Chapters 28 and 29: Quantum Physics and Atoms Solutions

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1 Chapters 8 and 9: Quantum Physics and Atoms Solutions Chapter 8: Questions: 3, 8, 5 Exercises & Problems:, 6, 0, 9, 37, 40, 48, 6 Chapter 9: Questions, 6 Problems 3, 5, 8, 9 Q8.3: How does Einstein's explanation account for each of these characteristics of the photoelectric effect? A. The photoelectric current is zero for frequencies below some threshold. B. The photoelectric current increases with increasing light intensity. C. The photoelectric current is independent of ΔV for ΔV > 0. D. The photoelectric current decreases slowly as ΔV becomes more negative. E. The stopping potential is independent of the light intensity. Which of these cannot be explained by classical physics? Explain. Reason: (a) For all materials that emit photo-electrons, the electrons need a minimum amount of energy in order to be released (called the work function). Since a single photon gives all its energy to a single electron, photons must have a minimum energy and hence frequency (called the threshold frequency). (b) Increasing the light intensity increases the number of incident photons. Increasing the number of incident photons increases the number of electrons emitted. Since the moving electrons amount to an electric current, the more electrons, the more current. (c) If the anode is positive, it attracts all electrons to the anode. A further increase in V does not cause any more electrons to reach the anode and thus does not cause a further increase in the current I. (d) As V becomes more negative, fewer electrons have the energy to reach the anode. Hence a decrease in current. (e) The greater the light intensity, the greater the number of photons reaching the photoelectron emitting surface. Since a single photon interacts with a single electron, the energy of the emitted electron depends on the energy of the single incident photon it interacts with, not the number of photons it can only interact with one photon. Classical physics can explain some aspects of the photoelectric effect, but, as mentioned in the text, not all of the aspects we see. Classically, we do not expect a sharp threshold frequency f. 0 We also do not expect the current to appear instantly when the light is turned on. Classically, we expect that as the light intensity is increased, so is the energy absorbed by the surface electrons. Hence, the number of photoelectrons emitted, or the photocurrent, is expected to increase proportionately with light intensity. Classically, we expect the more intense light will give more energy to the photoelectrons and hence their stopping potential will be increases this does not happen. Classical physics cannot explain A and E. Assess: Only the quantum hypothesis can account for all the characteristics of the photoelectric effect. Einstein s explanation of the photoelectric effect won him a well-deserved Nobel Prize.

2 Q8.8: Metal has a larger work function than metal. Both are illuminated with the same short-wavelength ultraviolet light. Do electrons from metal have a higher speed, a lower speed, or the same speed as electrons from metal? Explain. Reason: The kinetic energy of the emitted electrons is related to the energy of the incident photons and the work function of the metal by K = E E. photon o The kinetic energy of the electron is related to its speed by K = mv /, and the energy of the photon is related to its wavelength by / Ephoton = hc/. Combining these and solving for v we obtain v = (/ m)[( hc/ ) Eo]. Examining this function we see that if the wavelength is constant, then as the work function increases, the speed of the electrons decreases. So electrons from metal have lower speed. Assess: It makes sense that if the electron requires more energy to get out of the metal, it will have less energy once it is out. Q8.5: An electron and a proton are accelerated from rest through potential differences of the same magnitude. Afterward, which particle has the larger de Broglie wavelength? Explain. Reason: Both particles accelerate through the same potential difference and so have the same kinetic energy. Combine Equation 8.8 with v= Km /. h h h = = = mv m K/ m Km This shows that the particle with the smaller mass corresponds to the larger wavelength. Thus, the electron has the larger de Broglie wavelength. Assess: Equation 8.8 shows that the wavelength is inversely proportional to the momentum. Q9.: The two spectra shown in the figure below belong to the same element, a fictional Element X. Explain why they are different. Reason: The spectrum at the top is an emission spectrum and the one at the bottom is an absorption spectrum. All lines in the absorption spectrum are in the emission spectrum but not all lines in the emission spectrum are in the absorption spectrum. To see why this is the case, consider a number of identical atoms all in the n = 4 excited state. These atoms may undergo four different transitions to reach the ground state and in the process emit as many as six different

3 wavelengths of light. Each atom will undergo only one transition, but different atoms may use different transition routes to get to the ground state. The possible transitions are as follows: Transition A: n = 4 3. Four different wavelengths of light are emitted. Transition B: n = 4. Two different wavelengths of light are emitted; however the second wavelength from this transition is identical to the third wavelength from transition A. Transition C: n = 4 3. Two different wavelengths of light are emitted; however the first wavelength in this transition is identical to the first wavelength in transition A. Transition D: n = 4. One wavelength of light is emitted. The above shows that from these four transitions we can expect six different wavelengths of light to be emitted and appear in the emission spectrum. If the atoms are in the ground state ( n = ) and light is absorbed to excite the atoms, we expect spectral lines associated with transitions such as n =, n = 3, n = 4. However since there are no electrons in the n = 3 state, we do not expect to see a wavelength associated with the n = 3 4 transition. It is, however, possible to excite an electron to the n = 3 and then again from the n = 3 4 before it can de-excite to the ground state. In general, since for atoms in the ground or unexcited state the upper levels are not populated, we do not expect to see wavelengths associated with all transitions in the emission spectrum in the absorption spectrum. Assess: The absorption spectrum is a subset of the emission spectrum. Q9.6: If an electron is in a stationary state of an atom, is the electron at rest? If not, what does the term mean? Reason: An electron in a stationary state is not at rest, rather it might be in the n = state and not making a transition to another state. Assess: It is important to understand the terminology of any area of the sciences. P 8.: Zinc has a work function of 4.3 ev. a. What is the longest wave length of light that will release an electron from a zinc surface? b. A 4.7 ev photon strikes the surface and an electron is emitted. What is the maximum possible speed of the electron? Prepare: We know the energy of a photon of frequency f and wavelength is E = hf = hc/. Solve: (a) The longest wavelength is the photon with the minimum energy; longer wavelengths would not have enough energy to eject a photon. Solve the equation for hc ( J s)( m/s) ev = = 9 =. = E 4. 3 ev J m 90 nm (b) Review Example 8.4. If 4.3 ev of the energy is used just to get the electron ejected, the remaining 0.4 ev will become the kinetic energy of the electron, that is, K max = 0. 4 ev. v K.. = = = m 9. 0 kg ev 9 max (0 4 ev) 60 0 J max m/s Assess: Our data is a little different from that in Example 8.4, so we don t expect to get exactly the same answer; however we do get an answer that is in the same ballpark, so ours is probably correct. 3

4 P 8.6: A firefly glows by the direct conversion of chemical energy to light. The light emitted by a firefly has peak intensity at a wavelength of 550 nm. a. What is the minimum chemical energy, in ev, required to generate each photon? b. One molecule of ATP provides 0.30 ev of energy when it is metabolized in a cell. What is the minimum number of ATP molecules that must be consumed in the reactions that lead to the emission of one photon of 550 nm light? Prepare: We are given = 550 nm. We also recall that for photons E = hf. Solve: (a) The minimum chemical energy required to generate a photon of wavelength = 550 nm is 34 8 hc ( J s)( m/s) ev E = hf = = 3 ev 9 =. 550 nm J (b) The number of ATP molecules needed to produce that much energy is.3 ev 7.5 molecules of ATP 0.30 ev/molecule of ATP = However, the number of ATP molecules must be an integer, so the minimum number is 8. Assess: The firefly must be using quite a few molecules of ATP since it will emit many photons of light. P 8.0: The wavelengths of light emitted by a firefly span the visible spectrum but have maximum intensity near 550 nm. A typical flash lasts for 00 ms and has a power of. mw. If we assume that all of the light is emitted at the peak-intensity wavelength of 550 nm, how many photons are emitted in one flash? Prepare: Power is the energy per unit time ( P= E / t). The total energy comes from a number of photons ( E = NE ) and the energy of each photon is determined by its wavelength ( photon E = hc / ). Solve: Combine these expressions and solve for the number of photons. E NE N( hc / ) Pt (. 0 W)( m)(0.00 s) P= = = = = = t t t hc J m 3 7 photon 6 N photons Assess: While this seems like a large number, to help keep it in perspective, compare it to the number of molecules in a mole. P 8.9: What is the kinetic energy, in ev, of an electron with a de Broglie wavelength of.0 nm? Prepare: The kinetic energy of a particle of mass m is related to its momentum by E = p /( m). The momentum of the particle is related to its wavelength through = hp /. Solve: Using = hp /, for an electron we have 4

5 9 In ev, the energy is (.4 0 J) Assess: 34 p h ( J s) E = = = = 3 9 m m (9. 0 kg)(.0 0 m) ev (.6 0 J ) 9 =.5 ev J This is a small but reasonable amount of energy for an electron. P 8.37: The nucleus of a typical atom is 5.0 fm ( fm = 0-5 m) in diameter. A very simple model of the nucleus is a one-dimensional box in which protons are confined. Estimate the energy of a proton in the nucleus by finding the first three allowed energies of a proton in a 5.0-fm-long box. Prepare: We will model the 5.0-fm-diameter nucleus as a box of length L = 5.0 fm. Equation 8.3 tells us that the energy of the confined proton is E h 8mL n = n n =,, 3, 4, Solve: The proton s energy is restricted to the discrete values h ( J s) n En = n = = n 8mL 8(.67 0 kg)(5.0 0 m) 34 (.36 0 J) 7 5 where n =,, 3, For n =, E =.36 0 J =.3 0 J, for n =, E = (.36 0 J)4 = J, and for n = 3, E3= 9E=. 0 J. Assess: Energies increase with n as we calculated. P 8.40: The allowed energies of a quantum system are 0.0 ev, 4.0 ev, and 6.0 ev. a. Draw the system's energy-level diagram. Label each level with the energy and the quantum number. b. What wavelengths appear in the system's emission spectrum? Prepare: To conserve energy, the emission spectra must have exactly the energy lost by the system in the appropriate quantum jumps. Solve: (a) 5

6 (b) From Equation 8.3, the energy of a light quantum is E = hf =hc/. We can use this equation to find the emission spectra wavelengths. The emission energies from the energy-level diagram are: E = 4.0 ev, E3 = 6.0 ev, and E 3 =.0 ev. The wavelength corresponding to the transition is 5 8 hc (4.4 0 ev s)(3.0 0 m/s) = = = 30 nm E 4.0 ev Likewise, 3 = hc/ E3 = 0 nm, and 3 = 60 nm. Assess: These are reasonable wavelengths for the three allowed transitions. P 8.48: Potassium and gold cathodes are used in a photoelectric effect experiment. For each cathode, find: a. The threshold frequency b. The threshold wavelength c. The maximum electron ejection speed if the light has a wavelength of 0 nm d. The stopping potential if the wavelength is 0 nm Prepare: The threshold energies for potassium and gold are given in Table 8.. Solve: (a) The threshold frequency is f 0 = E 0 /h. The threshold frequency for potassium and gold are given in the table in part (b). (b) The corresponding threshold wavelength is 0 = c/f 0. The results of the calculations are in the following table. Metal E 0 (ev) f 0 (Hz) 0 (nm) Potassium Gold (c) When light of wavelength is incident on the metal, the maximum kinetic energy of the photoelectrons is given by Equation 8.6: hc hc Kmax = mvmax = hf E0 = E0 v max = E 0 m E 0 must be converted to SI units of joules before this formula can be used. v max for potassium and gold are given in the table in part (d). (d) The stopping potential from Equation 8.7 is V K hc = = E e e max stop 0 where again E 0 has to be joules. The results of the calculations are in the following table. Metal E 0 (J) v max (m/s) Potassiu m Gold V stop (V) Assess: Values obtained for the threshold frequency, threshold wavelength, maximum speed of the emitted photoelectrons, and stopping potential are reasonable. 6

7 P 8.6: Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema a sunburn. The amount of exposure necessary to produce this reddening depends on the wavelength. For a.0 cm patch of skin, 3.7 mj of ultraviolet light at a wavelength of 54 nm will produce reddening; at 300 nm wavelength, 3 mj are required. a. What is the photon energy corresponding to each of these wavelengths? b. How many total photons does each of these exposures correspond to? c. Explain why there is a difference in the number of photons needed to provoke a response in the two cases. Prepare: The energy of a photon is related to its wavelength by E = hc /. The total energy is related to the number of photons and the energy per photon by E = NE. photon Solve: (a) The energy of the photon with wavelength m is E = hc = = / J m / (.54 0 m) J In like manner, the energy of the photon with wavelength m is J. (b) The number of photons for the first case is N E E = / photon = J / ( J) = photons In like manner for the second case obtain N = photons. (c) A sunburn will require more of the less energetic photons. Assess: We expect a large number of photons because the energy of each photon is very small. P 9.3: The Paschen series is analogous to the Balmer series, but with m = 3. Calculate the wavelengths of the first three members in the Paschen series. What part(s) of the electromagnetic spectrum are these in? Prepare: We ll look at the emission spectrum, so the transitions we are interested in are 4 3, 5 3, and 6 3. Equation 9. gives the generalized Balmer formula. Solve: Plug in 4, 5, and 6 for n nm = n 3 ( ) 9 6 ( ) 9 5 ( ) 9 36 ( 9 ) 9. 8 nm = = 876 nm 9. 8 nm = = 8 nm 9. 8 nm = = 094 nm These wavelengths are in the infrared (IR) region of the spectrum. Assess: The Lyman series wavelengths are in the UV region of the spectrum, the Balmer series in the visible, and the Paschen series in the IR region. n 7

8 P 9.5: How many electrons, protons, and neutrons are contained in the following atoms or ions: (a) 9 Be+, (b) C. and (c) 5 N +++? Prepare: For a neutral atom, the number of electrons is the same as the number of protons. The atomic number of Be, C, and N is 4, 6, and 7, respectively. Solve: (a) Since Z = 4, the 9 Be + ion has 3 electrons, 4 protons, and 9 4 = 5 neutrons. (b) Since Z = 6, C has 6 electrons, 6 protons, and 6 = 6 neutrons. (c) Since Z = 7, the triply charged 5 N +++ ion has 4 electrons, 7 protons, and 5 7 = 8 neutrons. Assess: We need a good understanding of the atomic model to label atoms and ions. P 9.8: The figure below is an energy-level diagram for a simple atom. What wavelengths appear in the atom 's (a) emission spectrum and (b) absorption spectrum? Prepare: Please refer to Figure P9.8. The energy of a light quantum is E = hf = hc/. To conserve energy, the emission and the absorption photons must have exactly the energy lost or gained by the atom in the appropriate quantum jumps. Solve: (a) The wavelength of the emission photon from the n = to n = transition is 5 8 (4.4 0 ev s)(3.0 0 m/s) hc = = = 830 nm E E.5 ev Likewise, = 500 nm for the 3 transition with E =.5 ev, and = 30 nm for the 3 transition with E = 4.0 ev. (b) Because the atom in the ground state is in the n = state, the absorption lines correspond to the and 3 transitions. The absorption wavelengths are 830 nm and 30 nm. The 3 transition is not seen in absorption. P 9.9: Determine all possible wavelengths of photons that can be emitted from the n = 4 state of a hydrogen atom. 8

9 Prepare: We will use Equation 9. or 9.8 with 0 = 9.8 nm. Solve: Photons emitted from the n = 4 state start in energy level n = 4 and undergo a quantum jump to a lower energy level with m < 4. The possibilities are 4, 4, and 4 3. According to Equation 9.8, the transition 4 m emits a photon of wavelength These values are given in the following table. Assess: 9.8 nm ( / m / n ) ( / m 6 / ) 0 = = Transition Wavelength 97.3 nm 486 nm 876 nm Transitions are in the ultraviolet, visible, and infrared regions. 9

Chapters 28 and 29: Quantum Physics and Atoms Questions & Problems

Chapters 28 and 29: Quantum Physics and Atoms Questions & Problems Chapters 8 and 9: Quantum Physics and Atoms Questions & Problems hc = hf = K = = hf = ev P = /t = N h h h = = n = n, n = 1,, 3,... system = hf photon p mv 8 ml photon max elec 0 0 stop total photon 91.1nm