Sales Toll Free No: 1-855-666-7446

# Compton Scattering

Top
 Sub Topics The development of quantum mechanics forms as the foundation of the modern atomic model. In 1922, the American physicist and researcher Arthur H. Compton (1892–1962) conducts x-ray scattering experiments that confirmed and advanced Einstein's theory of the dual nature of light. Lets see what exactly it is and how it works!The compton Scattering effect is a collision between a photon and a loosely bound outer orbital electron. After the interaction, the scattered photon undergoes a change in direction and the recoil electron is ejected from the atom. The scattered photon is deflected through an angle ($\theta_c$) proportional to the amount of energy lost. The maximum energy loss occurs when the angle $\theta_c$ is 180o or when the photon is back scattered. Compton scattering example: The annihilation photon (511 kev) after back scatter will have an energy of 170 kev. The energy lost by the scattered photon is divided between the small binding energy of the orbital electron and the kinetic energy of the recoil electron. The relative probability of the compton scattering increases slightly as the energy of the incident photon increases and as the energy of the incident photon increases and as the effective atomic number of the interacting medium decreases.

## Definition

When a photon of energy h $nu$ collides the electron assuming it at rest the electron gains the energy that results in loss of photon energy. The electron at rest becomes a recoil electron due to the energy it gains and the scattered photon will be having lower energy and hence a lower frequency compared to that of incident photons. This is what we call Compton Scattering.
Compton Scattering

## Equation

The Compton scattering equation is given by
$\lambda_1$ - $\lambda_2$ = $\frac{h}{m_o c}$ (1 - cos $\theta$)where
$\lambda_1$ is the wavelength of incident rays after scattering,
$\lambda_2$ is initial wavelength of the incident ray,
h is the planck's constant,
mo is electron mass,
c is speed of light,
$\theta$ is the scattering angle of incident ray
Here $\lambda_1$ - $\lambda_2$ is known as Compton shift.

## Derivation

If $\lambda_1$ and $\lambda_2$ are the wavelength of the incident and scattered x rays. The momenta corresponding is given by

p1 = $\frac{E_1}{c}$ = $\frac{h}{\lambda_1}$
and
p2 = $\frac{E_2}{c}$ = $\frac{h}{\lambda_2}$

The conservation of momentum gives
p1 = pe + p2
or
pe2 = p12 + p22 - 2p1.p2
= p12 + p22 - 2p1p2cos $\theta$

we have $\lambda$ = $frac{h}{p}$.Hence the compton equation is given by
$\lambda_2$ - $\lambda_1$ = $\frac{h}{mc}$ (1 - cos $\theta$).

## Experiment

The experiment consists of source covered by a sheild (it carries photon), target material (electron), detector (to detect scattered photon).A x-ray beam is made to fall on the target material and the scattered wavelength of x-rays depending on the scattering angle are determined by its angle $\theta$.

The intensity distributed by the scattered x-rays are plotted versus wavelength that depends on the angle scattered. For each scattering angle graph we could see two peaks one has initial wavelength $\lambda$ and other has greater wavelength $\lambda^{bar}$.

## Cross Section

$\frac{d \sigma}{d \Omega}$ = $\frac{\alpha^{2}h^{2}}{m_{o}^{2} c^{2}}$ ( P (E, $\theta$)2 ) [ P (E, $\theta$) +P (E, $\theta$)-1 – 1 + cos $\theta$]
P (E, $\theta$) is energy of photon before and after the collision ratio
$\alpha$ is a constant equal to $\frac{1}{137.04}$
$\theta$ is scattering angle
$\frac{d \sigma}{d \Omega}$ is Compton Scattering Cross Section.