ch01 - SOLUTIONS MANUAL CHAPTER 1 1 The energy contained in...

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SOLUTIONS MANUAL CHAPTER 1 1. The energy contained in a volume dV is U ( ν , T ) dV = U ( ν , T ) r 2 d r s in θ d θ d ϕ 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 dA cos θ 4 π r 2 The total energy emitted is . dE ( ν , T ) = dr d θ d ϕ U ( ν , T )sin θ cos θ dA 4 π 0 2 π 0 π /2 0 c t = dA 4 π 2 π c tU ( ν , T ) d θ sin θ cos θ 0 π /2 = 1 4 c tdAU ( ν , T ) By definition of the emissivity, this is equal to E tdA . Hence E ( ν , T ) = c 4 U ( ν , T ) 2. We have w ( λ , T ) = U ( ν , T ) | d ν / d λ | = U ( c λ ) c λ 2 = 8 π hc λ 5 1 e hc / λ kT - 1 This density will be maximal when dw ( λ , T )/ d λ = 0 . What we need is d d λ 1 λ 5 1 e A / λ - 1 = ( - 5 1 λ 6 - 1 λ 5 e A / λ e A / λ - 1 ( - A λ 2 )) 1 e A / λ - 1 = 0 Where A = hc / kT . The above implies that with x = A / λ , we must have 5 - x = 5 e - x A solution of this is x = 4.965 so that
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λ m ax T = hc 4 .965 k = 2 .898 × 10 - 3 m In example 1.1 we were given an estimate of the sun’s surface temperature as 6000 K. From this we get λ m ax sun = 28 .98 × 10 - 4 mK 6 × 10 3 K = 4 .83 × 10 - 7 m = 483 nm 3 . The relationship is h ν = K + W where K is the electron kinetic energy and W is the work function. Here h ν = hc λ = (6 .626 × 10 - 34 J . s )(3 × 10 8 m / s ) 350 × 10 - 9 m = 5 .68 × 10 - 19 J = 3 .55 eV With K = 1.60 eV, we get W = 1.95 eV 4 . We use hc λ 1 - hc λ 2 = K 1 - K 2 since W cancels. From ;this we get h = 1 c λ 1 λ 2 λ 2 - λ 1 ( K 1 - K 2 ) = = (200 × 10 - 9 m )(258 × 10 - 9 m ) (3 × 10 8 m / s )(58 × 10 - 9 m ) × (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 h ν , and the backward- scattered photon energy be h ν ' . Let the energy of the recoiling proton be E. Then its recoil momentum is obtained from E = p 2 c 2 + m 2 c 4 . The energy conservation equation reads h ν + m c 2 = h ν ' + E and the momentum conservation equation reads h ν c =- h ν ' c + p
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that is h ν =- h ν ' + pc We get E + pc - m c 2 = 2 h ν from which it follows that p 2 c 2 + m 2 c 4 = (2 h ν - pc + m c 2 ) 2 so that pc = 4 h 2 ν 2 + 4 h ν m c 2 4 h ν + 2 m c 2 The energy loss for the photon is the kinetic energy of the proton K = E - m c 2 . Now h ν = 100 MeV and mc 2 = 938 MeV, so that pc = 182
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