Compton Scattering

Figure 3.14: Compton Scattering.

Compton scattering occurs when X-rays scatter off electrons in ordinary matter. The result is an increase in the wavelength of the scattered X-rays. This increase is inexplicable within the context of classical physics, which predicts that radiation that scatters off a stationary target should suffer no change in wavelength. In fact, as we shall explain, this effect can be explained in terms of the scattering of individual X-ray photons by individual electrons.

Consider the situation, illustrated in Figure 3.14, in which an X-ray photon of momentum ${\bf p}_\gamma$ collides with a stationary electron of rest mass $m_e$. After the collision, the momentum of the photon is ${\bf p}_\gamma'$, and the recoil momentum of the electron is ${\bf p}_e'$. Conservation of momentum in the collision requires that

$${\bf p}_\gamma = {\bf p}_\gamma' +{\bf p}_e'.$$ (3.224)

However, we know that

$${\bf p}_\gamma =\hbar\,{\bf k},$$ (3.225)

$${\bf p}_\gamma' =\hbar\,{\bf k}',$$ (3.226)

$${\bf p}_e' =\gamma\,m_e\,v,$$ (3.227)

where ${\bf k}$ and ${\bf k}'$ are the photon's initial and final wavevector, respectively, $v$ is the electron's recoil speed, and $\gamma=(1-v^2/c^2)^{-1/2}$. [See Equations (3.162) and (3.200).] Thus, we obtain

$${\bf k}- {\bf k}' = \frac{m_e\,c}{\hbar}\,\gamma\,\frac{v}{c}.$$ (3.228)

The previous equation yields

$$\vert{\bf k}-{\bf k}'\vert^2 = \left(\frac{m_e\,c}{\hbar}\right)^... ...\,\frac{v^2}{c^2} = \left(\frac{m_e\,c}{\hbar}\right)^2\left(\gamma^2-1\right),$$ (3.229)

or

$$k^2 - 2\,k\,k'\,\cos\theta+k'^{\,2} = \left(\frac{m_e\,c}{\hbar}\right)^2\left(\gamma^2-1\right).$$ (3.230)

Here, $\theta$ is the angle through which the photon is scattered (i.e., the angle subtended between ${\bf k}$ and ${\bf k}'$). See Figure 3.14.

Let $E_\gamma$, $E_\gamma'$, $E_e$, and $E_e'$ be the initial photon energy, the final photon energy, the initial electron energy, and the final electron energy, respectively. Energy conservation in the collision requires that

$$E_\gamma + E_e = E_\gamma'+E_e'.$$ (3.231)

However, we know that

$$E_\gamma = \hbar\,c\,k,$$ (3.232)

$$E_\gamma' =\hbar\,c\,k',$$ (3.233)

$$E_e = m_e\,c^2,$$ (3.234)

$$E_e' = \gamma\,m_e\,c^2.$$ (3.235)

[See Equations (3.174), (3.197), and (3.199).] Hence, we get

$$\gamma = \frac{k-k'+m_e\,c/\hbar}{m_e\,c/\hbar}.$$ (3.236)

Equations (3.230) and (3.236) can be combined to give

$$k^2 - 2\,k\,k'\,\cos\theta+k'^{\,2} = \left(k-k'+\frac{m_e\,c}{\hbar}\right)^2 - \left(\frac{m_e\,c}{\hbar}\right)^2,$$ (3.237)

or

$$k^2 - 2\,k\,k'\,\cos\theta+k'^{\,2}= k^2-2\,k\,k'+k'^{\,2} + 2\,(k-k')\,\frac{m_e\,c}{\hbar},$$ (3.238)

which can be rearranged to produce

$$\frac{1}{k'} - \frac{1}{k} = \frac{\hbar}{m_e\,c}\,(1-\cos\theta).$$ (3.239)

Finally, if $\lambda =2\pi/k$ and $\lambda'=2\pi/k'$ are the initial and final wavelengths of the photon then we obtain

$$\lambda'-\lambda = \frac{h}{m_e\,c}\,(1-\cos\theta).$$ (3.240)

The previous equation relates the increase in wavelength of the scattered photon to its scattering angle in a simple manner. Here, $h/(m_e\,c) =2.43\times 10^{-12}\,{\rm m}$ is known as the Compton wavelength of the electron. The previous formula was verified experimentally by Arthur Compton in 1923.