Question

A beam of photons of energy 184 keV is incident on a target. Scattered photons are observed at an angle of 60° relative to the direction of the incident beam. Assume that the photons scatter from free electrons in the target.

(a) What is the energy of the scattered photons?
(b) Find the kinetic energy that is acquired by the scattered electrons.
(c) Find the magnitude of the electron’s momentum and the component of the electron’s momentum perpendicular to the direction of the incident beam.

Answer 1

The scattering of photons from charged particles is called Compton scattering after When the incoming photon gives part of its energy to the electron, then the scattered photon has lower energy and according to the Planck relationship has lower frequency and longer wavelength. The wavelength change in such scattering depends only upon the angle of scattering for a given target particle. The constant in the Compton formula above can be written

$\frac{h}{m_ec} = \frac{hc}{m_ec^2} = \frac{1240eV}{0.511MeV} = 0.00243nm$

Given is:-

Energy of the beam of photns is = 184keV

Angle at which scattered photons are observed is  $\theta = 60^\circ$

Thus the energy of the scattered photons is

$E_f = \frac{hc}{\lambda _f}$ (we assume unspecified parameters to the case of molybdenum K-alpha xrays )

thus by putting values in above equation ($\lambda _f = 1.213 \times 10^{-12}m$), we get the energy of scattered photons

$E_f = 1021.99799 keV$ or   $E_f = 1.02199 MeV$

This implies that the energy transferred to the electron is

$KE_{electron} = 1.14 \times 10^{21}eV$ or   $KE_{electron} =184 J$