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 Compton scattering Feynman diagrams s-channel220px u-channel220px Light-matter interaction Low energy phenomena Photoelectric effect Mid-energy phenomena Compton scattering High energy phenomena Pair production

In physics, Compton scattering or the Compton effect is the decrease in energy (increase in wavelength) of an X-ray or gamma ray photon, when it interacts with matter. Inverse Compton scattering also exists, where the photon gains energy (decreasing in wavelength) upon interaction with matter. The amount the wavelength changes by is called the Compton shift. Although nuclear compton scattering exists[1], Compton scattering usually refers to the interaction involving only the electrons of an atom. The Compton effect was observed by Arthur Holly Compton in 1923 and further verified by his graduate student Y. H. Woo in the years following. Arthur Compton earned the 1927 Nobel Prize in Physics for the discovery.

The effect is important because it demonstrates that light cannot be explained purely as a wave phenomenon. Thomson scattering, the classical theory of an electromagnetic wave scattered by charged particles, cannot explain any shift in wavelength. Light must behave as if it consists of particles in order to explain the Compton scattering. Compton's experiment convinced physicists that light can behave as a stream of particles whose energy is proportional to the frequency.

The interaction between electrons and high energy photons(~keV) results in the electron being given part of the energy (making it recoil), and a photon containing the remaining energy being emitted in a different direction from the original, so that the overall momentum of the system is conserved. If the photon still has enough energy left, the process may be repeated. In this scenario, the electron is treated as free or loosely bound. Experimental verification of momentum conservation in individual Compton scattering processes by Bothe and Geiger as well as by Compton and Simon has been important in falsifying the BKS theory.

If the photon is of lower energy, but still has sufficient energy (in general a few eV, right around the energy of visible light), it can eject an electron from its host atom entirely (a process known as the Photoelectric effect), instead of undergoing Compton scattering. Higher energy photons(~MeV) may be able to bombard the nucleus and cause an electron and a positron to be formed, a process called pair production.

## The Compton shift formulaEdit

Template:See also Compton used a combination of three fundamental formulas representing the various aspects of classical and modern physics, combining them to describe the quantum behavior of light.

The final result gives us the Compton scattering equation:

$\lambda' - \lambda = \frac{h}{m_e c}(1-\cos{\theta})$

where

$\lambda\,$ is the wavelength of the photon before scattering,
$\lambda'\,$ is the wavelength of the photon after scattering,
$m_e$ is the mass of the electron,
$\theta\,$ is the angle by which the photon's heading changes,
$h$ is Planck's constant, and
$c$ is the speed of light in vaccum or not.
$\frac{h}{m_e c} = 2.43 \times 10^{-12}\,m$ is known as the Compton wavelength.

### NOW:-Edit

$E_\gamma + E_e = E_{\gamma^\prime} + E_{e^\prime} \quad \quad (1) \,$
$\vec p_\gamma = \vec{p}_{\gamma^\prime} + \vec{p}_{e^\prime} \quad \quad \quad \quad \quad (2) \,$
where
$E_\gamma \,$ and $p_\gamma \,$ are the energy and momentum of the photon and
$E_e \,$ and $p_e \,$ are the energy and momentum of the electron.

#### Solving (Part 1)Edit

Now we fill in for the energy part:

$E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,$
$hf + mc^2 = hf' + \sqrt{(p_{e'}c)^2 + (mc^2)^2}\,$

The square of the second equation gives an equation for pe':

$p_{e'}^2c^2 = (hf + mc^2-hf')^2-m^2c^4 \quad \quad \quad \quad \quad (3) \,$

#### Solving (Part 2)Edit

Rearrange equation (2)

$\vec{p}_{e'} = \vec{p}_\gamma - \vec{p}_{\gamma'} \,$

and square it to see

$p_{e'}^2 = (\vec{p}_\gamma - \vec{p}_{\gamma'}) \cdot (\vec{p}_\gamma - \vec{p}_{\gamma'})$
$p_{e'}^2 = p_{\gamma}^2 + p_{\gamma'}^2 - 2\vec{p_{\gamma}} \cdot \vec{p_{\gamma'}}$
$p_{e'}^2 = p_\gamma^2 + p_{\gamma'}^2 - 2|p_{\gamma}||p_{\gamma'}|\cos(\theta) \,$

Energy and momentum of photons are connected by the relativistic equation $p_{\gamma}=\frac{E_{\gamma}}{c}\,.$

Therefore we have also

$p_{e'}^2c^2 = (h f)^2 + (h f')^2 - 2(hf)(h f')\cos{\theta} \quad \quad \quad (4)$

#### Putting it togetherEdit

Now we have the two equations (3 & 4) for $p_{e'}^2c^2$, which we equate:

$\left(h f\right)^2 + \left(h f'\right)^2 - 2h^2 ff'\cos{\theta} = (hf + mc^2-hf')^2 -m^2c^4 \,$

Next we multiply out the right-hand term $(hf + mc^2-hf')^2$ and cancel square terms on both sides and get:

$-2h^2ff'\cos{\theta} = -2h^2ff'+2h(f-f')mc^2 .\,$

Then divide both sides by '$-2h$' to see

$hff'\cos{\theta} = hff'-(f-f')mc^2 \,$
$(f-f')mc^2 = hff'(1-\cos{\theta}) .\,$

After dividing both sides by $mc^2$ and $ff^\prime$ we get:

$\frac{f-f^\prime}{f f^\prime} = \frac{h}{mc^2}\left(1-\cos \theta \right) . \,$

The left-hand side can be rewritten as simply

 $\frac{1}{f^\prime} - \frac{1}{f} = \frac{h}{mc^2}\left(1-\cos \theta \right) \,$

This is equivalent to the Compton scattering equation, but it is usually written in terms of wavelength rather than frequency. To make that switch use

$f=\frac{c}{\lambda} \,$

so that finally,

 $\lambda'-\lambda = \frac{h}{mc}(1-\cos{\theta}) \,$