The Compton effect

Figure 17.1: X-ray scattering

A nearly monochromatic X-ray beam of frequency $\nu$ is incident on a graphite sample as shown in Figure 17.1. The oscillating electric field of the incident electromagnetic wave causes the electrons in the graphite to oscillate at the same frequency $\nu$. These oscillating electrons will emit radiation in all directions, the frequency of this radiation is expected to also be $\nu$, same as the incident frequency. This process where the incident X-ray is scattered in different directions, its frequency being unchanged is called Thomson scattering. In addition to this, it is found that there is a component of scattered X-ray which has a smaller frequency $\nu^{'}$ or larger wavelength $\lambda^{'}$. The situation where there is a change in the frequency of the incoming light is referred to as the Compton effect. It is not possible to explain the Compton effect if we think of the incident X-ray as a wave.

To explain the Compton effect it is necessary to associate a particle called a photon ($\gamma$) with the incident electromagnetic wave,

(17.1)

The momentum $\vec{p}$ and energy $E$ of the photon are related to the wave number and angular frequency respectively of the wave as

$$\vec{p}=\hbar \vec{k} \hspace{2cm} E=\hbar \omega\,.$$ (17.2)

where $\hbar=h/2 \pi$ and $h=6.63 \, \times 10^{-34} \, {\rm J \, s}$ is the Planck's constant.

Figure 17.2: The Compton scattering

It is possible to explain the Compton effect if we think of it as the elastic scattering of a photon ($\gamma$) and an electron ($e$) as shown in Figure 17.2. The electron's energy is related to its momentum, $\vec{p}_e$, as

$$E^2 = p_e^2 c^2 + m_e^2 c^4 \,.$$ (17.3)

This relativistic formula taken into account the rest mass energy $m_e c^2$ of the electron and is valid even if the electron moves at a high speed approaching the speed of light. Applying the conversation of energy to the $\gamma$, $e$ collision we have,

$$\hbar \omega + m_e c^2 = \hbar \omega^{'} + \sqrt{p_e^2 c^2 + m_e^2 c^4} \,.$$ (17.4)

The $x$ and $y$ components of the conservation of momentum are respectively,

$$\hbar k - \hbar k^{'} \cos \theta = p_e \cos \alpha$$ (17.5)

and

$$\hbar k^{'} \sin \theta = p_e \sin \alpha.$$ (17.6)

Squaring and adding equations (17.5) and (17.6) we have

$$p^2_e=\hbar^2 (k^2 + k^{'2} - 2 k k^{'} \, \cos \theta) \,.$$ (17.7)

Multiplying this by $c^2$ and writing it in terms of wavelengths we have,

Squaring the conservation of energy (eq. 17.5) and writing it in terms of wavelength gives us

$$c^2 p_e^2=\left( \frac{c h}{\lambda}-\frac{c h}{\lambda^{'}} +m_e c^2 \right)^2 - m_e^2 c^4 \,.$$ (17.9)

Subtracting equations (17.8) and (17.9) and rearranging the terms we have the difference in wavelength,

$$\lambda^{'}-\lambda=\lambda_c (1 - \cos \theta),$$ (17.10)

where $\lambda_c=h/m_e c=2.4 \times 10^{-12} \, {\rm m}$ is the Compton wavelength. The difference, $\Delta \lambda=\lambda^{'}-\lambda$, is known as the Compton shift. It is interesting to note that the Compton shift is independent of initial wavelength of the light and depends only on the scattering angle $\theta$ of the light. The maximum difference in the wavelength $\Delta \lambda=\lambda^{'}-\lambda$ is in the backwards direction $\theta=180^{\circ}$ where $\Delta \lambda=2 \lambda_c$.

In this picture we think of the incident X-ray as particles called photons which lose energy when they collide with the electrons. This results in the increase in wavelength observed in the Compton effect. The change in wavelength is very small, of the order of $\lambda_c$. This change will be significant only when the incident wavelength $\lambda$ is comparable to $\lambda_c$, which is the case in X-rays where $\lambda \sim 10^{-10} \, {\rm m}$.

The photoelectric effect and the Compton effect require us to think of electromagnetic radiation in terms of a particle called the photon. This does not mean that we can abandon the wave theory. We cannot explain interference or diffraction without this. This basically tells us that light has a dual nature. It is sometimes necessary to think of it as a wave and sometimes as a particle, depending on the phenomenon that we are trying to explain. This dual wave-particle behaviour is not restricted to light alone, and it actually extends to the whole of nature.