InterviewSolution
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InterviewSolution
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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 to`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_0C^2` = energy equivalent of 1.007276 a.m.u. `APPROX` 938 MeV . [`because` 1.007276 x 931 `approx` 938 MeV] THUS K.E. of each particle =`1/2`[1896 MeV-2 x 938 MeV] =10 MeV |
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