1134_Physics ProblemsTechnical Physics

1134_Physics ProblemsTechnical Physics - Chapter 40 P40.29...

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Chapter 40 475 P40.29 −= λλ θ h mc e 1c o s af ′′ − =− ′′ − + πθ h h h h h e ee o s cos cos Now cos cos , so ′′ − = = 20 0 0 4 8 6 h e . n m . FIG. P40.29 P40.30 Maximum energy loss appears as maximum increase in wavelength, which occurs for scattering angle 180°. Then λ ° F H G I K J = 11 8 0 2 cos h mc h mc where m is the mass of the target particle. The fractional energy loss is EE E hc hc hc hmc 0 0 0 0 0 00 2 2 = −′ = ′ − = + = + . Further, 0 0 = hc E , so E hc E h mc E mc E 0 0 2 0 2 2 2 2 = + = + . (a) For scattering from a free electron, mc 2 0511 = . MeV, so E 0 0 20511 0667 = + = . .. . MeV MeV MeV a f . (b) For scattering from a free proton, mc 2 938 = MeV, and E 0 0 938 2 0 511 000109 = + = . . . MeV MeV MeV a f . Section 40.4 Photons and Electromagnetic Waves
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