Question

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 ____________

(a) 4.1 MeV

(b) 2.05 MeV

(c) 1.02 MeV

(d) 0.51 MeV

The question was posed to me in an interview.

My question comes from Pair Productions topic in portion Quantum Physics of Engineering Physics I

Answer 1

Correct option is (b) 2.05 MeV

To elaborate: We Know, in pair production momentum and energy is conserved.

By conserving energy, Etotal = Ee + EP

4.1 MeV = Ee + EP

Now, by conserving momentum in y-direction we get

Pe – PP = 0

Pe = PP

Hence, if momentum is same the energy should be same

Therefore, 4.1 MeV = 2Ee

Ee = 2.05 MeV.