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

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

Sol: Working on the same lines as an electron-positron pair production, we notice that the reaction.

γ → Proton + antiproton, has the energy balance

E = m0(proton) c2 + K.E(proton) + m0(antiproton) c2 + K.E(antiproton)

But m0 c2 = energy equivalent of 1.007276 a.m.u ≈ 938 MeV

[∵ 1.007276 x 931 ≈ 938 MeV]

Thus K.E of each particle = (1/2 )[1896 MeV – 2 × 938 MeV ] = 10 MeV.