In Physics 2 you saw that X-rays bouncing off electrons come away with longer wavelengths, because a photon carries momentum $p = h/\lambda$. In 1923 Arthur Compton derived exactly how much longer, treating the collision like two billiard balls but with relativistic energy for the electron. His formula matched the measurements, and it convinced most physicists that photons are real particles.
🔑 Key idea
A photon scattered by a free electron through an angle $\theta$ comes away with a longer wavelength: $$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_ec}(1 - \cos\theta),\qquad \frac{h}{m_ec} = 2.426\ \text{pm}.$$ The shift depends only on the angle, not on the starting wavelength. The electron recoils with the energy the photon lost: $K_e = \frac{hc}{\lambda} - \frac{hc}{\lambda'}$.
💡 Why it works
Conserve momentum and energy. Momentum (a vector): the electron gets $\vec{p}_e = \vec{p} - \vec{p}\,'$, so $p_e^2 = p^2 + p'^2 - 2pp'\cos\theta$. Energy: $pc + m_ec^2 = p'c + E_e$, with $E_e^2 = (p_ec)^2 + (m_ec^2)^2$. Square the energy equation written as $E_e = (p - p')c + m_ec^2$: $$E_e^2 = (p - p')^2c^2 + 2(p - p')m_ec^3 + m_e^2c^4.$$ Set it equal to $p_e^2c^2 + m_e^2c^4$ and cancel: $-2pp' + 2(p - p')m_ec = -2pp'\cos\theta$, so $\frac{p - p'}{pp'} = \frac{1 - \cos\theta}{m_ec}$. With $p = h/\lambda$, the left side is $\frac{\lambda' - \lambda}{h}$. Multiply by $h$: done. The scale $h/m_ec$ is the electron's Compton wavelength.
A 50 pm X-ray scattering from an electron. Change the angle: the shift grows from zero (straight through) to $2h/m_ec = 4.85$ pm (straight back).
Worked examples
✏️ Worked example: A right-angle scatter
X-rays of wavelength 50.0 pm scatter from electrons at $90^\circ$. Find the scattered wavelength.
Answer: About one part in 100,000, too small to notice; for 50 pm X-rays the same shift is about 10%
Compton's own experiment scattered 71 pm X-rays from molybdenum off a block of graphite. At each angle he found two peaks: one shifted by exactly $\frac{h}{m_ec}(1 - \cos\theta)$, and one not shifted at all. The unshifted peak comes from photons that bounce off a whole atom rather than a single electron; the same formula applies with the atom's mass in place of $m_e$, and that mass is so large that the shift is far too small to see.
⚠️ Common mistake
Putting the photon's wavelength into the factor in front, as in $\Delta\lambda = \lambda(1 - \cos\theta)$. The factor is the electron's Compton wavelength $h/m_ec = 2.426$ pm, the same for every photon. A check: the formula says the shift is the same in picometers for X-rays and for gamma rays; only the fraction differs.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.