p39_060 - 60. (a) We use the result of problem 3: Ephoton =...

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60. (a) We use the result of problem 3: E photon = hc λ = 1240 nm · eV 1 . 00 nm =1 . 24 keV and for the electron K = p 2 2 m e = ( h/λ ) 2 2 m e = ( hc/λ ) 2 2 m e c 2 = 1 2(0 . 511 MeV) µ 1240 eV · nm 1 . 00 nm 2 . 50 eV . (b) In this case, we Fnd E photon = 1240 nm · eV 1 . 00 × 10 6 nm . 24 × 10 9 eV = 1 . 24 GeV , and for the electron (recognizing that 1240 eV · nm = 1240 MeV · fm) K = p p 2 c 2 +( m e c 2 ) 2 m e c 2 = p ( hc/λ ) 2 m e c 2 ) 2 m e c 2 = s µ 1240 MeV · fm 1 . 00 fm 2 +(0 . 511 MeV) 2 0 . 511 MeV . 24 × 10 3 MeV = 1 .
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