The Compton effect — the photon has momentum

What just happened?

An X-ray photon struck a free electron like a billiard ball. It gave up part of its momentum and energy, so it bounced off “weaker” — with lower energy, and therefore a longer wavelength. This is the Compton effect.

The Compton shift:

— the increase in wavelength depends only on the scattering angle $\theta$, not on the incident wavelength. The constant $\lambda_C = h/(m_e c) \approx 2.43\ \text{pm}$ is the Compton wavelength of the electron. Since $1-\cos\theta \ge 0$, always $\Delta\lambda \ge 0$ — the photon never gains energy.

Extreme cases:

For $\theta = 0^\circ$ (the photon flies straight on) $\Delta\lambda = 0$ — no collision. For $\theta = 180^\circ$ (back-scattering) the shift is largest: $\Delta\lambda = 2\lambda_C$, because the electron receives maximum momentum.

Why does this matter?

A wave cannot “collide” and transfer momentum like a particle. The fact that the collision calculation works out proves the photon has momentum $p = h/\lambda$ and behaves like a particle. The Compton effect is one of the pillars of wave–particle duality.

Why you need this for the exam:

The formula $\Delta\lambda = \lambda_C(1-\cos\theta)$ plus $p = h/\lambda$ and conservation of momentum solve Compton scattering problems. The trick: the angle $\theta$ gives Δλ, hence $\lambda'$ and the energy of the scattered photon $E' = hc/\lambda'$.

Remember: the scattered photon always has a longer wavelength (lower energy) — Δλ = λ_C(1−cosθ) ≥ 0. That is proof that light has momentum and a corpuscular nature.