Question

Show that a Fermi energy of 25 MeV lowers the threshold incident kinetic energy for antiproton production by proton incident on nucleus to 4.3 GeV.

Answer 1

Use the invariance of E² − P² = E^∗2 − P^∗2 = E^∗2 − 0 = E^∗2 

E^∗2 = (4m)² = (T + m + m + 0.025)² − (P₁ − 0.218)² 

Putting m = 0.938, P₁ = (T² + 2Tm)^1/2 and 

solving for T , we find that T (threshold) = 4.3 GeV