p38_049 - 49. We interpret the given 2.50 MeV = 2500 keV to...

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49. We interpret the given 2 . 50 MeV = 2500 keV to be the kinetic energy of the electron. Using Table 38-3 and Eq. 38-49, we Fnd γ = K m e c 2 +1= 2500 keV 511 keV +1=5 . 892 , and (from Eq. 38-8) β = r 1 1 γ 2 =0 . 9855 . Therefore, using the equation introduced in problem 46 (with “
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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