Q: An electron – positron pair is produced when a γ-ray photon of energy 2.36 MeV passed close to a
heavy nucleus. Find the kinetic energy carried by each particle produced, as well as the total energy
with each.
Sol: the reaction is represented by
γ → -1e0 + -1e0 , so that
E = m0 c2 + K.E(electron) + m0 c2 + K.E(positron)
2.36 MeV = 2 m0 c2 + K.E(electron) + K.E(positron)
2.36 MeV = 1.02 MeV + K.E(electron) + K.E(positron)
K.E(electron) = K.E(positron) = (2.36-1.02)/2
(K.E. carried each) = 0.67 MeV (motional energy)
Total energy shared by each particle is obviously ,
m0 c2 + K.E = 0.51 MeV + 0.67 MeV = 1.81 MeV.