A photon has no mass, but it carries momentum. Turn the de Broglie relation around: a photon of wavelength $\lambda$ has $p = h/\lambda$. In 1923 Arthur Compton bounced X-rays off electrons and found that the photons behave like billiard balls in a collision, conserving both energy and momentum.
Key idea
A photon has momentum $p = \dfrac{h}{\lambda} = \dfrac{E}{c}$. A photon that scatters off a free electron at angle $\theta$ comes away with a longer wavelength: $$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta),$$ where $\dfrac{h}{m_e c} = 2.43\ \text{pm}$ (the Compton wavelength of the electron).
Why it works
The electron recoils and carries off some kinetic energy. The photon must lose that much energy, and for a photon less energy means a lower frequency and a longer wavelength, since $E = hc/\lambda$. Its speed cannot change: every photon moves at $c$. A head-on bounce ($\theta = 180^\circ$) gives the electron the biggest kick, so the shift is largest there, $2(2.43) = 4.86\ \text{pm}$. A grazing scatter ($\theta \to 0$) changes almost nothing.
Common mistake
Thinking the scattered photon slows down. It still moves at $c$; its lost energy shows up as a longer wavelength (lower frequency). Check: the shift depends only on the angle, and $\lambda'$ is always longer than $\lambda$.
Worked example: Scattering at 90°
X-rays of $71.0\ \text{pm}$ scatter at $90^\circ$ from free electrons. Find the scattered wavelength.
- $\cos 90^\circ = 0$, so $\Delta\lambda = 2.43\ \text{pm}$.
- $\lambda' = 71.0 + 2.43 = 73.4\ \text{pm}$.
Answer: $73.4\ \text{pm}$
Worked example: Energy given to the electron
For the same scattering, how much kinetic energy does the electron get? Use $hc = 1240\ \text{eV·nm} = 1.240 \times 10^{6}\ \text{eV·pm}$.
- Before: $E = \dfrac{1.240 \times 10^{6}}{71.0} = 17{,}460\ \text{eV}$.
- After: $E' = \dfrac{1.240 \times 10^{6}}{73.43} = 16{,}890\ \text{eV}$.
- The electron gets the difference: about $580\ \text{eV}$, so the photon gave up about $3\%$ of its energy.
Answer: About $580\ \text{eV}$
Worked example: Momentum of visible light
Find the momentum of a $500\ \text{nm}$ photon.
- $$p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{5.00 \times 10^{-7}} = 1.33 \times 10^{-27}\ \text{kg·m/s}.$$
- Tiny, but sunlight's push on a large thin sail can steer a spacecraft.
Answer: $1.33 \times 10^{-27}\ \text{kg·m/s}$