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An 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 the total energy with each. |
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Answer» Solution :The REACTION is represented by `gamma to (._(-1)E^0)+(._(+)e^0)` , so that `E=m_0 C^2+K.E_"electron"+m_0C^2+K.E_"positron"` 2.36 MeV= `2m_0. C_2+ K.E_"(electron)" + K.E_"(positron)"` = 1.02 MeV+ `K.E_((e)) + K.E_((e))` `THEREFORE` Kinetic energy 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_0C^2`+K.E=0.51 MeV+0.67 MeV= 1.18 MeV |
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