A photon of energy 4.1 MeV is incident on a lead nucleus, causing the creation of electron-positron pair. They travel perpendicular to the initial direction of the photon. The energy of the electron is?
Answer: 2.05 MeV
Explanation
Given:
Energy of photon (E) = 4.1 MeV
The total energy available for the electron and positron is:
E - 2 × 0.511 MeV = 4.1 MeV - 1.022 MeV = 3.078 MeV
Since the electron and positron share the energy equally, the kinetic energy of the electron is:
3.078 MeV / 2 = 1.539 MeV
Adding the rest mass energy of the electron (0.511 MeV), we get:
Energy of electron = 1.539 MeV + 0.511 MeV = 2.05 MeV
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