If you shine a green laser beam on a mirror, it reflects as a green laser beam. Neither the wavelength nor the color of visible light is changed by reflection or by scattering. But in 1922, the American physicist Arthur Compton made the remarkable discovery that x rays scattered from a solid target have a longer wavelength than the incident x rays. It is as if a green laser beam were to reflect as a red laser beam.

Compton was soon able to show that what we now call Compton scattering, the shifted wavelength of scattered x rays, is exactly what the photon model of light—but not the wave model—predicts. His experiment was perhaps the most convincing demonstration of the reality of photons.

FIGURE 38.11a shows Compton’s experiment: A collimated beam of x rays of known wavelength ${\lambda }_{\text{i}}$ is incident on a target, and the wavelength ${\lambda }_{\text{s}}$ is measured of x rays that are scattered at the scattering angle $\theta$. X rays, like visible light, interact with the electrons of the target atoms. If x rays are photons, rather than classical waves, then, Compton reasoned, scattering is really an elastic collision of a photon with an electron, not unlike the collision of two billiard balls.

FIGURE 38.11 Compton’s x-ray scattering experiment.

Elastic collisions conserve both energy and momentum. We’ve seen that photons have energy, but what about momentum? You learned in Chapter 31 that light beams do carry momentum, and thus the photons that make up the light beam must have momentum. Specifically, we found that an object that absorbs electromagnetic wave energy $\text{Δ}E$ gains momentum $\text{Δ}p=\text{Δ}E/c$. If the energy absorbed is that of one photon, $hf$, and if the photon vanishes, then momentum conservation dictates that the photon had momentum

Compton’s recognition that photons have momentum was key to his analysis of the experiment.

FIGURE 38.11b is a before-and-after representation of a scattering event in which an x-ray photon is incident along the $x$-axis. Scattering sends the photon at angle $\theta$ and the electron at angle $\varphi$. (X-ray energies are much greater than the ionization energies of atoms, so the influence of the nucleus is small and the electron acts very much like a free electron.) We need to write equations for both momentum and energy conservation.

Momentum is a vector, and both its $x$- and $y$-components are conserved. This gives, respectively, the two equations

where $p_{\text{e}}$ is the momentum of the scattered electron. The scattered electron is moving at very high speed, so we need to use the Chapter 36 relativistic equation for energy. Initially the electron has only rest energy, $m{c}^2$, where $m$ is the electron mass. The scattered electron’s energy, from Equation 36.45, is $({\left(m{c}^2\right)}^2+(p_{\text{e}}c{)}^2{)}^{1/2}$. Consequently, the energy-conservation equation is

where we used $f=c\text{/}\lambda$ to write the photon energies in terms of wavelength rather than frequency.

Equations 38.12 and 38.13 are three simultaneous equations. After a lot of algebra, which we’ll skip, they lead to the quite simple result that x rays scattered at angle $\theta$ have a wavelength shift

This result is in perfect agreement with Compton’s measurements.