f5ch39 - CHAPTER 39 PHOTONS AND MATTER WAVES 1059 CHAPTER...

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Unformatted text preview: CHAPTER 39 PHOTONS AND MATTER WAVES 1059 CHAPTER 39 Answer to Checkpoint Questions 1. 2. 3. 4. 5. (b), (a), (d), (c) (a) lithium, sodium, potassium, cesium; (b) all tie (a) same; (b) { (d) x rays (a) proton; (b) same; (c) proton same Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. (a) microwave; (b) x ray; (c) x ray (a) true; (b) false; (c) false; (d) true; (e) true; (f) false potassium only (b) Positive charge builds up on the plate, inhibiting further electron emission only (e) none The fractional wavelength change for visible light is too small (a) greater; (b) less (a) B ; (b) { (d) A no essential change electron p (a) decreases by a factor of 1= 2; (b) decreases by a factor of 1/2 electron, neutron, alpha particle extremely small (d) (a) increasing; (b) decreasing; (c) same; (d) same 1060 CHAPTER 39 PHOTONS AND MATTER WAVES 18. 19. 20. 21. 22. proton (a) amplitude of re ected wave is less than that of incident wave (a) zero; (b) yes all tie Solutions to Exercises & Problems The energy of a photon is given by E = hf , where h is the Planck constant and f is the frequency. The wavelength is related to the frequency by f = c, so E = hc=. Since h = 6:63 10 34 J s and c = 3:00 108 m/s, 1E hc = 1:99 10 Thus 25 J m = 1:99 10 25 J m (1:602 10 19 J/eV)(10 9 m/nm) = 1240 eV nm : E = 1240 eV nm : 2E From the result of 1E E = 1240 eV nm = 1240 eV nm = 2:11 eV : 589 nm 6:626 10 34 J s = 4:14 10 1:60 10 19 J/eV 3E h = 6:626 10 4E 34 J s = 15 eV s : Let E = 1240 eV nm=min = 0:6 eV to get = 2:1 103 nm = 2:1 m. It is in the infrared region. 5E 1240 E = 1240 eV nm = 21 eV7 nm = 5:9 10 6 eV = 5:9 eV : 10 nm CHAPTER 39 PHOTONS AND MATTER WAVES 1061 6E Let and solve for v: r s 1 m v2 = E = hc ph 2 e 2 v = mhc = (m2hc ) c e e c2 s = 2(1240 eV nm) 8 5 (0:511 106 eV)(590 nm) (3:00 10 m/s) = 8:6 10 m/s : Since v c the non-relativistic formula K = 1 mv2 may be used. 2 7E Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon. Then the power output of the Sun is given by P = RE . Now E = hf = hc=, where h is the Planck constant, f is the frequency of the light emitted, and is the wavelength. Thus P = Rhc= and 9 R = P = (6:626(550 nm)(3:s)(3:10 W)8 m/s) = 1:0 1045 photons/s : hc 10 34 J 00 10 26 8E Let the diameter of the laser beam be d, then the corss-sectional area of the beam is A = d2 =4. From the formula obtained in 7E the rate in question is given by R = P A hc(d2 =4) 3 0 = (6:626 10 4(633 nm)(5: 10 W):5 10 3 m)2 34 J s)(3:00 108 m/s)(3 = 1:7 1021 photons/m2 s : 9E Since = (1; 650; 763:73) 1 m = 6:0578021 10 7 m = 605:78021 nm, the energy is 1240 nm E = hc = 605:78021eV = 2:047 eV : nm 1062 CHAPTER 39 PHOTONS AND MATTER WAVES 10P P = E = 1240 eV nm (100=s) = 1240 eV nm (1:60 10 t 550 nm = 3:6 10 17 W : 11P 19 J/eV)(100=s) (a) Use the formula obtained in 7E: R = P=hc. The bulb emitting light with the longer wavelength (the 700 nm bulb) emits more photons per unit time. The energy of each photon is less so it must emit photons at a greater rate. (b) Let R` be the rate of photon production at the longer wavelength (` ) and let Rs be the rate of production at the shorter wavelength (s ). Then (700 J/s) R` Rs = (` hc s )P = (1:60 nm 19 400 nm)(400eV nm) = 6:0 1020 photons/s : 10 J/eV)(1240 The result hc = 1240 eV nm, developed in 1E, was used. 12P (a) The rate at which solar energy strikes the panel is P = (1:39 kW/m2 )(2:60 m2 ) = 3:61 kW : (b) The rate at which solar photons are absorbed by the panel is :61 kW P R = E = (1240 eV nm=5503nm)(1:60 10 ph (c) The time in question is given by 19 J/eV) = 1:00 1022 =s : NA = 6:02 1023 = 60:2 s : t = R 1:00 1022 =s 13P The total energy emitted by the bulb is E = 93%Pt, where P = 60 W and t = 730 h = (730 h)(3600 s/h) = 2:628 106 s. The energy of each photon emitted is Eph = hc=. Thus the number of photons emitted is 6 (93%)(60 E N = E = 93%Pt = (1240 eV nm=630W)(2:628 10 s) J/eV) = 4:7 1026 : hc= nm)(1:60 10 19 ph CHAPTER 39 PHOTONS AND MATTER WAVES 1063 The rate at which photons are emitted from the argon laser source is given by R = P=Eph , where P = 1:5 W is the power of the laser beam and Eph = hc= is the energy of each photon of wavelength . Since = 84% of the energy of the laser beam falls with in the central disk, the rate of photon absorption of the central disk is 14P P (0: R0 = R = hc= = (1240 eV nm=51584)(1:5:W) 10 nm)(1 60 = 3:3 1018 photons/s : 19 J/eV) (a) Assume all the power results in photon production at the wavelength = 589 nm. Let R be the rate of photon production and E be the energy of a single photon. Then P = RE = Rhc=, where E = hf and f = c= were used. Here h is the Planck constant, f is the frequency of the emitted light, and is its wavelength. Thus 15P P = (589 10 9 m)(100 W) R = hc (6:63 10 34 J s)(3:00 108 m/s) = 2:96 1020 photons/s : (b) Let I be the photon ux a distance r from the source. Since photons are emitted uniformly in all directions R = 4r2 I and 2:96 1020 photons/s = 4:85 107 m : r = 4R = I 4(1:00 104 photons/m2 s) (c) Let n be the photon density and consider a narrow column with cross-sectional area A and length `, with its axis along the direction of photon motion. At any instant the number of photons in the column is N = nA`. All of them will move through one end in time t = `=c, so the photon ux through the end is I = N=At = nA`c=A` = nc. Now I = R=4r2 , so R=4r2 = nc and r s R 2:96 1020 photons/s r = 4nc = = 280 m : 4(1:00 106 photons/m3 )(3:00 108 m/s) (d) The photon ux is 20 photons/s R = 5:89 1018 photons/m2 s : I = 4r2 = 2:96 10 :00 m)2 4 (2 r s According to Eq. 39-5 the condition for photoelectric eect to occur is hf . 16E 1064 CHAPTER 39 PHOTONS AND MATTER WAVES (a) For = 565 nm sium at this wavelength. (b) Now = 518 nm so hc = (4:14 10 15 eV s)(3:00 108 m/s) = 2:20 eV : hf = 565 10 9 m Since potassium > hf > cesium photoelectric eect can occur in cesium but not in potas15 eV s)(3: 8 hf = hc = (4:14 10 518 10 900 10 m/s) = 2:40 eV : m This is greater than both cesium and potassium so photoelectric eect can now occur for both metals. 17E The energy of the most energetic photon in the visible light range (with wavelength of about 400 nm) is about E = (1240 eV nm=400 nm) = 3:1 eV. So barium and lithium can be used, since their work functions are both lower than 3:1 eV. 18E (a) Since Eph = h= = 1240 eV nm=680 nm = 1:82 eV < = 2:28 eV, there is no photoelectric emission. (b) The longest (cuto) wavelength of photons which will cause photoelectric emission in sodium is given by Eph = h=max = , or max = h= = (1240 eV nm)=2:28 eV = 544 nm. This corresponds to the green color. Use Eq. 39-5 to nd Kmax : 19E Kmax = hf = (4:14 10 20E 15 eV s)(3:0 1015 Hz) 2:3 eV = 10 eV : The energy of an incident photon is Eph = hf = hc=, where h is the Planck constant, f is the frequency of the electromagnetic radiation, and is its wavelength. The kinetic energy of the most energetic electron emitted is Kmax = Eph = (hc=) , where is the work function for sodium. The stopping potential Vstop is related to the maximum kinetic energy by eVstop = Kmax , so eVstop = (hc=) and = eV hc+ = 5:1240 eV2:nm = 170 nm : 0 eV + 2 eV stop CHAPTER 39 PHOTONS AND MATTER WAVES 1065 Here eVstop = 5:0 eV and hc = 1240 eV nm were used. 21E The speed v of the electron sati es Kmax = 1 me v2 = Eph , or 2 2 v = 2(Eph ) = 2(Eph c2)c me me r 8 2 50 = 2(5:80 eV 04::511eV)(36:00 10 m/s) = 6:76 105 m/s : 10 eV s s (a) The kinetic energy Kmax of the fastest electron emitted is given by Kmax = hf = (hc=) , where is the work function of aluminum, f is the frequency of the incident radiation, and is its wavelength. The relationship f = c= was used to obtain the second form. Thus Kmax = 1240 eV nm 4:20 eV = 2:00 eV : 200 nm (b) The slowest electron just breaks free of the surface and so has zero kinetic energy. (c) The stopping potential Vstop is given by Kmax = eVstop , so V = Kmax = 2:00 eV = 2:00 V : stop 22P (d) The value of the cuto wavelength is such that Kmax = 0. Thus hc= = or eV = hc = 1240:2 eVnm = 295 nm : 4 If the wavelength is longer the photon energy is less and a photon does not have sucient energy to knock even the most energetic electron out of the aluminum sample. (a) Use Eq. 39-6: (b) Use the formula obtained in 21E: s 23P e e Vstop = hf e = hc= = (1240 eV nm=400 nm) 1:8 eV = 1:3 V : e e 2 v = 2(Eph ) = 2eVstop = 2eVstop c me me me c2 r 2 V)(3 00 8 = 2e(1:30:511: 10610 m/s) = 6:8 105 m/s : eV r s 1066 CHAPTER 39 PHOTONS AND MATTER WAVES 24P From Eq. 39-6 Kmax = Eph = hc hc = 1240 eV nm 1240 eV nm = 1:07 eV : 254 nm 325 nm max To nd the longest possible wavelength max (corresponding to the lowest possible energy) of a photon which can cause photoelectric eect in platinum, put Kmax = 0 in Eq. 39-5 and use hf = hc=. Thus hc=max = . Solve for max : eV nm max = hc = 124032 eV = 233 nm : 5: 25P (a) Use the photoelectric eect equation (Eq. 39-5) in the form hc= = + Kmax . The work function depends only on the material and the condition of the surface and not on the wavelength of the incident light. Let 1 be rst wavelength described and 2 be the second. Let Km1 (= 0:710 eV) be the maximum kinetic energy of electrons ejected by light with the rst wavelength and Km2 (= 1:43 eV) be the maximum kinetic energy of electron ejected by light with the second wavelength. Then 26P hc = + K m1 1 hc = + K : m2 2 Solve these equations simultaneously for 2 . The rst equation yields = (hc=1 ) Km1 . When this is used to substitute for in the second equation the result is (hc=2 ) = (hc=1 ) Km1 + Km2 . The solution for 2 is (1240 eV nm)(491 nm) 2 = hc + (hc1 K ) = 1240 eV nm + (491 nm)(1:43 eV 0:710 eV) 1 Km2 m1 = 382 nm : (b) The rst equation displayed above yields and hc = Km1 = 1240 eV nm 0:710 eV = 1:82 eV : 491 nm 1 CHAPTER 39 PHOTONS AND MATTER WAVES 1067 For the rst and second case (labeled 1 and 2) we have eV01 = hc=1 and eV02 = hc=2 , from which h and can be solved. (a) 1:85 eV :820 1 h = e(V1 V2 )1 = (3:00 1017 nm/s)[(3000nm) eV (400 nm) 1 ] 1 c(1 2 ) = 4:12 10 15 eV s : 27P V (1: = 3(V2 2 1 1 ) = (0:820 eV)(400 nm) 40085 eV)(300 nm) = 2:27 eV : 300 nm nm 1 2 (c) Let = hc=max to obtain eV nm max = hc = 124027 eV = 545 nm : 2: 28P (b) Vstop (V) 3.0 2.0 a 1.0 c b 0 4 6 8 10 12 f (10 14 Hz) (a) According to Eq. 39-6 the slope of the straight line plotted above is equal to h=e. Measure the slope directly from the plot to obtain h = ab = 2:0 V 0:75 V 15 e cb (10:5 1014 7:5 1014 ) Hz = 4:17 10 V s ; 1068 CHAPTER 39 PHOTONS AND MATTER WAVES so h = (4:17 10 15 V s)(1:6 10 19 C) = 6:7 10 34 J s : (b) From the plot we nd the lowest photon frequency for photoelectric eect to occur to be f0 = 5:7 1014 Hz. Thus 34 7 14 = hf0 = (6:63 10:6 J s)(5:J/C 10 Hz) = 2:4 eV ; 1 10 19 which compares favourably with the actual value of 2:3 eV. 29P The number of photons emitted from the laser per unit time is 2:00 10 3 W P = R= E ph (1240 eV nm=600 nm)(1:60 10 19 J/eV) = 6:05 1015 =s ; of which (1:0 10 16 )(6:05 1015 =s) = 0:605=s actually cause photoelectric emissions. Thus the current is i = (0:605=s)(1:60 10 19 C) = 9:68 10 20 A: 30P (a) Find the speed v of the electron from r = me v=eB : v = rBe=me . Thus 2 2 2 1 1 Kmax = 2 me v2 = 2 me rBe = (rB ) e me 2me 4 T m)2 19 C)2 (1: = 2(988 10 31 kg)(1(1:60 1019 J/eV) = 3:10 keV : :11 10 :60 10 (b) The work done is 1240 eV W = Eph Kmax = 71 10 3nm 3:10 keV = 14 keV : nm 31E (a) (b) c 3: 108 m/s f = = 3500 10 12 m = 8:57 1018 Hz : :0 E = hf = (4:14 10 15 eV s)(8:57 1018 Hz) = 3:55 104 eV : 34 J s 12 m (c) From Eq. 37-9 h :626 10 p = = 635:0 10 = 1:89 10 23 kg m/s : CHAPTER 39 PHOTONS AND MATTER WAVES 1069 (a) The rest energy of an electron is given by E = me c2 . Thus the momentum of the photon in question is given by 2 p = E = me c = me c c c = (9:11 10 31 kg)(3:00 108 m/s) = 2:73 10 22 kg m/s : 32E We may also express the momentum in terms of MeV/c: p = me c2 =c = 0:511 MeV=c. (b) 34 s = h = 2:6:626 10 10 kg Jm/s = 2:43 10 12 m = 2:43 pm : p 73 22 (c) c 3 00 108 m/s f = = 2::43 10 12 m = 1:24 1020 Hz : 33E where me is the mass of an electron and is the scattering angle. Now h=me c = 2:43 10 12 m = 2:43 pm, so = (2:43 pm)(1 cos 30 ) = 0:326 pm. The nal wavelength is 0 = + = 2:4 pm + 0:326 pm = 2:7 pm. (b) Now = (2:43 pm)(1 cos 120 ) = 3:645 pm and 0 = 2:4 pm + 3:645 pm = 6:05 pm. (a) (b) 34P (a) When a photon scatters from an electron initially at rest the change in wavelength is given by h = m c (1 cos ) ; e = hc = 1240 nm eV = 2:43 10 3 nm = 2:43 pm : E 0:511 MeV h 0 = + = + m c (1 cos ) e = 2:43 pm + (2:43 pm)(1 cos 90:0 ) = 4:86 pm : (c) 2 43 pm E 0 = E 0 = (0:511 MeV) 4::86 pm = 0:255 MeV : 1070 CHAPTER 39 PHOTONS AND MATTER WAVES Since the electron is free it is undergoing no acceleration. Therefore we may, without loss of generality, consider a stationary electron of rest mass me which collides with a photon of linear momentum p. Prior to the collision the energy of the electron is its rest energy Ee = me c2 , while the energy of the photon is simply Eph = pc. Thus the total energy of the electron-photon system before the collision is Ei = Ee + Eph = me c2 + pc : After the collision, the photon has been entirely absorbed by the electron which, according to the conservation of linear momentum, must have acquired a linear momentum equal to p, the initial linear momentum of the photon. Thus the nal energy of the system after the collision is p Ef = (pc)2 + m2 c4 : e Equate the initial and the nal energies to obtain p me c2 + pc = (pc)2 + m2 c4 : e Now square both sides: (me c2 + pc)2 = (me c2 )2 + 2me pc3 + (pc)2 = (pc)2 + m2 c4 ; e which yields 2me pc3 = 0, or p = 0: This means that the only photon that can participate in the collision is the one with zero linear momentum. The energy of the photon is thus Eph = pc = 0, which means that the photon is nonexistent and the collision is impossible. 36P 35P (a) 1 1 E = hc hc = (1240 nm eV) 0:01 nm + 4:86 pm 0:01 nm = 41 keV : 0 (c) From conservation of energy K = E = 41 keV. (b) h = m c (1 cos ) = (2:43 pm)(1 cos 180 ) = +4:86 pm : e (d) The electron will move straight ahead after the collision, since it has acquired some of the foward linear momentum from the photon. 37P (a) Since the mass of an electron is me = 9:109 10 31 kg, its Compton wavelength is 6: 10 34 J s h C = m c = (9:109 10626 kg)(2:998 108 m/s) = 2:426 10 12 m = 2:43 pm : 31 e CHAPTER 39 PHOTONS AND MATTER WAVES 1071 (b) Since the mass of a proton is mp = 1:673 10 27 kg, its Compton wavelength is 15 m = 1:32 fm : 6: 10 34 J s C = (1:673 10626 kg)(2:998 108 m/s) = 1:321 10 27 (c) Use the formula developed in 1E: E = (1240 eVnm)=, where E is the energy and is the wavelength. Thus for the electron 1240 E = 2:426 eV nm = 5:11 105 eV = 0:511 MeV : 10 3 nm (d) For the proton 1240 E = 1:321 eV nm = 9:39 108 eV = 939 MeV : 10 6 nm 38P The fractional change is E = (hc=) = 1 = 1 1 = 1 = E hc= 0 0 + 1 = =1 + 1 = (= )(1 1 ) 1 + 1 : cos C (a) Now = 3:0 cm = 3:0 1010 pm and = 90 so 1 E = 10 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 8:1 10 E (3:0 10 (b) Now = 500 nm = 5:00 105 pm and = 90 so E = 1 6 5 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 4:9 10 : E (5:00 10 (c) Now = 25 pm and = 90 so 1 E = 2 E (25 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 8:9 10 : (d) Now = hc=E = 1240 nm eV=1:0 MeV = 1:24 10 3 nm = 1:24 pm and = 90 so 1 E = E (1:24 pm=2:43 pm)(1 cos 90 ) 1 + 1 = 0:66 : 11 : 1072 CHAPTER 39 PHOTONS AND MATTER WAVES (e) From the calculation above we see that the shorter the wavelength the greater the fractinal energy change for the photon as a result of the Compton scattering. Since E=E is virtually zero for microwave and visible light, Compton eect is signi cant only in the x-ray to gamma ray range of the electromagnetic spectrum. If E is the original energy of the photon and E 0 is the energy after scattering, then the fractional energy loss is 0 frac = E E E : Sample Problem 39-4 shows that this is frac = + : Thus = frac = 0:75 = 3 : 1 frac 1 0:75 A 300% increase in the wavelength leads to a 75% decrease in the energy of the photon. 40P 39P h 2 max = m c (1 cos ) = mhc p p max 15 eV s)(3:00 108 m/s) = 2(4:14 10 = 2:65 10 15 m = 2:65 fm : 938 MeV 41P The dierence between the electron-photon scattering process in this problem and the one studied in the text (the Compton shift, see Eq. 39-11) is that the electron is in motion relative with speed v to the laboratory frame. To utilize the result in Eq. 39-11, shift to a new reference frame in which the electron is at rest before the scattering. Denote the quantities measured in this new frame with a prime ('), and apply Eq. 39-11 to yield 0 = 0 0 = h (1 cos ) = 2h ; 0 where we noted that = (since the photon is scattered back in the direction of incidence). 0 Now, from the Doppler shift formula (Eq. 38-25) the frequency f0 of the photon prior to the scattering in the new reference frame satis es me c me c f00 = c0 = f0 1 + ; 1 0 s CHAPTER 39 PHOTONS AND MATTER WAVES 1073 where = v=c. Also, as we switch back from the new reference frame to the original one after the scattering s s c f = f 0 1 + = 0 1 + : 1 1 Solve the two Doppler-shift equations above for 0 and 00 and substitute the results into the Compton shift formula for 0 : 1 0 = f 1 + 1 Some simple algebra then leads to s f0 1 s 1 = 2h : 1 + me c2 s ! 1 E = hf = hf0 where we used E = hf . (a) From Eq. 39-11 42P 2 1 + m h2 1 + ec 1 ; h = m c (1 cos ) = (2:43 pm)(1 cos 90 ) = 2:43 pm : e (b) 2 15 8 = (4:14 10 eV s)(3:00 210 m/s)(2:43 pm) (590 nm) 6 eV : = 8:67 10 For an x ray photon of energy Eph = 50 keV, remains the same (= 2:43 pm) since it is independent of Eph . The fractional change in wavelength is now = = (50 103 eV)(2:43 pm) 2 hc=Eph (4:14 10 15 eV s)(3:00 108 m/s) = 9:78 10 ; and the change in photon energy is now 1 1 = Eph = hc + = 2:425 pm = 4:11 10 6 : 590 nm (c) The change in energy for a photon with = 590 nm is given by E = hc hc ph hc = E ph 1 + ; + 1074 CHAPTER 39 PHOTONS AND MATTER WAVES where = =. Plug in Eph = 50 keV and = 9:78 10 2 to obtain Eph = 4:45 keV. (Note that in this case 0:1 is not close enough to zero so the approximation Eph hc=2 is not as accurate as in the rst case, in which = 4:12 10 6 . In fact if you were to use this approximation here you would get Eph 4:89 keV, which does not amount to a satisfactory approximation.) (a) From Eq. 39-11 = (h=me c)(1 cos ): In this case = 180 so cos = 1, and the change in wavelength for the photon is given by = 2h=me c. The energy E 0 of the scattered photon (whose initial energy is E = hc=) is then 43P E E hc E 0 = + = 1 + E = = 1 + (2h=m c)(E=hc) = 1 + 2E=m c2 e e 50:0 keV = 1 + 2(50:0 keV)=0:511 MeV = 41:8 keV : (b) From conservation of energy the kinetic energy K of the electron is given by K = E E 0 = 50:0 keV 41:8 keV = 8:2 keV. 44P From the result of 38P E = 1 = 0 = = (h=mc)(1 cos ) = hf 0 (1 cos ) : E 0 0 0 c=f 0 mc2 Here the minus sign indicates that part of the photon energy is lost during the scattering. The magnitude of the fractional energy change for the photon is given by Eph E 45P ph = (hc=) hc= = 1 1 = 1 = + = ; + where = 10%. Thus = =(1 ). Plug in this expression for into Eq. 39-11 and solve for cos : 2 mc cos = 1 mc = 1 h(1 ) = 1 (1 (mcE) h ) ph = 1 (1(10%)(511 keV) = 0:716 : 10%)(200 keV) The angle is = 44 . CHAPTER 39 PHOTONS AND MATTER WAVES 1075 The maximum energy of the electron is attained when the loss of energy of the photon is the greatest: E = 1 = =21 + 1 : 1+1 E (=C )(1 cos ) C max Conservation of energy then gives 2 E Kmax = Emax = =2E + 1 = (hc=E )(m c=2h) + 1 = m c2E2 + E : = 46P C e e 47P Use the formula obtained in 46P: e 2 2 (17 Kmax = E + E c2 =2 = 17:5 keV:5 keV)keV=2 = 1:12 keV : m + 511 48P Rewrite Eq. 39-9 as and Eq. 39-10 as h h cos = p v cos m m0 1 (v=c)2 h sin = p v sin : m0 1 (v=c)2 1 cos 0 2 Square both equations and add up the two sides: h m 2 " 1 1 + 0 sin 2 # 2 = 1 vv=c)2 ; ( where we used sin2 + cos2 = 1 to eliminate . Now the RHS can be written as v2 = c2 1 1 1 (v=c)2 1 (v=c)2 ; so # " v2 = h 2 1 1 cos 2 + 1 sin 2 + 1 : 1 (v=c)2 mc 0 0 Now rewrite Eq. 39-8 as h 1 1 +1= p 1 mc 0 1 (v=c)2 1076 CHAPTER 39 PHOTONS AND MATTER WAVES and compare with the previous equation we obtained for [1 (v=c)2 ] 1 . This yields h 1 1 +1 2 = h mc 0 mc 2 " 1 1 cos 0 2 1 + 0 sin 2 # + 1: We have so far eliminated and v. Working out the squares on both sides and noting that sin2 + cos2 = 1, we get h 0 = = mc (1 cos ) : 49E (a) (b) The de Broglie wavelength of the bullet is much too short for the wave nature of the bullet to reveal itself through diraction eects. (a) Substitute the classical relationship between momentum p and velocity v, v = p=me into the classical de nition of kinetic energy, K = 1 me v2 , to obtain K = p2 =2me . Here p2 me is the mass of an electron. Solve for p: p = 2me K . The relationship between the momentum and the de Broglie wavelength is = h=p, where h is the Planck constant. Thus = p h : 2me K If K is given in electron volts then 6:626 10 34 J s p = p 2(9:109 10 31 kg) (1:602 10 19 J/eV)K 1=2 1=2 9 p p = 1:226 10 m (eV) = 1:226 nm (eV) : 50E h 6 63 10 34 J = h = mv = (40 :10 kg)(1000sm/s) = 1:7 10 3 p 35 m : K K 51E Use the result of 50E and K = eV , where V = 25:0 kV. Thus 1=2 1=2 1=2 :226 p p = 1:226 nm (eV) = 1:226 nm (eV) = 1p nm (eV) K eV e(25 103 V) = 7:75 10 3 nm = 7:75 pm ; CHAPTER 39 PHOTONS AND MATTER WAVES 1077 where we noted that 1 eV = 1e 1 V. 52E (a) (b) (c) = 38:7 pm : = h = p h = p hc 2 = p 1240 eV nm p 2me K 2me c K 2(0:511 MeV)(1:00 keV) = hc = 1240 eV nm = 1:24 nm : E 1:00 keV = 9:04 10 4 nm = 904 fm : = p hc 2 = p 1240 eV nm mn c K 2(939 MeV)(1:00 keV) 53P Use the result of 50E: = (1:226 nmeV1=2 )= K , where K is the kinetic energy. Solve for K: " # # 2 " 2 1:226 nm (eV)1=2 = 1:226 nm (eV)1=2 = 4:32 10 6 eV : K= 590 nm p 54P (a) and (b) The momenta of the electron and the photon are the same: 34 h : p = = 6063 10 9 J s = 3:3 10 :20 10 m 24 kg m/s : The kinetic energy of the electron is 2 :3 10 24 2 Ke = 2p = (32(911 10kg31m/s) = 6:0 10 me : kg) 18 J = 38 eV ; while that for the photon is Kph = pc = (3:3 10 55P 24 kg m/s)(3:00 108 m/s) = 9:9 10 16 J = 6:2 keV : (a) 38 3 K = 2 kT = 3(1:2(1: 10 10 J/K)(300 K) = 3:88 10 2 eV = 38:8 meV : 60 19 J/eV) 23 1078 CHAPTER 39 PHOTONS AND MATTER WAVES (b) 1240 nm eV = p h = p hc 2 = p 2mn K 2(mn c )K 2(939 MeV)(3:88 10 2 eV) = 1:46 10 10 m = 146 pm : 56P (a) Solve v from = h=p = h=(mp v): 34 J s 6 v = mh = (1:675 10:63 10 :100 10 27 kg)(0 p 12 m) = 3:96 106 m/s : (b) Let eV = K = 1 mp v2 and solve for V : 2 27 kg)(3 6 2 p v2 V = m2e = (1:67 10 :60 10:96 10 m/s) = 8:18 103 V : 19 C) 2(1 57P (a) The average de Broglie wavelength is = h = p h = p h = p hc2 p 2m(3kT=2) 2(mc )kT 2mK 1240 eV nm =p = 7:3 10 3(4)(938 MeV)(8:62 10 5 eV/K)(300 K) 11 m = 73 pm ; while the average separation is 31 d = pn = (b) Yes, since d. (a) The momentum of the photon is given by p = Eph =c, where Eph is its energy. Its wavelength is eV nm hc = h = E = 124000 eV = 1240 nm = 1:24 nm : p 1: ph 58P 1 p 3 p=kT = r 3 (1:38 10 23 J/K)(300 K) = 3:4 nm : 1:01 105 Pa CHAPTER 39 PHOTONS AND MATTER WAVES 1079 The momentum of the electron is given by p = 2me K , where K is its kinetic energy and me is its mass. Its wavelength is According to 50E if K is in electron volts this is 226 p = 1:226 nm = 1:p nm = 1:23 nm : 1:00 K (b) For the photon nm hc = E = 11240 eV 9 eV = 1:24 10 6 nm = 1:24 fm : :00 10 ph Relativity theory must be used to calculate the wavelength for the electron. The momentum p and kinetic energy K of an electron are related by (pc)2 = K 2 + 2Kme c2 . Thus p = h = p h : p 2me K pc = K 2 + 2Kme c2 = (1:00 109 eV)2 + 2(1:00 109 eV)(0:511 106 eV) = 1:00 109 eV : The wavelength is nm = h = hc = 11240 eV 9 eV = 1:24 10 6 nm = 1:24 fm : p pc :00 10 p p 59P (a) For the photon and for the electron Eph = hc = 1240 nm eV = 1:24 keV 1:00 nm 2 2 2 hc= 2 1 K = 2p = (h=) = (2m c) = 2(0:511 MeV) 1240 nm eV 2 me 2me 1:00 nm e = 1:50 eV : (b) Now for the photon and for the electron p p K = p2 c2 + (me c2 )2 me c2 = (hc=)2 + (me c2 )2 me c2 s 1240 fm MeV 2 + (0:511 MeV)2 0:511 MeV = 1:00 fm = 1:24 103 MeV = 1:24 GeV : nm Eph = 124000 fmeV = 1:24 GeV 1: 1080 CHAPTER 39 PHOTONS AND MATTER WAVES Note that at short (large K ) the kinetic energy of the electron, calculated with relativistic formula, is about the same as that of the photon. This is expected since now K E pc for the electron, which is the same as E = pc for the photon. (a) The kinetic energy acquired is K = qV , where q is the charge on an ion and V is the accelerating potential. Thus K = (1:60 10 19 C)(300 V) = 4:80 10 17 J. The mass of a single sodium atom is, from Appendix F, m = (22:9898 g/mol)=(6:02 1023 atom/mol) = 3:819 10 23 g = 3:819 10 26 kg. Thus the momentum of an ion is 60P p = 2mK = 2(3:819 10 (b) The de Broglie wavelength is p p 26 kg)(4:80 10 17 J) = 1:91 10 21 kg m/s : 34 s 6:626 = h = 1:915 10 kgJm/s = 3:46 10 p 10 21 13 m = 346 fm : 61P We need to use the relativistic formula p = (E=c)2 m2 c2 E=c K=c (since E e me c2 ). So eV nm = h hc = 1240 GeV = 0:025 fm ; p K 50 which is about 200 times smaller than the radius of an average nucleus. 62P p (a) Since K = 7:5 MeV m c2 = 4(932 MeV), we may use the non-relativistic formula p p = 2m K . So 1240 nm eV = 5:2 fm : = h = p hc 2 = p p 2m c K 2(4)(932 MeV)(7:5 MeV) (b) Since = 5:2 fm 30 fm, to a fairly good approximation the wave nature of the particle dose not need to be taken into consideration. The wavelength associated with the unknown particle is = h=p = h=(mv), where p is its momentum, m is its mass, and v is its speed. The classical relationship p = mv was used. Similarly, the wavelength associated with the electron is e = h=(me ve ), where me is its mass and ve is its speed. The ratio of the wavelengths is =e = (me ve )=(mv), so 31 e m = vve me = 9:109 10 10 kg = 1:675 10 3(1:813 4 ) 27 kg : 63P CHAPTER 39 PHOTONS AND MATTER WAVES 1081 According to Appendix B this is the mass of a neutron. 64P (a) Let = h=p = h= (E=c)2 m2 c2 and solve for K = E me c2 : e p K= s s 1240 eV nm 2 + (0:511 MeV)2 0:511 MeV = 10 10 3 nm = 0:015 MeV = 15 keV : (b) 1240 eV E = hc = 10 10 3nm = 1:2 105 eV = 120 keV : nm hc 2 + m2 c4 me c2 e (c) The electron microscope is more suitable, as the required energy of the electrons is much less than that of the photons. The same resolution requires the same wavelength and since the wavelength and particle momentum are related by p = h=, this means the same particle momentum. The momentum of a 100-keV photon is p = E=c = (100 103 eV)(1:60 10 19 J/eV)=(3:00 108 m/s) = 5:33 10 23 kg m/s. This is also the magnitude of the momentum of the electron. In the classical approximation, the kinetic energy of the electron is 2 23 2 33 K = 2p = (5:2(9 10 10 kg m/s) = 1:56 10 me :11 31 kg) 15 J : 65P The accelerating potential is thus 1 56 10 V = K = 1::60 10 e 15 J 19 C = 9:76 103 V : p Relativistically, the kinetic energy of the electron satis es eV = K = (me c2 )2 + (pc)2 me c2 , which we may solve for V : V=1 e 1 = hp hp (me c2 )2 + (pc)2 me c2 i e = 9:69 103 V : (5:11 105 eV)2 + (100 103 eV)2 5:11 105 eV i 1082 CHAPTER 39 PHOTONS AND MATTER WAVES (a) 66E nn = (a + ib)(a + ib) = (a + ib)(a + i b ) = (a + ib)(a ib) = a2 + iba iab + (ib)( ib) = a2 + b2 ; which is always real since both a and b are real. (b) jnmj = j(a + ib)(c + id)j = jac + iad + ibc + ( i)2 bdj = j(ac bd) + i(ad + bc)j p = (ac bd)2 + (ad + bc)2 p = a2 c2 + b2 d2 + a2 d2 + b2 c2 : p p But p jnj jmj = ja + ibj jc + idj = a2 + b2 c2 + d2 = a2 c2 + b2 d2 + a2 d2 + b2 c2 ; so jnmj = jnj jmj. Plug Eq. 39-17 into Eq. 39-16. Note that 67P d = d Aeikx + Be ikx = ikAeikx ikBe ikx dx dx and Thus d2 = d ikAeikx ikBe ikx = k2 Aeikx k2 Beikx : dx2 dx d2 + k2 = k2 Aeikx k2 Beikx + k2 Aeikx + Be ikx = 0 : dx2 Use the Euler formula ei = cos + i sin to re-write (x) as (x) = 0 eikx = 0 (cos kx + i sin kx) = ( 0 cos kx) + i( 0 sin kx) = a + ib ; where a = 0 cos kx 68P and b = 0 sin kx are both real quantities. CHAPTER 39 PHOTONS AND MATTER WAVES 1083 (b) (x; t) = (x)e i!t = 0 eikx e i!t = 0 ei(kx !t) = [ 0 cos(kx !t)] + i [ 0 sin(kx !t)] : The angular wave number k is related to the wavelength by k = 2= and the wavelength is related to the particle momentum p by = h=p, so k = 2p=h. Now the kinetic energy K and the momentum are related by K = p2 =2m, where m is the mass of the particle. p Thus p = 2mK and p 2 2mK : k= h 70P 69P For Epot = E0 , Schrdinger's equation becomes o d2 + 82 m (E E ) = 0 : 0 dx2 h2 Substitute result is = 0 eikx . The second derivative is d2 =dx2 = k2 0 eikx = k2 . The 2 k2 + 8 2m (E E0 ) = 0 : h Solve for k and obtain r 2 p k = 8 2m [E E0 ] = 2h 2m(E E0 ) : h From Eq. 39-14 71P j(x; y; z; t)j = =j =j =j Here we used the result of 66E. (x; y; z) e i!t = j (x; y; z)j e i!t (x; y; z)j jcos( !t) + i sin( !t)j p (x; y; z)j (cos !t)2 + (sin !t)2 (x; y; z)j : 1084 CHAPTER 39 PHOTONS AND MATTER WAVES The wave function is now given by (x; t) = 0 e i(kx+!t) : This function describes a plane matter wave traveling in the negative x direction. An example of the actual particles that t this description is a free electron with linear momentum p = (hk=2) i and kinetic energy K = p2 =2me = h2 k2 =82 me . (a) The wave function is now given by (x; t) = Thus 0 h 73P 72P i ei(kx !t) + e i(kx+!t) = 0e i!t eikx + e ikx : j(x; t)j2 = 0 e (b) 4 = 0e 2 = 0 eikx + e ikx 2 2 = 0 j(cos kx + i sin kx) + (cos kx i sin kx)j2 2 = 4 0 (cos kx)2 2 = 2 0 (1 + cos 2kx) : i!t eikx + e ikx 2 i!t 2 eikx + e ikx 2 3 |(x, t)/0|2 2 1 0 0 2 4 6 8 kx Consider two plane matter waves, each with the same amplitude 0 = 2 and travels in opposite directions along the x axis. The combined wave is a standing wave: (x; t) = 0 ei(kx !t) + 0 e i(kx+!t) = 0 eikx + e ikx e i!t = (2 0 cos kx) e i!t : p CHAPTER 39 PHOTONS AND MATTER WAVES 1085 Thus the squared amplitude of the matter wave is j(x; t)j2 = (2 0 cos kx)2 e i!t 2 = 2 2 (1 + cos 2kx) ; 0 which is the same as the function plotted above. 2 (c) Set j(x; t)j2 = 2 0 (1 + cos 2kx) = 0 to obtain cos(2kx) = 1. This gives 2kx = 2 2 = (2n + 1) ; Solve for x: (n = 0; 1; 2; 3; . . . ) (d) The most probable positions for nding the particle are where j(x; t)j / (1 + cos 2kx) reaches its maximum. Thus cos 2kx = 1, or x = 1 (2n + 1) : 4 2kx = 2 2 = 2n ; Solve for x: 1 x = 2 n : (n = 0; 1; 2; 3; . . . ) (a) Since px = py = 0, px = py = 0. Thus from Eq. 39-20 both x and y are in nite. It is therefore impossible to assign a y or z coordinate to the position of an electron. (b) Since it is independent of y and z the wave function (x) should describe a plane wave that extends in nitely in both the y and z directions. Also from Fig. 39-12 we see that j(x)j2 extends inifnitely along the x axis. Thus the matter wave described by (x) extends throughout the entire thhree-dimensional space. 75E 74E h 10 34 p x = 6:63(50 pm)J s = 2:1 10 2 24 kg m/s : 76E Denote the Plank's constant in the imaginary universe as h0 . Then the uncertainty in the position of the baseball would be 0= h0 0 60 J x h2 = 2mv = 2(0:50:kg)(1s0 m/s) = 0:19 m : p : 1086 CHAPTER 39 PHOTONS AND MATTER WAVES It would be hard to catch such a baseball, since its size is less than the uncertainty in its position. 77P If x = =2 then from Eq. 39-20 2h h p = x = =h = h=p = p ; 2 were we used h = 2h and p = h=. For p = 0 the formula above gives p = 0. This is understandable, since now = h=p ! 1 so x = =2 ! 1. So while the measurement of momentum can be exact, it is now impossible to locate the position x of the particle. As the measured momentum p increases, so does p. This is also understandable because as p increases x = =2 = h=p / 1=p decreases, keeping xp = h. 78P (a) (b) eV E = hc = 101240 nm 3 nm = 124 keV : :0 10 1 E = hc = hc +1 = hc + = 1 + E = E = 1 + (= )(1 cos ) 1 = 1 + (10:0 pm=2:124 keV cos 180 ) 43 pm)(1 C = 40:5 keV : 1 (c) It is impossible to \view" an atomic electron with such a high-energy photon, because with the energy imparted to the electron the photon would have knocked the electron out of its orbit. The probability T that a particle of mass m and energy E tunnels through a barrier of height U and width L is given by T = e 2kL ; where r 2 k = 8 m(U E ) : h2 79P CHAPTER 39 PHOTONS AND MATTER WAVES 1087 For the proton 2 k = 8 (1:6726 10 s 27 kg)(10 Mev = 5:8082 1014 m 1 ; kL = (5:8082 1014 m 1 )(10 10 15 m) = 5:8082, and T = e 25:8082 = 9:02 10 6 : The value of k was computed to a greater number of signi cant digits than usual because an exponential is quite sensitive to the value of the exponent. The mass of a deuteron is 2:0141 u = 3:3454 10 27 kg, so 2 k = 8 (3:3454 10 s 3:0 Mev)(1:6022 10 (6:6261 10 34 J s)2 13 J/MeV) 27 kg)(10 MeV = 8:2143 1014 m 1 ; kL = (8:2143 1014 m 1 )(10 10 80P 3:0 MeV)(1:6022 10 (6:6261 10 34 J s)2 13 J/MeV) 15 m) = 8:2143, and T = e 28:2143 = 7:33 10 8 : r # Let T = exp( 2kL) = exp " and solve for E : 2 2L 8 m(U E ) h2 2 E = U 21 h4ln T m L (1240 eV nm)(ln 0:001) 2 1 = 6:0 eV 2(0:511 MeV) 4(0:70 nm) = 5:1 eV : (a) The rate at which incident protons arrive at the barrier is n = 1:0 kA=1:60 10 6:25 1023 =s. Let nTt = 1 to nd the waiting time t: 81P 19 C = t = (nT ) 1= 70 p = 6:25 11023 =s exp 2(0:eVnm) 8(938 MeV)(6:0 eV 5:0 eV) 1240 nm 111 s 10104 y ; = 3:37 10 1 exp 2L 82 mp (U E ) n h2 " r # 1088 CHAPTER 39 PHOTONS AND MATTER WAVES which is much, much longer than the age of the universe. (b) Replace the mass of the proton with that of the electron to obtain the corresponding waiting time for an electron: # " r 1 exp 2L 82 me (U E ) t = (nT ) 1 = 70 p = 6:25 11023 =s exp 2(0:eVnm) 8(0:511 MeV)(6:0 eV 5:0 eV) 1240 nm = 2:1 10 19 s : The enormous dierence between the two waiting times is the result of the dierence between the masses of the two kinds of particles. 82P n h2 (a) If m is the mass of the particle and E is its energy then the probability it will tunnel through a barrier of height U and width L is given by T = e 2kL ; where r 2 k = 8 m(U E ) : h2 If the change U in U is small (as it is) the change in the tunneling probability is given by Now Thus dT dk T = dU U = 2LT dU U : r 82 m = 82 m(U E ) = 1 k : dk = p 1 dU 2 U E h2 2(U E ) h2 2(U E ) r T = LTk U E : For the data of Sample Problem 39-7, 2kL = 10:0, so kL = 5:0 and T = kL U = (5:0) (0:0100)(6:8 eV) = 0:20 : T U E 6:8 eV 5:1 eV There is a 20% decrease in the tunneling probability. (b) The change in the tunneling probability is given by and T = dT L = 2ke 2kL L = 2kT L dL T = 2k L = 2(6:67 109 m 1 )(0:0100)(750 10 T There is a 10% decrease in the tunneling probability. U 12 m) = 0:10 : CHAPTER 39 PHOTONS AND MATTER WAVES 1089 (c) The change in the tunneling probability is given by dT dk dk T = dE E = 2Le 2kL dE E = 2LT dE E : Now dk=dE = dk=dU = k=2(U E ), so T = kL E = (5:0) (0:01000)(5:1 eV) = 0:15 : T U E 6:8 eV 5:1 eV There is a 15% increase in the tunneling probability. The kinetic energy of the car of mass m moving at speed v is given by E = 1 mv2 ; while 2 the potential barrier it has to tunnel through is Epot = mgh, where h = 24 m. According to Eq. 39-21 and 39-22 the tunneling probability is given by T e 2kL , where 83P k= r 2 = 6:626(1500 kg)J s 2 (9:80 m/s2 )(24 m) 1 (20 m/s)2 10 34 2 = 1:2 1038 m 1 : 82 m(Epot h2 E ) = 82 m(mgh h2 s s 1 mv 2 ) 2 Thus 2kL = 2(1:2 1038 m 1 )(30 m) = 7:2 1039 . You can see that T e 2kL is almost zero. (a) Set Epot = 0 in Eq. 39-15: 84P d2 + 82 mE = 0 : dx2 h2 For the wave function (x) = Aeikx + Be ikx d2 = (ik)2 Aeikx + ( ik)2 Be ikx = k2 (Aeikx + B ikx ) = k2 : dx2 Plug this into the Schrdinger's equation to obtain o k2 2 (x) + 8hmE (x) = 2 k2 + 82 mE h2 (x) = 0 : 1090 CHAPTER 39 PHOTONS AND MATTER WAVES Since (x) 6= 0, Thus 2 k2 + 8hmE = 0 : 2 82 mE = 82 m p2 = 42 p2 ; h2 h2 2m h2 or k = 2p=h. Here k is arbitrarily set to be positive, which does not aect the generality of the solution since it contains both e+ikx and e ikx . (b) k2 = = [(A + B ) cos kx]2 + [(A B ) sin kx]2 = (A2 + 2AB + B 2 ) cos2 kx + (A2 2AB + B 2 ) sin2 kx = (A2 + B 2 )(cos2 kx + sin2 kx) + 2AB (cos2 kx sin2 kx) = A2 + B 2 + 2AB cos 2kx ; where to reach the last step we used the trigonometric identities cos2 + sin2 = 1 and cos2 sin2 = cos 2. (c) Since both A and B are constants j (x)j2 is a function of cos 2kx, which varies between 1 and +1. Thus j (x)j2 varies between A2 + B 2 + 2AB ( 1) = (A B )2 and A2 + B 2 + 2AB (+1) = (A + B )2 . (a) For Epot = const. > E Eq. 39-15 can be written as d2 + 82 m(Epot E ) = d2 + k2 = 0 ; dx2 h2 dx2 r 85P j (x)j2 = Aeikx + Be ikx 2 = jA cos kx + iA sin kx + B cos kx iB sin kxj2 = j(A + B ) cos kx + i(A B ) sin kxj2 where This solution, however, should be discarted in our case becasue it leads to the unphysical limit (x) ! 1 at x ! 1. (b) Since (x) is real, 2 k = 8 m(Epot E ) : h2 Plug the trial solution (x) = Ce kx into the Schrdinger's equation: o ( k)2 Ce kx + k2 Ce kx = 0 : Thus (x) = Ce kx is indeed a solution to the equation in this region. You can verify that in general there can be another solution in the form De+kx , where D is another constant. j (x)j2 = [ (x)]2 = Ce kx 2 = C 2 e 2kx : ...
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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