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A gamma ray photon of energy 1896 MeV 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 each particle will carry ? |
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Answer» Solution :Working on the same liens as an ELECTRON - positron PAIR PRODUCTION , wenotice that the reaction. `gamma rarr`PROTON + antiproton, has the energy balance `E=m_(0)` (proton) `c^2+K.E`(antiproton) But `m_(0).c^2`= energy equivalent of 1.007276 a.m.u `~~938 MeV [ :. 1.00726 xx931 ~~938]` Thus K.E of each particle = `1/2[1896 MeV -2xx938 ] MeV = 10 MeV` |
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