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Unformatted text preview: SOLUTIONS MANUAL CHAPTER 1 1. The energy contained in a volume dV is U ( , T ) dV = U ( , T ) r 2 dr sin d d when the geometry is that shown in the figure. The energy from this source that emerges through a hole of area dA is dE ( , T ) = U ( , T ) dV dA cos 4 r 2 The total energy emitted is . dE ( , T ) = dr d d U ( , T )sin cos dA 4 2 /2 c t = dA 4 2 c tU ( , T ) d sin cos / 2 = 1 4 c tdAU ( , T ) By definition of the emissivity, this is equal to E tdA . Hence E ( , T ) = c 4 U ( , T ) 2. We have w ( , T ) = U ( , T )  d / d  = U ( c ) c 2 = 8 hc 5 1 e hc / kT 1 This density will be maximal when dw ( , T ) / d = . What we need is d d 1 5 1 e A /  1 = ( 5 1 6 1 5 e A / e A /  1 ( A 2 )) 1 e A /  1 = Where A = hc / kT . The above implies that with x = A / , we must have 5 x = 5 e x A solution of this is x = 4.965 so that max T = hc 4.965 k = 2.898 10 3 m In example 1.1 we were given an estimate of the suns surface temperature as 6000 K. From this we get max sun = 28.98 10 4 mK 6 10 3 K = 4.83 10 7 m = 483 nm 3 . The relationship is h = K + W where K is the electron kinetic energy and W is the work function. Here h = hc = (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 1 hc 2 = K 1 K 2 since W cancels. From ;this we get h = 1 c 1 2 2 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 headon collision, with the photon 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 its recoil momentum is obtained from E = p 2 c 2 + m 2 c 4 . The energy conservation equation reads h + mc 2 = h ' + E and the momentum conservation equation reads h c =  h ' c + p that is h =  h ' + pc We get E + pc...
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 Spring '05
 mokhtari
 Energy

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