1129_Physics ProblemsTechnical Physics

1129_Physics ProblemsTechnical Physics - 470 Introduction...

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470 Introduction to Quantum Physics P40.22 (a) λθ =− h mc e 1c o s a f : λ = × ×× −° = × 6626 10 9 11 10 3 00 10 13 7 04 8 8 1 0 34 31 8 13 . .. cos . . ej e j a f m (b) E hc 0 0 = : 300 10 1 60 10 300 10 31 9 34 8 0 = eV J eV ms e j e j . 0 12 414 10 . m and ′= + = × λλ λ 0 12 463 10 m = ×⋅× × = E hc 430 10 268 34 8 12 14 . . Js m J k e V e j (c) KEE e = = 0 300 268 5 31 5 keV keV keV P40.23 With KE e =′ , e 0 gives − ′ EEE 0 E E 0 2 and hc E = = λλ hc E hc E 00 0 2 22 =+ θ 0 1 C cos af 21 λλλ C cos 1 000160 000243 0 −= = cos . . C 70 0 . P40.24 This is Compton scattering through 180°: E hc h e 0 0 34 8 91 9 12 12 0110 10 160 10 11 3 1 2 43 10 1 180 4 86 10 == = = × = × −− . cos . cos . m J e V keV m e j e j a f a f FIG. P40.24 + = 0 0115 . nm so ′ = = E hc 10 8 . k eV . By conservation of momentum for the photon-electron system, hh p e 0 ±± ± ii i = −+ and ph e F H G I K J 11 0 p c c e × × F H G G I K J J × + × F H G I K J = 1 1 0115 10 22 1 34 8 19 9 9 . . . . JeV m m keV . By conservation of system energy, 11 3 10 8
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