Exam (elaborations) TEST BANK FOR Quantum Physics 3rd Edition By Stephen Gasiorowicz (Solution manual)

1. The energy contained in a volume dV is U(ν,T)dV = U(ν ,T)r2drsinθ dθdϕ when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is dE(ν ,T) = U(ν ,T)dV dAcosθ 4πr2 The total energy emitted is . dE(ν ,T) = dr dθ dϕU(ν,T)sinθ cosθ dA 0 4π 2π ∫ 0 π /2 ∫ 0 cΔt ∫ = dA 4π 2π cΔtU(ν ,T) dθ sinθ cosθ 0 π / 2 ∫ = 1 4 cΔtdAU (ν ,T) By definition of the emissivity, this is equal to EΔtdA . Hence E(ν,T) = c 4 U(ν,T) 2. We have w(λ ,T) = U(ν ,T) | dν / dλ |= U( c λ ) c λ2 = 8πhc λ 5 1 ehc/λkT −1 This density will be maximal when dw(λ ,T) / dλ = 0. What we need is d dλ 1 λ 5 1 eA /λ −1 ⎛ ⎝ ⎞ ⎠ = (−5 1 λ6 − 1 λ5 eA /λ eA/λ −1 (− A λ 2 )) 1 eA /λ −1 = 0 Where A = hc / kT . The above implies that with x = A /λ , we must have 5 − x = 5e−x A solution of this is x = 4.965 so that λ maxT = hc 4.965k = 2.898 ×10−3m In example 1.1 we were given an estimate of the sun’s surface temperature as 6000 K. From this we get λ max sun = 28.98 ×10−4mK 6 ×103K = 4.83 ×10−7m = 483nm 3. The relationship is hν = K +W where K is the electron kinetic energy and W is the work function. Here hν = hc λ = (6.626 ×10−34 J .s)(3×108m / s) 350 ×10−9m = 5.68 ×10−19J = 3.55eV With K = 1.60 eV, we get W = 1.95 eV 4. We use hc λ1 − hc λ2 = K1 − K2 since W cancels. From ;this we get h = 1 c λ1 λ2 λ2 − λ 1 (K1 − K2) = = (200 ×10−9m)(258 ×10−9m) (3×108m / s)(58 ×10−9m) × (2.3− 0.9)eV × (1.60 ×10−19)J / eV = 6.64 ×10−34 J .s 5. The maximum energy loss for the photon occurs in a head-on collision, with the photon scattered backwards. Let the incident photon energy be hν , and the backwardscattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its recoil momentum is obtained from E = p2c 2 + m2c 4 . The energy conservation equation reads hν + mc2 = hν '+E and the momentum conservation equation reads hν c = − hν ' c + p that is hν = −hν '+ pc We get E + pc − mc2 = 2hν from which it follows that p2c2 + m2c4 = (2hν − pc + mc2)2 so that pc = 4h2ν 2 + 4hνmc2 4hν + 2mc2 The energy loss for the photon is the kinetic energy of the proton K = E − mc2 . Now hν = 100 MeV and mc2 = 938 MeV, so that pc = 182MeV and E − mc2 = K = 17.6MeV 6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing electron momentum. Energy conservation reads hν + mc2 = hν '+ p2c2 + m2c4 We write the equation for momentum conservation, assuming that the initial photon moves in the x –direction and the final photon in the y-direction. When multiplied by c it read i(hν ) = j(hν ') + (ipxc + jpyc) Hence pxc = hν ; pyc = −hν '. We use this to rewrite the energy conservation equation as follows: (hν + mc 2 − hν ')2 = m2c 4 + c 2(px 2 + py 2) = m2c4 + (hν )2 + (hν ')2 From this we get hν'= hν mc2 hν + mc2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ We may use this to calculate the kinetic energy of the electron K = hν − hν '= hν 1− mc2 hν + mc2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = hν hν hν + mc2 = (100keV )2 100keV + 510keV = 16.4keV Also pc = i(100keV ) + j(−83.6keV) which gives the direction of the recoiling electron. 7. The photon energy is hν = hc λ = (6.63×10−34 J.s)(3 ×108m / s) 3×106 ×10−9m = 6.63×10−17J = 6.63×10−17 J 1.60 ×10−19 J / eV = 4.14 ×10−4 MeV The momentum conservation for collinear motion (the collision is head on for maximum energy loss), when squared, reads hν c ⎛ ⎝ ⎞ ⎠ 2 + p2 + 2 hν c ⎛ ⎝ ⎞ ⎠ p η i = hν ' c ⎛ ⎝ ⎞ ⎠ 2 + p'2 +2 hν ' c ⎛ ⎝ ⎞ ⎠ p' η f Here η i = ±1, with the upper sign corresponding to the photon and the electron moving in the same/opposite direction, and similarly for η f . When this is multiplied by c2 we get (hν )2 + (pc)2 + 2(hν ) pc η i = (hν ')2 + ( p'c)2 + 2(hν ') p'c η f The square of the energy conservation equation, with E expressed in terms of momentum and mass reads (hν )2 + (pc)2 + m2c 4 + 2Ehν = (hν ')2 + ( p'c)2 + m2c4 + 2E' hν ' After we cancel the mass terms and subtracting, we get hν(E − η i pc) = hν '(E'− η f p'c) From this can calculate hν' and rewrite the energy conservation law in the form