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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 these will carry? |
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Answer» Solution :WORKING on the same lines as an electron - positron pair production ,we notice that the reaction. `GAMMA rarr` proton + antiproton + antiproton , has the energy balance `E=m_(0("proton"))C^2+K.E._(("proton"))+m_(0("antiproton"))C^2K.E_(("antiproton"))` But `m_(0)C^2`= energy equivalent of 1.007276 a.m.u `~~` 938 MEV. `[ :. 1.007276xx931~~938"MeV"]` THUS K.E of each particle `=1/2` [ 1896 MeV `-2xx938` MeV] = 10 MeV |
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