Compton scattering

Compton observed that the wavelength of X-rays scattered by an electron depends on their scattering angle $\theta$ .

From conservation of momentum and energy of the electron-photon system we find

\begin{align*} E &= \frac{hc}{\lambda_0} + m_e c^2 = \frac{hc}{\lambda_f} + \sqrt{p^2 c^2 + m_e^2c^4} \\ p_x &= \frac{h}{\lambda_0} = \frac{h}{\lambda_f} \cos \theta + p \cos \phi \\ p_y &= 0 = \frac{h}{\lambda_f} \sin \theta - p \sin \phi, \end{align*}

where $\phi$ is the electron scattering angle and $p$ is the momentum of the electron after the collision.

We isolate $p$ to find

p^2 = \left(\frac{h}{\lambda_0} - \frac{h}{\lambda_f} \cos \theta \right)^2 + \left(\frac{h}{\lambda_f} \sin \theta \right)^2.

Using the energy equation we solve for

p^2 = h^2 \left( \frac{1}{\lambda_0^2} - \frac{1}{\lambda_f^2} \right)^2 + 2h\left(\frac{1}{\lambda_0} - \frac{1}{\lambda_f} \right) m_e c.

After some algebra we find

-\frac{2h^2}{\lambda_0 \lambda_f} + 2h \left( \frac{1}{\lambda_0} + \frac{1}{\lambda_f} \right) m_e c= -\frac{2h^2}{\lambda_0 \lambda_f} \cos \theta.

Finally, we find the Compton shift

\Delta\lambda = \lambda_f - \lambda_0 = \underbrace{\frac{h}{m_e c}}_{\lambda_c} (1-\cos \theta).

Where $\lambda_c$ is the Compton wavelength.