49. We interpret the given 2.50 MeV = 2500 keV to be the kinetic energy of the electron. Using Table 38-3and Eq. 38-49, we Fndγ=Kmec2+1=2500 keV511 keV+1=5.892,and (from Eq. 38-8)β=r1−1γ2=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.