Question

6. Photons from a 3 MeV gamma source are normally incident on plane aluminum foil of 1 mm thick.

a) What is the maximum electron kinetic energy expected from scattering interaction of photons with the foil?

b) What is the minimum photon energy expected

Answer 1

a) This is the case of Compton scattering. The kinetic energy of the scattered electron depends on the energy of the incident and the scattered photon.

If E₀ = Energy of incident Photon = 3MeV

E = Energy of scattered Photon

Kinetic energy of scattered electron is given by T_e =  E₀ - E

Thus kinectic energy of electron is maximum when the energy of scattered photon is minimum.

The kinetic energy of electron can range upto to few MeV depending on the scattering angle and hence the energy of scattered photon.

b) According to the formula for Compton scattering,

The photon energy is minimum when it is backscattered. i.e, it emerges at 180 degrees with respect to its original direction.

Using rest mass of electron =0.51 MeV,

E_min = 0.235 MeV

This is the minimum photon energy expected.

Using this minimum scattered photon energy, maximum kinetic energy of scattered electrons = (3-0.235) MeV = 2.765 MeV