Chapter 40475P40.29′−=−λλθhmce1cosaf′′ −′=−−′′−− +−πθhhhhheeeoscoscosNow coscos−, so ′′ −==2000486he. nm.FIG. P40.29P40.30Maximum energy loss appears as maximum increase in wavelength, which occurs for scatteringangle 180°. Then ∆λ°FHGIKJ=11802coshmchmcwhere mis the mass of the target particle. Thefractional energy loss isEEEhchchchmc000000022−′=−′=′ −′=+=+∆∆.Further, 00=hcE, so Ehc Eh mcEmcE00202222−′=+=+.(a)For scattering from a free electron, mc20511=.MeV, soE00205110667=+=....MeVMeVMeVaf.(b)For scattering from a free proton, mc2938=MeV, andE009382 0 511000109=+=...MeVMeVMeVaf.Section 40.4Photons and Electromagnetic Waves
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .