48. We interpret the given 10 GeV = 10000 MeV to be the kinetic energy of the proton. Using Table 38-3and Eq. 38-49, we Fndγ=Kmpc2+1=10000 MeV938 MeV+1=11.66,and (from Eq. 38-8)β=r1−1γ2=0.9963.Therefore, using the equation introduced in problem 46, we obtain
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