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?

2.05 MeV

4.1 MeV

1.02 MeV

None of these

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