The Compton effect

If light is really made of photons with energy $hf$ and momentum $h/\lambda$, then the collision between a photon and a free electron should behave like an ordinary elastic collision between two particles, with conservation of energy and momentum. The photon would give up part of its energy to the electron, and would come out of the collision with less energy — that is, with lower frequency and longer wavelength. Arthur Compton verified this prediction in 1923, obtaining a spectacular confirmation of the corpuscular nature of light.

The experiment consists of firing X-rays at a target (graphite) and measuring the wavelength of the scattered radiation at various angles. Compton found that the scattered radiation has a longer wavelength than the incident one, and that the increase depends only on the scattering angle $\theta$ — exactly as predicted by treating the photon-electron collision with relativistic kinematics.

Key formula

A photon of wavelength $\lambda$ strikes a stationary electron; after the collision it continues with a longer wavelength $\lambda'$, deflected by an angle $\theta$. We have $\ev{\lambda' - \lambda = \frac{h}{m_e c}\,(1 - \cos\theta)}$ where $m_e$ is the electron mass. The quantity $\lambda_C = h/(m_e c) \approx 2{,}43\cdot 10^{-12}\;\text{m}$ is the electron’s Compton wavelength.

The shift $\Delta\lambda = \lambda' - \lambda$ is zero in the forward direction ($\theta = 0$, no collision), maximum backwards ($\theta = 180°$, where it equals $2\lambda_C$), and — crucially — does not depend on the initial wavelength: it is always of the order of $\lambda_C \approx 2{,}43$ pm, a constant fixed by the electron mass. This is the unmistakable signature of the corpuscular mechanism: a classically scattered wave would change intensity, not wavelength.

Observable only with X-rays

Since $\Delta\lambda$ is always of the order of $\lambda_C \approx 2{,}43$ pm, the effect is visible only if $\lambda$ is comparable to this scale. For a visible-light photon ($\lambda \approx 500$ nm) the relative change is $\Delta\lambda/\lambda \sim 2\lambda_C/\lambda \sim 10^{-5}$, too small to be detected. For an X-ray ($\lambda \approx 50$ pm) instead $\Delta\lambda/\lambda \sim 10\%$: this is why Compton scattering is observed with X-rays and not with visible light.

Together with the photoelectric effect, the Compton effect was the decisive proof that the photon is not merely a convenient way of counting energy, but a genuine corpuscle with well-defined momentum that takes part in collisions following the rules of relativistic kinematics. A complete numerical example — a $0{,}8$ MeV photon that loses a third of its energy, with calculation of the scattering angle and of the (relativistic!) velocity of the recoiling electron — is worked out among the exercises of the chapter.

Links

Topics: Quantum physics Concepts: Compton effect · Photon · Momentum · Elastic collision

Related exercises: Worked exercise — Compton scattering with a 0.8 MeV photon · Compton scattering at 90 degrees with X-rays · Bowling ball versus pin