The Compton Effect

By 1923 physicists had accepted that light carries energy in discrete quanta — photons — but a subtler question remained: does a photon also carry momentum? Arthur Compton answered it by firing X-rays at a graphite target and measuring the scattered wavelength.

Photon momentum and the scattering geometry

A photon of wavelength $\lambda$ carries energy $E = hc/\lambda$ and, because it is massless, momentum

$p = \frac{E}{c} = \frac{h}{\lambda}.$

When such a photon collides with a nearly-free electron (one loosely bound in the outer shells of carbon), both total relativistic energy and momentum are conserved. Working out the kinematics — the photon deflects by angle $\theta$, the electron recoils — yields the Compton formula:

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

Here $\lambda$ is the incident wavelength, $\lambda'$ the scattered wavelength, $m_e$ the electron rest mass, and $c$ the speed of light. The combination

$\lambda_C = \frac{h}{m_e c} \approx 2.426 \times 10^{-12}\,\text{m}$

is called the Compton wavelength of the electron. Because $\lambda_C$ is a fixed length set only by fundamental constants, the shift $\Delta\lambda$ is entirely independent of the incident wavelength — a striking prediction that classical wave scattering (Thomson scattering) cannot reproduce.

Deriving the formula

We use two conservation laws in the reference frame where the electron is initially at rest.

Energy conservation: The photon loses energy to the recoiling electron.

$\frac{hc}{\lambda} + m_e c^2 = \frac{hc}{\lambda'} + \gamma m_e c^2,$

where $\gamma = (1 - v^2/c^2)^{-1/2}$ is the Lorentz factor of the recoiling electron.

Momentum conservation (two components): the photon's initial momentum $h/\lambda$ must equal the vector sum of the scattered photon momentum $h/\lambda'$ and the electron recoil momentum $\gamma m_e v$.

Eliminating the electron's final energy and momentum from these three equations (one energy + two momentum components) gives, after some algebra, exactly the Compton formula above.

Angular dependence

The shift ranges from zero (forward scattering, $\theta = 0$) to a maximum of $2\lambda_C$ (back scattering, $\theta = \pi$):

| Angle $\theta$ | $\Delta\lambda$ | |---|---| | $0°$ | $0$ | | $90°$ | $\lambda_C \approx 2.43 \times 10^{-12}\,\text{m}$ | | $180°$ | $2\lambda_C \approx 4.85 \times 10^{-12}\,\text{m}$ |

At $\theta = 90°$, $\cos 90° = 0$, so $\Delta\lambda = \lambda_C$ exactly.

Compton's experimental results matched these predictions to high precision, providing unambiguous evidence that photons carry quantized momentum — a cornerstone of quantum theory.

Try it

This is a numerical exercise — return a number. An X-ray photon scatters off a stationary electron at $\theta = 90°$. The Compton wavelength is $\lambda_C = 2.426 \times 10^{-12}\,\text{m}$. What is the wavelength shift $\Delta\lambda$ (in metres)?

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