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An electron-positron pair is produced when a gamma -ray photon of energy 2.36MeV passes close to a heavy nuclens . Find the kinetic energy carried by each particle produced , as well as the total energy with each. |
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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")` `2.36MeV = 2m_(0) .C_(2) +K.E_(("electron"))+K.E_("(positror)")` `= 1.02 MeV +K.E_((e^(-)))+K.E_((e^(+)))` `:.` Kinetic ENEGY of `(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=0.51MeV +0.67 MeV = 1.18MeV.` |
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