucleon number, while exoergic fuion reaction occur only for nuclei with mall nucleon number. Why i thi? Q43.0 When a large nucleu plit during nuclear fiion, the daughter nuclei of the fiion fly apart with enormou kinetic energy. Why doe thi happen? Q43.1 A tar age, they ue up their upply of hydrogen and eventually begin producing energy by a reaction that involve the fuion of three helium nuclei to form a carbon nucleu. Would you expect the interior of thee old tar to be hotter or cooler than the interior of younger tar? Explain. EXERCISES Section 43.1 Propertie of Nuclei How many proton and how many neutron are there in a 8 nucleu of the mot common iotope of (a) ilicon, 14Si; (b) rubidium, 37Rb; (c) thallium, Tl? CP Hydrogen atom are placed in an external 1.65-T magnetic field. (a) The proton can make tranition between tate where the nuclear pin component i parallel and antiparallel to the field by aborbing or emitting a photon. Which tate ha lower energy: the tate with the nuclear pin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic pectrum doe it lie? (b) The electron can make tranition between tate where the electron pin component i parallel and antiparallel to the field by aborbing or emitting a photon. Which tate ha lower energy: the tate with the electron pin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic pectrum doe it lie? Hydrogen atom are placed in an external magnetic field. The proton can make tranition between tate in which the nuclear pin component i parallel and antiparallel to the field by aborbing or emitting a photon. What magnetic-field magnitude i required for thi tranition to be induced by photon with frequency.7 MHz? Neutron are placed in a magnetic field with magnitude.30 T. (a) What i the energy difference between the tate with the nuclear pin angular momentum component parallel and antiparallel to the field? Which tate i lower in energy: the one with it pin component parallel to the field or the one with it pin component antiparallel to the field? How do your reult compare with the energy tate for a proton in the ame field (ee Example 43.)? (b) The neutron can make tranition from one of thee tate to the other by emitting or aborbing a photon with energy equal to the energy difference of the two tate. Find the frequency and wavelength of uch a photon. Section 43. Nuclear Binding and Nuclear Structure The mot common iotope of boron i 11 5 B. (a) Determine 11 the total binding energy of 5 B from Table 43. in Section (b) Calculate thi binding energy from Eq. (43.11). (Why i the fifth term zero?) Compare to the reult you obtained in part (a). What i the percent difference? Compare the accuracy of Eq. (43.11) for B to it accuracy for 8Ni (ee Example 43.4) The mot common iotope of uranium, 38 9U, ha atomic ma u. Calculate (a) the ma defect; (b) the binding energy (in MeV); (c) the binding energy per nucleon CP What i the maximum wavelength of a g ray that could break a deuteron into a proton and a neutron? (Thi proce i called photodiintegration.) Calculate (a) the total binding energy and (b) the binding energy per nucleon of 1 C. (c) What percent of the ret ma of thi nucleu i it total binding energy? CP A photon with a wavelength of 3.50 * m trike a deuteron, plitting it into a proton and a neutron. (a) Calculate the kinetic energy releaed in thi interaction. (b) Auming the two particle hare the energy equally, and taking their mae to be 1.00 u, calculate their peed after the photodiintegration Calculate the ma defect, the binding energy (in MeV), 14 and the binding energy per nucleon of (a) the nitrogen nucleu, 7 N, 4 and (b) the helium nucleu, He. (c) How doe the binding energy per nucleon compare for thee two nuclei? Ue Eq. (43.11) to calculate the binding energy per nucleon for the nuclei 36Kr and 73 Ta. Do your reult confirm what i hown in Fig. 43. that for A greater than 6 the binding energy per nucleon deceae a A increae? Section 43.3 Nuclear Stability and Radioactivity (a) I the decay n S p + b - + n e energetically poible? If not, explain why not. If o, calculate the total energy releaed. (b) I the decay p S n + b + + n e energetically poible? If not, explain why not. If o, calculate the total energy releaed What nuclide i produced in the following radioactive decay? (a) a decay of 39 94Pu; (b) b - decay of 4 11Na; (c) b + decay 15 of 8 O CP 38 U decay pontaneouly by a emiion to 34 Th. Calculate (a) the total energy releaed by thi proce and (b) the recoil velocity of the 34 Th nucleu. The atomic mae are u for 38 U and u for 34 Th The atomic ma of 14 C i u. Show that the b - decay of 14 C i energetically poible, and calculate the energy releaed in the decay What particle ( a particle, electron, or poitron) i 7 emitted in the following radioactive decay? (a) 14Si S 7 13Al; (b) 9U S 34 90Th; (c) 33A S 74 34Se (a) Calculate the energy releaed by the electron-capture 57 decay of 7Co (ee Example 43.7). (b) A negligible amount of thi 57 energy goe to the reulting 6Fe atom a kinetic energy. About 57 90% of the time, the 6Fe nucleu emit two ucceive gamma-ray photon after the electron-capture proce, of energie 0.1 MeV and MeV, repectively, in decaying to it ground tate. What i the energy of the neutrino emitted in thi cae? Tritium 1 3 1H i an untable iotope of hydrogen; it ma, including one electron, i u. (a) Show that tritium mut be untable with repect to beta decay becaue the decay product ( 3 He plu an emitted electron) have le total ma than the tritium. (b) Determine the total kinetic energy (in MeV) of the decay product, taking care to account for the electron mae correctly. Section 43.4 Activitie and Half-Live If a 6.13-g ample of an iotope having a ma number of 14 decay at a rate of Ci, what i it half-life?

155 Exercie BIO Radioactive iotope ued in cancer therapy have a helf-life, like pharmaceutical ued in chemotherapy. Jut after it ha been manufactured in a nuclear reactor, the activity of a ample of 60 Co i 5000 Ci. When it activity fall below 3500 Ci, it i conidered too weak a ource to ue in treatment. You work in the radiology department of a large hopital. One of thee 60 Co ource in your inventory wa manufactured on October 6, 004. It i now April 6, 007. I the ource till uable? The half-life of 60 Co i 5.71 year The common iotope of uranium, 38 U, ha a half-life of 4.47 * 10 9 year, decaying to 34 Th by alpha emiion. (a) What i the decay contant? (b) What ma of uranium i required for an activity of 1.00 curie? (c) How many alpha particle are emitted per econd by 10.0 g of uranium? BIO Radiation Treatment of Protate Cancer. In many cae, protate cancer i treated by implanting 60 to 100 mall eed of radioactive material into the tumor. The energy releaed from the decay kill the tumor. One iotope that i ued (there are other) i palladium ( 103 Pd), with a half-life of 17 day. If a typical grain contain 0.50 g of 103 Pd, (a) what i it initial activity rate in Bq, and (b) what i the rate 68 day later? A 1.0-g ample of carbon from living matter decay at the rate of decay>min due to the radioactive in it. What will be the decay rate of thi ample in (a) 1000 year and (b) 50,000 year? BIO Radioactive Tracer. Radioactive iotope are often introduced into the body through the bloodtream. Their pread through the body can then be monitored by detecting the appearance of radiation in different organ. a emitter with a half-life of 8.0 d, i one uch tracer. Suppoe a cientit introduce a ample with an activity of 375 Bq and watche it pread to the organ. (a) Auming that the ample all went to the thyroid gland, what will be the decay rate in that gland 4 d (about 3 1 week) later? (b) If the decay rate in the thyroid 4 d later i actually meaured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What iotope remain after the I-131 decay? 40 K 131 I, The untable iotope i ued for dating rock ample. It half-life i 1.8 * 10 9 y. (a) How many decay occur per econd in a ample containing 1.63 * 10-6 g of 40 K? (b) What i the activity of the ample in curie? A a health phyicit, you are being conulted about a pill in a radiochemitry lab. The iotope pilled wa 500 mci of 131 Ba, which ha a half-life of 1 day. (a) What ma of 131 Ba wa pilled? (b) Your recommendation i to clear the lab until the radiation level ha fallen 1.00 mci. How long will the lab have to be cloed? Meaurement on a certain iotope tell you that the decay rate decreae from 8318 decay>min to 3091 decay>min in 4.00 day. What i the half-life of thi iotope? The iotope 6 Ra undergoe a decay with a half-life of 160 year. What i the activity of 1.00 g of 6 Ra? Expre your anwer in Bq and in Ci The radioactive nuclide 199 Pt ha a half-life of 30.8 minute. A ample i prepared that ha an initial activity of 7.56 * Bq. (a) How many 199 Pt nuclei are initially preent in the ample? (b) How many are preent after 30.8 minute? What i the activity at thi time? (c) Repeat part (b) for a time 9.4 minute after the ample i firt prepared Radiocarbon Dating. A ample from timber at an archeological ite containing 500 g of carbon provide 3070 decay> min. What i the age of the ample? b - 14 C Section 43.5 Biological Effect of Radiation BIO (a) If a chet x ray deliver 0.5 msv to 5.0 kg of tiue, how many total joule of energy doe thi tiue receive? (b) Natural radiation and comic ray deliver about 0.10 msv per year at ea level. Auming an RBE of 1, how many rem and rad i thi doe, and how many joule of energy doe a 75-kg peron receive in a year? (c) How many chet x ray like the one in part (a) would it take to deliver the ame total amount of energy to a 75-kg peron a he receive from natural radiation in a year at ea level, a decribed in part (b)? BIO A peron expoed to fat neutron receive a radiation doe of 00 rem on part of hi hand, affecting 5 g of tiue. The RBE of thee neutron i 10. (a) How many rad did he receive? (b) How many joule of energy did thi peron receive? (c) Suppoe the peron received the ame rad doage, but from beta ray with an RBE of 1.0 intead of neutron. How many rem would he have received? BIO A nuclear chemit receive an accidental radiation doe of 5.0 Gy from low neutron (RBE = 4.0). What doe he receive in rad, rem, and J>kg? BIO To Scan or Not to Scan? It ha become popular for ome people to have yearly whole-body can (CT can, formerly called CAT can) uing x ray, jut to ee if they detect anything upiciou. A number of medical people have recently quetioned the adviability of uch can, due in part to the radiation they impart. Typically, one uch can give a doe of 1 msv, applied to the whole body. By contrat, a chet x ray typically adminiter 0.0 msv to only 5.0 kg of tiue. How many chet x ray would deliver the ame total amount of energy to the body of a 75-kg peron a one whole-body can? BIO Food Irradiation. Food i often irradiated with either x ray or electron beam to help prevent poilage. A low doe of 5 75 kilorad (krad) help to reduce and kill inactive paraite, a medium doe of krad kill microorganim and pathogen uch a almonella, and a high doe of krad terilize food o that it can be tored without refrigeration. (a) A doe of 175 krad kill poilage microorganim in fih. If x ray are ued, what would be the doe in Gy, Sv, and rem, and how much energy would a 0-g portion of fih aborb? (See Table 43.3.) (b) Repeat part (a) if electron of RBE 1.50 are ued intead of x ray BIO In an indutrial accident a 65-kg peron receive a lethal whole-body equivalent doe of 5.4 Sv from x ray. (a) What i the equivalent doe in rem? (b) What i the aborbed doe in rad? (c) What i the total energy aborbed by the peron body? How doe thi amount of energy compare to the amount of energy required to raie the temperature of 65 kg of water C? BIO A 67-kg peron accidentally inget 0.35 Ci of tritium. (a) Aume that the tritium pread uniformly throughout the body and that each decay lead on the average to the aborption of 5.0 kev of energy from the electron emitted in the decay. The half-life of tritium i 1.3 y, and the RBE of the electron i 1.0. Calculate the aborbed doe in rad and the equivalent doe in rem during one week. (b) The decay of tritium releae more than 5.0 kev of energy. Why i the average energy aborbed le than the total energy releaed in the decay? CP BIO In a diagnotic x-ray procedure, 5.00 * photon are aborbed by tiue with a ma of kg. The x-ray wavelength i nm. (a) What i the total energy aborbed by the tiue? (b) What i the equivalent doe in rem? b -

156 1476 CHAPTER 43 Nuclear Phyic Section 43.6 Nuclear Reaction, Section 43.7 Nuclear Fiion, and Section 43.8 Nuclear Fuion Conider the nuclear reaction 1H N S X B where X i a nuclide. (a) What are Z and A for the nuclide X? (b) Calculate the reaction energy Q (in MeV). (c) If the 1H nucleu 14 i incident on a tationary 7 N nucleu, what minimum kinetic energy mut it have for the reaction to occur? Energy from Nuclear Fuion. Calculate the energy releaed in the fuion reaction Conider the nuclear reaction where X i a nuclide. (a) What are the value of Z and A for the nuclide X? (b) How much energy i liberated? (c) Etimate the threhold energy for thi reaction The United State ue 1.0 * 10 0 J of electrical energy per year. If all thi energy came from the fiion of 35 U, which releae 00 MeV per fiion event, (a) how many kilogram of 35 U would be ued per year and (b) how many kilogram of uranium would have to be mined per year to provide that much 35 U? (Recall that only 0.70% of naturally occurring uranium i 35 U.) At the beginning of Section 43.7 the equation of a fiion proce i given in which 35 U i truck by a neutron and undergoe fiion to produce 144 Ba, 89 Kr, and three neutron. The meaured mae of thee iotope are u 1 35 U, u Ba, u 1 89 Kr, and u (neutron). (a) Calculate the energy (in MeV) releaed by each fiion reaction. (b) Calculate the energy releaed per gram of 35 U, in MeV>g Conider the nuclear reaction 8 14Si + g S 4 1Mg + X where X i a nuclide. (a) What are Z and A for the nuclide X? (b) Ignoring the effect of recoil, what minimum energy mut the 8 photon have for thi reaction to occur? The ma of a 14Si atom i u, and the ma of a 1Mg atom i u The econd reaction in the proton-proton chain (ee 3 3 Fig ) produce a He nucleu. A He nucleu produced in 4 thi way can combine with a He nucleu: Calculate the energy liberated in thi proce. (Thi i hared between the energy of the photon and the recoil kinetic energy of 7 the beryllium nucleu.) The ma of a 4Be atom i u Conider the nuclear reaction 4 He + 7 3Li S X + 1 0n where X i a nuclide. (a) What are Z and A for the nuclide X? (b) I energy aborbed or liberated? How much? CP In a cm 3 ample of water, 0.015% of the molecule are D O. Compute the energy in joule that i liberated if all the deuterium nuclei in the ample undergo the fuion reaction of Example PROBLEMS 3 He + 1H S 4 He + 1 1H 1H + 9 4Be S X + 4 He 3 He + 4 He S 7 4Be + g Comparion of Energy Releaed per Gram of Fuel. (a) When gaoline i burned, it releae 1.3 * 10 8 J of energy per gallon (3.788 L). Given that the denity of gaoline i 737 kg>m 3, expre the quantity of energy releaed in J>g of fuel. (b) During fiion, when a neutron i aborbed by a 35 U nucleu, about 00 MeV of energy i releaed for each nucleu that undergoe fiion. Expre thi quantity in J>g of fuel. (c) In the proton-proton chain that take place in tar like our un, the overall fuion reaction can be ummarized a ix proton fuing to form one 4 He nucleu with two leftover proton and the liberation of 6.7 MeV of energy. The fuel i the ix proton. Expre the energy produced here in unit of J>g of fuel. Notice the huge difference between the two form of nuclear energy, on the one hand, and the chemical energy from gaoline, on the other. (d) Our un produce energy at a meaured rate of 3.86 * 10 6 W. If it ma of 1.99 * kg were all gaoline, how long could it lat before conuming all it fuel? (Hitorical note: Before the dicovery of nuclear fuion and the vat amount of energy it releae, cientit were confued. They knew that the earth wa at leat many million of year old, but could not explain how the un could urvive that long if it energy came from chemical burning.) Ue conervation of ma-energy to how that the energy releaed in alpha decay i poitive whenever the ma of the original neutral atom i greater than the um of the mae of the final neutral atom and the neutral 4 He atom. (Hint: Let the parent nucleu have atomic number Z and nucleon number A. Firt write the reaction in term of the nuclei and particle involved, and then add Z electron mae to both ide of the reaction and allot them a needed to arrive at neutral atom.) Ue conervation of ma-energy to how that the energy releaed in b - decay i poitive whenever the neutral atomic ma of the original atom i greater than that of the final atom. (See the hint in Problem ) Ue conervation of ma-energy to how that the energy releaed in b + decay i poitive whenever the neutral atomic ma of the original atom i at leat two electron mae greater than that of the final atom. (See the hint in Problem ) (a) Calculate the minimum energy required to remove one proton from the nucleu 1 6C. Thi i called the proton-removal energy. (Hint: Find the difference between the ma of a 1 6C nucleu and the ma of a proton plu the ma of the nucleu formed when a proton i removed from 1.) (b) How doe the proton-removal energy for 1 6C compare to the binding energy per nucleon for 1 6C 6C, calculated uing Eq. (43.10)? (a) Calculate the minimum energy required to remove 17 one neutron from the nucleu 8 O. Thi i called the neutronremoval energy. (See Problem 43.5.) (b) How doe the neutronremoval energy for 8 O compare to the binding energy per nucleon for 8 O, calculated uing Eq. (43.10)? The neutral atomic ma of 14 6 C i u. Calculate 15 the proton removal energy and the neutron removal energy for 7 N. (See Problem 43.5 and ) What i the percentage difference between thee two energie, and which i larger? BIO Radioactive Fallout. One of the problem of in-air teting of nuclear weapon (or, even wore, the ue of uch weapon!) i the danger of radioactive fallout. One of the mot problematic nuclide in uch fallout i trontium-90 ( 90 Sr), which break down by b - decay with a half-life of 8 year. It i chemically imilar to calcium and therefore can be incorporated into bone and teeth, where, due to it rather long half-life, it remain for year a an internal ource of radiation. (a) What i the daughter nucleu of the 90 Sr decay? (b) What percentage of the original level of 90 Sr i left after 56 year? (c) How long would you have to wait for the original level to be reduced to 6.5% of it original value?

157 Problem CP Thorium 90Th decay to radium 88Ra by a emiion. The mae of the neutral atom are u for 90Th 30 6 and u for 88Ra. If the parent thorium nucleu i at ret, what i the kinetic energy of the emitted a particle? (Be ure to account for the recoil of the daughter nucleu.) The atomic ma of 5 1Mg i u, and the atomic 5 ma of 13Al i u. (a) Which of thee nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined thi. (c) How much energy (in MeV) i releaed in the decay? The polonium iotope 10 84Po ha atomic ma u. Other atomic mae are 8Pb, u; 83Bi, u; 83Bi, u; 84Po, u; and At, u. (a) Show that the alpha decay of 84Po i energetically poible, and find the energy of the emitted a particle. 10 (b) I 84Po energetically table with repect to emiion of a proton? Why or why not? (c) I 84Po energetically table with repect to emiion of a neutron? Why or why not? (d) I 84Po energetically table with repect to b - decay? Why or why not? (e) I 84Po 10 energetically table with repect to b + decay? Why or why not? BIO Irradiating Ourelve! The radiocarbon in our bodie i one of the naturally occurring ource of radiation. Let ee how large a doe we receive. 14 C decay via b - emiion, and 18% of our body ma i carbon. (a) Write out the decay cheme of carbon-14 and how the end product. (A neutrino i alo produced.) (b) Neglecting the effect of the neutrino, how much kinetic energy (in MeV) i releaed per decay? The atomic ma of 14 C i u. (c) How many gram of carbon are there in a 75-kg peron? How many decay per econd doe thi carbon produce? (Hint: Ue data from Example 43.9.) (d) Auming that all the energy releaed in thee decay i aborbed by the body, how many MeV> and J> doe the 14 C releae in thi peron body? (e) Conult Table 43.3 and ue the larget appropriate RBE for the particle involved. What radiation doe doe the peron give himelf in a year, in Gy, rad, Sv, and rem? BIO Pion Radiation Therapy. A neutral pion 1p 0 ha a ma of 64 time the electron ma and decay with a lifetime of 8.4 * to two photon. Such pion are ued in the radiation treatment of ome cancer. (a) Find the energy and wavelength of thee photon. In which part of the electromagnetic pectrum do they lie? What i the RBE for thee photon? (b) If you want to deliver a doe of 00 rem (which i typical) in a ingle treatment to 5 g of tumor tiue, how many p 0 meon are needed? Gold, Au, undergoe b - decay to an excited tate of Hg. If the excited tate decay by emiion of a g photon with energy 0.41 MeV, what i the maximum kinetic energy of the electron emitted in the decay? Thi maximum occur when the 198 antineutrino ha negligible energy. (The recoil energy of the 80Hg nucleu can be ignored. The mae of the neutral atom in their 198 ground tate are u for 79Au and u for Hg.) Calculate the ma defect for the decay of 11 b + 6 C. I thi decay energetically poible? Why or why not? The atomic ma 11 of 6 C i u Calculate the ma defect for the decay of 13 b + 7 N. I thi decay energetically poible? Why or why not? The atomic ma 13 of 7 N i u The reult of activity meaurement on a radioactive ample are given in the table. (a) Find the half-life. (b) How many radioactive nuclei were preent in the ample at t = 0? (c) How many were preent after 7.0 h? 6 Time (h) BIO A peron inget an amount of a radioactive ource with a very long lifetime and activity 0.63 mci. The radioactive material lodge in the lung, where all of the 4.0-MeV a particle emitted are aborbed within a 0.50-kg ma of tiue. Calculate the aborbed doe and the equivalent doe for one year Meauring Very Long Half-Live. Some radioiotope uch a amarium Sm and gadolinium 1 15 Gd have half-live that are much longer than the age of the univere, o we can t meaure their half-live by watching their decay rate decreae. Luckily, there i another way of calculating the half-life, uing Eq. (43.16). Suppoe a 1.0-g ample of i oberved to decay at a rate of.65 Bq. Calculate the half-life of the ample in year. (Hint: How many nuclei are there in the 1.0-g ample?) We Are Stardut. In 195 pectral line of the element technetium Tc were dicovered in a red giant tar. Red giant are very old tar, often around 10 billion year old, and near the end of their live. Technetium ha no table iotope, and the half-life of 99 Tc i 00,000 year. (a) For how many half-live ha the 99 Tc been in the red-giant tar if it age i 10 billion year? (b) What fraction of the original 99 Tc would be left at the end of that time? Thi dicovery wa extremely important becaue it provided convincing evidence for the theory (now eentially known to be true) that mot of the atom heavier than hydrogen and helium were made inide of tar by thermonuclear fuion and 99 Tc Decay/ 0 0, , , , ,00.5 4, , , , Sm other nuclear procee. If the had been part of the tar ince it wa born, the amount remaining after 10 billion year would have been o minute that it would not have been detectable. Thi knowledge i what led the late atronomer Carl Sagan to proclaim that we are tardut BIO A 70.0-kg peron experience a whole-body expoure to a radiation with energy 4.77 MeV. A total of 6.5 * 10 1 a particle are aborbed. (a) What i the aborbed doe in rad? (b) What i the equivalent doe in rem? (c) If the ource i g of 6 Ra (half-life 1600 y) omewhere in the body, what i the activity of thi ource? (d) If all the alpha particle produced are aborbed, what time i required for thi doe to be delivered? Meaurement indicate that 7.83% of all rubidium 87 Rb atom currently on the earth are the radioactive iotope. The ret are the table 85 Rb iotope. The half-life of 87 Rb i 4.75 * y. Auming that no rubidium atom have been formed ince, what percentage of rubidium atom were 87 Rb when our olar ytem wa formed 4.6 * 10 9 y ago? A O nucleu at ret decay by the emiion of a.76-mev a particle. Calculate the atomic ma of the daughter

158 1478 CHAPTER 43 Nuclear Phyic nuclide produced by thi decay, auming that it i produced in it 186 ground tate. The atomic ma of 76O i u BIO A 60 Co ource with activity.6 * 10-4 Ci i embedded in a tumor that ha ma 0.00 kg. The ource emit g photon with average energy 1.5 MeV. Half the photon are aborbed in the tumor, and half ecape. (a) What energy i delivered to the tumor per econd? (b) What aborbed doe (in rad) i delivered per econd? (c) What equivalent doe (in rem) i delivered per econd if the RBE for thee g ray i 0.70? (d) What expoure time i required for an equivalent doe of 00 rem? The nucleu 15 ha a half-life of 1. ; 19 8 O 8 O ha a halflife of 6.9. If at ome time a ample contain equal amount of O and 8 O, what i the ratio of 8 O to 8 O (a) after 4.0 minute and (b) after 15.0 minute? A bone fragment found in a cave believed to have been inhabited by early human contain 0.9 time a much 14 C a an equal amount of carbon in the atmophere when the organim containing the bone died. (See Example 43.9 in Section 43.4.) Find the approximate age of the fragment An Oceanographic Tracer. Nuclear weapon tet in the 1950 and 1960 releaed ignificant amount of radioactive tritium 1 3 1H, half-life 1.3 year into the atmophere. The tritium atom were quickly bound into water molecule and rained out of the air, mot of them ending up in the ocean. For any of thi tritium-tagged water that ink below the urface, the amount of time during which it ha been iolated from the urface can be calculated by meauring the ratio of the decay product, He, to the 3 remaining tritium in the water. For example, if the ratio of 3 He to 3 1H in a ample of water i 1:1, the water ha been below the urface for one half-life, or approximately 1 year. Thi method ha provided oceanographer with a convenient way to trace the movement of uburface current in part of the ocean. Suppoe 3 that in a particular ample of water, the ratio of He to 3 1H i 4.3 to 1.0. How many year ago did thi water ink below the urface? Conider the fuion reaction 1H + 1H S 3 He + 1 0n. (a) Etimate the barrier energy by calculating the repulive electrotatic potential energy of the two 1H nuclei when they touch. (b) Compute the energy liberated in thi reaction in MeV and in joule. (c) Compute the energy liberated per mole of deuterium, remembering that the ga i diatomic, and compare with the heat of combution of hydrogen, about.9 * 10 5 J>mol BIO In the 1986 diater at the Chernobyl reactor in the 1 Soviet Union (now Ukraine), about of the 137 preent in the reactor wa releaed. The iotope C C ha a half-life for b decay of y and decay with the emiion of a total of 1.17 MeV of energy per decay. Of thi, 0.51 MeV goe to the emitted electron and the remaining 0.66 MeV to a g ray. The radioactive 137 C i aborbed by plant, which are eaten by livetock and human. How many 137 C atom would need to be preent in each kilogram of body tiue if an equivalent doe for one week i 3.5 Sv? Aume that all of the energy from the decay i depoited in that 1.0 kg of tiue and that the RBE of the electron i CP (a) Prove that when a particle with ma m and kinetic energy K collide with a tationary particle with ma M, the total kinetic energy K cm in the center-of-ma coordinate ytem (the energy available to caue reaction) i M K cm = M + m K Aume that the kinetic energie of the particle and nuclei are much lower than their ret energie. (b) If K th i the minimum, or threhold, kinetic energy to caue an endoergic reaction to occur in the ituation of part (a), how that K th = - M + m M Q Calculate the energy releaed in the fiion reaction U + 1 0n S Xe + 38Sr + 1 0n. You can ignore the initial kinetic energy of the aborbed neutron. The atomic mae are U, u; 54Xe, u; and 38Sr, u. CHALLENGE PROBLEMS The reult of activity meaurement on a mixed ample of radioactive element are given in the table. (a) How many different nuclide are preent in the mixture? (b) What are their halflive? (c) How many nuclei of each type are initially preent in the ample? (d) How many of each type are preent at t = 5.0 h? Time (h) Decay/ Indutrial Radioactivity. Radioiotope are ued in a variety of manufacturing and teting technique. Wear meaurement can be made uing the following method. An automobile engine i produced uing piton ring with a total ma of 100 g, which include 9.4 mci of 59 Fe whoe half-life i 45 day. The engine i tet-run for 1000 hour, after which the oil i drained and it activity i meaured. If the activity of the engine oil i 84 decay>, how much ma wa worn from the piton ring per hour of operation? Anwer Chapter Opening Quetion? When an organim die, it top taking in carbon from atmopheric CO Some of thi carbon i radioactive 14. C, which decay with a half-life of 5730 year. By meauring the proportion of 14 C that remain in the pecimen, cientit can determine how long ago the organim died. (See Section 43.4.) Tet Your Undertanding Quetion 43.1 Anwer: (a) (iii), (b) (v) The radiu R i proportional to the cube root of the ma number A, while the volume i proportional to R 3 and hence to 1A 1>3 3 = A. Therefore, doubling the volume require increaing the ma number by a factor of ; doubling the radiu implie increaing both the volume and the ma number by a factor of 3 = 8.

159 Anwer Anwer: (ii), (iii), (iv), (v), (i) You can find the anwer by inpecting Fig The binding energy per nucleon i lowet for very light nuclei uch a 4 He, i greatet around A = 60, and then decreae with increaing A Anwer: (v) Two proton and two neutron are lot in an a decay, o Z and N each decreae by. A b + decay change a proton to a neutron, o Z decreae by 1 and N increae by 1. The net reult i that Z decreae by 3 and N decreae by Anwer: (iii) The activity -dn1t>dt of a ample i the product of the number of nuclei in the ample N1t and the decay contant l = 1ln >T 1>. Hence N(t) = 1-dN1t>dtT 1> >1ln. Taking the ratio of thi expreion for 40 Pu to thi ame expreion for the factor of ln cancel and we get N Pu N Am = 43 Am, 1-dN Pu>dtT 1> Pu mci16560 y = 1-dN Am >dtt 1> Am mci17370 y = Am The two ample contain equal number of nuclei. The ample ha a longer half-life and hence a lower decay rate, o it ha a lower activity than the 40 Pu ample Anwer: (ii) We aw in Section 43.3 that alpha particle can travel only a very hort ditance before they are topped. By contrat, x-ray photon are very penetrating, o they can eaily pa into the body Anwer: no The reaction 1 1 H + 7 3Li S 4 He + 4 He i a nuclear reaction, which can take place only if a proton (a hydrogen nucleu) come into contact with a lithium nucleu. If the hydrogen i in atomic form, the interaction between it electron cloud and the electron cloud of a lithium atom keep the two nuclei from getting cloe to each other. Even if iolated proton are ued, they mut be fired at the lithium atom with enough kinetic energy to overcome the electric repulion between the proton and the lithium nuclei. The tatement that the reaction i exoergic mean that more energy i releaed by the reaction than had to be put in to make the reaction occur Anwer: no Becaue the neutron ha no electric charge, it experience no electric repulion from a 35 U nucleu. Hence a low-moving neutron can approach and enter a 35 U nucleu, thereby providing the excitation needed to trigger fiion. By contrat, a low-moving proton (charge +e) feel a trong electric repulion from a 35 U nucleu (charge +9e). It never get cloe to the nucleu, o it cannot trigger fiion Anwer: no Fuion reaction between ufficiently light nuclei are exoergic becaue the binding energy per nucleon E B >A increae. If the nuclei are too maive, however, E B >A decreae and fuion i endoergic (i.e., it take in energy rather than releaing it). A an example, imagine fuing together two nuclei of A = 100 to make a ingle nucleu with A = 00. From Fig. 43., E B >A i more than 8.5 MeV for the A = 100 nuclei but i le than 8 MeV for the A = 00 nucleu. Such a fuion reaction i poible, but require a ubtantial input of energy. Bridging Problem 18 Xe Anwer: (a) (b) no; b + emiion would be endoergic (c) 3.5 * 10 9 atom, 1.50 * 10 6 Bq (d) N1t = 13.5 * 10 9 atom11 - e * t

160 44 PARTICLE PHYSICS AND COSMOLOGY LEARNING GOALS By tudying thi chapter, you will learn: The key varietie of fundamental ubatomic particle and how they were dicovered. How phyicit ue accelerator and detector to probe the propertie of ubatomic particle. The four way in which ubatomic particle interact with each other. How the tructure of proton, neutron, and other particle can be explained in term of quark. How phyicit probe the limit of the tandard model of particle and interaction. The evidence that the univere i expanding and that the expanion i peeding up. The hitory of the firt 380,000 year after the Big Bang.? Image made uing infrared and x-ray wavelength were combined to produce thi view of the dynamic center of our Milky Way galaxy. The image how atom in variou tate: a iolated atom in glowing, diffue cloud of ga (hown in blue), a clump of atom and molecule in immene, cold dut cloud (hown in red), and in dene accumulation that we call tar. What fraction of the ma and energy in the univere i compoed of normal matter that i, atom and their contituent? What i the world made of? What are the mot fundamental contituent of matter? Philoopher and cientit have been aking thee quetion for at leat 500 year. We till don t have the final anwer, but a we ll ee in thi chapter, we ve come a long way. The chapter title, Particle Phyic and Comology, may eem trange. Fundamental particle are the mallet thing in the univere, and comology deal with the bigget thing there i the univere itelf. Nonethele, we ll ee in thi chapter that phyic on the mot microcopic cale play an eential role in determining the nature of the univere on the larget cale. Fundamental particle, we ll find, are not permanent entitie; they can be created and detroyed. The development of high-energy particle accelerator and aociated detector ha been crucial in our emerging undertanding of particle. We can claify particle and their interaction in everal way in term of conervation law and ymmetrie, ome of which are abolute and other of which are obeyed only in certain kind of interaction. We ll conclude by dicuing our preent undertanding of the nature and evolution of the univere a a whole Fundamental Particle A Hitory The idea that the world i made of fundamental particle ha a long hitory. In about 400 B.C. the Greek philoopher Democritu and Leucippu uggeted that matter i made of indiviible particle that they called atom, a word derived from a- (not) and tomo (cut or divided). Thi idea lay dormant until about 1804, when the Englih cientit John Dalton ( ), often called the father of 1480

161 44.1 Fundamental Particle A Hitory 1481 modern chemitry, dicovered that many chemical phenomena could be explained if atom of each element are the baic, indiviible building block of matter. The Electron and the Proton Toward the end of the 19th century it became clear that atom are not indiviible. The exitence of characteritic atomic pectra of element uggeted that atom have internal tructure, and J. J. Thomon dicovery of the negatively charged electron in 1897 howed that atom could be taken apart into charged particle. Rutherford experiment in (ee Section 39.) revealed that an atom poitive charge reide in a mall, dene nucleu. In 1919 Rutherford made an additional dicovery: When alpha particle are fired into nitrogen, one of the product i hydrogen ga. He reaoned that the hydrogen nucleu i a contituent of the nuclei of heavier atom uch a nitrogen, and that a colliion with a fatmoving alpha particle can dilodge one of thoe hydrogen nuclei. Thu the hydrogen nucleu i an elementary particle, to which Rutherford gave the name proton. The following decade aw the blooming of quantum mechanic, including the Schrödinger equation. Phyicit were on their way to undertanding the principle that underlie atomic tructure. The Photon Eintein explained the photoelectric effect in 1905 by auming that the energy of electromagnetic wave i quantized; that i, it come in little bundle called photon with energy E = hƒ. Atom and nuclei can emit (create) and aborb (detroy) photon (ee Section 38.1). Conidered a particle, photon have zero charge and zero ret ma. (Note that any dicuion of a particle ma in thi chapter will refer to it ret ma.) In particle phyic, a photon i denoted by the ymbol g (the Greek letter gamma). The Neutron In 1930 the German phyicit Walther Bothe and Herbert Becker oberved that when beryllium, boron, or lithium wa bombarded by alpha particle, the target material emitted a radiation that had much greater penetrating power than the original alpha particle. Experiment by the Englih phyicit Jame Chadwick in 193 howed that the emitted particle were electrically neutral, with ma approximately equal to that of the proton. Chadwick chritened thee particle 1 neutron (ymbol n or 0n). A typical reaction of the type tudied by Bothe and Becker, uing a beryllium target, i 4 He Be S 1 6 C n (44.1) Elementary particle are uually detected by their electromagnetic effect for intance, by the ionization that they caue when they pa through matter. (Thi i the principle of the cloud chamber, decribed below.) Becaue neutron have no charge, they interact hardly at all with electron and produce little ionization when they pa through matter and o are difficult to detect directly. However, neutron can be lowed down by cattering from nuclei, and they can penetrate a nucleu. Hence low neutron can be detected by mean of a nuclear reaction in which a neutron i aborbed and an alpha particle i emitted. An example i 0 1 n B S 7 3 Li + 4 He (44.) The ejected alpha particle i eay to detect becaue it i charged. Later experiment howed that neutron, like proton and electron, are pin- 1 particle (ee Section 43.1). The dicovery of the neutron cleared up a mytery about the compoition of the nucleu. Before 1930 the ma of a nucleu wa thought to be due only to proton, but no one undertood why the charge-to-ma ratio wa not the ame for all nuclide. It oon became clear that all nuclide (except 1 H) contain both proton 1

162 148 CHAPTER 44 Particle Phyic and Comology 44.1 Photograph of the cloudchamber track made by the firt poitron ever identified. The photograph wa made by Carl D. Anderon in 193. The poitron follow a curved path owing to the preence of a magnetic field. The track i more trongly curved above the lead plate, howing that the poitron wa traveling upward and lot energy and peed a it paed through the plate. Lead plate (6 mm thick) Poitron track and neutron. Hence the proton, the neutron, and the electron are the building block of atom. One might think that would be the end of the tory. On the contrary, it i barely the beginning. Thee are not the only particle, and they can do more than build atom. The Poitron The poitive electron, or poitron, wa dicovered by the American phyicit Carl D. Anderon in 193, during an invetigation of particle bombarding the earth from pace. Figure 44.1 how a hitoric photograph made with a cloud chamber, an intrument ued to viualize the track of charged particle. The chamber contained a upercooled vapor; ion created by the paage of charged particle through the vapor erved a nucleation center, and liquid droplet formed around them, making a viible track. The cloud chamber in Fig i in a magnetic field directed into the plane of the photograph. The particle ha paed through a thin lead plate (which extend from left to right in the figure) that lie within the chamber. The track i more tightly curved above the plate than below it, howing that the peed wa le above the plate than below it. Therefore the particle had to be moving upward; it could not have gained energy paing through the lead. The thickne and curvature of the track uggeted that it ma and the magnitude of it charge equaled thoe of the electron. But the direction of the magnetic field and the velocity in the magnetic force equation F S qv S : B S howed that the particle had poitive charge. Anderon chritened thi particle the poitron. To theorit, the appearance of the poitron wa a welcome development. In 198 the Englih phyicit Paul Dirac had developed a relativitic generalization of the Schrödinger equation for the electron. In Section 41.5 we dicued how Dirac idea helped explain the pin magnetic moment of the electron. One of the puzzling feature of the Dirac equation wa that for a free electron it predicted not only a continuum of energy tate greater than it ret energy m e c, a hould be expected, but alo a continuum of negative energy tate le than -m e c (Fig. 44.a). That poed a problem. What wa to prevent an electron from emitting a photon with energy m e c or greater and hopping from a poitive tate to a negative tate? It wan t clear what thee negative-energy tate meant, and there wa no obviou way to get rid of them. Dirac ingeniou interpretation wa that all the negative-energy tate were filled with electron, and that thee electron were for ome reaon unobervable. The excluion principle (ee Section 41.6) would then forbid a tranition to a tate that wa already occupied. A vacancy in a negative-energy tate would act like a poitive charge, jut a a hole in the valence band of a emiconductor (ee Section 4.6) act like a poitive charge. Initially, Dirac tried to argue that uch vacancie were proton. But after 44. (a) Energy tate for a free electron predicted by the Dirac equation. (b) Raiing an electron from an E 6 0 tate to an E 7 0 tate correpond to electron poitron pair production. (c) An electron dropping from an E 7 0 tate to a vacant E 6 0 tate correpond to electron poitron pair annihilation. (a) Continuum of (b) (c) poitive-energy tate Electron.m e c m e c Photon 0 m e c m e c m e c m e c Photon Photon Continuum of negative-energy tate + Poitron m e c

163 44.1 Fundamental Particle A Hitory 1483 Anderon dicovery it became clear that the vacancie were oberved phyically a poitron. Furthermore, the Dirac energy-tate picture provide a mechanim for the creation of poitron. When an electron in a negative-energy tate aborb a photon with energy greater than m e c, it goe to a poitive tate (Fig. 44.b), in which it become obervable. The vacancy that it leave behind i oberved a a poitron; the reult i the creation of an electron poitron pair. Similarly, when an electron in a poitive-energy tate fall into a vacancy, both the electron and the vacancy (that i, the poitron) diappear, and photon are emitted (Fig. 44.c). Thu the Dirac theory lead naturally to the concluion that, like photon, electron can be created and detroyed. While photon can be created and detroyed ingly, electron can be produced or detroyed only in electron poitron pair or in aociation with other particle. (Creating or detroying an electron alone would mean creating or detroying an amount of charge e, which would violate the conervation of electric charge.) In 1949 the American phyicit Richard Feynman howed that a poitron could be decribed mathematically a an electron traveling backward in time. Hi reformulation of the Dirac theory eliminated difficult calculation involving the infinite ea of negative-energy tate and put electron and poitron on the ame footing. But the creation and detruction of electron poitron pair remain. The Dirac theory provide the beginning of a theoretical framework for creation and detruction of all fundamental particle. Experiment and theory tell u that the mae of the poitron and electron are identical, and that their charge are equal in magnitude but oppoite in ign. The S poitron pin angular momentum and magnetic moment M S are parallel; they S are oppoite for the electron. However, and M S have the ame magnitude for both particle becaue they have the ame pin. We ue the term antiparticle for a particle that i related to another particle a the poitron i to the electron. Each kind of particle ha a correponding antiparticle. For a few kind of particle (necearily all neutral) the particle and antiparticle are identical, and we can ay that they are their own antiparticle. The photon i an example; there i no way to ditinguih a photon from an antiphoton. We ll ue the tandard ymbol e - for the electron and e + for the poitron, and the generic term electron will often include both electron and poitron. Other antiparticle are often denoted by a bar over the particle ymbol; for example, an antiproton i p. We ll ee everal other example of antiparticle later. Poitron do not occur in ordinary matter. Electron poitron pair are produced during high-energy colliion of charged particle or g ray with matter. Thi proce i called e + e - pair production (Fig. 44.3). Enough energy E mut be available to account for the ret energy m e c of the two particle. The minimum energy for electron poitron pair production i 44.3 (a) Photograph of bubble-chamber track of electron poitron pair that are produced when 300-MeV photon trike a lead heet. A magnetic field directed out of the photograph made the electron and poitron curve in oppoite direction. (b) Diagram howing the pair-production proce for two of the photon. (a) Electron poitron pair E min = m e c = * kg1.998 * 10 8 m> = * J = 1.0 MeV The invere proce, e + e - pair annihilation, occur when a poitron and an electron collide (ee Example 38.6 in Section 38.3). Both particle diappear, and two (or occaionally three) photon can appear, with total energy of at leat m e c = 1.0 MeV. Decay into a ingle photon i impoible: Such a proce could not conerve both energy and momentum. Poitron alo occur in the decay of ome untable nuclei, in which they are called beta-plu particle 1b +. We dicued b + decay in Section It often convenient to repreent particle mae in term of the equivalent ret energy uing m = E>c. Then typical ma unit are MeV>c ; for example, m = MeV>c for an electron or poitron. We ll ue thee unit frequently in thi chapter. (b) g g B S e e 1 e e 1

164 1484 CHAPTER 44 Particle Phyic and Comology Application Pair Annihilation in Medical Diagnoi A technique called poitron emiion tomography (PET) can be ued to identify the early tage of Alzheimer dieae. A patient i adminitered a glucoe-like compound called FDG in which one of the oxygen atom i replaced by radioactive 18 F. FDG accumulate in active area of the brain, where glucoe metabolim i high. The 18 F undergoe b + decay (poitron emiion) with a half-life of 110 minute, and the emitted poitron immediately annihilate with an atomic electron to produce two gamma-ray photon. A canner detect both photon, then calculate where the annihilation took place and hence the ite of FDG accumulation. Thee PET image which how area of tronget emiion, and hence greatet glucoe metabolim, in red reveal change in the brain of patient with mild cognitive impairment and with Alzheimer dieae. Particle A Force Mediator In claical phyic we decribe the interaction of charged particle in term of electric and magnetic force. In quantum mechanic we can decribe thi interaction in term of emiion and aborption of photon. Two electron repel each other a one emit a photon and the other aborb it, jut a two kater can puh each other apart by toing a heavy ball back and forth between them (Fig. 44.4a). For an electron and a proton, in which the charge are oppoite and the force i attractive, we imagine the kater trying to grab the ball away from each other (Fig. 44.4b). The electromagnetic interaction between two charged particle i mediated or tranmitted by photon. If charged-particle interaction are mediated by photon, where doe the energy to create the photon come from? Recall from our dicuion of the uncertainty principle (ee Section 38.4 and 39.6) that a tate that exit for a hort time t ha an uncertainty E in it energy uch that E t Ú U (44.3) Thi uncertainty permit the creation of a photon with energy E, provided that it live no longer than the time t given by Eq. (44.3). A photon that can exit for a hort time becaue of thi energy uncertainty i called a virtual photon. It a though there were an energy bank; you can borrow energy, provided that you pay it back within the time limit. According to Eq. (44.3), the more you borrow, the ooner you have to pay it back An analogy for how particle act a force mediator. (a) Two kater exert repulive force on each other by toing a ball back and forth. F F Meon I there a particle that mediate the nuclear force? By the mid-1930 the nuclear force between two nucleon (neutron or proton) appeared to be decribed by a potential energy U1r with the general form U1r = -ƒ a e-r>r 0 b 1nuclear potential energy (44.4) r The contant ƒ characterize the trength of the interaction, and r 0 decribe it range. Figure 44.5 how a graph of the abolute value of thi function and compare it with the function ƒ >r, which would be analogou to the electric interaction of two proton: U1r = 1 e 4pP 0 r 1electric potential energy (44.5) (b) Two kater exert attractive force on each other when one trie to grab the ball out of the other hand. F F In 1935 the Japanee phyicit Hideki Yukawa uggeted that a hypothetical particle that he called a meon might mediate the nuclear force. He howed that the range of the force wa related to the ma of the particle. Yukawa argued that the particle mut live for a time t long enough to travel a ditance comparable to the range r 0 of the nuclear force. Thi range wa known from the ize of nuclei and other information to be about 1.5 * m = 1.5 fm. If we aume that an average particle peed i comparable to c and travel about half the range, it lifetime t mut be about t = r 0 c = 1.5 * m 13.0 * 10 8 m> =.5 * 10-4 From Eq. (44.3), the minimum neceary uncertainty E in energy i E = U t = 1.05 * J # 1.5 * 10-4 =.1 * J = 130 MeV

165 44. Particle Accelerator and Detector 1485 The ma equivalent m of thi energy i m = E c =.1 * J * 10 8 m> =.3 * 10-8 kg = 130 MeV>c Thi i about 50 time the electron ma, and Yukawa potulated that an a yet undicovered particle with thi ma erve a the meenger for the nuclear force. A year later, Carl Anderon and hi colleague Seth Neddermeyer dicovered in comic radiation two new particle, now called muon. The m - ha charge equal to that of the electron, and it antiparticle the m + ha a poitive charge with equal magnitude. The two particle have equal ma, about 07 time the electron ma. But it oon became clear that muon were not Yukawa particle becaue they interacted with nuclei only very weakly. In 1947 a family of three particle, called p meon or pion, were dicovered. Their charge are +e, -e, and zero, and their mae are about 70 time the electron ma. The pion interact trongly with nuclei, and they are the particle predicted by Yukawa. Other, heavier meon, the v and r, evidently alo act a horter-range meenger of the nuclear force. The complexity of thi explanation ugget that it ha impler underpinning; thee involve the quark and gluon that we ll dicu in Section Before dicuing meon further, we ll decribe ome particle accelerator and detector to ee how meon and other particle are created in a controlled fahion and oberved Graph of the magnitude of the Yukawa potential-energy function for nuclear force, ƒu1rƒ = ƒ e -r/r 0 >r. The function U1r = ƒ >r, proportional to the potential energy for Coulomb law, i alo hown. The two function are imilar at mall r, but the Yukawa potential energy drop off much more quickly at large r. 0U(r)0 4f /r 0 3f /r 0 f /r 0 f /r 0 O Coulomb potential energy f r Yukawa potential energy ( ) f er / r 0 r r 0 r 0 3r 0 r Tet Your Undertanding of Section 44.1 Each of the following particle can be exchanged between two proton, two neutron, or a neutron and a proton a part of the nuclear force. Rank the particle in order of the range of the interaction that they mediate, from larget to mallet range. (i) the p + (pi-plu) meon of ma 140 MeV>c ; (ii) the r + (rho-plu) meon of ma 776 MeV>c ; (iii) the h 0 (eta-zero) meon of ma 548 MeV>c ; (iv) the v 0 (omega-zero) meon of ma 783 MeV>c. 44. Particle Accelerator and Detector Early nuclear phyicit ued alpha and beta particle from naturally occurring radioactive element for their experiment, but they were retricted in energy to the few MeV that are available in uch random decay. Preent-day particle accelerator can produce preciely controlled beam of particle, from electron and poitron up to heavy ion, with a wide range of energie. Thee beam have three main ue. Firt, high-energy particle can collide to produce new particle, jut a a colliion of an electron and a poitron can produce photon. Second, a high-energy particle ha a hort de Broglie wavelength and o can probe the mall-cale interior tructure of other particle, jut a electron microcope (ee Section 39.1) can give better reolution than optical microcope. Third, they can be ued to produce nuclear reaction of cientific or medical ue. Linear Accelerator Particle accelerator ue electric and magnetic field to accelerate and guide beam of charged particle. A linear accelerator (linac) accelerate particle in a traight line. J. J. Thomon cathode-ray tube were early example of linac. Modern linac ue a erie of electrode with gap to give the particle a erie of boot. Mot preent-day high-energy linear accelerator ue a traveling electromagnetic wave; the charged particle ride the wave in more or le the way that a urfer ride an incoming ocean wave. In the highet-energy linac in the world today, at the SLAC National Accelerator Laboratory, electron and poitron can be accelerated to 50 GeV in a tube 3 km long. At thi energy their de Broglie wavelength are 0.05 fm, much maller than the ize of a proton or a neutron.

166 1486 CHAPTER 44 Particle Phyic and Comology 44.6 Layout and operation of a cyclotron. (a) Schematic diagram of a cyclotron (b) A the poitive particle reache the gap, it i accelerated by the electric-field force... D 1 (c)... and the next emicircular orbit ha a larger radiu. D 1 Dee High-frequency alternating voltage B S Dee + B S D F S E S + F S By the time the particle reache the gap again, the dee voltage D ha revered and the particle i again accelerated. B S E S The Cyclotron Many accelerator ue magnet to deflect the charged particle into circular path. The firt of thee wa the cyclotron, invented in 1931 by E. O. Lawrence and M. Stanley Livington at the Univerity of California (Fig. 44.6a). Particle with ma m and charge q move inide a vacuum chamber in a uniform magnetic field B S that i perpendicular to the plane of their path. In Section 7.4 we howed that in uch a field, a particle with peed v move in a circular path with radiu r given by r = mv ƒqƒb and with angular peed (angular frequency) v given by v = v r = ƒqƒb m (44.6) (44.7) An alternating potential difference i applied between the two hollow electrode D 1 and D (called dee), creating an electric field in the gap between them. The polarity of the potential difference and electric field i changed preciely twice each revolution (Fig. 44.6b and 44.6c), o that the particle get a puh each time they cro the gap. The puhe increae their peed and kinetic energy, booting them into path of larger radiu. The maximum peed v max and kinetic energy K max are determined by the radiu R of the larget poible path. Solving Eq. (44.6) for v, we find v = ƒqƒbr>m and v max = ƒqƒbr>m. Auming nonrelativitic peed, we have K max = 1 mv max = q B R m (44.8) Example 44.1 Frequency and energy in a proton cyclotron One cyclotron built during the 1930 ha a path of maximum radiu m and a magnetic field of magnitude 1.50 T. If it i ued to accelerate proton, find (a) the frequency of the alternating voltage applied to the dee and (b) the maximum particle energy. SOLUTION IDENTIFY and SET UP: The frequency ƒ of the applied voltage mut equal the frequency of the proton orbital motion. Equation (44.7) give the angular frequency v of the proton orbital motion; we find ƒ uing ƒ = v>p. The proton reache it maximum energy K max, given by Eq. (44.8), when the radiu of it orbit equal the radiu of the dee. EXECUTE: (a) For proton, q = 1.60 * C and m = 1.67 * 10-7 kg. From Eq. (44.7), f = v p = ƒ q ƒb pm * C11.50 T = p11.67 * 10-7 kg =.3 * 10 7 Hz = 3 MHz

167 44. Particle Accelerator and Detector 1487 (b) From Eq. (44.8) the maximum kinetic energy i K max = * C T m * 10-7 kg = 4.3 * 10-1 J =.7 * 10 7 ev = 7 MeV Thi proton kinetic energy i much larger than that available from natural radioactive ource. EVALUATE: From Eq. (44.6) or Eq. (44.7), the proton peed i v = 7. * 10 7 m>, which i about 5% of the peed of light. At uch peed, relativitic effect are beginning to become important. Since we ignored thee effect in our calculation, the above reult for ƒ and K max are in error by a few percent; thi i why we kept only two ignificant figure. The maximum energy that can be attained with a cyclotron i limited by relativitic effect. The relativitic verion of Eq. (44.7) i v = ƒqƒb m 1 - v >c A the particle peed up, their angular frequency v decreae, and their motion get out of phae with the alternating dee voltage. In the ynchrocyclotron the particle are accelerated in burt. For each burt, the frequency of the alternating voltage i decreaed a the particle peed up, maintaining the correct phae relationhip with the particle motion. Another limitation of the cyclotron i the difficulty of building very large electromagnet. The larget ynchrocyclotron ever built ha a vacuum chamber that i about 8 m in diameter and accelerate proton to energie of about 600 MeV. The Synchrotron To attain higher energie, another type of machine, called the ynchrotron, i more practical. Particle move in a vacuum chamber in the form of a thin doughnut, called the accelerating ring. The particle beam i bent to follow the ring by a erie of electromagnet placed around the ring. A the particle peed up, the magnetic field i increaed o that the particle retrace the ame trajectory over and over. The Large Hadron Collider (LHC) near Geneva, Switzerland, i the highet-energy accelerator in the world (Fig. 44.7). It i deigned to accelerate proton to a maximum energy of 7 TeV, or ev. (A we ll dicu in Section 44.3, hadron are a cla of elementary particle that include proton and neutron.) A we pointed out in Section 3.1, accelerated charge radiate electromagnetic energy. In an accelerator in which the particle move in curved path, thi radiation i often called ynchrotron radiation. High-energy accelerator are typically contructed underground to provide protection from thi radiation. From the accelerator tandpoint, ynchrotron radiation i undeirable, ince the energy given to an accelerated particle i radiated right back out. It can be minimized by making the accelerator radiu r large o that the centripetal acceleration v >r i mall. On the poitive ide, ynchrotron radiation i ued a a ource of wellcontrolled high-frequency electromagnetic wave. Available Energy When a beam of high-energy particle collide with a tationary target, not all the kinetic energy of the incident particle i available to form new particle tate. Becaue momentum mut be conerved, the particle emerging from the colliion mut have ome net motion and thu ome kinetic energy. The dicuion following Example (Section 43.6) preented a nonrelativitic example of thi principle. The maximum available energy i the kinetic energy in the frame of reference in which the total momentum i zero. We call thi the center-of-momentum ytem; it i the relativitic generalization of the center-of-ma ytem that we dicued in Section 8.5. In thi ytem the total kinetic energy after the colliion can be zero, o that the maximum amount of the initial kinetic energy become available to caue the reaction being tudied (a) The Large Hadron Collider at the European Organization for Nuclear Reearch (CERN). The underground accelerating ring (hown by the red circle) i 100 m underground and 8.5 km in diameter, o large that it pan the border between Switzerland and France. (Note the Alp in the background.) When accelerated to 7 TeV, proton travel around the ring more than 11,000 time per econd. (b) An engineer working on one of the 9593 uperconducting electromagnet around the LHC ring. (a) (b)

168 1488 CHAPTER 44 Particle Phyic and Comology Conider the laboratory ytem, in which a target particle with ma M i initially at ret and i bombarded by a particle with ma m and total energy (including ret energy) E m. The total available energy E a in the center-of-momentum ytem (including ret energie of all the particle) can be hown to be given by E a = Mc E m + 1Mc + 1mc 1available energy (44.9) When the mae of the target and projectile particle are equal, thi can be implified to E a = mc 1E m + mc 1available energy, equal mae (44.10) If in addition E i much greater than mc m, we can neglect the econd term in the parenthee in Eq (44.10). Then E a i E a = mc E m 1available energy, equal mae, E m W mc (44.11) The quare root in Eq. (44.11) i a diappointing reult for an accelerator deigner: Doubling the energy E m of the bombarding particle increae the available energy E a by only a factor of 1 = Example 44. and 44.3 explore the limitation of having a tationary target particle. Example 44. Threhold energy for pion production A proton (ret energy 938 MeV) with kinetic energy K collide with a proton at ret. Both proton urvive the colliion, and a neutral pion (p 0, ret energy 135 MeV) i produced. What i the threhold energy (minimum value of K) for thi proce? SOLUTION IDENTIFY and SET UP: The final tate include the two original proton (ma m) and the pion (ma m p ). The threhold energy correpond to the minimum-energy cae in which all three particle are at ret in the center-of-momentum ytem. The total available energy E a in that ytem mut be at leat the total ret energy, mc + m p c. We ue thi to olve Eq. (44.10) for the total energy E m of the bombarding proton; the kinetic energy K (our target variable) i then E minu the proton ret energy mc m. EXECUTE: We ubtitute E a = mc + m p c into Eq. (44.10), implify, and olve for E m : 4m c 4 + 4mm p c 4 + m p c 4 = mc E m + 1mc E m = mc + m p c a + m p m b = mc + K K = m p c a + m p m b We ee that the bombarding proton kinetic energy K mut be omewhat greater than twice the pion ret energy m p c. With mc = 938 MeV and m p c = 135 MeV, we have m p >m = 0.07 and K = 1135 MeV = 80 MeV EVALUATE: Compare thi reult with the reult of Example (Section 37.8), where we found that a pion can be produced in a head-on colliion of two proton, each with only 67.5 MeV of kinetic energy. We dicu the energy advantage of uch colliion in the next ubection. Example 44.3 Increaing the available energy The Fermilab accelerator in Illinoi wa deigned to bombard tationary target with 800-GeV proton. (a) What i the available energy E a in a proton-proton colliion? (b) What i E a if the beam energy i increaed to 980 GeV? SOLUTION IDENTIFY and SET UP: Our target variable i the available energy E a in a tationary-target colliion between identical particle. In both part (a) and (b) the beam energy E m i much larger than the proton ret energy mc = 938 MeV = GeV, o we can afely ue the approximation of Eq. (44.11). EXECUTE: (a) For E m = 800 GeV, Eq. (44.11) give E a = GeV1800 GeV = 38.7 GeV (b) For E m = 980 GeV, E a = GeV1980 GeV = 4.9 GeV EVALUATE: With a tationary-proton target, increaing the proton beam energy by 180 GeV increae the available energy by only 4. GeV! Thi how a major limitation of experiment in which one of the colliding particle i initially at ret. Below we decribe how phyicit can overcome thi limitation.

169 44. Particle Accelerator and Detector 1489 Colliding Beam The limitation illutrated by Example 44.3 i circumvented in colliding-beam experiment. In thee experiment there i no tationary target; intead, beam of particle moving in oppoite direction are tightly focued onto one another o that head-on colliion can occur. Uually the two colliding particle have momenta of equal magnitude and oppoite direction, o the total momentum i zero. Hence the laboratory ytem i alo the center-of-momentum ytem, and the available energy i maximized. The highet-energy colliding beam available are thoe at the Large Hadron Collider (ee Fig. 44.7). In operation, 808 bunche of 7-TeV proton circulate around the ring, half in one direction and half in the oppoite direction. Each bunch contain about proton. Magnet teer the oppoitely moving bunche to collide at interaction point. The available energy E a in the reulting head-on colliion i the total energy of the two colliding particle: E a = * 7 TeV = 14 TeV. (Strictly, E a i 14 TeV minu the ret energy of the two colliding proton. But thi ret energy i only mc = 1938 MeV = * 10-3 TeV, which i o mall compared to 14 TeV that it can be ignored.) Phyicit expect that the very large available energy at the Large Hadron Collider will make it poible to produce particle that have never been een before. Detector Ordinarily, we can t ee or feel individual ubatomic particle or photon. How, then, do we meaure their propertie? A wide variety of device have been deigned. Many detector ue the ionization caued by charged particle a they move through a ga, liquid, or olid. The ion along the particle path act a nucleation center for droplet of liquid in the uperaturated vapor of a cloud chamber (Fig. 44.1) or caue mall volume of vapor in the uperheated liquid of a bubble chamber (Fig. 44.3a). In a emiconducting olid the ionization can take the form of electron hole pair. We dicued their detection in Section 4.7. Wire chamber contain array of cloely paced wire that detect the ion. The charge collected and time information from each wire are proceed uing computer to recontruct the particle trajectorie. The detector at the Large Hadron Collider ue an array of device to follow the track of particle produced by colliion between proton (Fig. 44.8). The giant olenoid in the photo that open Chapter 8 i at the heart of one thee detector array. The intene magnetic field of the olenoid help identify newly produced particle, which curve in different direction and along path of different radii depending on their charge and energy Thi computer-generated image how the reult of a imulated colliion between two proton (not hown) in one of the interaction region at the Large Hadron Collider. The view i along the beampipe. The different color track how different type of particle emerging from the colliion. A variety of different detector urround the colliion region. (Note the drawing of woman in a red dre, hown for cale.) Comic-Ray Experiment Large number of particle called comic ray continually bombard the earth from ource both within and beyond our galaxy. Thee particle conit motly of neutrino, proton, and heavier nuclei, with energie ranging from le than 1 MeV to more than 10 0 ev. The earth atmophere and magnetic field protect u from much of thi radiation. Thi mean that comic-ray experimentation often mut be carried out above all or mot of the atmophere by mean of rocket or high-altitude balloon. In contrat, neutrino detector are buried below the earth urface in tunnel or mine or ubmerged deep in the ocean. Thi i done to creen out all other type of particle o that only neutrino, which interact only very weakly with matter, reach the detector. It would take a light-year thickne of lead to aborb a izable fraction of a beam of neutrino. Thu neutrino detector conit of huge amount of matter: The Super-Kamiokande detector look for flahe of light produced when a neutrino interact in a tank containing 5 * 10 7 kg of water (ee Section 44.5).

170 1490 CHAPTER 44 Particle Phyic and Comology Comic ray were important in early particle phyic, and their tudy currently bring u important information about the ret of the univere. Although comic ray provide a ource of high-energy particle that doe not depend on expenive accelerator, mot particle phyicit ue accelerator becaue the high-energy comic-ray particle they want are too few and too random. Tet Your Undertanding of Section 44. In a colliding-beam experiment, a 90-GeV electron collide head-on with a 90-GeV poitron. The electron and the poitron annihilate each other, forming a ingle virtual photon that then tranform into other particle. Doe the virtual photon obey the ame relationhip E = pc a real photon do? 44.3 Particle and Interaction We have mentioned the array of ubatomic particle that were known a of 1947: photon, electron, poitron, proton, neutron, muon, and pion. Since then, literally hundred of additional particle have been dicovered in accelerator experiment. The vat majority of known particle are untable and decay pontaneouly into other particle. Particle of all kind, whether table or untable, can be created or detroyed in interaction between particle. Each uch interaction involve the exchange of virtual particle, which exit on borrowed energy allowed by the uncertainty principle. Although the world of ubatomic particle and their interaction i complex, ome key reult bring order and implicity to the eeming chao. One key implification i that there are only four fundamental type of interaction, each mediated or tranmitted by the exchange of certain characteritic virtual particle. Furthermore, not all particle repond to all four kind of interaction. In thi ection we will examine the fundamental interaction more cloely and ee how phyicit claify particle in term of the way in which they interact The tie that bind u together originate in the fundamental interaction of nature. The nuclei within our bodie are held together by the trong interaction. The electromagnetic interaction bind nuclei and electron together to form atom, bind atom together to form molecule, and bind molecule together to form u. Four Force and Their Mediating Particle In Section 5.5 we firt decribed the four fundamental type of force or interaction (Fig. 44.9). They are, in order of decreaing trength: 1. The trong interaction. The electromagnetic interaction 3. The weak interaction 4. The gravitational interaction The electromagnetic and gravitational interaction are familiar from claical phyic. Both are characterized by a 1>r dependence on ditance. In thi cheme, the mediating particle for both interaction have ma zero and are table a ordinary particle. The mediating particle for the electromagnetic interaction i the familiar photon, which ha pin 1. (That mean it pin quantum number i = 1, o the magnitude of it pin angular momentum i S = 1 + 1U = U. That particle for the gravitational force i the pin- graviton 1 =, S = 1 + 1U =6U. The graviton ha not yet been oberved experimentally becaue the gravitational force i very much weaker than the electromagnetic force. For example, the gravitational attraction of two proton i maller than their electrical repulion by a factor of about The gravitational force i of primary importance in the tructure of tar and the large-cale behavior of the univere, but it i not believed to play a ignificant role in particle interaction at the energie that are currently attainable. The other two force are le familiar. One, uually called the trong interaction, i reponible for the nuclear force and alo for the production of pion and everal other particle in high-energy colliion. At the mot fundamental level, the mediating particle for the trong interaction i called a gluon. However, the force between nucleon i more eaily decribed in term of meon a the mediating particle. We ll dicu the pin-1, male gluon in Section 44.4.

171 44.3 Particle and Interaction 1491 Equation (44.4) i a poible potential-energy function for the nuclear force. The trength of the interaction i decribed by the contant ƒ, which ha unit of energy time ditance. A better bai for comparion with other force i the dimenionle ratio ƒ >Uc, called the coupling contant for the interaction. (We invite you to verify that thi ratio i a pure number and o mut have the ame value in all ytem of unit.) The oberved behavior of nuclear force ugget that ƒ >Uc L 1. The dimenionle coupling contant for electromagnetic interaction i 1 e 4pP 0 Uc = 7.97 * 10-3 = (44.1) Thu the trong interaction i roughly 100 time a trong a the electromagnetic interaction; however, it drop off with ditance more quickly than 1>r. The fourth interaction i called the weak interaction. It i reponible for beta decay, uch a the converion of a neutron into a proton, an electron, and an antineutrino. It i alo reponible for the decay of many untable particle (pion into muon, muon into electron, and o on). It mediating particle are the hort-lived particle W +, W -, and Z 0. The exitence of thee particle wa confirmed in 1983 in experiment at CERN, for which Carlo Rubbia and Simon van der Meer were awarded the Nobel Prize in The W and Z 0 have pin 1 like the photon and the gluon, but they are not male. In fact, they have enormou mae, 80.4 GeV>c for the W and 91. GeV>c for the Z 0. With uch maive mediating particle the weak interaction ha a much horter range than the trong interaction. It alo live up to it name by being weaker than the trong interaction by a factor of about Table 44.1 compare the main feature of thee four fundamental interaction. ActivPhyic 19.5: Particle Phyic More Particle In Section 44.1 we mentioned the dicoverie of muon in 1937 and of pion in The electric charge of the muon and the charged pion have the ame magnitude e a the electron charge. The poitive muon m + i the antiparticle of the negative muon m- 1. Each ha pin like the electron, and a ma of about 07m e = 106 MeV>c,. Muon are untable; each decay with a lifetime of. * 10-6 into an electron of the ame ign, a neutrino, and an antineutrino. There are three kind of pion, all with pin 0; they have no pin angular momentum. The p + and p - have mae of 73m e = 140 MeV>c. They are untable; each p decay with a lifetime of.6 * 10-8 into a muon of the ame ign along with a neutrino for the p + and an antineutrino for the p -. The p 0 i omewhat le maive, 64m e = 135 MeV>c, and it decay with a lifetime of 8.4 * into two photon. The p + and p - are antiparticle of one another, while the p 0 i it own antiparticle. (That i, there i no ditinction between particle and antiparticle for the p 0. The exitence of the antiproton p had been upected ever ince the dicovery of the poitron. The p wa found in 1955, when proton antiproton 1pp pair were created by ue of a beam of 6-GeV proton from the Bevatron at the Univerity of Table 44.1 Four Fundamental Interaction Mediating Particle Relative Interaction Strength Range Name Ma Charge Spin Strong 1 Short Gluon ' 1 fm Electromagnetic Long Photon Weak 10-9 Short W, Z , 91. e, >r 1 ' fm GeV>c Gravitational Long Graviton >r

172 149 CHAPTER 44 Particle Phyic and Comology California, Berkeley. The antineutron n wa found oon afterward. After 1960, a higher-energy accelerator and more ophiticated detector were developed, a veritable blizzard of new untable particle were identified. To decribe and claify them, we need a mall blizzard of new term. Initially, particle were claified by ma into three categorie: (1) lepton ( light one uch a electron); () meon ( intermediate one uch a pion); and (3) baryon ( heavy one uch a nucleon and more maive particle). But thi cheme ha been upereded by a more ueful one in which particle are claified in term of their interaction. For intance, hadron (which include meon and baryon) have trong interaction, and lepton do not. In the following dicuion we will alo ditinguih between fermion, which have half-integer pin, and boon, which have zero or integer pin. Fermion obey the excluion principle, on which the Fermi-Dirac ditribution function (ee Section 4.5) i baed. Boon do not obey the excluion principle and have a different ditribution function, the Boe-Eintein ditribution. Lepton The lepton, which do not have trong interaction, include ix particle; the electron 1e - and it neutrino 1n the muon 1m - e, and it neutrino 1n m, and the tau particle 1t - and it neutrino 1n t. Each of the ix particle ha a ditinct 1 antiparticle. All lepton have pin and thu are fermion. The family of lepton i hown in Table 44.. The tau have ma 3478m e = 1777 MeV>c. Tau and muon are untable; a t - decay into a m - plu a tau neutrino and a muon antineutrino, or an electron plu a tau neutrino and an electron antineutrino. A m - decay into a electron plu a muon neutrino and an electron antineutrino. They have relatively long lifetime becaue their decay are mediated by the weak interaction. Depite their zero charge, a neutrino i ditinct from an antineutrino; the pin angular momentum of a neutrino ha a component that i oppoite it linear momentum, while for an antineutrino that component i parallel to it linear momentum. Becaue neutrino are o eluive, phyicit have only been able to place upper limit on the ret mae of the n e, the n m, and the n t. Until recently, it wa thought that the ret mae of the neutrino were zero; compelling evidence now indicate that they have mall but nonzero mae. We ll return to thi point and it implication later. Lepton obey a conervation principle. Correponding to the three pair of lepton are three lepton number L and The electron e - e, L m, L t. and the electron neutrino n are aigned L and their antiparticle e + e e = 1, and n e are given L e = -1. Correponding aignment of L m and L t are made for the m and t particle and their neutrino. In all interaction, each lepton number i eparately conerved. For example, in the decay of the m -, the lepton number are S + n e + n m L m = 1 L e = 1 L e = -1 L m = 1 m - e - Thee conervation principle have no counterpart in claical phyic. Table 44. The Six Lepton Principal Particle Anti- Ma Lifetime Decay Name Symbol particle 1MeV/c L e L M L T () Mode Electron e - e Stable Electron neutrino n e n e 63 * Stable Muon m - m * 10-6 e - n e n m Muon neutrino n m n m Stable Tau t - t * m - n m n t Tau neutrino n t n t Stable or e - n e n t

173 44.3 Particle and Interaction 1493 Example 44.4 Lepton number conervation Check conervation of lepton number for thee decay cheme: (a) m + S e + + n e + n m (b) p - S m - + n m (c) p 0 S m - + e + + n e SOLUTION IDENTIFY and SET UP: Lepton number conervation require that L e, L m, and L t (given in Table 44.) eparately have the ame um after the decay a before. L e EXECUTE: We tabulate and L m for each decay cheme. An antiparticle ha the oppoite lepton number from it correponding particle lited in Table 44.. No t particle or t neutrino appear in any of the cheme, o L t = 0 both before and after each decay and L t i conerved. (a) m + S e + + n e + n m L e : 0 = L m : -1 = (b) p - S m - + n m L e : 0 = L m : 0 = (c) p 0 S m - + e + + n e L e : 0 = L m : 0 Z EVALUATE: Decay (a) and (b) are conitent with lepton number conervation and are oberved. Decay (c) violate the conervation of L m and ha never been oberved. Phyicit ued thee and other experimental reult to deduce the principle that all three lepton number mut eparately be conerved. Hadron Hadron, the trongly interacting particle, are a more complex family than lepton. Each hadron ha an antiparticle, often denoted with an overbar, a with the antiproton p. There are two ubclae of hadron: meon and baryon. Table 44.3 how ome of the many hadron that are currently known. (We ll explain trangene and quark content later in thi ection and in the next one.) Meon include the pion that have already been mentioned, K meon or kaon, h meon, and other that we will dicu later. Meon have pin 0 or 1 and therefore are all boon. There are no table meon; all can and do decay to le maive particle, obeying all the conervation law for uch decay. Table 44.3 Some Hadron and Their Propertie Charge Baryon Typical Ma Ratio, Number, Strangene, Mean Decay Quark Particle 1MeV/c Q/e Spin B S Lifetime () Mode Content Meon p * gg uu, dd p * 10-8 m + n m ud p * 10-8 m - n m ud K * 10-8 m + n m u K * 10-8 m - n m u h L10-18 gg uu, dd, Baryon p Stable uud * pp - or 0 np ud n pe n e udd * pp 0 or np uu * 10-0 g ud * np dd * p 0 - u * p - d L10-3 pp uuu Æ - + c * K * pk - p + udc

174 1494 CHAPTER 44 Particle Phyic and Comology Baryon include the nucleon and everal particle called hyperon, including the,,, and Æ. Thee reemble the nucleon but are more maive. Baryon have half-integer pin, and therefore all are fermion. The only table baryon i the proton; a free neutron decay to a proton, and the hyperon decay to other hyperon or to nucleon by variou procee. Baryon obey the principle of conervation of baryon number, analogou to conervation of lepton number, again with no counterpart in claical phyic. We aign a baryon number B = 1 to each baryon (p, n,,, and o on) and B = -1 to each antibaryon 1p, n,,, and o on). In all interaction, the total baryon number i conerved. Thi principle i the reaon the ma number nuclear reaction that we tudied in Chapter 43. A wa conerved in all of the Example 44.5 Baryon number conervation Check conervation of baryon number for thee reaction: (a) n + p S n + p + p + p (b) n + p S n + p + n SOLUTION IDENTIFY and SET UP: Thi example i imilar to Example We compare the total baryon number before and after each reaction, uing data from Table EXECUTE: We tabulate the baryon number, noting that a baryon ha B = 1 and an antibaryon ha B = -1: (a) n + p S n + p + p + p: = (b) n + p S n + p + n: Z EVALUATE: Reaction (a) i conitent with baryon number conervation. It can occur if enough energy i available in the n + p colliion. Reaction (b) violate baryon number conervation and ha never been oberved. Example 44.6 Antiproton creation What i the minimum proton energy required to produce an antiproton in a colliion with a tationary proton? SOLUTION IDENTIFY and SET UP: The reaction mut conerve baryon number, charge, and energy. Since the target and bombarding proton are of equal ma and the target i at ret, we determine the minimum energy E m of the bombarding proton uing Eq. (44.10). EXECUTE: Conervation of charge and conervation of baryon number forbid the creation of an antiproton by itelf; it mut be created a part of a proton antiproton pair. The complete reaction i p + p S p + p + p + p For thi reaction to occur, the minimum available energy E a in Eq. (44.10) i the final ret energy 4mc of three proton and an antiproton. Equation (44.10) then give 14mc = mc 1E m + mc E m = 7mc EVALUATE: The energy E m of the bombarding proton include it ret energy mc, o it minimum kinetic energy mut be 6mc = MeV = 5.63 GeV. The earch for the antiproton wa a principal reaon for the contruction of the Bevatron at the Univerity of California, Berkeley, with beam energy of 6 GeV. The earch ucceeded in 1955, and Emilio Segrè and Owen Chamberlain were later awarded the Nobel Prize for thi dicovery. Strangene The K meon and the and hyperon were dicovered during the late Becaue of their unuual behavior they were called trange particle. They were produced in high-energy colliion uch a p - + p, and a K meon and a hyperon were alway produced together. The relatively high rate of production of thee particle uggeted that it wa a trong-interaction proce, but their relatively long lifetime uggeted that their decay wa a weak-interaction proce. The K 0 appeared to have two lifetime, one about 9 * and another nearly 600 time longer. Were the K meon trongly interacting hadron or not?

175 44.3 Particle and Interaction 1495 The earch for the anwer to thi quetion led phyicit to introduce a new quantity called trangene. The hyperon 0 and,0 were aigned a trangene quantum number S = -1, and the aociated K 0 and K + meon were aigned S = +1. The correponding antiparticle had oppoite trangene, S = +1 for 0 and,0 and S = -1 for K 0 and K -. Then trangene wa conerved in production procee uch a The proce p + p - S - + K + p + p - S 0 + K 0 p + p - S p + K - doe not conerve trangene and it doe not occur. When trange particle decay individually, trangene i uually not conerved. Typical procee include + S n + p + 0 S p + p - K - S p + + p - + p - In each of thee decay, the initial trangene i 1 or -1, and the final value i zero. All obervation of thee particle are conitent with the concluion that trangene i conerved in trong interaction but it can change by zero or one unit in weak interaction. There i no counterpart to the trangene quantum number in claical phyic. CAUTION Strangene v. pin Take care not to confue the ymbol S for trangene with the identical ymbol for the magnitude of the pin angular momentum. Conervation Law The decay of trange particle provide our firt example of a conditional conervation law, one that i obeyed in ome interaction and not in other. By contrat, everal conervation law are obeyed in all interaction. Thee include the familiar conervation law; energy, momentum, angular momentum, and electric charge. Thee are called abolute conervation law. Baryon number and the three lepton number are alo conerved in all interaction. Strangene i conerved in trong and electromagnetic interaction but not in all weak interaction. Two other quantitie, which are conerved in ome but not all interaction, are ueful in claifying particle and their interaction. One i iopin, a quantity that i ued to decribe the charge independence of the trong interaction. The other i parity, which decribe the comparative behavior of two ytem that are mirror image of each other. Iopin i conerved in trong interaction, which are charge independent, but not in electromagnetic or weak interaction. (The electromagnetic interaction i certainly not charge independent.) Parity i conerved in trong and electromagnetic interaction but not in weak one. The Chinee-American phyicit T. D. Lee and C. N. Yang received the Nobel Prize in 1957 for laying the theoretical foundation for nonconervation of parity in weak interaction. Thi dicuion how that conervation law provide another bai for claifying particle and their interaction. Each conervation law i alo aociated with a ymmetry property of the ytem. A familiar example i angular momentum. If a ytem i in an environment that ha pherical ymmetry, there can be no torque acting on it becaue the direction of the torque would violate the ymmetry. In uch a ytem, total angular momentum i conerved. When a conervation law i violated, the interaction i often decribed a a ymmetry-breaking interaction.

176 1496 CHAPTER 44 Particle Phyic and Comology Tet Your Undertanding of Section 44.3 From conervation of energy, a particle of ma m and ret energy mc can decay only if the decay product have a total ma le than m. (The remaining energy goe into the kinetic energy of the decay product.) Can a proton decay into le maive meon? 44.4 Quark and the Eightfold Way The lepton form a fairly neat package: three particle and three neutrino, each with it antiparticle, and a conervation law relating their number. Phyicit believe that lepton are genuinely fundamental particle. The hadron family, by comparion, i a me. Table 44.3 contain only a ample of well over 100 hadron that have been dicovered ince 1960, and it ha become clear that thee particle do not repreent the mot fundamental level of the tructure of matter. Our preent undertanding of the tructure of hadron i baed on a propoal made initially in 1964 by the American phyicit Murray Gell-Mann and hi collaborator. In thi propoal, hadron are not fundamental particle but are compoite tructure whoe contituent are pin- 1 fermion called quark. (The name i found in the line Three quark for Muter Mark! from Finnegan Wake by Jame Joyce.) Each baryon i compoed of three quark 1qqq, each antibaryon of three antiquark 1qqq, and each meon of a quark antiquark pair 1qq. Table 44.3 of the preceding ection give the quark content of many hadron. No other compoition eem to be neceary. Thi cheme require that 1 quark have electric charge with magnitude 3 and 3 of the electron charge e, which had previouly been thought to be the mallet unit of charge. Each quark 1 alo ha a fractional value for it baryon number and each antiquark ha a baryon-number value B, In a meon, a quark and antiquark combine with net baryon number 0 and can have their pin angular momentum component parallel to form a pin-1 meon or antiparallel to form a pin-0 meon. Similarly, the three quark in a baryon combine with net baryon number 1 and can form a pin- 1 baryon or a pin- 3 baryon Quark content of four different hadron. The variou color combination that are needed for color neutrality are not hown. u 1 e 3 u u d 1 e 1 1 e e d 1 d e 1 3 e 3 Proton (p) Neutron (n) The Three Original Quark The firt (1964) quark theory included three type (called flavor) of quark, labeled u (up), d (down), and (trange). Their principal propertie are lited in Table The correponding antiquark u, d, and have oppoite value of charge Q, B, and S. Proton, neutron, p and K meon, and everal hyperon can be contructed from thee three quark. For example, the proton quark content i uud. Checking Table 44.4, we ee that the value of Q>e add to 1 and that the value of the baryon number B alo add to 1, a we hould expect. The neutron i udd, with total Q = 0 and B = 1. The p + meon i ud, with Q>e = 1 and B = 0, and the K + meon i u. Checking the value of S for the quark content, we ee that the proton, neutron, and p + have trangene 0 and that the K + ha trangene 1, in agreement with Table The antiproton i p = uud, the negative pion i p - = ud, and o on. The quark content can alo be ued to explain hadron excited tate and magnetic moment. Figure how the quark content of two baryon and two meon. u d e 1 e 3 3 Poitive pion (p 1 ) u e 1 e 3 3 Poitive kaon (K 1 ) Table 44.4 Propertie of the Three Original Quark Baryon Strange- Bottom- Topne, Symbol Q/e Spin Number, B ne, S Charm, C ne, B œ T u d

177 44.4 Quark and the Eightfold Way 1497 Example 44.7 Determining the quark content of baryon, Given that they contain only u, d,, u, d, and> or find the quark (a) The + ha Q>e = +1, o the other two quark mut both content of (a) + and (b) 0. The + and 0 (the antiparticle of be u quark (each of which ha Q>e = + ). Hence the quark content of + 3 the 0 ) are both baryon with trangene S = -1. i uu. (b) Firt we find the quark content of the 0. To yield zero total SOLUTION charge, the other two quark mut be u 1Q>e = + d 1Q>e = - 1, o the quark content of the IDENTIFY and SET UP: We ue the idea that the total charge of 3 3 and i ud. The quark content of the each baryon i the um of the individual quark charge, and imilarly for the baryon number and trangene. We ue the quark EVALUATE: Although the 0 and 0 are both electrically neutral i therefore u d. propertie given in Table EXECUTE: Baryon contain three quark. If S = -1, exactly one of and have the ame ma, they are different particle: 0 ha B = 1 and S = -1, while 0 ha B = -1 and S = 1. the three mut be an quark, which ha S = -1 and Q>e = Motivating the Quark Model What caued phyicit to upect that hadron were made up of omething maller? The magnetic moment of the neutron (ee Section 43.1) wa one of the firt reaon. In Section 7.7 we learned that a magnetic moment reult from a circulating current (a motion of electric charge). But the neutron ha no charge, or, to be more accurate, no total charge. It could be made up of maller particle whoe charge add to zero. The quantum motion of thee particle within the neutron would then give it urpriing nonzero magnetic moment. To verify thi hypothei by eeing inide a neutron, we need a probe with a wavelength that i much le than the neutron ize of about a femtometer. Thi probe hould not be affected by the trong interaction, o that it won t interact with the neutron a a whole but will penetrate into it and interact electromagnetically with thee uppoed maller charged particle. A probe with thee propertie i an electron with energy above 10 GeV. In experiment carried out at SLAC, uch electron were cattered from neutron and proton to help how that nucleon are indeed made up of fractionally charged, pin- 1 pointlike particle. The Eightfold Way Symmetry conideration play a very prominent role in particle theory. Here are two example. Conider the eight pin- 1 baryon we ve mentioned: the familiar p and n; the trange and -,,, ; and the doubly trange 0 and -. For each we plot the value of trangene S veru the value of charge Q in Fig The reult i a hexagonal pattern. A imilar plot for the nine pin-0 meon (ix hown in Table 44.3 plu three other not included in that table) i hown in Fig. 44.1; the particle fall in exactly the ame hexagonal pattern! In each plot, all the particle have mae that are within about 00 MeV>c of the median ma value of that plot, with variation due to difference in quark mae and internal potential energie. (a) n p S L 0 S 0 S 1 S 5 0 S 51 (b) dd udd ud ud uud uu S 5 0 S (a) Plot of S and Q value for pin- 1 baryon, howing the ymmetry pattern of the eightfold way. (b) Quark content of each pin- 1 baryon. The quark content of the 0 and 0 are the ame; the 0 i an excited tate of the 0 and can decay into it by photon emiion. J J 0 S 5 d u S 5 Q 5e Q 5 0 Q 51e Q 5e Q 5 0 Q 51e

178 1498 CHAPTER 44 Particle Phyic and Comology 44.1 (a) Plot of S and Q value for nine pin-0 meon, howing the ymmetry pattern of the eightfold way. Each particle i on the oppoite ide of the hexagon from it antiparticle; each of the three particle in the center i it own antiparticle. (b) Quark content of each pin-0 meon. The particle in the center are different mixture of the three quark antiquark pair hown. (a) K 0 K K 1 h 0 p 0 p p 1 h9 0 K 0 S 511 S 5 0 S 51 (b) ud d u dd uu ud u d S 511 S 5 0 S 51 Q 5 e Q 5 0 Q 5 e Q 5e Q 5 0 Q 51e (a) A pion containing a blue quark and an antiblue antiquark. (b) The blue quark emit a blue antired gluon, changing to a red quark. (c) The gluon i aborbed by the antiblue antiquark, which become an antired antiquark. The pion now conit of a red antired quark antiquark pair. The actual quantum tate of the pion i an equal uperpoition of red antired, green antigreen, and blue antiblue pair. (a) (b) (c) Blue quark Antiblue antiquark Blue antired gluon Red quark Red quark Antiblue antiquark Antired antiquark The ymmetrie that lead to thee and imilar pattern are collectively called the eightfold way. They were dicovered in 1961 by Murray Gell-Mann and independently by Yu val Ne eman. (The name i a lightly irreverent reference to the Noble Eightfold Path, a et of principle for right living in Buddhim.) A imilar pattern for the pin- 3 baryon contain ten particle, arranged in a triangular pattern like pin in a bowling alley. When thi pattern wa firt dicovered, one of the particle wa miing. But Gell-Mann gave it a name anyway 1Æ -, predicted the propertie it hould have, and told experimenter what they hould look for. Three year later, the particle wa found during an experiment at Brookhaven National Laboratory, a pectacular ucce for Gell-Mann theory. The whole erie of event i reminicent of the way in which Mendeleev ued gap in the periodic table of the element to predict propertie of undicovered element and to guide chemit in their earch for thee element. What bind quark to one another? The attractive interaction among quark are mediated by male pin-1 boon called gluon in much the ame way that photon mediate the electromagnetic interaction or that pion mediated the nucleon nucleon force in the old Yukawa theory. Color 1 Quark, having pin, are fermion and o are ubject to the excluion principle. Thi would eem to forbid a baryon having two or three quark with the ame flavor and ame pin component. To avoid thi difficulty, it i aumed that each quark come in three varietie, which are whimically called color. Red, green, and blue are the uual choice. The excluion principle applie eparately to each color. A baryon alway contain one red, one green, and one blue quark, o the baryon itelf ha no net color. Each gluon ha a color anticolor combination (for example, blue antired) that allow it to tranmit color when exchanged, and color i conerved during emiion and aborption of a gluon by a quark. The gluon-exchange proce change the color of the quark in uch a way that there i alway one quark of each color in every baryon. The color of an individual quark change continually a gluon are exchanged. Similar procee occur in meon uch a pion. The quark antiquark pair of meon have canceling color and anticolor (for example, blue and antiblue), o meon alo have no net color. Suppoe a pion initially conit of a blue quark and an antiblue antiquark. The blue quark can become a red quark by emitting a blue antired virtual gluon. The gluon i then aborbed by the antiblue antiquark, converting it to an antired antiquark (Fig ). Color i conerved in each emiion and aborption, but a blue antiblue pair ha become a red antired pair. Such change occur continually, o we have to think of a pion a a uperpoition of three quantum tate: blue antiblue, green antigreen, and red antired. On a larger cale, the trong interaction between nucleon wa decribed in Section 44.3 a due to the exchange of virtual meon. In term of quark and gluon, thee mediating virtual meon are quark antiquark ytem bound together by the exchange of gluon.

179 44.5 The Standard Model and Beyond 1499 The theory of trong interaction i known a quantum chromodynamic (QCD). No one ha been able to iolate an individual quark, and indeed QCD predict that quark are bound in uch a way that it i impoible to obtain a free quark. An impreive body of experimental evidence upport the correctne of the quark model and the idea that quantum chromodynamic i the key to undertanding the trong interaction. Three More Quark Before the tau particle were dicovered, there were four known lepton. Thi fact, together with ome puzzling decay rate, led to the peculation that there might be a fourth quark flavor. Thi quark i labeled c (the charmed quark); it ha Q>e = B = 1 3, 3, S = 0, and a new quantum number charm C = +1. Thi wa confirmed in 1974 by the obervation at both SLAC and the Brookhaven National Laboratory of a meon, now named c, with ma 3097 MeV>c. Thi meon wa found to have everal decay mode, decaying into e + e -, m + m -, or hadron. The mean lifetime wa found to be about Thee reult are conitent with c being a pin-1 cc ytem. Almot immediately after thi, imilar meon of greater ma were oberved and identified a excited tate of the cc ytem. A few year later, individual meon with a nonzero net charm quantum number, D 0 1cu and D + + 1cd, and a charmed baryon, c 1udc, were alo oberved. In 1977 a meon with ma 9460 MeV>c, called upilon 1, wa dicovered at Brookhaven. Becaue it had propertie imilar to c, it wa conjectured that the meon wa really the bound ytem of a new quark, b (the bottom quark), and it antiquark, b. The bottom quark ha the value 1 of a new quantum number B (not to be confued with baryon number B) called bottomne. Excited tate of the were oon oberved, a were the B + 1bu and B 0 1bd meon. With the five flavor of quark 1u, d,, c, and b) and the ix flavor of lepton (e, m, t, n e, n m, and n t it wa an appealing conjecture that nature i ymmetric in it building block and that therefore there hould be a ixth quark. Thi quark, labeled t (top), would have Q>e = B = 1 3, 3, and a new quantum number, T = 1. In 1995, group uing two different detector at Fermilab Tevatron announced the dicovery of the top quark. The group collided 0.9-TeV proton with 0.9-TeV antiproton, but even with 1.8 TeV of available energy, a top antitop 1tt pair wa detected in fewer than two of every colliion! Table 44.5 lit ome propertie of the ix quark. Each ha a correponding antiquark with oppoite value of Q, B, S, C, B, and T. Table 44.5 Propertie of the Six Quark Baryon Strange- Bottom- Topne, Symbol Q/e Spin Number, B ne, S Charm, C ne, B œ T u d c b t Tet Your Undertanding of Section 44.4 I it poible to have a baryon with charge Q = +e and trangene S = -? 44.5 The Standard Model and Beyond The particle and interaction that we ve dicued in thi chapter provide a reaonably comprehenive picture of the fundamental building block of nature. There i enough confidence in the baic correctne of thi picture that it i called the tandard model.

180 1500 CHAPTER 44 Particle Phyic and Comology The tandard model include three familie of particle: (1) the ix lepton, which have no trong interaction; () the ix quark, from which all hadron are made; and (3) the particle that mediate the variou interaction. Thee mediator are gluon for the trong interaction among quark, photon for the electromagnetic interaction, the W and Z 0 particle for the weak interaction, and the graviton for the gravitational interaction. Electroweak Unification Theoretical phyicit have long dreamed of combining all the interaction of nature into a ingle unified theory. A a firt tep, Eintein pent much of hi later life trying to develop a field theory that would unify gravitation and electromagnetim. He wa only partly ucceful. Between 1961 and 1967, Sheldon Glahow, Abdu Salam, and Steven Weinberg developed a theory that unifie the weak and electromagnetic force. One outcome of their electroweak theory i a prediction of the weak-force mediator particle, the Z 0 and W boon, including their mae. The baic idea i that the ma difference between photon (zero ma) and the weak boon 1L100 GeV>c make the electromagnetic and weak interaction behave quite differently at low energie. At ufficiently high energie (well above 100 GeV), however, the ditinction diappear, and the two merge into a ingle interaction. Thi prediction wa verified in 1983 in experiment with proton-antiproton colliion at CERN. The weak boon were found, again with the help provided by the theoretical decription, and their oberved mae agreed with the prediction of the electroweak theory, a wonderful convergence of theory and experiment. The electroweak theory and quantum chromodynamic form the backbone of the tandard model. Glahow, Salam, and Weinberg received the Nobel Prize in A remaining difficulty in the electroweak theory i that photon are male but the weak boon are very maive. To account for the broken ymmetry among thee interaction mediator, a particle called the Higg boon ha been propoed. It ma i expected to be le than 1 TeV>c, but to produce it in the laboratory may require a much greater available energy. The earch for the Higg boon i an important miion of the Large Hadron Collider at CERN. Grand Unified Theorie Perhap at ufficiently high energie the trong interaction and the electroweak interaction have a convergence imilar to that between the electromagnetic and weak interaction. If o, they can be unified to give a comprehenive theory of trong, weak, and electromagnetic interaction. Such cheme, called grand unified theorie (GUT), are till peculative. One intereting feature of ome grand unified theorie i that they predict the decay of the proton (in violation of conervation of baryon number), with an etimated lifetime of more than 10 8 year. (For comparion the age of the univere i known to be 1.37 * year.) With a lifetime of 10 8 year, ix metric ton of proton would be expected to have only one decay per day, o huge amount of material mut be examined. Some of the neutrino detector that we mentioned in Section 44. originally looked for, and failed to find, evidence of proton decay. Neverthele, experimental work continue, with current etimate etting the proton lifetime well over year. Some GUT alo predict the exitence of magnetic monopole, which we mentioned in Chapter 7. At preent there i no confirmed experimental evidence that magnetic monopole exit. In the tandard model, the neutrino have zero ma. Nonzero value are controverial becaue experiment to determine neutrino mae are difficult both to perform and to analyze. In mot GUT the neutrino mut have nonzero mae. If neutrino do have ma, tranition called neutrino ocillation can occur, in which one type of neutrino 1n e,, or n t change into another type. n m

181 44.6 The Expanding Univere 1501 In 1998, cientit uing the Super-Kamiokande neutrino detector in Japan (Fig ) reported the dicovery of ocillation between muon neutrino and tau neutrino. Subequent meaurement at the Sudbury Neutrino Obervatory in Canada have confirmed the exitence of neutrino ocillation. Thi dicovery i evidence for exciting phyic beyond that predicted by the tandard model. The dicovery of neutrino ocillation ha cleared up a long-tanding mytery about the un. Since the 1960, phyicit have been uing enitive detector to look for electron neutrino produced by nuclear fuion reaction in the un core (ee Section 43.8). However, the oberved flux of olar electron neutrino i only one-third of the predicted value. The explanation wa provided in 00 by the Sudbury Neutrino Obervatory, which can detect neutrino of all three flavor. The reult howed that the combined flux of olar neutrino of all flavor i equal to the theoretical prediction for the flux of electron neutrino. The explanation i that the un i indeed producing electron neutrino at the rate predicted by theory, but that two-third of thee electron neutrino are tranformed into muon or tau neutrino during their flight from the un core to a detector on earth Thi photo how the interior of the Super-Kamiokande neutrino detector in Japan. When in operation, the detector i filled with 5 * 10 7 kg of water. A neutrino paing through the detector can produce a faint flah of light, which i detected by the 13,000 photomultiplier tube lining the detector wall. Data from thi detector were the firt to indicate that neutrino have ma. Superymmetric Theorie and TOE The ultimate dream of theorit i to unify all four fundamental interaction, adding gravitation to the trong and electroweak interaction that are included in GUT. Such a unified theory i whimically called a Theory of Everything (TOE). It turn out that an eential ingredient of uch theorie i a pace-time continuum with more than four dimenion. The additional dimenion are rolled up into extremely tiny tructure that we ordinarily do not notice. Depending on the cale of thee tructure, it may be poible for the next generation of particle accelerator to reveal the preence of extra dimenion. Another ingredient of many theorie i uperymmetry, which give every boon and fermion a uperpartner of the other pin type. For example, the propoed uperymmetric partner of the pin- 1 electron i a pin-0 particle called the electron, and that of the pin-1 photon i a pin- 1 photino. A yet, no uperpartner particle have been dicovered, perhap becaue they are too maive to be produced by the preent generation of accelerator. Within a few year, new data from the Large Hadron Collider and other accelerator will help u decide whether thee intriguing theorie have merit. Tet Your Undertanding of Section 44.5 One apect of the tandard model i that a d quark can tranform into a u quark, an electron, and an antineutrino by mean of the weak interaction. If thi happen to a d quark inide a neutron, what kind of particle remain afterward in addition to the electron and antineutrino? (i) a proton; (ii) a - ; (iii) a + ; (iv) a 0 or a 0 ; (v) any of thee The Expanding Univere In the lat two ection of thi chapter we ll explore briefly the connection between the early hitory of the univere and the interaction of fundamental particle. It i remarkable that there are uch cloe tie between phyic on the mallet cale that we ve explored experimentally (the range of the weak interaction, of the order of m and phyic on the larget cale (the univere itelf, of the order of at leat 10 6 m. Gravitational interaction play an eential role in the large-cale behavior of the univere. One of the great achievement of Newtonian mechanic, including the law of gravitation, wa the undertanding it brought to the motion of planet in the olar ytem. Atronomical evidence how that gravitational

182 150 CHAPTER 44 Particle Phyic and Comology (a) The galaxy M101 i a larger verion of the Milky Way galaxy of which our olar ytem i a part. Like all galaxie, M101 i held together by the mutual gravitational attraction of it tar, ga, dut, and other matter, all of which orbit around the galaxy center of ma. M101 i 5 million light-year away. (b) Thi image how part of the Coma cluter, an immene grouping of over 1000 galaxie that lie 300 million light-year from u. The galaxie within the cluter are all in motion. Gravitational force between the member galaxie of thi cluter prevent them from ecaping. (a) (b) force alo dominate in larger ytem uch a galaxie and cluter of galaxie (Fig ). Until early in the 0th century it wa uually aumed that the univere wa tatic; tar might move relative to each other, but there wa not thought to be any overall expanion or contraction. But if everything i initially itting till in the univere, why doen t gravity jut pull it all together into one big clump? Newton himelf recognized the erioune of thi troubling quetion. Meaurement that were begun in 191 by Veto Slipher at Lowell Obervatory in Arizona, and continued in the 190 by Edwin Hubble with the help of Milton Humaon at Mount Wilon in California, indicated that the univere i not tatic. The motion of galaxie relative to the earth can be meaured by oberving the hift in the wavelength of their pectra. For ditant galaxie thee hift are alway toward longer wavelength, o they appear to be receding from u and from each other. Atronomer firt aumed that thee were Doppler hift and ued a relationhip between the wavelength l 0 of light meaured now from a ource receding at peed v and the wavelength l S meaured in the ret frame of the ource when it wa emitted. We can derive thi relationhip by inverting Eq. (37.5) for the Doppler effect, making ubcript change, and uing l = c>ƒ; the reult i l 0 = l SA c + v c - v (44.13) Wavelength from receding ource are alway hifted toward longer wavelength; thi increae in l i called the redhift. We can olve Eq. (44.13) for v; the reult i v = 1l 0>l S - 1 (44.14) 1l 0 >l S + 1 c CAUTION Redhift, not Doppler hift Equation (44.13) and (44.14) are from the pecial theory of relativity and refer to the Doppler effect. A we ll ee, the redhift from ditant galaxie i caued by an effect that i explained by the general theory of relativity and i not a Doppler hift. However, a the ratio v>c and the fractional wavelength change 1l 0 - l S >l S become mall, the general theory equation approach Eq. (44.13) and (44.14), and thoe equation may be ued. Example 44.8 Receion peed of a galaxy The pectral line of variou element are detected in light from a galaxy in the contellation Ura Major. An ultraviolet line from ingly ionized calcium 1l S = 393 nm i oberved at wavelength l 0 = 414 nm, redhifted into the viible portion of the pectrum. At what peed i thi galaxy receding from u? SOLUTION IDENTIFY and SET UP: Thi example ue the relationhip between redhift and receion peed for a ditant galaxy. We can ue the wavelength l S at which the light i emitted and l 0 that we detect on earth in Eq. (44.14) to determine the galaxy receion peed v if the fractional wavelength hift i not too great. EXECUTE: The fractional redhift i l 0 >l S = 1414 nm> 1393 nm = Thi i only a 5.3% increae, o we can ue Eq. (44.14) with reaonable accuracy: v = c = c = 1.55 * 107 m> EVALUATE: The galaxy i receding from the earth at 5.16% of the peed of light. Rather than going through thi calculation, atronomer often jut tate the redhift z = 1l 0 - l S >l S = 1l 0 >l S - 1. Thi galaxy ha redhift z =

183 44.6 The Expanding Univere 1503 The Hubble Law Analyi of redhift from many ditant galaxie led Edwin Hubble to a remarkable concluion: The peed of receion v of a galaxy i proportional to it ditance r from u (Fig ). Thi relationhip i now called the Hubble law; expreed a an equation, v = H 0 r (44.15) where H 0 i an experimental quantity commonly called the Hubble contant, ince at any given time it i contant over pace. Determining H 0 ha been a key goal of the Hubble Space Telecope, which can meaure ditance to galaxie with unprecedented accuracy. The current bet value i.3 * , with an uncertainty of 5%. Atronomical ditance are often meaured in parec (pc); one parec i the ditance at which there i a one-arcecond 11>3600 angular eparation between two object 1.50 * m apart (the average ditance from the earth to the un). A ditance of 1 pc i equal to 3.6 light-year (ly), where 1 ly = 9.46 * 10 1 km i the ditance that light travel in one year. The Hubble contant i then commonly expreed in the mixed unit 1km>>Mpc (kilometer per econd per megaparec), where 1 Mpc = 10 6 pc: H 0 = 1.3 * a 9.46 * 101 km 3.6 ly pc ba 1 ly 1 pc ba106 1 Mpc b = 71 km> Mpc Graph of receion peed veru ditance for everal galaxie. The bet-fit traight line illutrate Hubble law. The lope of the line i the Hubble contant, H 0. v (10 3 km/) Slope 5 H r (megaparec) Example 44.9 Determining ditance with the Hubble law Ue the Hubble law to find the ditance from earth to the galaxy in Ura Major decribed in Example SOLUTION IDENTIFY and SET UP: The Hubble law relate the redhift of a ditant galaxy to it ditance r from earth. We olve Eq. (44.15) for r and ubtitute the receion peed v from Example EXECUTE: Uing H 0 = 71 1km>>Mpc = 7.1 * m>>Mpc, EVALUATE: A ditance of 0 million parec (710 million lightyear) i truly tupendou, but many galaxie lie much farther away. To appreciate the immenity of thi ditance, conider that our farthet-ranging unmanned pacecraft have traveled only about ly from our planet. r = v 1.55 * 10 7 m> = = 0 Mpc H * m>>Mpc =. * 10 8 pc = 7.1 * 10 8 ly = 6.7 * 10 4 m Another apect of Hubble obervation wa that, in all direction, ditant galaxie appeared to be receding from u. There i no particular reaon to think that our galaxy i at the very center of the univere; if we lived in ome other galaxy, every ditant galaxy would till eem to be moving away. That i, at any given time, the univere look more or le the ame, no matter where in the univere we are. Thi important idea i called the comological principle. There are local fluctuation in denity, but on average, the univere look the ame from all location. Thu the Hubble contant i contant in pace although not necearily contant in time, and the law of phyic are the ame everywhere. The Big Bang The Hubble law ugget that at ome time in the pat, all the matter in the univere wa far more concentrated than it i today. It wa then blown apart in an immene exploion called the Big Bang, giving all obervable matter more or le the velocitie that we oberve today. When did thi happen? According to the

184 1504 CHAPTER 44 Particle Phyic and Comology Hubble law, matter at a ditance r away from u i traveling with peed v = H 0 r. The time t needed to travel a ditance r i t = r v = r H 0 r = 1 H 0 = 4.3 * = 1.4 * y By thi hypothei the Big Bang occurred about 14 billion year ago. It aume that all peed are contant after the Big Bang; that i, it neglect any change in the expanion rate due to gravitational attraction or other effect. We ll return to thi point later. For now, however, notice that the age of the earth determined from radioactive dating (ee Section 43.4) i 4.54 billion * 10 9 year. It encouraging that our hypothei tell u that the univere i older than the earth! An inflating balloon a an analogy for an expanding univere. (a) Point (repreenting galaxie) on the urface of a balloon are decribed by their latitude and longitude coordinate. R R R (b) The radiu R of the balloon ha increaed. The coordinate of the point are the ame, but the ditance between them ha increaed. The rate of receion for any two point i proportional to the ditance between them. R Expanding Space The general theory of relativity take a radically different view of the expanion jut decribed. According to thi theory, the increaed wavelength i not caued by a Doppler hift a the univere expand into a previouly empty void. Rather, the increae come from the expanion of pace itelf and everything in intergalactic pace, including the wavelength of light traveling to u from ditant ource. Thi i not an eay concept to grap, and if you haven t encountered it before, it may ound like doubletalk. Here an analogy that may help to develop ome intuition on thi point. Imagine we are all bug crawling around on a horizontal urface. We can t leave the urface, and we can ee in any direction along the urface, but not up or down. We are then living in a two-dimenional world; ome writer have called uch a world flatland. If the urface i a plane, we can locate our poition with two Carteian coordinate 1x, y. If the plane extend indefinitely in both the x- and y-direction, we decribed our pace a having infinite extent, or a being unbounded. No matter how far we go, we never reach an edge or a boundary. An alternative habitat for u bug would be the urface of a phere with radiu R. The pace would till eem infinite in the ene that we could crawl forever and never reach an edge or a boundary. Yet in thi cae the pace i finite or bounded. To decribe the location of a point in thi pace, we could till ue two coordinate: latitude and longitude, or the pherical coordinate u and f hown in Fig Now uppoe the pherical urface i that of a balloon (Fig ). A we inflate the balloon more and more, increaing the radiu R, the coordinate of a point don t change, yet the ditance between any two point get larger and larger. Furthermore, a R increae, the rate of change of ditance between two point (their receion peed) i proportional to their ditance apart. The receion peed i proportional to the ditance, jut a with the Hubble law. For example, the ditance from Pittburgh to Miami i twice a great a the ditance from Pittburgh to Boton. If the earth were to begin to well, Miami would recede from Pittburgh twice a fat a Boton would. We ee that although the quantity R in t one of the two coordinate giving the poition of a point on the balloon urface, it neverthele play an eential role in any dicuion of ditance. It i the radiu of curvature of our two-dimenional pace, and it i alo a varying cale factor that change a thi two-dimenional univere expand. Generalizing thi picture to three dimenion in t o eay. We have to think of our three-dimenional pace a being embedded in a pace with four or more dimenion, jut a we viualized the two-dimenional pherical flatland a being embedded in a three-dimenional Carteian pace. Our real three-pace i not Carteian; to decribe it characteritic in any mall region require at leat one additional parameter, the curvature of pace, which i analogou to the radiu of the phere. In a ene, thi cale factor, which we ll continue to call R, decribe the ize of the univere, jut a the radiu of the phere decribed the ize of our

185 44.6 The Expanding Univere 1505 two-dimenional pherical univere. We ll return later to the quetion of whether the univere i bounded or unbounded. Any length that i meaured in intergalactic pace i proportional to R, o the wavelength of light traveling to u from a ditant galaxy increae along with every other dimenion a the univere expand. That i, l 0 l = R 0 R (44.16) The zero ubcript refer to the value of the wavelength and cale factor now, jut a H 0 i the current value of the Hubble contant. The quantitie l and R without ubcript are the value at any time pat, preent, or future. In the ituation decribed in Example 44.8, we have l 0 = 414 nm and l = l S = 393 nm, o Eq. (44.16) give R 0 >R = That i, the cale factor now 1R 0 i 5.3% larger than it wa 710 million year ago when the light wa emitted from that galaxy in Ura Major. Thi increae of wavelength with time a the cale factor increae in our expanding univere i called the comological redhift. The farther away an object i, the longer it light take to get to u and the greater the change in R and l. The current larget meaured wavelength ratio for galaxie i about 7, meaning that the volume of pace itelf i about 7 3 L 340 time larger than it wa when the light wa emitted. Do not attempt to ubtitute l 0 >l S = 7 into Eq. (44.14) to find the receion peed; that equation i accurate only for mall comological redhift and v V c. The actual value of v depend on the denity of the univere, the value of H 0, and the expanion hitory of the univere. Here a urprie for you: If the ditance from u in the Hubble law i large enough, then the peed of receion will be greater than the peed of light! Thi doe not violate the pecial theory of relativity becaue the receion peed i not caued by the motion of the atronomical object relative to ome coordinate in it region of pace. Rather, we can have v 7 c when two et of coordinate move apart fat enough a pace itelf expand. In other word, there are object whoe coordinate have been moving away from our coordinate o fat that light from them han t had enough time in the entire hitory of the univere to reach u. What we ee i jut the obervable univere; we have no direct evidence about what lie beyond it horizon. CAUTION The univere in t expanding into emptine The balloon hown in Fig i expanding into the empty pace around it. It a common miconception to picture the univere in the ame way a a large but finite collection of galaxie that expanding into unoccupied pace. The reality i quite different! All the accumulated evidence how that our univere i infinite: It ha no edge, o there i nothing outide it and it in t expanding into anything. The expanion of the univere imply mean that the cale factor of the univere i increaing. A good two-dimenional analogy i to think of the univere a a flat, infinitely large rubber heet that tretching and expanding much like the urface of the balloon in Fig In a ene, the infinite univere i imply becoming more infinite! Critical Denity We ve mentioned that the law of gravitation in t conitent with a tatic univere. We need to look at the role of gravity in an expanding univere. Gravitational attraction hould low the initial expanion, but by how much? If thee attraction are trong enough, the univere hould expand more and more lowly, eventually top, and then begin to contract, perhap all the way down to what been called a Big Crunch. On the other hand, if gravitational force are much weaker, they low the expanion only a little, and the univere hould continue to expand forever. The ituation i analogou to the problem of ecape peed of a projectile launched from the earth. We tudied thi problem in Example 13.5 (Section 13.3); now would be an excellent time to review that dicuion. The total energy

186 1506 CHAPTER 44 Particle Phyic and Comology E = K + U when a projectile of ma m and peed v i at a ditance r from the center of the earth (ma m E ) i An imaginary phere of galaxie. The net gravitational force exerted on our galaxy (at the urface of the phere) by the other galaxie i the ame a if all of their ma were concentrated at the center of the phere. (Since the univere i infinite, there alo an infinity of galaxie outide thi phere.) v v Sphere of galaxie, total ma M Radiu R v v Our galaxy, ma m E = 1 mv - Gmm E r If E i poitive, the projectile ha enough kinetic energy to move infinitely far from the earth 1r S q and have ome kinetic energy left over. If E i negative, the kinetic energy K = 1 mv become zero and the projectile top when r = -Gmm E >E. In that cae, no greater value of r i poible, and the projectile can t ecape the earth gravity. We can carry out a imilar analyi for the univere. Whether the univere continue to expand indefinitely hould depend on the average denity of matter. If matter i relatively dene, there i a lot of gravitational attraction to low and eventually top the expanion and make the univere contract again. If not, the expanion hould continue indefinitely. We can derive an expreion for the critical denity r c needed to jut barely top the expanion. Here a calculation baed on Newtonian mechanic; it in t relativitically correct, but it illutrate the idea. Conider a large phere with radiu R, containing many galaxie (Fig ), with total ma M. Suppoe our own galaxy ha ma m and i located at the urface of thi phere. According to the comological principle, the average ditribution of matter within the phere i uniform. The total gravitational force on our galaxy i jut the force due to the ma M inide the phere. The force on our galaxy and potential energy U due to thi pherically ymmetric ditribution are the ame a though m and M were both point, o U = -GmM>R, jut a in Section The net force from all the uniform ditribution of ma outide the phere i zero, o we ll ignore it. The total energy E (kinetic plu potential) for our galaxy i E = 1 mv - GmM R (44.17) If E i poitive, our galaxy ha enough energy to ecape from the gravitational attraction of the ma M inide the phere; in thi cae the univere hould keep expanding forever. If E i negative, our galaxy cannot ecape and the univere hould eventually pull back together. The croover between thee two cae occur when E = 0, o that 1 mv = GmM R (44.18) The total ma M inide the phere i the volume 4pR 3 >3 time the denity r c : M = 4 3 pr3 r c We ll aume that the peed v of our galaxy relative to the center of the phere i given by the Hubble law: v = H 0 R. Subtituting thee expreion for m and v into Eq. (44.18), we get 1 m1h 0R = Gm R A 4 3 pr3 r c B or r c = 3H 0 1critical denity of the univere (44.19) 8pG Thi i the critical denity. If the average denity i le than r c, the univere hould continue to expand indefinitely; if it i greater, the univere hould eventually top expanding and begin to contract. Putting number into Eq. (44.19), we find 31.3 * r c = 8p16.67 * N # m >kg = 9.5 * 10-7 kg>m 3

187 44.6 The Expanding Univere 1507 The ma of a hydrogen atom i 1.67 * 10-7 kg, o thi denity i equivalent to about ix hydrogen atom per cubic meter. Dark Matter, Dark Energy, and the Accelerating Univere Atronomer have made extenive tudie of the average denity of matter in the univere. One way to do o i to count the number of galaxie in a patch of ky. Baed on the ma of an average tar and the number of tar in an average galaxy, thi effort give an etimate of the average denity of luminou matter in the univere that i, matter that emit electromagnetic radiation. (You are made of luminou matter becaue you emit infrared radiation a a conequence of your temperature; ee Section 17.7 and 39.5.) It alo neceary to take into account other luminou matter within a galaxy, including the tenuou ga and dut between the tar. Another technique i to tudy the motion of galaxie within cluter of galaxie (Fig ; ee alo Fig b). The motion are o low that we can t actually ee galaxie changing poition within a cluter. However, obervation how that different galaxie within a cluter have omewhat different redhift, which indicate that the galaxie are moving relative to the center of ma of the cluter. The peed of thee motion are related to the gravitational force exerted on each galaxy by the other member of the cluter, which in turn depend on the total ma of the cluter. By meauring thee peed, atronomer can determine the average denity of all kind of matter within the cluter, whether or not the matter emit electromagnetic radiation. Obervation uing thee and other technique how that the average denity of all matter in the univere i 7.4% of the critical denity, but the average denity of luminou matter i only 4.6% of the critical denity. In other word, mot of the matter in the univere i not luminou: It doe not emit electromagnetic radiation of any kind. At preent, the nature of thi dark matter remain an outtanding mytery. Some propoed candidate for dark matter are WIMP (weakly interacting maive particle, which are hypothetical ubatomic particle far more maive than thoe produced in accelerator experiment) and MACHO (maive compact halo object, which include object uch a black hole that might form halo around galaxie). Whatever the true nature of dark matter, it i by far the dominant form of matter in the univere. For every kilogram of the ordinary matter that ha been our ubject for mot of thi book including electron, proton, atom, molecule, block on inclined plane, planet, and tar there are five kilogram of dark matter. Since the average denity of matter in the univere i le than the critical denity, it might eem fair to conclude that the univere will continue to expand indefinitely, and that gravitational attraction between matter in different part of the univere hould low the expanion down (albeit not enough to top it). One way to tet thi prediction i to examine the redhift of extremely ditant object. When atronomer look at a galaxy 10 9 light-year away, the light they receive ha been in tranit for 10 9 year, o they are eeing 10 9 year into the pat. If the expanion of the univere ha been lowing down, the expanion mut have been more rapid in the ditant pat. Thu we would expect very ditant galaxie to have greater redhift than predicted by the Hubble law, Eq. (44.15). Only ince the 1990 ha it become poible to accurately meaure both the ditance and the redhift of extremely ditant galaxie. The reult have been totally urpriing: Very ditant galaxie actually have maller redhift than predicted by the Hubble law! The implication i that the expanion of the univere wa lower in the pat than it i now, o the expanion ha been peeding up rather than lowing down. If gravitational attraction hould make the expanion low down, why i it peeding up intead? The explanation generally accepted by atronomer and phyicit i that pace i uffued with a kind of energy that ha no gravitational The bright pot in thi image are not tar, but entire galaxie. They are part of a cluter of galaxie about 10. billion ly (3.13 billion pc, or 3130 Mpc) away. (The blue glow i x-ray emiion from hot ga within the cluter.) When the galaxie emitted the light ued to make thi image, the cale factor of the univere wa only about 35% a large a it i now. By comparion, we ee the relatively nearby Coma cluter (ee Fig b) a it wa 300 million year ago, when the cale factor wa 98% of the preent-day value.

188 1508 CHAPTER 44 Particle Phyic and Comology Application A Foil Both Ancient and Recent Thi foil trilobite i an example of a group of marine arthropod that flourihed in earth ocean from 540 to 50 million year ago. (By comparion, the firt dinoaur did not appear until 30 million year ago.) From our perpective, thi make trilobite almot unfathomably ancient. But compared to the time that ha elaped ince the Big Bang, 13.7 billion year, even trilobite are a very recent phenomenon: They firt appeared when the univere wa already 96% of it preent age. effect and emit no electromagnetic radiation, but rather act a a kind of antigravity that produce a univeral repulion. Thi inviible, immaterial energy i called dark energy. A the name ugget, the nature of dark energy i poorly undertood but i the ubject of very active reearch. Obervation how that the energy denity of dark energy (meaured in, ay, joule per cubic meter) i 7.6% of the critical denity time c; that i, it i equal to 0.76r c c. A decribed above, the average denity of matter of all kind i 7.4% of the critical denity. From the Eintein relationhip E = mc, the average energy denity of matter in the univere i therefore 0.74r c c. Becaue the energy denity of dark energy i nearly three time greater than that of matter, the expanion of the univere will continue to accelerate. Thi expanion will never top, and the univere will never contract. If we account for energy of all kind, the average energy denity of the univere i equal to 0.76r c c r c c = 1.00r c c. Of thi, 7.6% i? the myteriou dark energy,.8% i the no le myteriou dark matter, and a mere 4.6% i well-undertood conventional matter. How little we know about the content of our univere! When we take account of the denity of matter in the univere (which tend to low the expanion of pace) and the denity of dark energy (which tend to peed up the expanion), the age of the univere turn out to be 13.7 billion year. What i the ignificance of the reult that within obervational error, the average energy denity of the univere i equal to r c c? It tell u that the univere i infinite and unbounded, but jut barely o. If the average energy denity were even lightly larger than r c c, the univere would be finite like the urface of the balloon depicted in Fig A of thi writing, the obervational error in the average energy denity i till large enough (about 1%) that we can t be totally ure that the univere i unbounded. Improving thee meaurement will be an important tak for phyicit and atronomer in the year ahead. Tet Your Undertanding of Section 44.6 i made of ordinary matter? I it accurate to ay that your body 44.7 The Beginning of Time What an odd title for the very lat ection of a book! We will decribe in general term ome of the current theorie about the very early hitory of the univere and their relationhip to fundamental particle interaction. We ll find that an atonihing amount happened in the very firt econd. A lot of looe end will be left untied, and many quetion will be left unanwered. Thi i, after all, one of the frontier of phyic. Temperature The early univere wa extremely dene and extremely hot, and the average particle energie were extremely large, all many order of magnitude beyond anything that exit in the preent univere. We can compare particle energy E and abolute temperature T uing the equipartition principle (ee Section 18.4): E = 3 kt (44.0) In thi equation k i Boltzmann contant, which we ll often expre in ev>k: k = * 10-5 ev>k Thu we can replace Eq. (44.0) by E L ev>kt = GeV>KT when we re dicuing order of magnitude.

189 44.7 The Beginning of Time 1509 Example Temperature and energy (a) What i the average kinetic energy E (in ev) of particle at room temperature 1T = 90 K and at the urface of the un 1T = 5800 K? (b) What approximate temperature correpond to the ionization energy of the hydrogen atom and to the ret energie of the electron and the proton? SOLUTION IDENTIFY and SET UP: In thi example we are to apply the equipartition principle. We ue Eq. (44.0) to relate the target variable E and T. EXECUTE: (a) At room temperature, from Eq. (44.0), E = 3 kt = * 10-5 ev>k190 K = ev The temperature at the un urface i higher than room temperature by a factor of K>190 K = 0, o the average kinetic energy there i ev = 0.75 ev. (b) The ionization energy of hydrogen i 13.6 ev. Uing the approximation E L ev>kt, we have T L E 10-4 ev>k = 13.6 ev 10-4 ev>k L 105 K Repeating thi calculation for the ret energie of the electron 1E = MeV and proton 1E = 938 MeV give temperature of K and K, repectively. EVALUATE: Temperature in exce of 10 5 K are found in the un interior, o mot of the hydrogen there i ionized. Temperature of K or K are not found anywhere in the olar ytem; a we will ee, temperature were thi high in the very early univere. Uncoupling of Interaction We ve characterized the expanion of the univere by a continual increae of the cale factor R, which we can think of very roughly a characterizing the ize of the univere, and by a correponding decreae in average denity. A the total gravitational potential energy increaed during expanion, there were correponding decreae in temperature and average particle energy. A thi happened, the baic interaction became progreively uncoupled. To undertand the uncoupling, recall that the unification of the electromagnetic and weak interaction occur at energie that are large enough that the difference in ma among the variou pin-1 boon that mediate the interaction become inignificant by comparion. The electromagnetic interaction i mediated by the male photon, and the weak interaction i mediated by the weak boon W and Z 0 with mae of the order of 100 GeV>c. At energie much le than 100 GeV the two interaction eem quite different, but at energie much greater than 100 GeV they become part of a ingle interaction. The grand unified theorie (GUT) provide a imilar behavior for the trong interaction. It become unified with the electroweak interaction at energie of the order of GeV, but at lower energie the two appear quite ditinct. One of the reaon GUT are till very peculative i that there i no way to do controlled experiment in thi energy range, which i larger by a factor of than energie available with any current accelerator. Finally, at ufficiently high energie and hort ditance, it i aumed that gravitation become unified with the other three interaction. The ditance at which thi happen i thought to be of the order of m. Thi ditance, called the Planck length l P, i determined by the peed of light c and the fundamental contant of quantum mechanic and gravitation, h and G, repectively. The Planck length l P i defined a l P = A UG c 3 (44.1) You hould verify that thi combination of contant doe indeed have unit of length. The Planck time t P = l P >c i the time required for light to travel a ditance l P : t P = l P c = UG A c 5 = * m = * (44.)

190 1510 CHAPTER 44 Particle Phyic and Comology If we mentally go backward in time, we have to top when we reach t = becaue we have no adequate theory that unifie all four interaction. So a yet we have no way of knowing what happened or how the univere behaved at time earlier than the Planck time or when it ize wa le than the Planck length. The Standard Model of the Hitory of the Univere The decription that follow i called the tandard model of the hitory of the univere. The title indicate that there are ubtantial area of theory that ret on olid experimental foundation and are quite generally accepted. The figure on page i a graphical decription of thi hitory, with the characteritic ize, particle energie, and temperature at variou time. Referring to thi figure frequently will help you to undertand the following dicuion. In thi tandard model, the temperature of the univere at time t = (the Planck time) wa about 10 3 K, and the average energy per particle wa approximately E L GeV>K110 3 K = GeV In a totally unified theory thi i about the energy below which gravity begin to behave a a eparate interaction. Thi time therefore marked the tranition from any propoed TOE to the GUT period. During the GUT period, roughly t = to 10-35, the trong and electroweak force were till unified, and the univere conited of a oup of quark and lepton tranforming into each other o freely that there wa no ditinction between the two familie of particle. Other, much more maive particle may alo have been freely created and detroyed. One important characteritic of GUT i that at ufficiently high energie, baryon number i not conerved. (We mentioned earlier the propoed decay of the proton, which ha not yet been oberved.) Thu by the end of the GUT period the number of quark and antiquark may have been unequal. Thi point ha important implication; we ll return to it at the end of the ection. By t = the temperature had decreaed to about 10 7 K and the average energy to about GeV. At thi energy the trong force eparated from the electroweak force (Fig. 44.0), and baryon number and lepton number began to be eparately conerved. Thi eparation of the trong force wa analogou to a phae 44.0 Schematic diagram howing the time and energie at which the variou interaction are thought to have uncoupled. The energy cale i backward becaue the average energy decreaed a the age of the univere increaed. Age of the univere min 500,000 y y Age increaing Strong force Electroweak force Gravitation Electromagnetic force Weak force v ( e v ) ( m v ) ( t ) Energy decreaing GeV GeV 100 GeV 100 MeV 1 MeV 1 ev 1 mev Lepton e m Quark u c ( d) ( ) t t ( b ) v, n e Hadron n, p K neutrino background e Nuclei Atom

191 44.7 The Beginning of Time 1511 change uch a boiling a liquid, with an aociated heat of vaporization. Think of it a being imilar to boiling a heavy nucleu, pulling the particle apart beyond the hort range of the nuclear force. A a reult, the univere underwent a dramatic expanion (far more rapid than the preent-day expanion rate) called comic inflation. In one model, the cale factor R increaed by a factor of in At t = 10-3 the univere wa a mixture of quark, lepton, and the mediating boon (gluon, photon, and the weak boon W and Z 0. It continued to expand and cool from the inflationary period to t = 10-6, when the temperature wa about K and typical energie were about 1 GeV (comparable to the ret energy of a nucleon; ee Example 44.11). At thi time the quark began to bind together to form nucleon and antinucleon. Alo there were till enough photon of ufficient energy to produce nucleon antinucleon pair to balance the proce of nucleon antinucleon annihilation. However, by about t = 10 -, mot photon energie fell well below the threhold energy for uch pair production. There wa a light exce of nucleon over antinucleon; a a reult, virtually all of the antinucleon and mot of the nucleon annihilated one another. A imilar equilibrium occurred later between the production of electron poitron pair from photon and the annihilation of uch pair. At about t = 14 the average energy dropped to around 1 MeV, below the threhold for e + e - pair production. After pair production ceaed, virtually all of the remaining poitron were annihilated, leaving the univere with many more proton and electron than the antiparticle of each. Up until about t = 1, neutron and neutrino could be produced in the endoergic reaction e - + p S n + n e After thi time, mot electron no longer had enough energy for thi reaction. The average neutrino energy alo decreaed, and a the univere expanded, equilibrium reaction that involved aborption of neutrino (which occurred with decreaing probability) became inoperative. At thi time, in effect, the flux of neutrino and antineutrino throughout the univere uncoupled from the ret of the univere. Becaue of the extraordinarily low probability for neutrino aborption, mot of thi flux i till preent today, although cooled greatly by expanion. The tandard model of the univere predict a preent neutrino temperature of about K, but no experiment ha yet been able to tet thi prediction. Nucleoynthei At about t = 1, the ratio of proton to neutron wa determined by the Boltzmann ditribution factor e - E>kT, where E i the difference between the neutron and proton ret energie: E = 1.94 MeV. At a temperature of about K, thi ditribution factor give about 4.5 time a many proton a neutron. However, a we have dicued, free neutron (with a half-life of 887 ) decay pontaneouly to proton. Thi decay caued the proton neutron ratio to increae until about t = 5. At thi time, the temperature wa about 10 9 K, and the average energy wa well below MeV. Thi energy ditribution wa critical becaue the binding energy of the deuteron (a neutron and a proton bound together) i. MeV (ee Section 43.). A neutron bound in a deuteron doe not decay pontaneouly. A the average energy decreaed, a proton and a neutron could combine to form a deuteron, and there were fewer and fewer photon with. MeV or more of energy to diociate the deuteron again. Therefore the combining of proton and neutron into deuteron halted the decay of free neutron. The formation of deuteron tarting at about t = 5 marked the beginning of the period of formation of nuclei, or nucleoynthei. At thi time, there were about even proton for each neutron. The deuteron 1 H can aborb a neutron and form a triton 1 3 H, or it can aborb a proton and form 3 He. Then 3 H can aborb a proton, and 3 He can aborb a neutron, each yielding 4 He (the alpha particle).

192 151 CHAPTER 44 Particle Phyic and Comology AGE OF QUARKS AND AGE OF NUCLEONS AND GLUONS (GUT Period) ANTINUCLEONS Dene concentration of matter and Quark bind together to form antimatter; gravity a eparate force; nucleon and antinucleon; energy more quark than antiquark. too low for nucleon antinucleon Inflationary period (10 35 ): rapid expanion, pair production at 10. trong force eparate from electroweak force. AGE OF LEPTONS Lepton ditinct from quark; W and Z 0 boon mediate weak force (10 1 ). 5 AGE OF NUCLEOSYNTHESIS Stable deuteron; matter 74% H, 5% He, 1% heavier nuclei. BIG BANG Neutrino Quark Proton Neutron Antineutrino Antiquark Antineutron Antiproton g 1 g n p H g g 4 He e 3 H TOE GUT Electroweak unification Force eparate Matter domination ( 1 H K y 10 3 y 10 6 y 10 9 y 10 5 K 10 0 K K Nuclear K Atomic 10 5 K Solar 1K binding binding ytem energy energy form GeV GeV 10 1 GeV 10 9 GeV 10 6 GeV 1TeV 1 GeV 1 MeV 1 kev 1eV 1 mev t T E Size Nucleoynthei Logarithmic cale how characteritic temperature, energy, and ize of the univere a function of time.

193 44.7 The Beginning of Time 1513 AGE OF IONS Expanding, cooling ga of ionized H and He. A Brief Hitory of the Univere AGE OF ATOMS Neutral atom form; univere become tranparent to mot light AGE OF STARS AND GALAXIES Thermonuclear fuion begin in tar, forming heavier nuclei. NOW 1 H 1 H 4 He 4 He 8 Be 1 H 4 He 4 He 1 C 4 He H 16 O H 4 He He

194 1514 CHAPTER 44 Particle Phyic and Comology A few 7 Li nuclei may alo have been formed by fuion of 3 H and 4 He nuclei. According to the theory, eentially all the 1 H and 4 He in the preent univere wa formed at thi time. But then the building of nuclei almot ground to a halt. The reaon i that no nuclide with ma number A = 5 ha a half-life greater than Alpha particle imply do not permanently aborb neutron or proton. The nuclide 8 Be that i formed by fuion of two 4 He nuclei i untable, with an extremely hort half-life, about 7 * Note alo that at thi time, the average energy wa till much too large for electron to be bound to nuclei; there were not yet any atom. Conceptual Example The relative abundance of hydrogen and helium in the univere Nearly all of the proton and neutron in the even-to-one ratio at t = 5 either formed 4 He or remained a 1 H. After thi time, what wa the reulting relative abundance of 1 H and 4 He, by ma? SOLUTION The 4 He nucleu contain two proton and two neutron. For every two neutron preent at t = 5 there were 14 proton. The two neutron and two of the 14 proton make up one 4 He nucleu, leaving 1 proton ( 1 H nuclei). So there were eventually 1 1 H nuclei for every 4 He nucleu. The mae of 1 H and 4 He are about 1 u and 4 u, repectively, o there were 1 u of 1 H for every 4 u of 4 He. Therefore the relative abundance, by ma, wa 75% 1 H and 5% 4 He. Thi reult agree very well with etimate of the preent H He ratio in the univere, an important confirmation of thi part of the theory The Veil Nebula in the contellation Cygnu i a remnant of a upernova exploion that occurred more than 0,000 year ago. The ga ejected from the upernova i till moving very rapidly. Colliion between thi fat-moving ga and the tenuou material of intertellar pace excite the ga and caue it to glow. The portion of the nebula hown here i about 40 ly (1 pc) in length. Further nucleoynthei did not occur until very much later, well after t = (about 380,000 y). At that time, the temperature wa about 3000 K, and the average energy wa a few tenth of an electron volt. Becaue the ionization energie of hydrogen and helium atom are 13.6 ev and 4.5 ev, repectively, almot all the hydrogen and helium wa electrically neutral (not ionized). With the electrical repulion of the nuclei canceled out, gravitational attraction could lowly pull the neutral atom together to form cloud of ga and eventually tar. Thermonuclear reaction in tar then produced all of the more maive nuclei. In Section 43.8 we dicued one cycle of thermonuclear reaction in which 1 H become 4 He. For tar whoe ma i 40% of the un ma or greater, a the hydrogen i conumed the tar core begin to contract a the inward gravitational preure exceed the outward ga and radiation preure. The gravitational potential energy decreae a the core contract, o the kinetic energy of nuclei in the core increae. Eventually the core temperature become high enough to begin another proce, helium fuion. Firt two 4 He nuclei fue to form 8 Be, which i highly untable. But becaue a tar core i o dene and colliion among nuclei are o frequent, there i a nonzero probability that a third 4 He nucleu will fue with the 8 Be nucleu before it can decay. The reult i the table nuclide 1 C. Thi i called the triple-alpha proce, ince three 4 He nuclei (that i, alpha particle) fue to form one carbon nucleu. Then ucceive fuion with 4 He give 16 O, 0 Ne, and 4 Mg. All thee reaction are exoergic. They releae energy to heat up the tar, and 1 C and 16 O can fue to form element with higher and higher atomic number. For nuclide that can be created in thi manner, the binding energy per nucleon peak at ma number A = 56 with the nuclide 56 Fe, o exoergic fuion reaction top with Fe. But ucceive neutron capture followed by beta decay can continue the ynthei of more maive nuclei. If the tar i maive enough, it may eventually explode a a upernova, ending out into pace the heavy element that were produced by the earlier procee (Fig. 44.1; ee alo Fig. 37.7). In pace, the debri and other intertellar matter can gravitationally bunch together to form a new generation of tar and planet. Our own un i one uch econd-generation tar. Thi mean that the un planet and everything on them (including you) contain matter that wa long ago blated into pace by an exploding upernova.

195 44.7 The Beginning of Time 1515 Background Radiation In 1965 Arno Penzia and Robert Wilon, working at Bell Telephone Laboratorie in New Jerey on atellite communication, turned a microwave antenna kyward and found a background ignal that had no apparent preferred direction. (Thi ignal produce about 1% of the hah you ee on a TV creen when you turn to an unued channel.) Further reearch ha hown that the radiation that i received ha a frequency pectrum that fit Planck blackbody radiation law, Eq. (39.4) (Section 39.5). The wavelength of peak intenity i mm (in the microwave region of the pectrum), with a correponding abolute temperature T =.75 K. Penzia and Wilon contacted phyicit at nearby Princeton Univerity who had begun the deign of an antenna to earch for radiation that wa a remnant from the early evolution of the univere. We mentioned above that neutral atom began to form at about t = 380,000 year when the temperature wa 3000 K. With far fewer charged particle preent than previouly, the univere became tranparent at thi time to electromagnetic radiation of long wavelength. The 3000-K blackbody radiation therefore urvived, cooling to it preent.75-k temperature a the univere expanded. The comic background radiation i among the mot clear-cut experimental confirmation of the Big Bang theory. Figure 44. how a modern map of the comic background radiation. 44. Thi fale-color map how microwave radiation from the entire ky mapped onto an oval. When thi radiation wa emitted 380,000 year after the Big Bang, the region hown in blue were lightly cooler and dener than average. Within thee cool, dene region formed galaxie, including the Milky Way galaxy of which our olar ytem, our earth, and our elve are part. Example 44.1 Expanion of the univere By approximately what factor ha the univere expanded ince t = 380,000 y? SOLUTION IDENTIFY and SET UP: We ue the idea that a the univere ha expanded, all intergalactic wavelength have expanded with it. The Wien diplacement law, Eq. (39.1), relate the peak wavelength l m in blackbody radiation to the temperature T. Given the temperature of the comic background radiation today (.75 K) and at t = 380,000 y K we can determine the factor by which wavelength have changed and hence determine the factor by which the univere ha expanded. EXECUTE: We rewrite Eq. (39.1) a l m =.90 * 10-3 m # K T Hence the peak wavelength l m i inverely proportional to T. A the univere expand, all intergalactic wavelength (including l m ) increae in proportion to the cale factor R. The temperature ha decreaed by the factor K>1.75 K L 1100, o l m and the cale factor mut both have increaed by thi factor. Thu, between t = 380,000 y and the preent, the univere ha expanded by a factor of about EVALUATE: Our reult how that ince t = 380,000 y, any particular intergalactic volume ha increaed by a factor of about = 1.3 * They alo how that when the comic background radiation wa emitted, it peak wavelength wa of the preent-day value of mm, or 967 nm. Thi i in the infrared region of the pectrum.

196 1516 CHAPTER 44 Particle Phyic and Comology Matter and Antimatter One of the mot remarkable feature of our univere i the aymmetry between matter and antimatter. One might think that the univere hould have equal number of proton and antiproton and of electron and poitron, but thi doen t appear to be the cae. Theorie of the early univere mut explain thi imbalance. We ve mentioned that mot GUT include violation of conervation of baryon number at energie at which the trong and electroweak interaction have converged. If particle antiparticle ymmetry i alo violated, we have a mechanim for making more quark than antiquark, more lepton than antilepton, and eventually more matter than antimatter. One eriou problem i that any aymmetry that i created in thi way during the GUT era might be wiped out by the electroweak interaction after the end of the GUT era. If o, there mut be ome mechanim that create particle antiparticle aymmetry at a much later time. The problem of the matter antimatter aymmetry i till very much an open one. There are till many unanwered quetion at the interection of particle phyic and comology. I the energy denity of the univere preciely equal to r c c, or are there mall but important difference? What i dark energy? Ha the denity of dark energy remained contant over the hitory of the univere, or ha the denity changed? What i dark matter? What happened during the firt after the Big Bang? Can we ee evidence that the trong and electroweak interaction undergo a grand unification at high energie? The earch for the anwer to thee and many other quetion about our phyical world continue to be one of the mot exciting adventure of the human mind. Tet Your Undertanding of Section 44.7 Given a ufficiently powerful telecope, could we detect photon emitted earlier than t = 380,000 y?

197 CHAPTER 44 SUMMARY Fundamental particle: Each particle ha an antiparticle; ome particle are their own antiparticle. Particle can be created and detroyed, ome of them (including electron and poitron) only in pair or in conjunction with other particle and antiparticle. Particle erve a mediator for the fundamental interaction. The photon i the mediator of the electromagnetic interaction. Yukawa propoed the exitence of meon to mediate the nuclear interaction. Mediating particle that can exit only becaue of the uncertainty principle for energy are called virtual particle. B S e e e e Particle accelerator and detector: Cyclotron, ynchrotron, and linear accelerator are ued to accelerate charged particle to high energie for experiment with particle interaction. Only part of the beam energy i available to caue reaction with target at ret. Thi problem i avoided in colliding-beam experiment. (See Example ) Highfrequency alternating voltage B S Particle and interaction: Four fundamental interaction are found in nature: the trong, electromagnetic, weak, and gravitational interaction. Particle can be decribed in term of their interaction and of quantitie that are conerved in all or ome of the interaction. Fermion have half-integer pin; boon have integer pin. Lepton, which are fermion, have no trong interaction. Strongly interacting particle are called hadron. They include meon, which are alway boon, and baryon, which are alway fermion. There are conervation law for three different lepton number and for baryon number. Additional quantum number, including trangene and charm, are conerved in ome interaction and not in other. (See Example ) Quark: Hadron are compoed of quark. There are thought to be ix type of quark. The interaction between quark i mediated by gluon. Quark and gluon have an additional attribute called color. (See Example 44.7.) Proton (p) u 1 e 3 d u e 3 u e Poitive pion (p + ) Symmetry and the unification of interaction: Symmetry conideration play a central role in all fundamental-particle theorie. The electromagnetic and weak interaction become unified at high energie into the electroweak interaction. In grand unified theorie the trong interaction i alo unified with thee interaction, but at much higher energie. Age of the univere min 500,000 y y Age increaing Strong force Electroweak force Gravitation Electromagnetic force Weak force Energy decreaing GeV GeV 100 GeV 100 MeV 1 MeV 1 ev 1 mev The expanding univere and it compoition: The Hubble law how that galaxie are receding from each other and that the univere i expanding. Obervation how that the rate of expanion i accelerating due to the preence of dark energy, which make up 7.6% of the energy in the univere. Only 4.6% of the energy in the univere i in the form of ordinary matter; the remaining.8% i dark matter, whoe nature i poorly undertood. (See Example 44.8 and 44.9.) v (10 3 km/) Slope 5 H 0 r (megaparec) The hitory of the univere: In the tandard model of the univere, a Big Bang gave rie to the firt fundamental particle. They eventually formed into the lightet atom a the univere expanded and cooled. The comic background radiation i a relic of the time when thee atom formed. The heavier element were manufactured much later by fuion reaction inide tar. (See Example ) 1517