﻿ Compton Scattering - College Physics

# Compton Scattering

## Introduction

In 1922, Arthur Compton investigated the scattering of high-energy X-rays on graphite. This element has a very low work function, which means that the electrons are bound loosely in metal and therefore behave similar to free electrons. Arthur Compton found that the photons of X-rays have a lower frequency and thus a higher wavelength after scattering at the free electrons. This is because they have given energy and momentum to the electrons. The change in direction $$\beta$$ of the photon determines how much its wavelength increases and its energy and momentum decreases.

Compton's measurements showed that the wavelength of the scattered radiation depends on the scattering angle and the process behaves like a elastic collision of particles, the photon and the electron.

The change in wavelength can be derived using the energy and momentum conservation law (see below).

$$\Delta \lambda = \lambda ' - \lambda = \dfrac{h}{m_e \cdot c} \cdot (1 - \cos \beta)$$

The constant $$\dfrac{h}{m_e \cdot c}$$ is called compton wavelength $$\lambda_c$$ and has the value:

$$\lambda_c = 2,426 \cdot 10^{-12} m$$

Substituting it in the above formular leads to:

$$\Delta \lambda = \lambda_c \cdot (1 - \cos \beta)$$

## Derivation

Below the Compton formula for the change of direction $$\ beta$$ is derived. Here, the electron is assumed to be a free electron at rest.

Moment of the photon $$p = \dfrac{E}{c} \qquad \mathrm{(1.1)}$$ $$p^{\prime} = \dfrac{E^{\prime}}{c} \qquad \mathrm{(1.2)}$$ Energy-momentum relationship $$\left( E^{\prime}_e \right) ^2 = \left( E_e \right) ^2 + c^2 \cdot \left( p_e^{\prime} \right) ^2$$

Umgestellt nach $$\left( p_e^{\prime} \right) ^2$$ :

$$\left( p_e^{\prime} \right) ^2 = \dfrac{ \left( E^{\prime}_e \right) ^2 - \left( E_e \right) ^2}{c^2} \qquad \mathrm{(2)}$$ Conservation of momentum $$\vec p = \vec p^{\,\prime} + \vec p^{\,\prime}_e$$
(vectorial)
$$\left( p_e^{\prime} \right) ^2 = p^2 + p^{\prime \, 2} - 2 \cdot p \cdot p' \cdot \cos \beta \qquad \mathrm{(3)}$$
(By cosine rule)

We substitute into the previous formular $$\mathrm{(3)}$$ the formulas $$\mathrm{(1.1)}$$, $$\mathrm{(1.2)}$$ and $$\mathrm{(2)}$$:

$$\dfrac{ \left( E^{\prime}_e \right) ^2 - \left( E_e \right) ^2}{c^2 \cdot } = \left( \dfrac{E}{c} \right) ^2 + \left( \dfrac{E^{\prime}}{c} \right) ^2 - 2 \cdot \dfrac{E}{c} \cdot \dfrac{E^{\prime}}{c} \cdot \cos \beta \qquad \mathrm{(4)}$$

Solving a bit:

$$\dfrac{ \left( E^{\prime}_e \right) ^2 - \left( E_e \right) ^2}{c^2} = \dfrac{E^2}{c^2} + \dfrac{E^{\prime \, 2}}{c^2} - 2 \cdot \dfrac{E \cdot E^{\prime}}{c^2} \cdot \cos \beta \qquad \mathrm{(5)}$$

Mal $$c^2$$:

$$\left( E^{\prime}_e \right) ^2 - \left( E_e \right) ^2 = E^2 + E^{\prime \, 2} - 2 \cdot E \cdot E^{\prime} \cdot \cos \beta \qquad \mathrm{(5)}$$
Conservation of energy $$E + E_e = E^{\,\prime}+ E^{\,\prime}_e$$

Solved for $$E^{\,\prime}_e$$ :

$$E^{\,\prime}_e = E + E_e - E^{\,\prime}$$

Substituting:

$$\left( E + E_e - E^{\,\prime} \right) ^2 - \left( E_e \right) ^2 = E^2 + E^{\prime \, 2} - 2 \cdot E \cdot E^{\prime} \cdot \cos \beta$$ $$\cancel{E^2} + \cancel{ \left( E_e \right) ^2 } + \cancel{ E^{\prime \, 2} } + 2 \cdot E \cdot E_e - 2 \cdot E \cdot E^{\,\prime} - 2 \cdot E^{\prime} \cdot E_e - \cancel{ \left( E_e \right) ^2 } = \cancel{E^2} + \cancel{ E^{\prime \, 2} } - 2 \cdot E \cdot E^{\prime} \cdot \cos \beta$$ $$2 \cdot E \cdot E_e - 2 \cdot E \cdot E^{\,\prime} - 2 \cdot E^{\prime} \cdot E_e = - 2 \cdot E \cdot E^{\prime} \cdot \cos \beta \qquad | \cdot \dfrac{1}{2 \cdot E \cdot E^{\prime} \cdot E_e}$$ \begin{aligned} \dfrac{1}{E^{\prime}} - \dfrac{1}{E_e} - \dfrac{1}{E} & = - \dfrac{\cos \beta}{E_e} \enspace \,\, \qquad \qquad \qquad | + \dfrac{1}{E_e} \\ \dfrac{1}{E^{\prime}} - \dfrac{1}{E} & = \dfrac{1}{E_e} - \dfrac{\cos \beta}{E_e} \qquad \qquad | \dfrac{1}{E_e} \mathrm{ausklammern} \\ \dfrac{1}{E^{\prime}} - \dfrac{1}{E} & = \dfrac{1}{E_e} \cdot (1 - \cos \beta) \\ \end{aligned} Energies of the photon $$E = \dfrac{h \cdot c}{\lambda}$$ $$E^{\prime} = \dfrac{h \cdot c}{\lambda^{\prime}}$$

Substituting:

\begin{aligned} \dfrac{\lambda^{\prime}}{h \cdot c} - \dfrac{\lambda}{h \cdot c} & = \dfrac{1}{E_e} \cdot (1 - \cos \beta) \qquad | \cdot h \cdot c\\ \lambda^{\prime} - \lambda & = \dfrac{h \cdot c}{E_e} \cdot (1 - \cos \beta) \\ \end{aligned}

With $$E_e = m_e \cdot c^2$$ :

$$\lambda^{\prime} - \lambda = \dfrac{h}{m_e \cdot c} \cdot (1 - \cos \beta)$$

With $$\lambda_c = \dfrac{h}{m_e \cdot c}$$ :

$$\Delta \lambda = \lambda_c \cdot (1 - \cos \beta)$$

### Sources

College-Physics © 2018, Partner: Abi-Mathe, Abi-Chemie, Deutsche website: Abi-Physik Lernportal