Question

An electron-positron pair is produced when a `gamma`-ray photon of energy `2.36 MeV` passes close to a heavy nucleus. Find the kinetic energy carried by each particle produced, as well as the total energy with each.

Answer 1

The reaction is represented by
`gamma rarr(._(-1)e^(0))+(._(+1)e^(0))`, so that
`E=m_(0)C^(2)+K.E_("electron")+m_(0)C^(2)+K.E._("positron")`
`=1.02 MeV+K.E_((e^(-)))+K.E_((e^(+)))`
`K.E` of `(e^(-))=K.E_((e^(-)))+(1)/(2)(2.36-1.02)MeV`
`(K.E.` carried each )` =0.67 MeV`(motional energy)
Total energy shared by each particle is obviously
`m_(0)C^(2)+K.E.=0.51 MeV+0.67MeV=1.18MeV`.