Gasiorowicz quantum physics 3rd ed solutions (2)

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  • 1. SOLUTIONS MANUAL CHAPTER 11. The energy contained in a volume dV is U(,T )dV = U ( ,T )r 2 dr sin ddwhen the geometry is that shown in the figure. The energy from this source that emergesthrough a hole of area dA is dAcos dE( ,T ) = U ( ,T )dV 4r 2The total energy emitted is /2 2 dA dE ( ,T ) = dr d dU (,T )sin cos ct 0 0 0 4 dA /2 = 2 ctU ( ,T ) d sin cos. 4 0 1 = ctdAU ( ,T ) 4 By definition of the emissivity, this is equal to EtdA . Hence c E (, T ) = U (, T ) 42. We have c c 8 hc 1 w( ,T ) = U ( ,T ) | d / d |= U ( ) 2 = 5 e hc/kT 1This density will be maximal when dw( ,T ) / d = 0 . What we need is A / d 1 1 1 1 e A 1 = (5 6 5 A / ( 2 )) A / =0 d e 1 5 A / e 1 e 1Where A = hc / kT . The above implies that with x = A / , we must have 5 x = 5e xA solution of this is x = 4.965 so that
  • 2. hc maxT = = 2.898 10 3 m 4.965kIn example 1.1 we were given an estimate of the suns surface temperatureas 6000 K. From this we get 28.98 10 4 mK sun = = 4.83 107 m = 483nm max 6 10 3 K3. The relationship is h = K + Wwhere K is the electron kinetic energy and W is the work function. Here hc(6.626 1034 J .s)(3 10 8 m / s)h = = = 5.68 10 19 J = 3.55eV 350 10 m 9With K = 1.60 eV, we get W = 1.95 eV4. We use hc hc = K1 K2 1 2since W cancels. From ;this we get 1 1 2h= (K K 2 ) = c 2 1 1 (200 10 9 m)(258 109 m)= (2.3 0.9)eV (1.60 10 19 )J / eV (3 10 8 m / s)(58 109 m) 34= 6.64 10 J .s5. The maximum energy loss for the photon occurs in a head-on collision, with thephoton scattered backwards. Let the incident photon energy be h , and the backward-scattered photon energy be h . Let the energy of the recoiling proton be E. Then itsrecoil momentum is obtained from E = p 2c 2 + m 2c 4 . The energy conservationequation reads h + mc 2 = h + Eand the momentum conservation equation reads h h = +p c c
  • 3. that is h = h + pcWe get E + pc mc 2 = 2h from which it follows that p2 c 2 + m2 c 4 = (2h pc + mc 2 ) 2so that 4 h 2 2 + 4 hmc 2 pc = 4 h + 2mc 2The energy loss for the photon is the kinetic energy of the protonK = E mc 2 . Now h = 100 MeV and mc 2 = 938 MeV, so that pc = 182MeVand E mc 2 = K = 17.6 MeV6. Let h be the incident photon energy, h the final photon energy and p the outgoing electron momentum. Energy conservation reads h + mc = h + p c + m c 2 2 2 2 4We write the equation for momentum conservation, assuming that the initial photonmoves in the x direction and the final photon in the y-direction. When multiplied by c itread i(h ) = j(h ) + (ipx c + jpy c)Hence px c = h ; pyc = h . We use this to rewrite the energy conservation equation asfollows: (h + mc 2 h )2 = m 2c 4 + c 2 ( px2 + py2 ) = m2 c 4 + (h ) 2 + (h ) 2From this we get 2 h= h mc h + mc 2We may use this to calculate the kinetic energy of the electron
  • 4. 1 mc 2 h K = h h = h 2 = h h + mc h + mc 2 (100keV )2 = = 16.4 keV 100keV + 510 keVAlso pc = i(100keV ) + j(83.6keV)which gives the direction of the recoiling electron.7. The photon energy is hc (6.63 10 34 J.s)(3 108 m / s) h = = 9 = 6.63 10 17 J 6 3 10 10 m 17 6.63 10 J = 19 = 4.14 10 4 MeV 1.60 10 J / eVThe momentum conservation for collinear motion (the collision is head on for maximumenergy loss), when squared, reads 2 2 h h h h c + p + 2 c pi = c + p +2 c p f 2 2Here i = 1, with the upper sign corresponding to the photon and the electron moving inthe same/opposite direction, and similarly for f . When this is multiplied by c2 we get (h ) 2 + (pc) 2 + 2(h ) pci = (h )2 + ( pc) 2 + 2(h ) pc fThe square of the energy conservation equation, with E expressed in terms ofmomentum and mass reads (h ) 2 + (pc) 2 + m 2c 4 + 2Eh = (h )2 + ( pc)2 + m2 c 4 + 2E h After we cancel the mass terms and subtracting, we get h(E i pc) = h (E f pc)From this can calculate h and rewrite the energy conservation law in the form
  • 5. E i pc E E= h 1 E pc f The energy loss is largest if i = 1; f = 1. Assuming that the final electron momentum is (mc 2 )2not very close to zero, we can write E + pc = 2E and E p c = so that 2 E 2 E 2 E E E = h (mc 2 )2 1 1It follows that = + 16 h with everything expressed in MeV. This leads to E E E =(100/1.64)=61 MeV and the energy loss is 39MeV.8.We have = 0.035 x 10-10 m, to be inserted into h h 6.63 1034 J .s = (1 cos600 ) = = = 1.23 10 12 m me c 2 mec 2 (0.9 10 30 kg)(3 10 8 m / s)Therefore = = (3.50-1.23) x 10-12 m = 2.3 x 10-12 m.The energy of the X-ray photon is therefore 34 8 hc(6.63 10 J .s)(3 10 m / s) h = 5 = = 5.4 10 eV (2.3 10 m)(1.6 10 J / eV ) 12 199. With the nucleus initially at rest, the recoil momentum of the nucleus must be equaland opposite to that of the emitted photon. We therefore have its magnitude given by p = h / c , where h = 6.2 MeV . The recoil energy is p2 h 6.2 MeV E= = h 2 = (6.2 MeV ) = 1.5 103 MeV 2M 2 Mc 2 14 (940 MeV )10. The formula = 2asin / n implies that / sin 2a / 3 . Since = h/p this leads to p 3h / 2asin , which implies that the kinetic energy obeys p2 9h2 K= 2m 8 ma2 sin2 Thus the minimum energy for electrons is 9(6.63 10 34 J .s)2 K= = 3.35eV 8(0.9 1030 kg)(0.32 109 m)2 (1.6 1019 J / eV )
  • 6. For Helium atoms the mass is 4(1.67 1027 kg) / (0.9 10 30 kg) = 7.42 103 larger, sothat 33.5eV K= = 4.5