Answer to Question #113965 in Quantum Mechanics for Nabeel

Question

A photon of energy E is scattered by a particle of rest energy Eo. Find the maximum kinetic energy of the recoiling particle in terms of E and Eo.

Answer 1

Energy of the photon before scattering:

$E=\frac{hc}{\lambda}.$

Energy of the photon after scattering:

$E_s=\frac{hc}{\lambda_s}.$

According to the law of conservation of energy, the energy of the photon before scattering and the rest energy of the particle are spent to give kinetic energy to the particle and to change the photon's energy to E₂:

$E+E_0=K+E_s,\\ K=E+E_o-E_s=\\ \space\\=\frac{h}{c}\bigg(\frac{1}{\lambda}-\frac{1}{\lambda_s}\bigg)+E_0.$

We know that in Compton scattering, the wavelength can be expressed as

$\lambda_s=\lambda+\frac{h}{m_ec}(1-\text{cos}\theta).$

The scattered wavelength is maximum when cosine theta is -1:

$\lambda_s=\lambda+\frac{2h}{m_ec}.$

Substitute this to the equation for kinetic energy:

$K=E-\frac{hc}{\frac{hc}{E}+\frac{2h}{m_ec}}+E_0=\\ \space\\=E_0+E\bigg(1-\frac{m_ec^2}{hc^2m_e+2E}\bigg).$