Compton Scattering

Light carries momentum even though it has no mass, with p = h/lambda. When an X-ray photon hits a loosely bound electron, treat it like a collision between two billiard balls. The photon knocks the electron into motion, handing over some of its energy and momentum. A photon that loses energy must have a lower frequency, and since c = f·lambda, a lower frequency means a longer wavelength. That is why the scattered X-ray comes out with more wavelength than it went in with: the shift is direct proof the photon behaved as a particle with its own momentum, not just a wave.

Working the collision means conserving both energy and momentum in two dimensions, exactly like any other particle collision, except one particle is a photon whose energy is hc/lambda and whose momentum is h/lambda. Write momentum conservation along and perpendicular to the original direction, conserve total energy (photon plus electron, using relativistic electron energy), and combine the equations. The result collapses to a clean formula: delta_lambda = (h/mc)(1 - cos(theta)), where theta is the angle the photon scatters through and m is the electron's mass. The constant h/mc, about 2.426 picometers, is the Compton wavelength.

The angle controls everything. At theta = 0, the photon barely scatters and delta_lambda = 0: no collision, no shift. As theta grows toward 180 degrees, 1 - cos(theta) grows toward 2, so the shift maxes out at twice the Compton wavelength, about 4.85 pm, for a direct backscatter. This sits next to the photoelectric effect in showing light's particle side, and it sets up pair production and nuclear physics, where mass, energy, and momentum keep trading places.