Question

A gamma ray photon of energy `1896 MeV` annihilates to produce a photon-antiproton pair. If the rest mass of each of the particles involved be `1.007276 a.m.u` aapproximately, find how much `K.E` these will carry?

Answer 1

Working on the same lines as an electron-position pair production, we notice that the reaction. `gamma rarr` Proton + antiproton, has the energy balance
`E=m_(0("Proton"))C^(2)+K.E_(("Proton"))+`
`m_(0("antiproton"))C^(2)+K.E_(("antiproton"))`
But `m_(0)C^(2)=` energy equivalent of `1.007276 a.m.u`
`~~ 938 MeV.[ because 1.007276xx931~~938 MeV]`
Thus `K.E`. of each particle
`=(1)/(2)[1896MeV-2xx938MeV]=10 MeV`.