# The Compton Effect

Wednesday, February 2, 2022

## What is the Compton Effect?

When a incident photon interacts with a loosely bound free electron, both the photon and electron scatter, in what is called the Compton effect. The photon has some initial energy, $E$, and momentum, $p$, which is converted into energy and momentum of the scattered photon, $E'$ and $p'$, and energy and momentum for the electron, $E_e$ and $p_e$ (we assume the electron to be at rest).

$E=hf=\frac{hc}{\lambda} \newline p=\frac{E}{c}$

The electron has rest energy $m_ec^2$, and is scattered at an angle $\phi$ below the line of incidence. The photon scatters with energy $E'=hc/\lambda'$ and momentum $p'=E'/c$ at an angle of $\theta$ above the line of incidence.

$E_{initial}=E_{final} : E+m_ec^2=E'+E_e \newline p_{x, initial}=p_{x,final} : p = p_ecos{\phi}+p'cos{\theta} \newline p_{y, initial}=p_{y,final} : 0 = p_esin{\phi}-p'sin{\theta}$

This system of equations is not enough alone to be solved uniquely, so the scattered photon's energy and direction are measured to eliminate $E_e$ and $\phi$. After some algebra and rearranging, the relationship between the energy is as follows:

$\frac{1}{E'} - \frac{1}{E}=\frac{1}{m_ec^2}\left(1-\cos{\theta}\right)$

Or, equivalently:

$\lambda'-\lambda=\frac{h}{m_ec}\left(1-\cos{\theta}\right)$

$h/m_ec$ is known as the *Compton wavelength of the electron*, and is about $0.002426$nm. This is simply the change in wavelength the electron causes, and is not an actual wavelength.

Since the scattered photon always has less energy than the incident photon (the difference being the kinetic energy of the electron), the following equation is true:

$K_e=E-E'$

## Compton Effect Experiments

Compton used an x-ray source radiating on a target atom with loosely attached valence electrons, then measured the wavelength of the scattered photons at a variety of angles around the target.