# Compton Scattering

Electromagnetic radiation interacts with matter in different ways. One way is Compton Scattering where an incident ray ejects an electron by elastic scattering. The scattered photon must have lower energy as the electron obtains kinectic energy from the collision. Convervation of energy and momentum show that the scattered photon has energy:

$$E^{'} = \cfrac{EE_e}{E(1-cos(\theta))+E_e}$$

Since the energy can be expressed as $$E=hf=\frac{hc}{\lambda}$$, the scattered photon wavelength is:

$$\lambda^{'} = \cfrac{hc}{E^{'}} = \cfrac{hc(E(1-cos(\theta))+E_e)}{EE_e}$$

$$\lambda^{'} = \cfrac{hcE(1-cos(\theta))}{EE_e} + \cfrac{hcE_e}{EE_e}$$

$$\lambda^{'} = \lambda_c(1-cos(\theta)) + \lambda$$

Where $$\lambda_c=\frac{hc}{E_e} = 2.426\cdot10^{-10} \; [cm]$$ is the Compton Wavelength.

A $$2 \; [MeV]$$ photon interacts with an atom, ejecting an electron. A photon is observed $$45^{\circ}$$ from the line of sight of the incident ray. What is the energy of the ejected photon?