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Course Number: PHYSICS 31, Spring 2007

College/University: Lehigh

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Physics 31 Spring, 2007 Solution to HW #2 Problem A Use the first order Taylor approximation (1 + )n 1 + n (for 1) to simplify the following expressions for the indicated limiting cases: (a) (b) (c) M -m M +m 1 1 - + p2 c2 + m2 c4 for m M mc 217 A metal surface illuminated by 8.5 1014 Hz light emits electrons whose maximum energy is 0.52 eV. The same surface illuminated by 12.0 1014 Hz light emits...

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31 Physics Spring, 2007 Solution to HW #2 Problem A Use the first order Taylor approximation (1 + )n 1 + n (for 1) to simplify the following expressions for the indicated limiting cases: (a) (b) (c) M -m M +m 1 1 - + p2 c2 + m2 c4 for m M mc 217 A metal surface illuminated by 8.5 1014 Hz light emits electrons whose maximum energy is 0.52 eV. The same surface illuminated by 12.0 1014 Hz light emits electrons whose maximum energy is 1.97 eV. From these data, find Planck's constant and the work function of the surface. Use Kmax = h - to set up two equations: 0.52 eV = h(8.5 1014 Hz) - 1.97 eV = h(12.0 1014 Hz) - Subracting the first equation from the second gives for for p 1.45 eV = h(3.5 1014 Hz) h = 4.14 10-15 eV s. and substituting this value of h back into the first equation gives -1 For part (a), m M 1- M -m m M = 1- m = 1+ m M +m M M M 1+ M m m m 1- 1-2 1- M M M For part (b), 1 1 1 - = - + 1 1- 1+ 1 1+ -1 = h(8.5 1014 Hz) - 0.52 eV = 3.00 eV. 222 The smallest angle of Bragg scattering in potassium chloride (KCl) is 28.4 for 0.30 nm x-rays. Find the distance between atomic planes in potassium chloride. Use 2d sin = n (n = 1) to get 1 1- 1- = 2 d= 0.3 nm = = 0.315 nm = 3.15 A 2 sin 2 sin 28.4 = 226 How much energy must a photon have if it is to have the momentum of a 10 MeV proton? The momentum of the proton and that of the photon are the same, so call that value p. For the proton, Eproton = p2 p= 2mp 2mp Eproton For part (c), as we did in class, p2 c2 + m2 c4 = = mc2 p2 1+ 2 2 m c m2 c4 1 + 1/2 p2 c2 m2 c4 p2 p2 1 + 1 2 2 = mc2 + 2m c 2m mc2 and for the photon, p = Ephoton /c. Equating the expressions for p, Ephoton = c 2mp Eproton = c 2mp (107 eV)(1.602 10-19 J/eV = 2.196 10-11 J = 137 MeV 228 A monochromatic x-ray beam whose wavelength 55.8 is pm is scattered through 46 . Find the wavelength of the scattered beam. - = C (1 - cos ) = + C (1 - cos ) . Substituting C = 2.426 pm leads to = 55.8 pm + 0.74 pm = 56.5 pm 216 Light of wavelength 400 nm is shone on a metal surface in an apparatus like that of Fig. 2.9. The work function of the metal is 2.50 eV. (a) Find the extinction voltage, that is, the retarding voltage at which the photoelectron current disappears. (b) Find the speed of the fastest photoelectrons. The basic equation is Kmax = h - = hc - , and we just need to substitute = 400 nm and = 2.50 eV to obtain 1240 eV nm - 2.50 eV = 0.60 eV Kmax = 400 nm When an electron is accelerated through a voltage 0.60 V, its energy is 0.60 eV, so the extinction voltage is 0.6 V. For part (b), we get the speed by equating Kmax to 1 mv 2 : 2 v= 2 0.6 eV 1.602 10-19 J/eV 9.1095 10-31 kg 1/2 = 4.59 105 m/s 234 (a) Find the change in wavelength of 80 pm x-rays that are scattered 120 by a target. (b) Find the angle between the directions of the recoil electron and the incident photon. (c) Find the energy of the recoil electron. For part (a), - = C (1 - cos 120 ) = 3.64 pm For part (b), we write the equations for conservation of momentum. Let pe be the final momentum of the electron, and be the angle of recoil of the electron (as in Fig. 2.22). h h = cos 120 + pe cos h sin 120 - pe sin 0 = Rearrange these equations to get pe cos = h h - cos 120 h pe sin = sin 120 Now divide the second equation by the first and substitute = 80 pm and = 83.64 pm. pe cancels out, and we get tan = 0.5604 = 29.3 . For part (c), the energy lost by the photon is gained by the electron, so the electron's energy E is E = hc hc - = 1240eV nm 1 1 - 0.080 nm 0.08364 nm = 674 eV = 1.08 10-16 J January 31, 2008

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