CHAPTER 39. Answer to Checkpoint Questions

Transcription

1 CHAPTR 39 PHOTONS AND MATTR WAVS 059 CHAPTR 39 Answer to Ceckoint Questions. (b), (a), (d), (c). (a) litium, sodium, otassium, cesium; (b) all tie 3. (a) same; (b) { (d) x rays 4. (a) roton; (b) same; (c) roton 5. same Answer to Questions. (a) microwave; (b) x ray; (c) x ray. (a) true; (b) false; (c) false; (d) true; (e) true; (f) false 3. otassium 4. only (b) 5. Positive carge builds u on te late, inibiting furter electron emission 6. only (e) 7. none 8. Te fractional wavelengt cange for visible ligt is too small 9. (a) greater; (b) less 0. (a) B; (b) { (d) A. no essential cange. electron 3. (a) decreases by a factor of ; (b) decreases by a factor of / 4. electron, neutron, ala article 5. extremely small 6. (d) 7. (a) increasing; (b) decreasing; (c) same; (d) same

2 060 CHAPTR 39 PHOTONS AND MATTR WAVS 8. roton 9. (a) 0. amlitude of reected wave is less tan tat of incident wave. (a) zero; (b) yes. all tie Solutions to xercises & Problems Te energy of a oton is given by f, were is te Planck constant and f is te frequency. Te wavelengt is related to te frequency by f c, so c. Since 6: Js and c 3: m/s, c : Jm : Jm (: J/eV)(0 9 m/nm) : Tus : From te result of 589 nm : ev : 3 6: Js 6: Js : J/eV 4:4 0 5 evs : 4 Let min 0:6 ev to get : 0 3 nm : m. It is in te infrared region nm 5:9 0 6 ev 5:9 ev :

3 CHAPTR 39 PHOTONS AND MATTR WAVS 06 6 Let and solve for v: m ev c r s c v m e c (m e c ) c s () (0:5 0 6 ev)(590 nm) (3:00 08 m/s) 8:6 0 5 m/s : Since v c te non-relativistic formula K mv may be used. 7 Let R be te rate of oton emission (number of otons emitted er unit time) of te Sun and let be te energy of a single oton. Ten te ower outut of te Sun is given by P R. Now f c, were is te Planck constant, f is te frequency of te ligt emitted, and is te wavelengt. Tus P Rc and R P c (550 nm)(3:9 0 6 W) (6: Js)(3: m/s) :0 045 otons/s : 8 Let te diameter of te laser beam be d, ten te corss-sectional area of te beam is A d 4. From te formula obtained in 7 te rate in question is given by R A P c(d 4) 4(633 nm)(5:0 0 3 W) (6: Js)(3: m/s)(3:5 0 3 m) :7 0 otons/m s : 9 Since (; 650; 763:73) m 6: m 605:780 nm, te energy is c 40 nmev 605:780 nm :047 ev :

4 06 CHAPTR 39 PHOTONS AND MATTR WAVS 0P P t (00s) 3:6 0 7 W : (: J/eV)(00s) 550 nm P (a) Use te formula obtained in 7: R Pc. Te bulb emitting ligt wit te longer wavelengt (te 700 nm bulb) emits more otons er unit time. Te energy of eac oton is less so it must emit otons at a greater rate. (b) Let R` be te rate of oton roduction at te longer wavelengt (`) and let R s be te rate of roduction at te sorter wavelengt ( s ). Ten R` R s (` s )P c (700 nm 400 nm)(400 J/s) (: J/eV)() 6:0 00 otons/s : Te result c, develoed in, was used. P (a) Te rate at wic solar energy strikes te anel is P (:39 kw/m )(:60 m ) 3:6 kw : (b) Te rate at wic solar otons are absorbed by te anel is R P (c) Te time in question is given by 3:6 kw (550 nm)(: J/eV) :00 0 s : t N A R 6:0 03 :00 0 s 60: s : 3P Te total energy emitted by te bulb is 93%P t, were P 60 W and t 730 (730 )(3600 s/) : s. Te energy of eac oton emitted is c. Tus te number of otons emitted is N 93%P t c (93%)(60 W)(: s) (630 nm)(: J/eV) 4:7 06 :

5 CHAPTR 39 PHOTONS AND MATTR WAVS 063 4P Te rate at wic otons are emitted from te argon laser source is given by R P, were P :5 W is te ower of te laser beam and c is te energy of eac oton of wavelengt. Since 84% of te energy of te laser beam falls wit in te central disk, te rate of oton absortion of te central disk is R 0 R P c (0:84)(:5 W) (55 nm)(: J/eV) 3:3 0 8 otons/s : 5P (a) Assume all te ower results in oton roduction at te wavelengt 589 nm. Let R be te rate of oton roduction and be te energy of a single oton. Ten P R Rc, were f and f c were used. Here is te Planck constant, f is te frequency of te emitted ligt, and is its wavelengt. Tus R P c ( m)(00 W) (6: Js)(3: m/s) :96 00 otons/s : (b) Let I be te oton ux a distance r from te source. uniformly in all directions R 4r I and r Since otons are emitted r s R 4I : otons/s 4(: otons/m s) 4:85 07 m : (c) Let n be te oton density and consider a narrow column wit cross-sectional area A and lengt `, wit its axis along te direction of oton motion. At any instant te number of otons in te column is N na`. All of tem will move troug one end in time t `c, so te oton ux troug te end is I NAt na`ca` nc. Now I R4r, so R4r nc and r s R r 4nc : otons/s 4(: otons/m 3 )(3: m/s) 80 m : (d) Te oton ux is I R 4r :96 00 otons/s 5: otons/m s : 4(:00 m) 6 According to q te condition for otoelectric eect to occur is f.

6 064 CHAPTR 39 PHOTONS AND MATTR WAVS (a) For 565 nm f c (4:4 0 5 evs)(3: m/s) m :0 ev : Since otassium > f > cesium otoelectric eect can occur in cesium but not in otassium at tis wavelengt. (b) Now 58 nm so f c (4:4 0 5 evs)(3: m/s) m :40 ev : Tis is greater tan bot cesium and otassium so otoelectric eect can now occur for bot metals. 7 Te energy of te most energetic oton in te visible ligt range (wit wavelengt of about 400 nm) is about (400 nm) 3: ev. So barium and litium can be used, since teir work functions are bot lower tan 3: ev. 8 (a) Since 680 nm :8 ev < :8 ev, tere is no otoelectric emission. (b) Te longest (cuto) wavelengt of otons wic will cause otoelectric emission in sodium is given by max, or max ():8 ev 544 nm. Tis corresonds to te green color. 9 Use q to nd K max : K max f (4:4 0 5 evs)(3:0 0 5 Hz) :3 ev 0 ev : 0 Te energy of an incident oton is f c, were is te Planck constant, f is te frequency of te electromagnetic radiation, and is its wavelengt. Te kinetic energy of te most energetic electron emitted is K max (c), were is te work function for sodium. Te stoing otential V sto is related to te maximum kinetic energy by ev sto K max, so ev sto (c) and c ev sto + 5:0 ev + : ev 70 nm :

7 CHAPTR 39 PHOTONS AND MATTR WAVS 065 Here ev sto 5:0 ev and c were used. Te seed v of te electron saties K max m ev s s ( ) ( )c v m e m e c r (5:80 ev 4:50 ev)(3:00 08 m/s) 0:5 0 6 ev, or 6: m/s : P (a) Te kinetic energy K max of te fastest electron emitted is given by K max f (c), were is te work function of aluminum, f is te frequency of te incident radiation, and is its wavelengt. Te relationsi f c was used to obtain te second form. Tus K max 4:0 ev :00 ev : 00 nm (b) Te slowest electron just breaks free of te surface and so as zero kinetic energy. (c) Te stoing otential V sto is given by K max ev sto, so V sto K max e :00 ev e :00 V : (d) Te value of te cuto wavelengt is suc tat K max 0. Tus c or c 4: ev 95 nm : If te wavelengt is longer te oton energy is less and a oton does not ave sucient energy to knock even te most energetic electron out of te aluminum samle. 3P (a) Use q. 39-6: V sto f e c e (b) Use te formula obtained in : s ( ) v m e (400 nm) :8 ev r evsto m e e s ev sto c m e c :3 V : r e(:3 V)(3:00 08 m/s) 6:8 0 5 m/s : 0:5 0 6 ev

8 066 CHAPTR 39 PHOTONS AND MATTR WAVS 4P From q K max c c max 54 nm 35 nm :07 ev : 5P To nd te longest ossible wavelengt max (corresonding to te lowest ossible energy) of a oton wic can cause otoelectric eect in latinum, ut K max 0 in q and use f c. Tus c max. Solve for max : max c 5:3 ev 33 nm : 6P (a) Use te otoelectric eect equation (q. 39-5) in te form c + K max. Te work function deends only on te material and te condition of te surface and not on te wavelengt of te incident ligt. Let be rst wavelengt described and be te second. Let K m ( 0:70 ev) be te maximum kinetic energy of electrons ejected by ligt wit te rst wavelengt and K m ( :43 ev) be te maximum kinetic energy of electron ejected by ligt wit te second wavelengt. Ten and Solve tese equations simultaneously for. c + K m c + K m : Te rst equation yields (c ) K m. Wen tis is used to substitute for in te second equation te result is (c ) (c ) K m + K m. Te solution for is c c + (K m K m ) ()(49 nm) + (49 nm)(:43 ev 0:70 ev) 38 nm : (b) Te rst equation dislayed above yields c K m 49 nm 0:70 ev :8 ev :

9 CHAPTR 39 PHOTONS AND MATTR WAVS 067 7P For te rst and second case (labeled and ) we ave ev 0 c and ev 0 c, from wic and can be solved. (a) e(v V ) c( ) :85 ev 0:80 ev (3: nm/s)[(300 nm) (400 nm) ] 4: 0 5 evs : (b) 3(V V ) (c) Let c max to obtain (0:80 ev)(400 nm) (:85 ev)(300 nm) 300 nm 400 nm :7 ev : max c :7 ev 545 nm : 8P V sto (V) a.0 c b f (0 4 Hz) (a) According to q te sloe of te straigt line lotted above is equal to e. Measure te sloe directly from te lot to obtain e ab cb :0 V 0:75 V (0: :5 0 4 ) Hz 4:7 0 5 Vs ;

10 068 CHAPTR 39 PHOTONS AND MATTR WAVS so (4:7 0 5 Vs)(:6 0 9 C) 6: Js : (b) From te lot we nd te lowest oton frequency for otoelectric eect to occur to be f 0 5:7 0 4 Hz. Tus f 0 (6: Js)(5:7 0 4 Hz) :6 0 9 J/C :4 ev ; wic comares favourably wit te actual value of :3 ev. 9P Te number of otons emitted from te laser er unit time is R P : W (600 nm)(: J/eV) 6:05 05 s ; of wic (:0 0 6 )(6: s) 0:605s actually cause otoelectric emissions. Tus te current is i (0:605s)(: C) 9: A: 30P (a) Find te seed v of te electron from r m e veb: v rbem e. Tus (b) Te work done is K max m ev m e rbe (rb) e m e m e (: T m) (: C) (9: 0 3 kg)(: J/eV) 3:0 kev : W K max nm 3:0 kev 4 kev : 3 (a) (b) f c 3:00 08 m/s 35:0 0 m 8:57 08 Hz : f (4:4 0 5 evs)(8: Hz) 3: ev : (c) From q : Js 35:0 0 m : kgm/s :

11 CHAPTR 39 PHOTONS AND MATTR WAVS (a) Te rest energy of an electron is given by m e c. Tus te momentum of te oton in question is given by c m ec m e c c (9: 0 3 kg)(3: m/s) :73 0 kgm/s : We may also exress te momentum in terms of MeV/c: m e c c 0:5 MeVc. (b) 6: Js :73 0 kgm/s :43 0 m :43 m : (c) f c 3:00 08 m/s :43 0 m :4 00 Hz : 33 (a) Wen a oton scatters from an electron initially at rest te cange in wavelengt is given by ( cos ) ; m e c were m e is te mass of an electron and is te scattering angle. Now m e c :43 0 m :43 m, so (:43 m)( cos 30 ) 0:36 m. Te nal wavelengt is 0 + :4 m + 0:36 m :7 m. (b) Now (:43 m)( cos 0 ) 3:645 m and 0 :4 m + 3:645 m 6:05 m. 34P (a) (b) c 40 nmev 0:5 MeV : nm :43 m : ( m e c cos ) :43 m + (:43 m)( cos 90:0 ) 4:86 m : (c) :43 m 0 (0:5 MeV) 0:55 MeV : 0 4:86 m

12 070 CHAPTR 39 PHOTONS AND MATTR WAVS 35P Since te electron is free it is undergoing no acceleration. Terefore we may, witout loss of generality, consider a stationary electron of rest mass m e wic collides wit a oton of linear momentum. Prior to te collision te energy of te electron is its rest energy e m e c, wile te energy of te oton is simly c. Tus te total energy of te electron-oton system before te collision is i e + m e c + c : After te collision, te oton as been entirely absorbed by te electron wic, according to te conservation of linear momentum, must ave acquired a linear momentum equal to, te initial linear momentum of te oton. Tus te nal energy of te system after te collision is f (c) + m e c4 : quate te initial and te nal energies to obtain Now square bot sides: wic yields m e c 3 0, or m e c + c (c) + m e c4 : (m e c + c) (m e c ) + m e c 3 + (c) (c) + m e c4 ; 0 : Tis means tat te only oton tat can articiate in te collision is te one wit zero linear momentum. Te energy of te oton is tus c 0, wic means tat te oton is nonexistent and te collision is imossible. 36P (a) (b) c 0 c m e c ( cos ) (:43 m)( cos 80 ) +4:86 m : (40 nmev) 0:0 nm + 4:86 m 4 kev : 0:0 nm (c) From conservation of energy K 4 kev. (d) Te electron will move straigt aead after te collision, since it as acquired some of te foward linear momentum from te oton. 37P (a) Since te mass of an electron is m e 9: kg, its Comton wavelengt is C m e c 6: Js (9: kg)(: m/s) :46 0 m :43 m :

13 CHAPTR 39 PHOTONS AND MATTR WAVS 07 (b) Since te mass of a roton is m : kg, its Comton wavelengt is C 6: Js (: kg)(: m/s) :3 0 5 m :3 fm : (c) Use te formula develoed in : (), were is te energy and is te wavelengt. Tus for te electron (d) For te roton : nm 5: 05 ev 0:5 MeV : :3 0 6 nm 9:39 08 ev 939 MeV : 38P Te fractional cange is (c) c ( C )( cos ) + : + (a) Now 3:0 cm 3:0 0 0 m and 90 so (3:0 0 0 m:43 m)( cos 90 ) + 8: 0 : (b) Now 500 nm 5: m and 90 so (5: m:43 m)( cos 90 ) + 4:9 0 6 : (c) Now 5 m and 90 so (5 m:43 m)( cos 90 ) + 8:9 0 : (d) Now c 40 nmev:0 MeV :4 0 3 nm :4 m and 90 so (:4 m:43 m)( cos 90 ) + 0:66 :

14 07 CHAPTR 39 PHOTONS AND MATTR WAVS (e) From te calculation above we see tat te sorter te wavelengt te greater te fractinal energy cange for te oton as a result of te Comton scattering. Since is virtually zero for microwave and visible ligt, Comton eect is signicant only in te x-ray to gamma ray range of te electromagnetic sectrum. 39P If is te original energy of te oton and 0 is te energy after scattering, ten te fractional energy loss is frac 0 : Samle Problem 39-4 sows tat tis is frac + : Tus frac frac 0:75 0:75 3 : A 300% increase in te wavelengt leads to a 75% decrease in te energy of te oton. 40P max ( cos ) m c m max c (4:4 0 5 evs)(3: m/s) 938 MeV : m :65 fm : 4P Te dierence between te electron-oton scattering rocess in tis roblem and te one studied in te text (te Comton sift, see q. 39-) is tat te electron is in motion relative wit seed v to te laboratory frame. To utilize te result in q. 39-, sift to a new reference frame in wic te electron is at rest before te scattering. Denote te quantities measured in tis new frame wit a rime ('), and aly q. 39- to yield m e c ( cos ) m e c ; were we noted tat (since te oton is scattered back in te direction of incidence). Now, from te Doler sift formula (q. 38-5) te frequency f 0 0 of te oton rior to te scattering in te new reference frame satises f 0 0 c 0 0 f 0 s + ;

15 CHAPTR 39 PHOTONS AND MATTR WAVS 073 were vc. Also, as we switc back from te new reference frame to te original one after te scattering s s f f 0 + c 0 + : Solve te two Doler-sift equations above for 0 and 0 0 and substitute te results into te Comton sift formula for 0 : s s 0 f + + m e c : f 0 Some simle algebra ten leads to s f f m e c! ; were we used f. 4P (a) From q. 39- m e c ( cos ) (:43 m)( cos 90 ) :43 m : (b) :45 m 590 nm 4: 0 6 : (c) Te cange in energy for a oton wit 590 nm is given by c c (4:4 0 5 evs)(3: m/s)(:43 m) (590 nm) 8: ev : For an x ray oton of energy 50 kev, remains te same ( :43 m) since it is indeendent of. Te fractional cange in wavelengt is now c ( ev)(:43 m) (4:4 0 5 evs)(3: m/s) 9:78 0 ; and te cange in oton energy is now c + c + ; +

16 074 CHAPTR 39 PHOTONS AND MATTR WAVS were. Plug in 50 kev and 9:780 to obtain 4:45 kev. (Note tat in tis case 0: is not close enoug to zero so te aroximation c is not as accurate as in te rst case, in wic 4: 0 6. In fact if you were to use tis aroximation ere you would get 4:89 kev, wic does not amount to a satisfactory aroximation.) 43P (a) From q. 39- (m e c)( cos ): In tis case 80 so cos, and te cange in wavelengt for te oton is given by m e c. Te energy 0 of te scattered oton (wose initial energy is c) is ten 0 c (m e c)(c) + m e c 50:0 kev 4:8 kev : + (50:0 kev)0:5 MeV (b) From conservation of energy te kinetic energy K of te electron is given by K 0 50:0 kev 4:8 kev 8: kev. 44P From te result of 38P (mc)( cos ) f 0 ( cos ) : cf 0 mc Here te minus sign indicates tat art of te oton energy is lost during te scattering. 45P Te magnitude of te fractional energy cange for te oton is given by (c) c + + ; were 0%. Tus ( solve for cos : ). Plug in tis exression for into q. 39- and cos mc mc ( ) (mc ) ( ) (0%)(5 kev) ( 0%)(00 kev) 0:76 : Te angle is 44.

17 CHAPTR 39 PHOTONS AND MATTR WAVS P Te maximum energy of te electron is attained wen te loss of energy of te oton is te greatest: Conservation of energy ten gives K max max ( C )( cos ) + max C + C + : (c)(m e c) + m e c + : 47P Use te formula obtained in 46P: K max + m e c (7:5 kev) 7:5 kev + 5 kev : kev : 48P Rewrite q as and q as m m 0 cos m 0 sin v cos (vc) v (vc) sin : Square bot equations and add u te two sides: " # m cos + 0 sin v 0 (vc) ; were we used sin + cos to eliminate. Now te RHS can be written as v (vc) c ; (vc) so " v (vc) mc # cos + 0 sin + : 0 Now rewrite q as mc + 0 (vc)

18 076 CHAPTR 39 PHOTONS AND MATTR WAVS and comare wit te revious equation we obtained for [ (vc) ]. Tis yields " # + mc 0 mc cos + 0 sin + : 0 We ave so far eliminated and v. Working out te squares on bot sides and noting tat sin + cos, we get 0 ( cos ) : mc 49 (a) mv 6: J s ( kg)(000 m/s) : m : (b) Te de Broglie wavelengt of te bullet is muc too sort for te wave nature of te bullet to reveal itself troug diraction eects. 50 (a) Substitute te classical relationsi between momentum and velocity v, v m e into te classical denition of kinetic energy, K m ev, to obtain K m e. Here m e is te mass of an electron. Solve for : m e K. Te relationsi between te momentum and te de Broglie wavelengt is, were is te Planck constant. Tus If K is given in electron volts ten me K : 6: Js (9: kg) (: J/eV)K :6 0 9 m(ev) K :6 nm(ev) K : 5 Use te result of 50 and K ev, were V 5:0 kv. Tus :6 nm(ev) K :6 nm(ev) ev :6 nm(ev) e(5 03 V) 7: nm 7:75 m ;

19 CHAPTR 39 PHOTONS AND MATTR WAVS 077 were we noted tat ev e V. 5 (a) me K c me c K (0:5 MeV)(:00 kev) 38:7 m : (b) (c) c mn c K c :00 kev :4 nm : (939 MeV)(:00 kev) 9: nm 904 fm : 53P Use te result of 50: (:6 nmev ) K, were K is te kinetic energy. Solve for K: " # :6 nm(ev) " # :6 nm(ev) K 4:3 0 6 ev : 590 nm 54P (a) and (b) Te momenta of te electron and te oton are te same: Te kinetic energy of te electron is K e wile tat for te oton is 6: Js 0:0 0 9 m 3:3 0 4 kgm/s : (3:3 0 4 kgm/s) m e (9: 0 3 kg) 6:0 0 8 J 38 ev ; K c (3:3 0 4 kgm/s)(3: m/s) 9:9 0 6 J 6: kev : 55P (a) K 3 kt 3(: J/K)(300 K) (: J/eV) 3:88 0 ev 38:8 mev :

20 078 CHAPTR 39 PHOTONS AND MATTR WAVS (b) mn K c (mn c )K : m 46 m : 40 nmev (939 MeV)(3:88 0 ev) 56P (a) Solve v from (m v): v m (b) Let ev K m v and solve for V : 6: Js (: kg)(0:00 0 m) 3:96 06 m/s : V m v e (: kg)(3: m/s) (: C) 8:8 0 3 V : 57P (a) Te average de Broglie wavelengt is m K m(3kt) c (mc )kt 3(4)(938 MeV)(8:6 0 5 ev/k)(300 K) 7:3 0 m 73 m ; wile te average searation is d 3 n r (: J/K)(300 K) 3 kt :0 0 5 Pa 3:4 nm : (b) Yes, since d. 58P (a) Te momentum of te oton is given by c, were is its energy. Its wavelengt is c :00 ev 40 nm :4 nm :

21 CHAPTR 39 PHOTONS AND MATTR WAVS 079 Te momentum of te electron is given by m e K, were K is its kinetic energy and m e is its mass. Its wavelengt is According to 50 if K is in electron volts tis is me K : :6 nm K :6 nm :00 :3 nm : (b) For te oton c : ev :4 0 6 nm :4 fm : Relativity teory must be used to calculate te wavelengt for te electron. Te momentum and kinetic energy K of an electron are related by (c) K + Km e c. Tus c K + Km e c (: ev) + (: ev)(0:5 0 6 ev) : ev : Te wavelengt is c c : ev :4 0 6 nm :4 fm : 59P (a) For te oton and for te electron c 40 nmev :00 nm :4 kev K () m e m e (c) m e c (0:5 MeV) 40 nmev :50 ev : :00 nm (b) Now for te oton and for te electron 40 nmev :00 fm :4 GeV K c + (m e c ) m e c (c) + (m e c ) m e c s 40 fmmev + (0:5 MeV) :00 fm :4 0 3 MeV :4 GeV : 0:5 MeV

22 080 CHAPTR 39 PHOTONS AND MATTR WAVS Note tat at sort (large K) te kinetic energy of te electron, calculated wit relativistic formula, is about te same as tat of te oton. Tis is exected since now K c for te electron, wic is te same as c for te oton. 60P (a) Te kinetic energy acquired is K qv, were q is te carge on an ion and V is te accelerating otential. Tus K (: C)(300 V) 4: J. Te mass of a single sodium atom is, from Aendix F, m (:9898 g/mol)(6:0 0 3 atom/mol) 3: g 3: kg. Tus te momentum of an ion is mk (3: kg)(4: J) :9 0 kgm/s : (b) Te de Broglie wavelengt is 6: Js :95 0 kgm/s 3: m 346 fm : 6P We need to use te relativistic formula (c) m e c ). So c K 50 GeV m e c c Kc (since 0:05 fm ; wic is about 00 times smaller tan te radius of an average nucleus. 6P (a) Since K 7:5 MeV m c 4(93 MeV), we may use te non-relativistic formula m K. So c m c K 40 nmev (4)(93 MeV)(7:5 MeV) 5: fm : (b) Since 5: fm 30 fm, to a fairly good aroximation te wave nature of te article dose not need to be taken into consideration. 63P Te wavelengt associated wit te unknown article is (mv), were is its momentum, m is its mass, and v is its seed. Te classical relationsi mv was used. Similarly, te wavelengt associated wit te electron is e (m e v e ), were m e is its mass and v e is its seed. Te ratio of te wavelengts is e (m e v e )(mv), so m v e e v m e 9: kg 3(: ) : kg :

23 CHAPTR 39 PHOTONS AND MATTR WAVS 08 According to Aendix B tis is te mass of a neutron. 64P (a) Let (c) m e c and solve for K m e c : s c K + m e c4 m e c s + (0:5 MeV) nm 0:05 MeV 5 kev : 0:5 MeV (b) c nm : 05 ev 0 kev : (c) Te electron microscoe is more suitable, as te required energy of te electrons is muc less tan tat of te otons. 65P Te same resolution requires te same wavelengt and since te wavelengt and article momentum are related by, tis means te same article momentum. Te momentum of a 00-keV oton is c (000 3 ev)(:600 9 J/eV)(3:000 8 m/s) 5: kgm/s. Tis is also te magnitude of te momentum of te electron. In te classical aroximation, te kinetic energy of te electron is K Te accelerating otential is tus (5: kgm/s) m e (9: 0 3 kg) : J : V K e : J : C 9:76 03 V : Relativistically, te kinetic energy of te electron satises ev K (m e c ) + (c) m e c, wic we may solve for V : V i (me c ) + (c) m e c e (5: 05 ev) + ( ev) 5: 0 evi 5 e 9: V :

24 08 CHAPTR 39 PHOTONS AND MATTR WAVS 66 (a) nn (a + ib)(a + ib) (a + ib)(a + i b ) (a + ib)(a a + iba iab + (ib)( ib) a + b ; ib) wic is always real since bot a and b are real. (b) But so jnmj jnj jmj. jnmj j(a + ib)(c + id)j jac + iad + ibc + ( i) bdj j(ac bd) + i(ad + bc)j (ac bd) + (ad + bc) a c + b d + a d + b c : jnj jmj ja + ibj jc + idj a + b c + d a c + b d + a d + b c ; 67P Plug q into q Note tat d dx d dx Aeikx + Be ikx ikae ikx ikbe ikx and Tus d d dx d dx ikaeikx ikbe ikx k Ae ikx k Be ikx : dx + k k Ae ikx k Be ikx + k Ae ikx + Be ikx 0 : 68P Use te uler formula e i cos + i sin to re-write (x) as (x) 0 e ikx 0 (cos kx + i sin kx) ( 0 cos kx) + i( 0 sin kx) a + ib ; were a 0 cos kx and b 0 sin kx are bot real quantities.

25 CHAPTR 39 PHOTONS AND MATTR WAVS 083 (b) (x; t) (x)e i!t 0 e ikx e i!t 0 e i(kx!t) [ 0 cos(kx!t)] + i [ 0 sin(kx!t)] : 69P Te angular wave number k is related to te wavelengt by k and te wavelengt is related to te article momentum by, so k. Now te kinetic energy K and te momentum are related by K m, were m is te mass of te article. Tus mk and k mk : 70P For ot 0, Scrodinger's equation becomes d dx + 8 m ( 0 ) 0 : Substitute 0 e ikx. Te second derivative is d dx k 0 e ikx k. Te result is k + 8 m ( 0 ) 0 : Solve for k and obtain k r 8 m [ 0 ] m( 0 ) : 7P From q j (x; y; z; t)j (x; y; z) e i!t j (x; y; z)j e i!t j (x; y; z)j jcos(!t) + i sin(!t)j j (x; y; z)j (cos!t) + (sin!t) j (x; y; z)j : Here we used te result of 66.

26 084 CHAPTR 39 PHOTONS AND MATTR WAVS 7P Te wave function is now given by (x; t) 0 e i(kx+!t) : Tis function describes a lane matter wave traveling in te negative x direction. An examle of te actual articles tat t tis descrition is a free electron wit linear momentum (k) i and kinetic energy K m e k 8 m e. 73P (a) Te wave function is now given by (x; t) 0 e i(kx!t) + e i(kx+!t) i 0 e i!t e ikx + e ikx : Tus (b) j (x; t)j 0 e i!t e ikx + e ikx 4 0 e i!t e ikx + e ikx e ikx + e ikx 0 0 j(cos kx + i sin kx) + (cos kx i sin kx)j 4 0(cos kx) 0( + cos kx) : 3 Ψ(x, t)/ψ kx Consider two lane matter waves, eac wit te same amlitude 0 and travels in oosite directions along te x axis. Te combined wave is a standing wave: (x; t) 0 e i(kx!t) + 0 e i(kx+!t) 0 e ikx + e ikx e i!t ( 0 cos kx) e i!t :

27 CHAPTR 39 PHOTONS AND MATTR WAVS 085 Tus te squared amlitude of te matter wave is j (x; t)j ( 0 cos kx) e i!t 0( + cos kx) ; wic is te same as te function lotted above. (c) Set j (x; t)j 0( + cos kx) 0 to obtain cos(kx). Tis gives kx (n + ) ; (n 0; ; ; 3;... ) Solve for x: x (n + ) : 4 (d) Te most robable ositions for nding te article are were j (x; t)j / ( + cos kx) reaces its maximum. Tus cos kx, or kx n ; (n 0; ; ; 3;... ) Solve for x: x n : 74 (a) Since x y 0, x y 0. Tus from q bot x and y are innite. It is terefore imossible to assign a y or z coordinate to te osition of an electron. (b) Since it is indeendent of y and z te wave function (x) sould describe a lane wave tat extends innitely in bot te y and z directions. Also from Fig. 39- we see tat j (x)j extends inifnitely along te x axis. Tus te matter wave described by (x) extends trougout te entire tree-dimensional sace. 75 x 6: Js (50 m) : 0 4 kgm/s : 76 Denote te Plank's constant in te imaginary universe as 0. Ten te uncertainty in te osition of te baseball would be x 0 0 mv 0:60 Js (0:50 kg)(:0 m/s) 0:9 m :

28 086 CHAPTR 39 PHOTONS AND MATTR WAVS It would be ard to catc suc a baseball, since its size is less tan te uncertainty in its osition. 77P If x ten from q x ; were we used and. For 0 te formula above gives 0. Tis is understandable, since now! so x!. So wile te measurement of momentum can be exact, it is now imossible to locate te osition x of te article. As te measured momentum increases, so does. Tis is also understandable because as increases x / decreases, keeing x. 78P (a) (b) c 40 nmev 0:0 0 3 nm 4 kev : c c c ( C )( cos ) 4 kev + (0:0 m:43 m)( cos 80 ) 40:5 kev : (c) It is imossible to \view" an atomic electron wit suc a ig-energy oton, because wit te energy imarted to te electron te oton would ave knocked te electron out of its orbit. 79P Te robability T tat a article of mass m and energy tunnels troug a barrier of eigt U and widt L is given by T e kl ; were k r 8 m(u ) :

29 CHAPTR 39 PHOTONS AND MATTR WAVS 087 For te roton s 8 k (: kg)(0 Mev 3:0 Mev)(: J/MeV) (6: Js) 5: m ; kl (5: m )(0 0 5 m) 5:808, and T e 5:808 9:0 0 6 : Te value of k was comuted to a greater number of signicant digits tan usual because an exonential is quite sensitive to te value of te exonent. Te mass of a deuteron is :04 u 3: kg, so s 8 k (3: kg)(0 MeV 3:0 MeV)(: J/MeV) (6: Js) 8: m ; kl (8: m )(0 0 5 m) 8:43, and T e 8:43 7: : 80P Let " r # 8 m(u ) T ex( kl) ex L and solve for : U 6:0 ev m 5: ev : ln T 4L (0:5 MeV) ()(ln 0:00) 4(0:70 nm) 8P (a) Te rate at wic incident rotons arrive at te barrier is n :0 ka: C 6:5 0 3 s. Let nt t to nd te waiting time t: t (nt ) r # 8 "L n ex m (U ) 6:5 0 3 s ex 3:37 0 s 0 04 y ; (0:70 nm) 8(938 MeV)(6:0 ev 5:0 ev)

30 088 CHAPTR 39 PHOTONS AND MATTR WAVS wic is muc, muc longer tan te age of te universe. (b) Relace te mass of te roton wit tat of te electron to obtain te corresonding waiting time for an electron: t (nt ) r # 8 "L n ex m e (U ) 6:5 0 3 s : 0 9 s : ex (0:70 nm) 8(0:5 MeV)(6:0 ev 5:0 ev) Te enormous dierence between te two waiting times is te result of te dierence between te masses of te two kinds of articles. 8P (a) If m is te mass of te article and is its energy ten te robability it will tunnel troug a barrier of eigt U and widt L is given by T e kl ; were r 8 k m(u ) : If te cange U in U is small (as it is) te cange in te tunneling robability is given by Now Tus T dt du U r dk du 8 m U T dk LT du U : r 8 m(u ) (U ) U LT k : U k (U ) : For te data of Samle Problem 39-7, kl 0:0, so kl 5:0 and T U T kl (0:000)(6:8 ev) (5:0) U 6:8 ev 5: ev 0:0 : Tere is a 0% decrease in te tunneling robability. (b) Te cange in te tunneling robability is given by T dt dl L ke kl L kt L and T T k L (6:67 09 m )(0:000)(750 0 m) 0:0 : Tere is a 0% decrease in te tunneling robability.

31 CHAPTR 39 PHOTONS AND MATTR WAVS 089 (c) Te cange in te tunneling robability is given by T dt d kl dk Le d dk LT d : Now dkd dkdu k(u ), so T T kl U Tere is a 5% increase in te tunneling robability. (0:0000)(5: ev) (5:0) 6:8 ev 5: ev 0:5 : 83P Te kinetic energy of te car of mass m moving at seed v is given by mv ; wile te otential barrier it as to tunnel troug is ot mg, were 4 m. According to q. 39- and 39- te tunneling robability is given by T e kl, were r 8 k m( ot ) s (500 kg) 6: Js : 0 38 m : s 8 m(mg (9:80 m/s )(4 m) mv ) (0 m/s) Tus kl (: 0 38 m )(30 m) 7: You can see tat T e kl is almost zero. 84P (a) Set ot 0 in q. 39-5: d dx + 8 m 0 : For te wave function (x) Ae ikx + Be ikx d dx (ik) Ae ikx + ( ik) Be ikx k (Ae ikx + B ikx ) k : Plug tis into te Scrodinger's equation to obtain k (x) + 8 m (x) k + 8 m (x) 0 :

32 090 CHAPTR 39 PHOTONS AND MATTR WAVS Since (x) 6 0, Tus k 8 m k + 8 m 0 : 8 m 4 ; m or k. Here k is arbitrarily set to be ositive, wic does not aect te generality of te solution since it contains bot e +ikx and e ikx. (b) j (x)j Ae ikx + Be ikx ja cos kx + ia sin kx + B cos kx ib sin kxj j(a + B) cos kx + i(a B) sin kxj [(A + B) cos kx] + [(A B) sin kx] (A + AB + B ) cos kx + (A AB + B ) sin kx (A + B )(cos kx + sin kx) + AB(cos kx sin kx) A + B + AB cos kx ; were to reac te last ste we used te trigonometric identities cos + sin and cos sin cos. (c) Since bot A and B are constants j (x)j is a function of cos kx, wic varies between and +. Tus j (x)j varies between A + B + AB( ) (A B) and A + B + AB(+) (A + B). 85P (a) For ot const. > q can be written as d dx + 8 m( ot ) d dx + k 0 ; were r 8 k m( ot ) : Plug te trial solution (x) Ce kx into te Scrodinger's equation: ( k) Ce kx + k Ce kx 0 : Tus (x) Ce kx is indeed a solution to te equation in tis region. You can verify tat in general tere can be anoter solution in te form De +kx, were D is anoter constant. Tis solution, owever, sould be discarted in our case becasue it leads to te unysical limit (x)! at x!. (b) Since (x) is real, j (x)j [ (x)] Ce kx C e kx :