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En electron - positron pair is produced when a gamma - ray photon of energy 2.36MeV passes close to a heavy nucleus. Find the kinetic energy carried by each particle produced , as well as its total energy. |
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Answer» Solution :The REACTION is represented by `GAMMA rarr (""_(-1)e^(0))+(""_(+)e^(0)),` so that `E=m_(0).C^(2)+K.E.` (electron) `+m_(0)c^2+` K.E. (positron) `2.36 MEV = 2m_(0) .c^2+K.E` (electron) + K.E (positron) ` = 1.02MeV + K.E (e^(-))+K.E(e^(+))` `:. ` Kinetic energy of electron or position `KE (e^(-))=K.E(e^(+))=1/2(2.36 - 1.02)MeV` , K.E . carried each = 0.67 MeV (motional energy) Total energy shared by each particle is obviously `m_(0).c^(2)+K.E = 0.51 MeV + 0.67 MeV` = 1.18 MeV |
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