Chapter. Prblem Slutins. If the speed f light were smaller than it is, wuld relativistic phenmena be mre r less cnspicuus than they are nw? All else being the same, including the rates f the chemical reactins that gvern ur brains and bdies, relativisitic phenmena wuld be mre cnspicuus if the speed f light were smaller. If we culd attain the abslute speeds btainable t us in the universe as it is, but with the speed f light being smaller, we wuld be able t mve at speeds that wuld crrespnd t larger fractins f the speed f light, and in such instances relativistic effects wuld be mre cnspicuus.. An athlete has learned enugh physics t knw that if he measures frm the earth a time interval n a mving spacecraft, what he finds will be greater than what smebdy n the spacecraft wuld measure. He therefre prpses t set a wrld recrd fr the 00-m dash by having his time taken by an bserver n a mving spacecraft. Is this a gd idea? Even if the judges wuld allw it, the bservers in the mving spaceship wuld measure a lnger time, since they wuld see the runners being timed by clcks that appear t run slwly cmpared t the ship's clcks. Actually, when the effects f length cntractin are included (discussed in Sectin. and Appendix ), the runner's speed may be greater than, less than, r the same as that measured by an bserver n the grund.
5. Tw bservers, A n earth and B in a spacecraft whse speed is.00 x 0 8 m/s, bth set their watches t the same time when the ship is abreast f the earth. (a) Hw much time must elapse by A's reckning befre the watches differ by.00 s? (b) T A, B's watch seems t run slw. T B, des A's watch seem t run fast, run slw, r keep the same time as his wn watch? Nte that the nnrelativistic apprximatin is nt valid, as v/c /. (a) See Example.. In Equatin (.), with t representing bth the time measured by A and the time as measured in A's frame fr the clck in B's frame t advance by t, we need v t t0 t t 0. 55. 00 s t c frm which t.9 s. (b) A mving clck always seems t run slwer. In this prblem, the time t is the time that bserver A measures as the time that B's clck takes t recrd a time change f t.
7. Hw fast must a spacecraft travel relative t the earth fr each day n the spacecraft t crrespnd t d n the earth? Frm Equatin (.), fr the time t n the earth t crrespnd t twice the time t 0 elapsed n the ship s clck, v, s v c. 60 0 c relating three significant figures. 9. A certain particle has a lifetime f.00 x0-7 s when measured at rest. Hw far des it g befre decaying if its speed is 0.99c when it is created? The lifetime f the particle is t 0, and the distance the particle will travel is, frm Equatin (.), vt vt v ( 0. 99) 8 m/s, 8 7 0 ( 0. 99)(. 0 0 m/s)(. 00 0 s) /c t tw significant figures. 0 m
. A galaxy in the cnstellatin Ursa Majr is receding frm the earth at 5,000 km/s. If ne f the characteristic wavelengths f the light the galaxy emits is 550 nm, what is the crrespnding wavelength measured by astrnmers n the earth? See Example.; fr the intermediate calculatins, nte that c c ν v/ c λ λ, ν ν ν + v/ c where the sign cnventin fr v is that f Equatin (.8), which v psitive fr an appraching surce and v negative fr a receding surce. Fr this prblem, s that v c. 50 0 km/s 8. 0 0 m/s 7 0. 050, λ λ v/ c + v/ c ( 550 nm) + 0. 050 0. 050 578 nm
. A spacecraft receding frm the earth emits radi waves at a cnstant frequency f 0 9 Hz. If the receiver n earth can measure frequencies t the nearest hertz, at what spacecraft speed can the difference between the relativistic and classical Dppler effects be detected? Fr the classical effect, assume the earth is statinary. This prblem may be dne in several ways, all f which need t use the fact that when the frequencies due t the classical and relativistic effects are fund, thse frequencies, while differing by Hz, will bth be sufficiently clse t v 0 9 Hz s that v culd be used fr an apprximatin t either. In Equatin (.), we have v 0 and V -u, where u is the speed f the spacecraft, mving away frm the earth (V < 0). In Equatin (.6), we have v u (r v -u in Equatin (.8)). The classical and relativistic frequencies, v c and v r respectively, are ν c ν ( u / c) ( u/ c) 0, νr ν ν + ( u/ c) + ( u / c) + ( u/ c) The last expressin fr v, is mtivated by the derivatin f Equatin (.6), which essentially incrprates the classical result (cunting the number f ticks), and allws expressin f the rati ν ν c r ( u/ c).
Use f the abve frms fr the frequencies allws the calculatin f the rati ν ν Hz 9 c νr ( u/ c) 0 ν ν + ( u/ c) 9 0 Hz Attempts t slve this equatin exactly are nt likely t be met with success, and even numerical slutins wuld require a higher precisin than is cmmnly available. Hwever, recgnizing that the numeratr ( u / c) is f the frm that can be apprximated using the methds utlined at the beginning f this chapter, we can use ( u/ c) ( / )( u/ c). The denminatr will be indistinguishable frm at lw speed, with the result u 9 0, c which is slved fr u 0 9 c. 0 m/s.km/s
5. If the angle between the directin f mtin f a light surce f frequency v and the directin frm it t an bserver is 0, the frequency v the bserver finds is given by where v is the relative speed f the surce. Shw that this frmula includes Eqs. (.5) t (.7) as special cases. The transverse Dppler effect crrespnds t a directin f mtin f the light surce that is perpendicular t the directin frm it t the bserver; the angle θ ±/ (r ±90 ), s cs θ 0, and which is Equatin (.5). Fr a receding surce, θ (r 80 ), and cs θ. The given expressin becmes /, c ν v ν, / / / / c v c v c v c v + + ν ν ν which is Equatin (.8). Fr an appraching surce, θ 0, cs θ, and the given expressin becmes, / / / / c v c v c v c v + ν ν ν which is Equatin (.8). θ ν ν )cs / ( / c v c v
7. An astrnaut whse height n the earth is exactly 6 ft is lying parallel t the axis f a spacecraft mving at 0.90c relative t the earth. What is his height as measured by an bserver in the same spacecraft? By an bserver n the earth? The astrnaut s prper length (height) is 6 ft, and this is what any bserver in the spacecraft will measure. Frm Equatin (.9), an bserver n the earth wuld measure L L v / c ( 6 ft) ( 0. 90). 6 ft 9. Hw much time des a meter stick mving at 0.00c relative t an bserver take t pass the bserver? The meter stick is parallel t its directin f mtin. The time will be the length as measured by the bserver divided by the speed, r t 8 L L 00 m 0 00 v / c (. ) (. ). 0 v v 8 ( 0. 00)(. 0 0 m/s) s
. A spacecraft antenna is at an angle f 0 relative t the axis f the spacecraft. If the spacecraft mves away frm the earth at a speed f 0.70c, what is the angle f the antenna as seen frm the earth? If the antenna has a length L' as measured by an bserver n the spacecraft (L' is nt either L r L O in Equatin (.9)), the prjectin f the antenna nt the spacecraft will have a length L'cs(0 ), and the prjectin nt an axis perpendicular t the spacecraft's axis will have a length L'sin(0 ). T an bserver n the earth, the length in the directin f the spacecraft's axis will be cntracted as described by Equatin (.9), while the length perpendicular t the spacecraft's mtin will appear unchanged. The angle as seen frm the earth will then be L sin( 0 ) arctan L cs( 0 ) v / c arctan tan( 0 ) ( 0. 70). The generalizatin f the abve is that if the angle is 00 as measured by an bserver n the spacecraft, an bserver n the earth wuld measure an angle θ given by tanθ tanθ v / c
. A wman leaves the earth in a spacecraft that makes a rund trip t the nearest star, lightyears distant, at a speed f 0.9c. The age difference will be the difference in the times that each measures the rund trip t take, r yr ( v /c ) ( 0 9 ) 5 yr. L t. v 0. 9 5. All definitins are arbitrary, but sme are mre useful than thers. What is the bjectin t defining linear mmentum as p mv instead f the mre cmplicated p gmv? It is cnvenient t maintain the relatinship frm Newtnian mechanics, in that a frce n an bject changes the bject's mmentum; symblically, F dp/dt shuld still be valid. In the absence f frces, mmentum shuld be cnserved in any inertial frame, and the cnserved quantity is p -γmv, nt mv 7. Dynamite liberates abut 5. x 0 6 J/kg when it expldes. What fractin f its ttal energy cntent is this? Fr a given mass M, the rati f the mass liberated t the mass energy is 6 M ( 5. 0 J/kg) 6. 0 0. 8 M (. 0 0 m/s)
9. At what speed des the kinetic energy f a particle equal its rest energy? If the kinetic energy K E mc, then E mc and Equatin (.) reduces t v / c (γ in the ntatin f Sectin.7). Slving fr v, v 8 c. 60 0 m/s. An electrn has a kinetic energy f 0.00 MeV. Find its speed accrding t classical and relativistic mechanics. Classically, v 9 K 0. 00 MeV. 60 0 J/eV 8. 88 0 m/s. 9. 0 kg m e Relativistically, slving Equatin (.) fr v as a functin f K, v c m ec E c m mec e c + K c + K /( m e c ).
With K/(m e c ) (0.00 MeV)/(0.5 MeV) 0.00/0.5, 8 8 v. 0 0 m/s. 6 0 m/s. + ( 0. 00)/( 0. 5) The tw speeds are cmparable, but nt the same; fr larger values f the rati f the kinetic and rest energies, larger discrepancies wuld be fund.. A particle has a kinetic energy 0 times its rest energy. Find the speed f the particle in terms f c. Using Equatin (.) in Equatin (.) and slving fr v/c, v c With E E, that is, E E + 0E, E E v c 0. 9989c.
5. Hw much wrk (in MeV) must be dne t increase the speed f an electrn frm. x 0 8 m/s t. X 0 8 m/s? The difference in energies will be, frm Equatin (.), MeV 9 0 0 0 5MeV 0. ). /. ( ). /. ( ). ( / / c v c v m e c 7. Prve that ½γmv, des nt equal the kinetic energy f a particle mving at relativistic speeds. Using the expressin in Equatin (.0) fr the kinetic energy, the rati f the tw quantities is. / c v c v c v K mv γ γ γ
9. An alternative derivatin f the mass-energy frmula E O mc, als given by Einstein, is based n the principle that the lcatin f the center f mass (CM) f an islated system cannt be changed by any prcess that ccurs inside the system. Figure.7 shws a rigid bx f length L that rests n a frictinless surface; the mass M f the bx is equally divided between its tw ends. A burst f electrmagnetic radiatin f energy E is emitted by ne end f the bx. Accrding t classical physics, the radiatin has the mmentum p E /c, and when it is emitted, the bx recils with the speed v E 0 Mc s that the ttal mmentum f the system remains zer. After a time t L/c the radiatin reaches the ther end f the bx and is absrbed there, which brings the bx t a stp after having mved the distance S. If the CM f the bx is t remain in its riginal place, the radiatin must have transferred mass frm ne end t the ther. Shw that this amunt f mass is m E O c. Measured frm the riginal center f the bx, s that the riginal psitin f the center f mass is 0, the final psitin f the center f mass is M L M L m + S + m S 0. Expanding the prducts and canceling similar terms [(M/)(L/), ms], the result MS ml is btained. The distance 5 is the prduct vt, where, as shwn in the prblem statement, v E/Mc (apprximate in the nnrelativistic limit M >> Elc ) and t L/c. Then, MS M E L E m L L Mc c c.
. In its wn frame f reference, a prtn takes 5 min t crss the Milky Way galaxy, which is abut 0 5 light-years in diameter. (a) What is the apprximate energy f the prtn in electrnvlts?. (b) Abut hw lng wuld the prtn take t crss the galaxy as measured by an bserver in the galaxy's reference frame? T crss the galaxy in a matter f minutes, the prtn must be highly relativistic, with v c (but v < c, f curse). The energy f the prtn will be E E γ, where E O is the prtn's rest energy and γ / v / c. Hwever, γ, frm Equatin (.9), is the same as the rati L O /L, where L is the diameter f the galaxy in the prtn's frame f reference, and fr the highly-relativistic prtn L ct, where t is the time in the prtn's frame that it takes t crss the galaxy. Cmbining, E L L 9 0 ly 7 E E E γ ( 0 ev) ( 0 s/yr ) 0 L ct c( 00 s). Find the mmentum (in MeV/c) f an electrn whse speed is 0.600c. Taking magnitudes in Equatin (.6), m v c c p e ( 0. 5 MeV / )( 0. 600 ) 0. 8 MeV / c v / c ( 0. 600) 5 9 ev
5. Find the mmentum f an electrn whse kinetic energy equals its rest energy f 5 kev When the kinetic energy f an electrn is equal t its rest energy, the ttal energy is twice the rest energy, and Equatin (.) becmes m e c m e c + p c, r p ( m c )/ c ( 5 kev / c). 9 GeV/ c The result f Prblem -9 culd be used directly; γ, v ( p m e c, as abve. e /)c, and Equatin (.7) gives 7. Find the speed and mmentum (in GeV/c) f a prtn whse ttal energy is.500 GeV Slving Equatin (.) fr the speed v in terms f the rest energy E O and the ttal energy E, v c ( E / E) c ( 0. 98/. 500) 0. 96c numerically.888 x 0 8 m/s. (The result f Prblem - des nt give an answer accurate t three significant figures.) The value f the speed may be substituted int Equatin (.6) (r the result f Prblem -6), r Equatin (.) may be slved fr the magnitude f the mmentum, p ( E/ c) ( E / c) (. 500 GeV/ c) ( 0. 98 GeV/ c). 7 GeV/ c
9. A particle has a kinetic energy f 6 MeV and a mmentum f 5 MeV/c. Find its mass (in MeV/c ) and speed (as a fractin f c). Frm E mc + K and Equatin (.), ( mc + K ) m c + p c Expanding the binmial, cancelling the m c term, and slving fr m, ( pc) K ( 5MeV) ( 6 MeV) m 87 MeV / c. c K c ( 6 MeV) The particle's speed may be fund any number f ways; a very cnvenient result is that f Prblem -6, giving p pc 5 MeV v c c c 0. 6c. E mc + K 87MeV + 6 MeV
5. An bserver detects tw explsins, ne that ccurs near her at a certain time and anther that ccurs.00 ms later 00 km away. Anther bserver finds that the tw explsins ccur at the, same place. What time interval separates the explsins t the secnd bserver? The given bservatin that the tw explsins ccur at the same place t the secnd bserver means that x' 0 in Equatin (.), and s the secnd bserver is mving at a speed 5 x. 00 0 m 7 v 5. 00 0 m/s t. 00 0 s with respect t the first bserver. Inserting this int Equatin (.), x x t ( x / t) t tc t c t t ( x / ct) x / c t c ( 5. 00 0 m/s) (. 00 ms). 97 ms. 8 (.998 0 m/s) (Fr this calculatin, the apprximatin is ( x / ct) ( x / c t ) valid t three significant figures.) An equally valid methd, and a gd cheek, is t nte that when the relative speed f the bservers (5.00 x 0 7 m/s) has been determined, the time interval that the secnd bserver measures shuld be that given by Equatin (.) (but be careful f which time it t, which is t). Algebraically and numerically, the different methds give the same result. 7
5. A spacecraft mving in the +x directin receives a light signal frm a surce in the xy plane. In the reference frame f the fixed stars, the speed f the spacecraft is v and the signal arrives at an angle θ t the axis f the spacecraft. (a) With the help f the Lrentz transfrmatin find the angle θ ' at which the signal arrives in the reference frame f the spacecraft. (b) What wuld yu cnclude frm this result abut the view f the stars frm a prthle n the side f the spacecraft? (a) A cnvenient chice fr the rigins f bth the unprimed and primed crdinate systems is the pint, in bth space and time, where the ship receives the signal. Then, in the unprimed frame (given here as the frame f the fixed stars, ne f which may be the surce), the signal was sent at a time t -r/c, where r is the distance frm the surce t the place where the ship receives the signal, and the minus sign merely indicates that the signal was sent befre it was received. Take the directin f the ship's mtin (assumed parallel t its axis) t be the psitive x-directin, s that in the frame f the fixed stars (the unprimed frame), the signal arrives at an angle 0 with respect t the psitive x-directin. In the unprimed frame, x r cs θ and y r sin θ. Frm Equatin (.), x vt r csθ ( r / c) csθ + ( v/ c) x r, v / c v / c v / c and y y r sin θ. Then,
tanθ 55. A man n the mn sees tw spacecraft, A and B, cming tward him frm ppsite directins at the respective speeds f 0.800c and 0.900c. (a) What des a man n A measure fr the speed with which he is appraching the mn? Fr the speed with which he is appraching B? (b) What des a man n B measure fr the speed with which he is appraching the mn? Fr the speed with which he is appraching A? (a) If the man n the mn sees A appraching with speed v 0.800 c, then the bserver n A will see the man in the mn appraching with speed v 0.800c. The relative velcities will have ppsite directins, but the relative speeds will be the same. The speed with which B is seen t apprach A, t an bserver in A, is then V v.. V x + 0 800 + 0 900 x c 0. 988c. + vv / c + ( 0. 800)( 0. 900) y x (csθ sinθ + ( v/ c))/ v / c, and sinθ v / c θ arctan csθ + ( v/ c) (b) Frm the frm f the result f part (a), it can be seen that the numeratr f the term in square brackets is less than sinθ, and the denminatr is greater than csθ, and s tan θ and θ < θ when v 0. Lking ut f a prthle, the surces, including the stars, will appear t be in the directins clse t the directin f the ship s mtin than they wuld fr a ship with v 0. As v c, θ 0, and all stars appear t be almst n the ship s axis(farther frward in the field f view). x.
(b) Similarly, the bserver n B will see the man n the mn appraching with speed 0.900 c, and the apparent speed f A, t an bserver n B, will be 0. 900 + 0. 800 c 0. 988c. + ( 0. 900)( 0. 800) (Nte that Equatin (.9) is unchanged if V x and v are interchanged.) S (mn) S B V x O v A
Chapter Prblem Slutins. If Planck's cnstant were smaller than it is, wuld quantum phenmena be mre r less cnspicuus than they are nw? Planck s cnstant gives a measure f the energy at which quantum effects are bserved. If Planck s cnstant had a smaller value, while all ther physical quantities, such as the speed f light, remained the same, quantum effects wuld be seen fr phenmena that ccur at higher frequencies r shrter wavelengths. That is, quantum phenmena wuld be less cnspicuus than they are nw.. Is it crrect t say that the maximum phtelectrn energy KE max is prprtinal t the frequency ν f the incident light? If nt, what wuld a crrect statement f the relatinship between KE max and ν be? N: the relatin is given in Equatin (.8) and Equatin (.9), KE max hν φ h( ν S that while KE max is a linear functin f the frequency ν f the incident light, KE max is nt prprtinal t the frequency. ν ),
5. Find the energy f a 700-nm phtn. Frm Equatin (.), 6. 0 ev m E. 77 ev. -9 700 0 m Or, in terms f jules, 8 (. 6 0 J s)(.0 0 m/s) E. 8 0 9 700 0 m 6 9 J 7. A.00-kW radi transmitter perates at a frequency f 880 khz. Hw many phtns per secnd des it emit? The number f phtns per unit time is the ttal energy per unit time(the pwer) divided by the energy per phtn, r P E P hν. 00 0 J/s 0. 7 0 phtns/s. ( 6. 6 0 J s)(880 0 Hz)
9. Light frm the sun arrives at the earth, an average f.5 x 0 m away, at the rate f. x 0 W/m f area perpendicular t the directin f the light. Assume that sunlight is mnchrmatic with a frequency f 5.0 x 0 Hz. (a) Hw many phtns fall per secnd n each square meter f the earth's surface directly facing the sun? (b) What is the pwer utput f the sun, and hw many phtns per secnd des it emit? (c) Hw many phtns per cubic meter are there near the earth? (a) The number f phtns per unit time per unit are will be the energy per unit time per unit area (the pwer per unit area, P/A), divided by the energy per phtn, r P / A. 0 W/m. 0 phtns/(s m ). hν - (6.6 0 J s)(5.0 0 Hz) (b) With the reasnable assumptin that the sun radiates unifrmly in all directins, all pints at the same distance frm the sun shuld have the same flux f energy, even if there is n surface t absrb the energy. The ttal pwer is then, R E S ( P / A) (. 0 W/m ) (. 5 0 m). 0 0 W, where R E-S is the mean Earth-Sun distance, cmmnly abbreviated as AU, fr astrnmical unit. The number f phtns emitted per secnd is this pwer divided by the energy per phtn, r 6. 0 0 J/s 5. 0 phtns/s. - (6.6 0 J s)( 5. 0 0 Hz) 6
(c) The phtns are all mving at the same speed c, and in the same directin (spreading is nt significant n the scale f the earth), and s the number f phtns per unit time per unit area is the prduct f the number per unit vlume and the speed. Using the result frm part (a),. 0 phtns/(s m ). 0 phtns/m 8. 0 0 m/s.. The maximum wavelength fr phtelectric emissin in tungsten is 0 nm. What wavelength f light must be used in rder fr electrns with a maximum energy f.5 ev t be ejected? Expressing Equatin (.9) in terms f λ c/ν and λ 0 c/ν 0, and perfrming the needed algebraic manipulatins, hc λ ( hc / λ ) + K max 9 (. 5eV)( 0 0 m) ( 0 nm) + 6. 0 ev m K λ0 + λ hc max 80 nm.
. What is the maximum wavelength f light that will cause phtelectrns t be emitted frm sdium? What will the maximum kinetic energy f the phtelectrns be if 00-nm light falls n a sdium surface? The maximum wavelength wuld crrespnd t the least energy that wuld allw an electrn t be emitted, s the incident energy wuld be equal t the wrk functin, and 6 hc. 0 ev m λmax 59 nm φ. ev where the value f φ fr sdium is taken frm Table.. Frm Equatin (.8), 6 hc. 0 ev m Kmax hν φ φ. ev.9ev. λ 9 00 0 m 5..5 mw f 00-nm light is directed at a phtelectric cell. If 0.0 percent f the incident phtns prduce phtelectrns, find the current in the cell. Because nly 0.0% f the light creates phtelectrns, the available pwer is (.0x0 - )(.5x0 - W).5x0-6 W. the current will be the prduct f the number f phtelectrns per unit time and the electrn charge, r I e P E P e hc / λ Pλ e hc 6 (. 5 0 J/s)( 00 0 ( e ) -6. 0 ev m 9 m) 0. 8 µ A
7. A metal surface illuminated by 8.5 x 0 Hz light emits electrns whse maximum energy is 0.5 ev The same surface illuminated by.0 x 0 Hz light emits electrns whse maximum energy is.97 ev Frm these data find Planck's cnstant and the wrk functin f the surface. Denting the tw energies and frequencies with subscripts and, K hν φ K ν φ. max,, max, h Subtracting t eliminate the wrk functin φ and dividing by ν - ν, Kmax, Kmax,. 7eV 0. 5eV h. 0 ν ν. 0 0 Hz 8. 5 0 Hz t the allwed tw significant figures. Keeping an extra figure gives 5 9 5 ev s h. 0 ev s 6.6 0 J s The wrk functin φ may be btained by substituting the abve result int either f the abve expressins relating the frequencies and the energies, yielding φ.0 ev t the same tw significant figures, r the equatins may be slved by rewriting them as Kmax, ν hνν φν, K max, ν hν ν φν, subtracting t eliminate the prduct hν ν and dividing by ν - ν t btain K max, ν Kmax, ν ( 9. 7 ev)(8.5 0 Hz) ( 0. 5 ev)(.0 0 Hz) φ. 0 ev ν ν (.0 0 Hz 8. 5 0 Hz) (This last calculatin, while pssibly mre cumbersme than direct substitutin, reflects the result f slving the system f equatins using a symblic-manipulatin prgram; using such a prgram fr this prblem is, f curse, a case f "swatting a fly with a sledgehammer".) -
9. Shw that it is impssible fr a phtn t give up all its energy and mmentum t a free electrn. This is the reasn why the phtelectric effect can take place nly when phtns strike bund electrns. Cnsider the prpsed interactin in the frame f the electrn initially at rest. The phtn's initial mmentum is p E /c, and if the electrn were t attain all f the phtn's mmentum and energy, the final mmentum f the electrn must be p e p p, the final electrn kinetic energy must be KE E pc, and s the final electrn energy is E e pc + m e c. Hwever, fr any electrn we must have E e (pc) + (m e c ). Equating the tw expressins fr E e r ( m c ) ( pc + m c ) ( pc) + ( pc) ( m c ) ( m c ), Ee ( pc) + e e e + 0 ( pc)( m c ). e This is nly pssible if p 0, in which case the phtn had n initial mmentum and n initial energy, and hence culd nt have existed. T see the same result withut using as much algebra, the electrn's final kinetic energy is p c + m e c m e c fr nnzer p. An easier alternative is t cnsider the interactin in the frame where the electrn is at rest after absrbing the phtn. In this frame, the final energy is the rest energy f the electrn, m e c, but befre the interactin, the electrn wuld have been mving (t cnserve mmentum), and hence wuld have had mre energy than after the interactin, and the phtn wuld have had psitive energy, s energy culd nt be cnserved. pc e
. Electrns are accelerated in televisin tubes thrugh ptential differences f abut 0 kv. Find the highest frequency f the electrmagnetic waves emitted when these electrns strike the screen f the tube. What kind f waves are these? Fr the highest frequency, the electrns will acquire all f their kinetic energy frm the accelerating vltage, and this energy will appear as the electrmagnetic radiatin emitted when these electrns strike the screen. The frequency f this radiatin will be E ev ( e)( 0 0 V) 8 ν. 0 Hz h h 5. 0 ev s which crrespnds t x-rays.. The distance between adjacent atmic planes in calcite (CaCO ) is 0.00 nm. Find the smallest angle f Bragg scattering fr 0.00-nm x-rays. Slving Equatin (.) fr θ with n, λ 0. 00 nm θ arcsin arcsin. 9 d 0.00 nm
5. What is the frequency f an x-ray phtn whse mmentum is. x 0 - kg m/s? Frm Equatin (.5), ν cp h 8 - (. 0 0 m/s)(. 0 kg m/s) 8 5. 0 0 Hz 6. 6 0 J s 7. In See..7 the x-rays scattered by a crystal were assumed t underg n change in wavelength. Shw that this assumptin is reasnable by calculating the Cmptn wavelength f a Na atm and cmparing it with the typical x-ray wavelength f 0. nm. Fllwing the steps that led t Equatin (.), but with a sdium atm instead f an electrn, λ C, Na 6 7 h. 6 0 J s 5. 8 0 cm 8-6 (.0 0 m/s)(.8 0 kg) Na r 5.8 x 0-8 nm, which is much less than. nm. (Here, the rest mass M Na.8 x 0-6 kg was taken frm Prblem -.) m,
9. A beam f x-rays is scattered by a target. At 5 frm the beam directin the scattered x-rays have a wavelength f. pm. What is the wavelength f the x-rays in the direct beam? Slving Equatin (.) fr λ, the wavelength f the x-rays in the direct beam, λ λ λ C ( csφ). pm (. 6 pm)( cs 5 ). 5 pm t the given tw significant figures.. An x-ray phtn f initial frequency.0 x 0 9 Hz cllides with an electrn and is scattered thrugh 90. Find its new frequency. Rewriting Equatin (.) in terms f frequencies, with λ c/ν and λ c/ν, and with cs 90 0, c c + λc ν ν and slving fr ν gives 9 λ. 0 m ν + C + 0 9 8. ν c. 0 0 Hz. 0 0 m/s The abve methd avids the intermediate calculatin f wavelengths. Hz
. At what scattering angle will incident 00-keV x-rays leave a target with an energy f 90 kev? Slving Equatin (.) fr cs φ, λ λ 5keV 5 kev cs mc mc φ + + + 0. λc λ C E E 00 kev 90 kev frm which φ 6 t tw significant figures. 5. A phtn f frequency ν is scattered by an electrn initially at rest. Verify that the maximum kinetic energy f the recil electrn is KE max (h ν /mc )/( + hν/mc ). Fr the electrn t have the maximum recil energy, the scattering angle must be 80 0, and Equatin (.0) becmes mc KE max (hv) (hv'), where KE max (hv - hv') has been used. T simplify the algebra smewhat, cnsider λ ν ν ν ν ν, λ + ( λ/ λ) + ( λc / λ) + ( νλ C / c) where λ λ C fr φ 80. With this expressin, ( hν)( hν ) ( hν) /( mc ) KEmax. mc + ( νλ C / c) Using λ C h/(mc) (which is Equatin (.)) gives the desired result.
7. A phtn whse energy equals the rest energy f the electrn underges a Cmptn cllisin with an electrn. If the electrn mves ff at an angle f 0 with the riginal phtn directin, what is the energy f the scattered phtn? As presented in the text, the energy f the scattered phtn is knwn in terms f the scattered angle, nt the recil angle f the scattering electrn. Cnsider the expressin fr the recil angle as given preceding the slutin t Prblem -5: sin φ sin φ sin φ tanθ. ( λ/ λ) + ( csφ) ( λc / λ)( csφ) + ( csφ) λ + C ( csφ) λ Fr the given prblem, with E mc, λ hc/e h/(mc) λ C, s the abve expressin reduces t sinφ tanθ. ( csφ) At this pint, there are many ways t prceed; a numerical slutin with θ 0 gives φ 6.6 0 t three significant figures. Fr an analytic slutin which avids the intermediate calculatin f the scattering angle φ, ne methd is t square bth sides f the abve relatin and use the trignmetric identity sin φ - cs φ ( + cs φ)( cs φ) t btain + csφ tan θ csφ (the factr - cs φ may be divided, as cs φ, φ 0, represents an undeflected phtn, and hence n interactin). This may be re-expressed as
( csφ)( tan θ ) + csφ ( csφ), csφ, csφ + tan θ Then with λ λ + λ C ( cs φ) λ C ( cs φ), + tan + tan λ + tan θ + tan ( 0 ) E E E ( 5keV) 5 ev λ + tan θ + tan ( 0 ) An equivalent but slightly mre cumbersme methd is t use the trignmetric identities φ φ φ sin φ sin cs, csφ sin in the expressin fr tan θ t btain φ tanθ ct, yielding the result θ 6.6 mre readily. φ arctan tan θ θ. θ r
9. A psitrn cllides head n with an electrn and bth are annihilated. Each particle had a kinetic energy f.00 MeV Find the wavelength f the resulting phtns. The energy f each phtn will he the sum f ne particle's rest and kinetic energies,.5 MeV (keeping an extra significant figure). The wavelength f each phtn will be 6 hc. 0 ev m λ 8. 0 m 0.8pm E 6. 5 0 ev. Shw that, regardless f its initial energy, a phtn cannt underg Cmptn scattering thrugh an angle f mre than 60 and still be able t prduce an electrn-psitrn pair. (Hint: Start by expressing the Cmptn wavelength f the electrn in terms f the maximum phtn wavelength needed fr pair prductin.) Fllwing the hint, λ h hc hc C, mc mc E min where E min mc is the minimum phtn energy needed fr pair prductin. The scattered wavelength (a maximum) crrespnding t this minimum energy is λ max (h/e min ), s λ C λ max. At this pint, it is pssible t say that fr the mst energetic incming phtns, λ ~ 0, and s - cs φ ½ fr λ ' λ C /, frm which cs φ ½ and φ 60. As an alternative, the angle at which the scattered phtns will have wavelength λ max can m be fund as a functin f the incming phtn energy E; slving Equatin (.) with λ ' λ' max )
csφ λmax λ λmax hc / E mc + + λ λ λ E C C This expressin shws that fr E >> mc, cs φ ½ and s φ 60, but it als shws that, because cs φ must always be less than, fr pair prductin at any angle, E must be greater than mc, which we knw t be the case.. (a) Shw that the thickness x /, f an absrber required t reduce the intensity f a beam f radiatin by a factr f is given by x / 0.69/µ. (b) Find the absrber thickness needed t prduce an intensity reductin f a factr f 0. (a) The mst direct way t get this result is t use Equatin (.6) with I /I, s that µ x ln 0. 69 I Ie x /. µ µ (b) Similarly, with I /I 0, ln0. 0 x / 0. µ µ C.
5. The linear absrptin cefficient fr -MeV gamma rays in lead is 78 m -. find the thickness f lead required t reduce by half the intensity f a beam f such gamma rays. Frm either Equatin (.6) r Prblem - abve, ln 0. 69 x / 8. 9 mm µ - 78 m 7. The linear absrptin cefficients fr.0-mev gamma rays are.9 m - in water and 5 in - in lead. What thickness f water wuld give the same shielding fr such gamma rays as 0 mm f lead? Rather than calculating the actual intensity ratis, Equatin (.6) indicates that the ratis will be the same when the distances in water and lead are related by µ x µ x, r H O Pb r cm tw significant figures. x x µ µ H H Pb O O H O ( 0 0 Pb Pb - 5 m m). 9 m - 0. 06 m
9. What thickness f cpper is needed t reduce the intensity f the beam in Exercise 8 by half. Either a direct applicatin f Equatin (.6) r use f the result f Prblem - gives ln 5 x/. 7 0 m, -. 7 0 m which is 0.05 mm t tw significant figures. 5. The sun's mass is.0 x 0 0 kg and its radius is 7.0 x 0 8 m. Find the apprximate gravitatinal red shift in light f wavelength 500 nm emitted by the sun. In Equatin (.9), the rati 0 6. 67 0 N m / kg)(. 0 0 kg) 6 0 8 - GM (. c R (. 0 0 m/s) ( 7. 0 0 m ) (keeping an extra significant figure) is s small that fr an apprximate red shift, the rati λ/λ will be the same as ν/ν, and λ GM λ c R ( 500 0 9 m)(. 0-6 ).06 0 - m.06 pm.
5. As discussed in Chap., certain atmic nuclei emit phtns in underging transitins frm "excited" energy states t their grund r nrmal states. These phtns cnstitute gamma 57 rays. When a nucleus emits a phtn, it recils in the ppsite directin. (a) The 7C nucleus 57 decays by K capture t 6 Fe, which then emits a phtn in lsing. kev t reach its grund 57 state. The mass f a 6 Fe atm is 9.5 x 0-6 kg. By hw much is the phtn energy reduced frm the full. kev available as a result f having t share energy and mmentum with the reciling atm? (b) In certain crystals the atms are s tightly bund that the entire crystal recils when a gamma-ray phtn is emitted, instead f the individual atm. This phenmenn is knwn as the Mössbauer effect. By hw much is the phtn energy reduced in this situatin if the ex- cited 576Fe nucleus is part f a.0-g crystal? (c) The essentially recil-free emissin f gamma rays in situatins like that f b means that it is pssible t cnstruct a surce f virtually mnenergetic and hence mnchrmatic phtns. Such a surce was used in the experiment described in See..9. What is the riginal frequency and the change in frequency f a.-kev gamma-ray phtn after it has fallen 0 m near the earth's surface? (a) The mst cnvenient way t d this prblem, fr cmputatinal purpses, is t realize that the nucleus will be mving nnrelativistically after the emissin f the phtn, and that the energy f the phtn will be very clse t E. kev, the energy that the phtn wuld have if the nucleus had been infinitely massive. S, if the phtn has an energy E, the recil mmentum f the nucleus is E/c, and its kinetic energy is p / M E /( Mc ), here M is the rest mass f the nucleus. Then, cnservatin f energy implies
E + E E Mc. This is a quadratic in E, and slutin might be attempted by standard methds, but t find the change in energy due t the finite mass f the nucleus, and recgnizing that E will be very clse t E, the abve relatin may be expressed as E E E E Mc Mc (. kev) (. 60 0 J/keV) 6. 9 0 kev -6 8 (9.5 0 kg)(.0 0 m/s) If the apprximatin E E, is nt made, the resulting quadratic is which is slved fr E E 6 + Mc E Mc E E Mc +. Mc 0,.9 0 Hwever, the dimensinless quantity E /(Mc ) is s small that standard calculatrs are nt able t determine the difference between E and E. The square rt must be expanded, using ( + x) / + (x/) - (x /8), and tw terms must be kept t find the difference between E and E. This apprximatin gives the previus result. ev.
It s happens that a relativistic treatment f the reciling nucleus gives the same numerical result, but withut intermediate apprximatins r slutin f a quadratic equatin. The relativistic frm expressing cnservatin f energy is, with pc E and befre, + E ( Mc ) + E Mc + E, r E + ( Mc ) Mc + E E. Squaring bth sides, canceling E and (Mc ), and then slving fr E, E + ( /( )). ( ) ( /( )) + Mc E E Mc E E + Mc E + E Mc Frm this frm, E E E, Mc + E /( Mc ) giving the same result. (b) Fr this situatin, the abve result applies, but the nnrelativistic apprximatin is by far the easiest fr calculatin; 9 E (. 0 ev) (. 6 0 J/eV) 5 E E. 8 0 ev. - 8 Mc (.0 0 kg)(.0 0 m/s) E 0 ev 8 (c) The riginal frequency is. ν 8 0 Hz. h 5. 0 ev s. Frm Equatin (.8), the change in frequency is gh ( 9. 8 m/s )( 0 m) 8 ν ν ν (. 8 0 Hz) 7. 6 Hz. ν 8 c (.0 0 m/s)
55. The gravitatinal ptential energy U relative t infinity f a bdy f mass m at a distance R frm the center f a bdy f mass M is U -GmM/R. (a) If R is the radius f the bdy f mass M, find the escape speed v, f the bdy, which is the minimum speed needed t leave it permanently. (b) Obtain a frmula fr the Schwarzschild radius f the bdy by setting v c c, the speed f light, and slving fr R. (Of curse, a relativistic calculatin is crrect here, but it is interesting t see what a classical calculatin prduces.) (a) T leave the bdy f mass M permanently, the bdy f mass m must have enugh kinetic energy s that there is n radius at which its energy is psitive. That is, its ttal energy must be nnnegative. The escape velcity v e is the speed (fr a given radius, and assuming M >> m) that the bdy f mass m wuld have fr a ttal energy f zer; GMm GM mve 0, r ve. R R (b) Slving the abve expressin fr R in terms f v e, GM R, and if v e c, Equatin (.0) is btained. v e
Chapter. Prblem Slutins. A phtn and a particle have the same wavelength. Can anything be said abut hw their linear mmenta cmpare? Abut hw the phtn's energy cmpares with the particle's ttal energy? Abut hw the phtn s energy cmpares with the particle's kinetic energy? Frm Equatin (.), any particle s wavelength is determined by its mmentum, and hence particles with the same wavelength have the same mmenta. With a cmmn mmentum p, the phtn s energy is pc, and the particle s energy is ( pc ) + ( mc ), which is necessarily greater than pc fr a massive particle. The particle s kinetic energy is K E mc ( pc) + ( mc ) mc Fr lw values f p (p<<mc fr a nnrelativistic massive particle), the kinetic energy is K p /m, which is necessarily less than pc. Fr a relativistic massive particle, K pc mc, and K is less than the phtn energy. The kinetic energy f a massive particle will always be less than pc, as can be seen by using E (pc) + (mc ) t btain ( pc ) K Kmc.
Chapter. Prblem Slutins. Find the de Brglie wavelength f a.0-mg grain f sand blwn by the wind at a speed f 0 m/s. Fr this nnrelativistic case, h 6. 6 0 J s 9 λ. 0 m; mv 6 (. 0 0 kg)(0 m/s) quantum effects certainly wuld nt be nticed fr such an bject. 5. By what percentage will a nnrelativistle calculatin f the de Brglie wavelength f a 00-keV electrn be in errr? Because the de Brglie wavelength depends nly n the electrn's mmentum, the percentage errr in the wavelength will be the same as the percentage errr in the reciprcal f the mmentum, with the nnrelativistic calculatin giving the higher wavelength due t a lwer calculated mmentum. The nnrelativistic mmentum is p nr mk. 7 0 and the relativistic mmentum is ( 9. 0 kg m/ s, kg)(00 0 ev)(.6 0-9 J/eV) p r ( K + mc ) ( mc ) ( 0. 00 + ( 0. 5) MeV / c. 79 0 kg m/s, c
Chapter. Prblem Slutins keeping extra figures in the intermediate calculatins. The percentage errr in the cmputed de Brglie wavelength is then ( h / pnr ) ( h / pr ) p. 79. 7 r pnr. 8 %. h / p p. 7 7. The atmic spacing in rck salt, NaCl, is 0.8 nm. Find the kinetic energy (in ev) f a neutrn with a de Brglie wavelength f 0.8 nm. Is a relativistic calculatin needed? Such neutrns can be used t study crystal structure. A nnrelativistic calculatin gives r ( hc / λ) nr p ( hc) (. 0 ev m) K. 0 0 ev m 6-9 mc mc λ ( 99. 6 0 ev)(0.8 0 m) (Nte that in the abve calculatin, multiplicatin f numeratr and denminatr by c and use f the prduct hc in terms f electrnvlts avided further unit cnversin.) This energy is much less than the neutrn's rest energy, and s the nnrelativistic calculatin is cmpletely valid. 6
Chapter. Prblem Slutins 9. Green light has a wavelength f abut 550 nm. Thrugh what ptential difference must an electrn be accelerated t have this wavelength? A nnrelativistic calculatin gives 6 p ( hc / λ) ( hc) (. 0 ev m) 6 K 5. 0 0 ev, m -9 mc ( mc ) λ ( 5 0 ev)(550 0 m) s the electrn wuld have t be accelerated thrugh a ptential difference f 5.0 x 0-6 V 5.0 µv. Nte that the kinetic energy is very small cmpared t the electrn rest energy, s the nnrelativistic calculatin is valid. (In the abve calculatin, multiplicatin f numeratr and denminatr by c and use f the prduct he in terms f electrnvlts avided further unit cnversin.). Shw that if the ttal energy f a mving particle greatly exceeds its rest energy, its de Brglie wavelength is nearly the same as the wavelength f a phtn with the same ttal energy. If E (pc) + (mc ) >> (mc ), then pc >> mc and E pc. Fr a phtn with the same energy, E pc, s the mmentum f such a particle wuld be nearly the same as a phtn with the same energy, and s the de Brglie wavelengths wuld be the same.
Chapter. Prblem Slutins. An electrn and a prtn have the same velcity Cmpare the wavelengths and the phase and grup velcities f their de Brglie waves. Fr massive particles f the same speed, relativistic r nnrelativistic, the mmentum will be prprtinal t the mass, and s the de Brglie wavelength will be inversely prprtinal t the mass; the electrn will have the lnger wavelength by a factr f (m p /m e ) 88. Frm Equatin (.) the particles have the same phase velcity and frm Equatin (.6) they have the same grup velcity. 5. Verify the statement in the text that, if the phase velcity is the same fr all wavelengths f a certain wave phenmenn (that is, there is n dispersin), the grup and phase velcities are the same. Suppse that the phase velcity is independent f wavelength, and hence independent f the wave number k; then, frm Equatin (.), the phase velcity v p (ω/k) u, a cnstant. It fllws that because ω uk, d v g ω u v p. dk
Chapter. Prblem Slutins 7. The phase velcity f cean waves is gλ/, where g is the acceleratin f gravity. Find the grup velcity f cean waves The phase velcity may be expressed in terms f the wave number k /λ as ω g v p, r ω gk r ω gk. k k Finding the grup velcity by differentiating ω(k) with respect t k, dω g ω vg g v p. dk k k k Using implicit differentiatin in the frmula fr ω (k), dω ω ωvg g, dk g gk ω ω s that vg v p, ω ωk ωk k the same result. Fr thse mre cmfrtable with calculus, the dispersin relatin may be expressed as ln( ω) ln( k ) + ln( g), dω dk ω frm which, and vg v p. ω k k
Chapter. Prblem Slutins 9. Find the phase and grup velcities f the de Brglie waves f an electrn whse kinetic energy is 500 kev. K + mc 500 + 5 Fr a kinetic energy f 500 kev, γ. 978. v / c mc 5 Slving fr v, v c ( / γ ) c ( /. 978) 0. 86c, and frm Equatin (.6), v g v 0.86c. The phase velcity is then v p c /v g.6 c.. (a) Shw that the phase velcity f the de Brglie waves f a particle f mass m and de Brglie wavelength λ is given by mcλ v p c + h (b) Cmpare the phase and grup velcities f an electrn whse de Brglie wavelength is exactly x 0 - m. (a) Tw equivalent methds will be presented here. Bth will assume the validity f Equatin (.6), in that v g v. First: Express the wavelength x in terms f v g, h λ p h mv γ g h mv g v c g.
Multiplying by mv g, squaring and slving fr v g gives. ) / ( ) ( + + h c m c c h m h v g λ λ Taking the square rt and using Equatin (.), v p c /v g, gives the desired result. Secnd: Cnsider the particle energy in terms f v p c lv g ; ( ) ( ) ( ) ( ). / ) ( mc hc v c mc mc mc pc E p + + λ γ Dividing by (mc ) leads t s that, /( λ) mc h v c p +, / ) ( ) ( ) ( ) /( h mc mc h mc h mc h v c p λ λ λ λ + + + which is an equivalent statement f the desired result. It shuld be nted that in the first methd presented abve culd be used t find λ in terms f v p directly, and in the secnd methd the energy culd be fund in terms f v g. The final result is, r curse, the same.
(b) Using the result f part (a), 8 - ( 9. 0 kg)(.0 0 m/s)(.0 0 m) v p c +. 00085c, 6. 6 0 J s and v g c /v p 0.9995c. Fr a calculatinal shrtcut, write the result f part (a) as - mc λ ( 5 0 ev)(.00 0 m) v p c + c. 00085c. hc + 6. 0 ev m In bth f the abve answers, the statement that the de Brglie wavelength is exactly 0 - m means that the answers can be given t any desired precisin.. What effect n the scattering angle in the Davissn-Germer experiment des increasing the electrn energy have? Increasing the electrn energy increases the electrn's mmentum, and hence decreases the electrn's de Brglie wavelength. Frm Equatin (.), a smaller de Brglie wavelength results in a smaller scattering angle.
Chapter. Prblem Slutins 5. In Sec..5 it was mentined that the energy f an electrn entering a crystal increase, which reduces its de Brglie wavelength. Cnsider a beam f 5-eV electrns directed at a nickel target. The ptential energy f an electrn that enters the target changes by 6 ev. (a) Cmpare the electrn speeds utside and inside the target. (b) Cmpare the respective de Brglie wavelengths. (a) Fr the given energies, a nnrelativistic calculatin is sufficient; v K m ( 5 ev)(.60 0-9 J/eV). 6 m/s 9. 0 kg utside the crystal, and (frm a similar calculatin, with K 80 ev), v 5.0 x 0 6 m/s inside the crystal (keeping an extra significant figure in bth calculatins). (b) With the speeds fund in part (a), the de Brgile wavelengths are fund frm 6 0 h h. 6 0 J s λ. 67 0 p mv 6 ( 9. 0 kg)(.6 0 m/s) r 0.67 nm utside the crystal, with a similar calculatin giving 0.7 nm inside the crystal. m,
Chapter. Prblem Slutins 7. Obtain an expressin fr the energy levels (in MeV) f a neutrn cnfined t a ne-dimensinal bx.00 x 0 - m wide. What is the neutrn's minimum energy? (The diameter f an atmic nucleus is f this rder f magnitude.) Frm Equatin (.8), h ( 6. 6 0 J s) E n n n n. 8 0 J n 7-8mL 8(. 67 0 kg)(.00 0 m) The minimum energy, crrespnding t n, is 0.5 MeV 9. A prtn in a ne-dimensinal bx has an energy f 00 kev in its first excited state. Hw wide is the bx? The first excited state crrespnds t n in Equatin (.8). Slving fr the width L, 0. 5 MeV. L n h 8mE 8(. 67 0 7 ( 6. 6 0 kg)(00 0 J s) ev)(.60 0-9 J/eV). 5 0 m 5. fm.
Chapter. Prblem Slutins. The atms in a slid pssess a certain minimum zer-pint energy even at 0 K, while n such restrictin hlds fr the mlecules in an ideal gas. Use the uncertainty principle t explain these statements. Each atm in a slid is limited t a certain definite regin f space - therwise the assembly f atms wuld nt be a slid. The uncertainty in psitin f each atm is therefre finite, and its mmentum and hence energy cannt be zer. The psitin f an ideal-gas mlecule is nt restricted, s the uncertainty in its psitin is effectively infinite and its mmentum and hence energy can be zer.. The psitin and mmentum f a.00-kev electrn are simultaneusly determined. If its psitin is lcated t within 0.00 nm, what is the percentage f uncertainty in its mmentum? The percentage uncertainty in the electrn's mmentum will be at least p h h hc p p x x mk x ( mc) K (. 00 0 0 (. 0 m) 6 (5 0 ev m) ev)(.00 0 ev). 0. %. Nte that in the abve calculatin, cnversin f the mass f the electrn int its energy equivalent in electrnvlts is purely ptinal; cnverting the kinetic energy int jules and using h 6.66 x 0 - J s will f curse give the same percentage uncertainty.
Chapter. Prblem Slutins 5. Hw accurately can the psitin f a prtn with v << c be determined withut giving it mre than.00 kev f kinetic energy? The prtn will need t mve a minimum distance h v t v, E where v can be taken t be K E v, s that m m K h h hc v t m E mk ( mc ) K. 0 ( 98 0 6 6 ev m MeV)(.00 0 ev). 0 m 0. pm. (See nte t the slutin t Prblem - abve). The result fr the prduct v t may be recgnized as v t h/p; this is nt incnsistent with Equatin (.), x p h/. In the current prblem, E was taken t be the (maximum) kinetic energy f the prtn. In such a situatin, ( p ) p E p v p m m, which is cnsistent with the previus result.
7. A marine radar perating at a frequency f 900 MHz emits grups f electrmagnetic waves 0.0800 µs in duratin. The time needed fr the reflectins f these grups t return indicates the distance t a target. (a) Find the length f each grup and the number f waves it cntains. (b) What is the apprximate minimum bandwidth (that is, spread f frequencies) the radar receiver must be able t prcess? (a) The length f each grup is c t (. 0 0 8 m/s)(8.0 0-5 s) m. The number f waves in each grup is the pulse duratin divided by the wave perid, which is the pulse duratin multiplied by the frequency, 8 6 ( 8. 0 0 s)(900 0 Hz) 75 waves. (b) The bandwidth is the reciprcal f the pulse duratin, 8 - ( 8. 0 0 s). 5 MHz.
Chapter. Prblem Slutins. / / ν m C 9. The frequency f scillatin f a harmnic scillatr f mass m and spring cnstant C is The energy f the scillatr is E p /m + Cx /, where p is its mmentum when its displacement frm the equilibrium psitin is x. In classical physics the minimum energy f the scillatr is E min 0. Use the uncertainty principle t find an expressin fr E in terms f x nly and shw that the minimum energy is actually E min hν/ by setting de/dx 0 and slving fr E min. T use the uncertainty principle, make the identificatin f p with p and x with x, s that p h/ (x), and. ) ( 8 x C x m h x E E + Differentiating with respect t x and setting, 0 E dx d 0, + Cx x m h which is slved fr. mc h x Substutin f this value int E(x) gives. min 8 ν h m C h mc h C h mc m h E +
Chapter. Prblem Slutins. The great majrity f alpha particles pass thrugh gases and thin metal fils with n deflectins. T what cnclusin abut atmic structure des this bservatin lead? The fact that mst particles pass thrugh undetected means that there is nt much t deflect these particles; mst f the vlume f an atm is empty space, and gases and metals are verall electrically neutral.. Determine the distance f clsest apprach f.00-mev prtns incident n gld nuclei. Fr a "clsest apprach", the incident prtn must be directed "head-n" t the nucleus, with n angular mmentum with respect t the nucleus (an "Impact parameter" f zer; see the Appendix t Chapter ). In this case, at the pint f clsest apprach the prtn will have n kinetic energy, and s the ptential energy at clsest apprach will be the initial kinetic energy, taking the ptential energy t be zer in the limit f very large separatin. Equating these energies, r min ε Ze K initial K initial ( 8. 99 0 9 N m / C Ze ε r min, r ( 79)(. 60 0 ). 60 0 9 J C). 0 m.
5. What is the shrtest wavelength present in the Brackett series f spectral lines? The wavelengths in the Brackett series are given in Equatin (.9); the shrtest wavelength (highest energy) crrespnds t the largest value f n. Fr n, 6 6 6 λ. 6 0 m.6 µ m R 7 -. 097 0 m 7. In the Bhr mdel, the electrn is in cnstant mtin. Hw can such an electrn have a negative amunt f energy? While the kinetic energy f any particle is psitive, the ptential energy f any pair f particles that are mutually attracted is negative. Fr the system t be bund, the ttal energy, the sum f the psitive kinetic energy and the ttal negative ptential energy, must be negative. Fr a classical particle subject t an inverse-square attractive frce (such as tw ppsitely charged particles r tw unifrm spheres subject t gravitatinal attractin in a circular rbit, the ptential energy is twice the negative f the kinetic energy.
9. The fine structure cnstant is defined as α e /ε hc. This quantity gt its name because it first appeared in a thery by the German physicist Arnld Smmerfeld that tried t explain the fine structure in spectral lines (multiple lines clse tgether instead f single lines) by assuming that elliptical as well as circular rbits are pssible in the Bhr mdel. Smmerfeld's apprach was n the wrng track, but α has nevertheless turned ut t be a useful quantity in atmic physics. (a) Shw that α v /c, where v, is the velcity f the electrn in the grund state f the Bhr atm. (b) Shw that the value f α is very clse t /7 and is a pure number with n dimensins. Because the magnetic behavir f a mving charge depends n its velcity, the small value f α is representative f the relative magnitudes f the magnetic and electric aspects f electrn behavir in an atm. (c) Shw that αa λ c /, where a is the radius f the grund-state Bhr rbit and λ c is the Cmptn wavelength f the electrn. (a) The velcity v, is given by Equatin (.), with r r a. Cmbining t find v, v (b) Frm the abve, α e ε ma e ε ε h m me e ε h v e, s α. c ε h c 9 ( 60 0 C) 8 (. 85 0 C / N m )( 6. 6 0 J s)(. 00 0 m/s) 8. 7. 0 0,
s that /α 7. t fur significant figures. A clse cheek f the units is wrthwhile; treating the units as algebraic quantities the units as given in the abve calculatin are [C ] [N][m [C ] [m] [ J][ s] ] [s] [N m] [J]. Thus, α is a dimensinless quantity, and will have the same numerical value in any system f units. The mst accurate (Nvember, 00) value f /α is 7.0599976, α accurate t better than parts per billin. (c) Using the abve expressin fr α and Equatin (.) with n fr a, e h ε α h a ε hc me mc where the Cmptn wavelength λ C is given by Equatin (.). λc,
. Find the quantum number that characterizes the earth's rbit arund the sun. The earth's mass is 6.0 x 0 kg, its rbital radius is.5 x 0 m, and its rbital speed is.0 x 0 m/s. With the mass, rbital speed and rbital radius f the earth knwn, the earth's rbital angular mmentum is knwn, and the quantum number that wuld characterize the earth's rbit abut the sun wuld be this angular mmentum divided by ; L mvr ( 6. 0 0 kg)(.0 0 m/s)(.5 0 m) 7 n. 6 0. h h. 06 0 J s (The number f significant figures nt f cncern.). Cmpare the uncertainty in the mmentum f an electrn cnfined t a regin f linear dimensin a with the mmentum f an electrn in a grund-state Bhr rbit. The uncertainty in psitin f an electrn cnfined t such a regin is, frm Equatin (.), p > /a, while the magnitude f the linear mmentum f an electrn in the first Bhr rbit is h h h p ; λ a a the value f p fund frm Equatin (.) is half f this mmentum.
5. What effect wuld yu expect the rapid randm mtin f the atms f an excited gas t have n the spectral lines they prduce? The Dppler effect shifts the frequencies f the emitted light t bth higher and lwer frequencies t prduce wider lines than atms at rest wuld give rise t. 7. A prtn and an electrn, bth at rest initially, cmbine t frm a hydrgen atm in the grund state. A single phtn is emitted in this prcess. What is its wavelength? It must assumed that the initial electrstatic ptential energy is negligible, s that the final energy f the hydrgen atm is E -.6 ev. The energy f the phtn emitted is then -E l, and the wavelength is hc λ E 6 8. 0 ev m. 6 ev 9. 0 m 9. nm, in the ultravilet part f the spectrum (see, fr instance, the back endpapers f the text).
9. Find the wavelength f the spectral line that crrespnds t a transitin in hydrgen frm the n 0 state t the grund state. In what part f the spectrum is this? Frm either Equatin (.7) with n 0 r Equatin (.8) with n f and n i 0, 00 00 8 λ 9. 0 m 9. nm, 99 R 99 7 -. 097 0 m which is in the ultravilet part f the spectrum (see, fr instance, the back endpapers f the text).. A beam f electrns bmbards a sample f hydrgen. Thrugh what ptential difference must the electrns have been accelerated if the first line f the Balmer series is t be emitted? The electrns energy must be at least the difference between the n and n levels, 8 E E E E (. 6 ev). ev 9 9 (this assumes that few r nne f the hydrgen atms had electrns in the n level). A ptential difference f. ev is necessary t accelerate the electrns t this energy.
. The lngest wavelength in the Lyman series is.5 nm and the shrtest wavelength in the Balmer series is 6.6 nm. Use the figures t find the lngest wavelength f light that culd inize hydrgen. The energy needed t inize hydrgen will be the energy needed t raise the energy frm the grund state t the first excited state plus the energy needed t inize an atm in the secnd excited state; these are the energies that crrespnd t the lngest wavelength (least energetic phtn) in the Lyman series and the shrtest wavelength (mst energetic phtn) in the Balmer series. The energies are prprtinal t the reciprcals f the wavelengths, and s the wavelength f the phtn needed t inize hydrgen is λ λ + λ As a check, nte that this wavelength is R -.. 5 nm + 6. 6 nm 9. nm. 5. An excited hydrgen atm emits a phtn f wavelength λ in returning t the grund state. (a) Derive a frmula that gives the quantum number f the initial excited state in terms f λ and R. (b) Use this frmula t find n i fr a 0.55-nm phtn. (a) Frm Equatin (.7) with n n i, R, λ n i which is slved fr
n i / λr λr. λr (b) Either f the abve frms gives n very clse (fur place) t ; specifically, with the prduct λr (0.55x0-9 m)(.097x0 7 m - ).5 runded t fur places as 9/8, n exactly. 7. When an excited atm emits a phtn, the linear mmentum f the phtn must be balanced by the recil mmentum f the atm. As a result, sme f the excitatin energy f the atm ges int the kinetic energy f its recil. (a) Mdify Eq. (.6) t include this effect. (b) Find the rati between the recil energy and the phtn energy fr the n n transitin in hydrgen, fr which E f - E i.9 ev. Is the effect a majr ne? A nnrelativistic calculatin is sufficient here. (a) A relativistic calculatin wuld necessarily invlve the change in mass f the atm due t the change in energy f the system. The fact that this mass change is t small t measure (that is, the change is measured indirectly by measuring the energies f the emitted phtns) means that a nnrelativistic calculatin shuld suffice. In this situatin, the kinetic energy f the reciling atm is p ( hν / c) K, M M where m is the ftequency f the emitted phtn and p h/λ hν/c is the magnitude f the mmentum f bth the phtn and the reciling atm. Equatin (.6) is then
( hν) hν Ei E f hν + K hν + hν +. Mc Mc This result is equivalent t that f Prblem -5, where hν E. and the term p /(M) crrespnds t E - E in that prblem. As in Prblem -5, a relativistic calculatin is manageable; the result wuld be Mc E f Ei hν +, + hν a frm nt ften useful; see part (b). (b) As indicated abve and in the prblem statement, a nnrelativistle calculatin is sufficient. As in part (a), ( E c ) p / K E. 9 ev 9 K, and. 0 0, M M E 6 Mc ( 99 0 ev) r.0 x 0-9 t tw significant figures. In the abve, the rest energy f the hydrgen atm is frm the frnt endpapers.
9. Shw that the frequency f the phtn emitted by a hydrgen atm in ging frm the level n + t the level n is always intermediate between the frequencies f revlutin f the electrn in the respective rbits. There are many equivalent algebraic methds that may be used t derive Equatin (.9), and that result will be cited here;. n h E f n The frequency v f the phtn emitted in ging frm the level n + t the level n is btained frm Equatin (.7) with n i n + and n f n;. ) ( ) ( + + + n n n h E n n h E ν This can be seen t be equivalent t the expressin fr v in terms f n and p that was fund in the derivatin f Equatin (.0), but with n replaced by n + and p. Nte that in this frm, ν is psitive because E l is negative. Frm this expressin n f n, n n n n f n n n n hn E < + + + + + + ν as the term in brackets is less than. Similarly,
ν ( )( ) ( E n + n + n + )( n + ) fn h( n + ) n n + > fn+ as the term in brackets is greater than.,. A µ - mun is in the n state f a munic atm whse nucleus is a prtn. Find the wavelength f the phtn emitted when the munic atm drps t its grund state. In what part f the spectrum is this wavelength? Fr a munic atm, the Rydberg cnstant is multiplied by the rati f the reduced masses f the muninc atm and the hydrgen atm, R' R (m'/m e ) 86R, as in Example.7; frm Equatin (.7), / / 0 λ 6. 5 0 m 0.65nm, R 7-86(. 097 0 m ) in the x-ray range. µ p µ 07 me, m p me m 86me mµ + m p m 86 m m
. A mixture f rdinary hydrgen and tritium, a hydrgen istpe whse nucleus is apprximately times mre massive than rdinary hydrgen, is excited and its spectrum bserved. Hw far apart in wavelength will the H α lines f the tw kinds f hydrgen be? The H α lines, crrespnding t n in Equatin (.6), have wavelengths f λ (6/5) (/R). Fr a tritium atm, the wavelength wuld be λ T (6/5) (/RT), where RT is the Rydberg cnstant evaluated with the reduced mass f the tritium atm replacing the reduced mass f the hydrgen atm. The difference between the wavelengths wuld then be λ T R λ λ λt λ λ λ. RT The values f R and RT are prprtinal t the respective reduced masses, and their rati is R memh /( me + mh ) mh( me + mt ). RT memt /( me + mt ) mt( me + mh ) Using this in the abve expressin fr λ, me( mt mh ) m λ λ e λ, me( me mh ) + m H where the apprximatins m e + rn H m H and m T m H have been used. Inserting numerical values, ( 6/ 5) ( 9. 0 kg) 0 λ. 8 0 7-7 (. 097 0 m ) (. 67 0 kg) m 0.8 nm.
5. (a) Derive a frmula fr the energy levels f a hydrgenic atm, which is an in such as He + r Li + whse nuclear charge is +Ze and which cntains a single electrn. (b) Sketch the energy levels f the He' in and cmpare them with the energy levels f the H atm. (c) An electrn jins a bare helium nucleus t frm a He + in. Find the wavelength f the phtn emitted in this prcess if the electrn is assumed t have had n kinetic energy when it cmbined with the nucleus. (a) The steps leading t Equatin (.5) are repeated, with Ze instead f e and Z e instead f e, giving m Z e E, n 8ε h n where the reduced mass m' will depend n the mass f the nucleus. (b) A plt f the energy levels is given belw. The scale is clse, but nt exact, and f curse there are many mre levels crrespnding t higher n. In the apprximatin that the reduced masses are the same, fr He +, with Z, the n level is the same as the n level fr Hydrgen, and the n level is the same as the n level fr hydrgen.
The energy levels fr H and He + : (c) When the electrn jins the Helium nucleus, the electrn-nucleus system lses energy; the emitted phtn will have lst energy E (-.6 ev) -5. ev, where the result f part (a) has been used. The emitted phtn's wavelength is λ hc E 6 8. 0 ev m 5. ev. 8 0 m.8 nm.
7. A certain ruby laser emits.00-j pulses f light whse wavelength is 69 nm. What is the minimum number f Cr + ins in the ruby? The minimum number f Cr + ins will he the minimum number f phtns, which is the ttal energy f the pulse divided by the energy f each phtn, E Eλ hc / λ hc -9 (. 00 J)(69 0 m) 8. 9 0 ins. 8 ( 6. 6 0 J s)(.0 0 m/s) 9. The Rutherfrd scattering frmula fails t agree with the data at very small scattering angles. Can yu think f a reasn? Small angles crrespnd t particles that are nt scattered much at all, and the structure f the atm des nt affect these particles. T these nnpenetrating particles, the nucleus is either partially r cmpletely screened by the atm's electrn clud, and the scattering analysis, based n a pintlike psitively charged nucleus, is nt applicable.
. A 5.0-MeV alpha particle appraches a gld nucleus with an impact parameter f.6 x 0 - m. Thrugh what angle will it be scattered? Frm Equatin (.9), using the value fr /ε given in the frnt endpapers, ct θ - ( 5. 0 ev)(.60 0 J/MeV) 9 (8.99 0 N m / C )( 79)(. 60 0 9 keeping extra significant figures. The scattering angle is then θ ct (. ) tan C). (. 6 0 0. m).,. What fractin f a beam f 7.7-MeV alpha particles incident upn a gld fil.0 x 0-7 m thick is scattered by less than? The fractin scattered by less than is - f, with f given in Equatin (.);
f Ze nt εk ( 5. 90 0 8 ct m - )(. 0 0 ( 79)(. 6 0 ( 7. 7MeV)(.6 0 θ nt ε 9-7 C) J/MeV) m)(9.0 0 Ze K ct 9 ( 0. 5 ct N m ) θ / C 0. 6, where n, the number f gld atms per unit vlume, is frm Example.8. The fractin scattered by less than is - f 0.8. ) 5. Shw that twice as many alpha particles are scattered by a fil thrugh angles between 60 and 90 as are scattered thrugh angles f 90 r mre. Regarding f as a functin f 0 in Equatin (.), the number f particles scattered between 60 and 90 is f (60 ) - f (90 ), and the number scattered thrugh angles greater than 90 is just f (90 ), and f ( 60 ) f ( 90 ) ct ( 0 ) ct ( 5 ), f ( 90 ) ct ( 5 ) s twice as many particles are scattered between 60 and 90 than are scattered thrugh angles greater than 90.
7. In special relativity, a phtn can be thught f as having a mass f m E ν /c. This suggests that we can treat a phtn that passes near the sun in the same way as Rutherfrd treated an alpha particle that passes near a nucleus, with an attractive gravitatinal frce replacing the repulsive electrical frce. Adapt Eq. (.9) t this situatin and find the angle f deflectin θ fr a phtn that passes b R sun frm the center f the sun. The mass and radius f the sun are respectively.0 x 0 0 kg and 7.0 x 0 8 m. In fact, general relativity shws that this result is exactly half the actual deflectin, a cnclusin supprted by bservatins made during slar clipses as mentined in Sec..0. If gravity acted n phtns as if they were massive bjects with mass m E v /c, the magnitude f the frce F in Equatin (.8) wuld be GM m F sun ; r the factrs f r wuld cancel, as they d fr the Culmb frce, and the result is θ θ θ c b mc b sin GMsunm cs and ct GM a result that is independent f the phtn s energy. Using b R sun, θ tan. 0 GM c R sun sun deg tan 0. 87. sun ( 6. 67 0 N m / kg )(. 0 0 8 8 (.0 0 m/s)(7.0 0 m), 0 kg)
Chapter 5 Prblem Slutins. Which f the wave functins in Fig. 5.5 cannt have physical significance in the interval shwn? Why nt? Figure (b) is duble valued, and is nt a functin at all, and cannt have physical significance. Figure (c) has discntinuus derivative in the shwn interval. Figure (d) is finite everywhere in the shwn interval. Figure (f) is discntinuus in the shwn interval.. Which f the fllwing wave functins cannt be slutins f Schrödinger's equatin fr all values f x? Why nt? (a) ψ A sec x; (b) ψ A tan x; (c) ψ A exp(x ); (d) ψ A exp(-x ). The functins (a) and (b) are bth infinite when cs x 0, at x ±/, ±/, ±(n+)/ fr any integer n, neither ψ A sec x r ψ A tan x culd be a slutin f Schrödinger's equatin fr all values f x. The functin (c) diverges as x ±, and cannt be a slutin f Schrödinger's equatin fr all values f x.
5. The wave functin f a certain particle is ψ A cs x fr -/ < x < /. (a) Find the value f A. (b) Find the prbability that the particle be fund between x 0 and x /. Bth parts invlve the integral cs xdx, evaluated between different limits fr the tw parts. Of the many ways t find this integral, including cnsulting tables and using symblicmanipulatin prgrams, a direct algebraic reductin gives [ ] [ ] [ ], cs cs ) cs ( cs ) ( cs cs ) cs ( ) (cs cs x x x x x x x x x 8 8 + + + + + + + + where the identity cs θ ½(+cs θ) has been used twice. (a) The needed nrmalizatin cnditin is [ ] 8 8 + + + + + + + / / / / / / / / / / cs cs cs ψ ψ xdx xdx dx A xdx A dx The integrals and / / / / / / / / sin cs sin cs + + + + x xdx x xdx are seen t be vanish, and the nrmalizatin cnditin reduces t., 8 r 8 A A
(b) Evaluating the same integral between the different limits, / 0 [ / x + sinx + sin x ] +, cs xdx 0 The prbability f the particle being fund between x 0 and x / is the prduct f this integral and A, r A 8 ( + ) ( + ) 0. 6 8 7. As mentined in Sec. 5., in rder t give physically meaningful results in calculatins a wave functin and its partial derivatives must be finite, cntinuus, and single-valued, and in additin must be nrmalizable. Equatin (5.9) gives the wave functin f a particle mving freely (that is, with n frces acting n it) in the +x directin as Ψ Ae ( i / h)( Etpc) where E is the particle's ttal energy and p is its mmentum. Des this wave functin meet all the abve requirements? If nt, culd a linear superpsitin f such wave functins meet these requirements? What is the significance f such a superpsitin f wave functins? The given wave functin satisfies the cntinuity cnditin, and is differentiable t all rders with respect t bth t and x, but is nt nrmalizable; specifically, Ψ Ψ A * A is cnstant in bth space and time, and if the particle is t mve freely, there can be n limit t its range, and s the integral f Ψ Ψ ver an infinite regin cannt be finite if A 0.
A linear superpsitin f such waves culd give a nrmalizable wave functin, crrespnding t a real particle. Such a superpsitin wuld necessarily have a nn- zer p, and hence a finite x; at the expense f nrmalizing the wave functin, the wave functin is cmpsed f different mmentum states, and is lcalized. 9. Shw that the expectatin values <px> and <xp>) are related by <px> - <xp> /i This result is described by saying that p and x d nt cmmute, and it is intimately related t the uncertainty principle. It's crucial t realize that the expectatin value <px> is fund frm the cmbined peratr pˆ ˆx, which, when perating n the wave functin Ψ(x, t), crrespnds t "multiply by x, differentiate with respect t x and multiply by /i," whereas the peratr xˆ ˆp crrespnds t "differentiate with respect t x, multiply by /i and multiply by x." Using these peratrs, h h ( px ˆˆ) Ψ pˆ( xˆ Ψ) ( xψ) Ψ + x Ψ, i x i x where the prduct rule fr partial differentiatin has been used. Als, h h ( xp ˆˆ) Ψ xˆ( pˆ Ψ) x Ψ x Ψ. i x i x
Thus Ψ Ψ i xp px h ˆˆ) ˆˆ ( and i dx i dx i xp px h h h Ψ Ψ Ψ Ψ > < * * fr Ψ(x, t) nrmalized.. Obtain Schrödinger s steady-state equatin frm Eq.(.5) with the help f de Brglie s relatinship λ h/mv by letting y ψ and finding ψ/ x. Using λν v p in Equatin (.5), and using ψ instead f y,. cs cs λ ν ψ x t A v x t A p Differentiating twice with respect t x using the chain rule fr partial differentiatin (similar t Example 5.),, sin sin λ ν λ λ λ ν ψ x t A x t A x ψ λ λ ν λ λ λ ν λ ψ x t A x t A x cs cs
The kinetic energy f a nnrelativistic particle is ) (, U E h m m h m p U E KE s that λ λ Substituting the abve expressin relating ψ λ ψ and x, ) ( ) ( ψ ψ ψ λ ψ U E m U E h m x 8 h which is Equatin (5.)
. One f the pssible wave functins f a particle in the ptential well f Fig. 5.7 is sketched there. Explain why the wavelength and amplitude f &P vary as they d. The wave functin must vanish at x 0, where V. As the ptential energy increases with x, the particle's kinetic energy must decrease, and s the wavelength increases. The amplitude increases as the wavelength increases because a larger wavelength means a smaller mmentum (indicated as well by the lwer kinetic energy), and the particle is mre likely t be fund where the mmentum has a lwer magnitude. The wave functin vanishes again where the ptential V ; this cnditin wuld determine the allwed energies.
5. An imprtant prperty f the eigenfunctins f a system is that they are rthgnal t ne anther, which means that + ψ n ψmdv 0 n m Verify this relatinship fr the eigenfunctins f a particle in a ne-dimensinal bx given by Eq. (5.6). The necessary integrals are f the frm + L nx mx ψnψ mdx dx L sin sin 0 L L fr integers n, m, with n m and n -m. (A mre general rthgnality relatin wuld invlve the integral f ψ n * ψ m, but as the eigenfunctins in this prblem are real, the distinctin need nt be made.) T d the integrals directly, a cnvenient identity t use is sin α sin β [cs( α β) cs( α + β)], as may be verified by expanding the csines f the sum and difference f α and β. T shw rthgnality, the stipulatin n m means that α β and α -β and the integrals are f the frm
, ) ( sin ) ( ) ( sin ) ( ) ( cs ) ( cs 0 0 + + + + L L m n L x m n m n L L x m n m n L dx L x m n L x m n L dx ψ ψ where sin(n - m) sin(n - m) sin 0 0 has been used. 7. As shwn in the text, the expectatin value <x> f a particle trapped in a bx L wide is L/, which means that its average psitin is the middle f the bx. Find the expectatin value <x >. Using Equatin (5.6), the expectatin value <x > is sin dx. L x n x L x L n > < 0 See the end f this chapter fr an alternate analytic technique fr evaluating this integral using Leibniz s Rule. Frm either a table r repeated integratin by parts, the indefinite integral is. sin cs sin sin sin + u u u u u u n L udu u n L dx L x n x 8 6 where the substitutin u (n/l)x has been made.
This frm makes evaluatin f the definite integral a bit simpler; when x 0 u 0, and when x L u n. Each f the terms in the integral vanish at u 0, and the terms with sin u vanish at u n, cs u cs n, and s the result is As a check, nte that which is the expectatin value f <x > in the classical limit, fr which the prbability distributin is independent f psitin in the bx.. ) ( > < 6 n L n n n L L x n, lim L x n n > < 9. Find the prbability that a particle in a bx L wide can be fund between x 0 and x L/n when it is in the nth state. This is a special case f the prbability that such a particle is between x and x, as fund in Example 5.. With x 0 and x L,. sin n L x n n L x P L L 0 0
. A particle is in a cubic bx with infinitely hard walls whse edges are L lng (Fig. 5. 8). The wave functins f the particle are given by K K K,,,,,,,,, sin sin sin z y x z z x n n n L z n L y n L x n A ψ Find the value f the nrmalizatin cnstant A. The nrmalizatin cnstant, assuming A t be real, is given by. sin sin sin * * dz L z n dy L y n dx L x n A dxdydz dv L z L y L x 0 0 0 ψ ψ ψ ψ Each integral abve is equal t L/ (frm calculatins identical t Equatin (5.)). The result is r / L A L A