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Students learn that a photon has energy hf and momentum h/λ, but the momentum part often feels like a formula to memorise. Arthur Compton's X-ray experiments are where that momentum became something you could measure. This post explains what he saw, derives the key result in words and works a full numerical example.
What did Compton observe?
When X-rays pass through a light element, some are scattered sideways. Classical wave theory says scattered radiation should have the same wavelength as the incoming beam: the electrons shake at the wave's frequency and re-radiate at that frequency. Compton found that the scattered X-rays had a longer wavelength, and the shift grew with the scattering angle. The Nobel Prize summary describes it this way: some of the incident photon energy was transferred to the electrons, so the X-rays' wavelength increased.
Why did a wavelength shift mean photons carry momentum?
Compton's 1923 paper in Physical Review states the hypothesis directly: "when an X-ray quantum is scattered it spends all of its energy and momentum upon some particular electron." The electron recoils, the scattered quantum has less energy, and less energy means a longer wavelength.
If you apply conservation of energy and of momentum to this photon-electron collision, as you would for two billiard balls, you get
Δλ = λ' − λ = (h / mc)(1 − cos θ)
where m is the electron mass and θ is the angle through which the X-ray is scattered. Compton wrote the same result as (2h/mc) sin²(θ/2), since 1 − cos θ = 2 sin²(θ/2). His abstract also notes the key surprise: the increase "is independent of the wave-length".
The constant h/mc is called the Compton wavelength of the electron. The NIST value is 2.426 × 10⁻¹² m, or 2.426 pm.
A worked example with numbers
Problem. X-rays of wavelength 71.0 pm scatter off nearly free electrons. Find the scattered wavelength at θ = 90°, the energy given to the electron and the momentum of an incoming photon. Use hc = 1240 eV nm and h = 6.626 × 10⁻³⁴ J s.
Shift. At 90°, cos θ = 0, so Δλ = h/mc = 2.43 pm.
Scattered wavelength. λ' = 71.0 + 2.43 = 73.43 pm.
Photon energies. E = hc/λ. With λ = 0.0710 nm, E = 1240 / 0.0710 ≈ 17,460 eV. With λ' = 0.07343 nm, E' ≈ 16,890 eV.
Electron's kinetic energy. K = E − E' ≈ 577 eV, about 0.58 keV. (Subtracting rounded values gives 570 eV; keep more digits until the end.)
Photon momentum. p = h/λ = 6.626 × 10⁻³⁴ / 7.10 × 10⁻¹¹ ≈ 9.33 × 10⁻²⁴ kg m/s.
Now compare visible light at 500 nm. The same 2.43 pm shift is only about 0.0005% of the wavelength, too small to notice. For 71 pm X-rays it is about 3.4%. That is why the effect was found with X-rays.
Common mistakes:
Mixing units. Keep both wavelengths in pm (or both in nm) before adding Δλ.
Thinking the shift depends on λ. Δλ depends only on the angle. The fractional shift, Δλ/λ, is what grows for short wavelengths.
Using the wrong angle. θ is the angle of the scattered photon, not the electron's recoil direction.
Subtracting rounded energies. E and E' are close, so small rounding errors become large in K.
On the syllabus: the NTA JEE (Main) 2026 syllabus lists the photoelectric effect, "particle nature of light" and the de Broglie relation in its Dual Nature unit, but does not name the Compton effect. Photon momentum still appears in radiation pressure and de Broglie questions. For practice, see our guide to photoelectric effect problems.
How was the discovery recognised?
Compton received half of the 1927 Nobel Prize in Physics "for his discovery of the effect named after him". The other half went to C.T.R. Wilson for the cloud chamber.
The idea also reached Kolkata. In his Nobel lecture, C.V. Raman said Compton's work had made familiar the idea that radiation's wavelength could change on scattering, and that he came to see his own puzzling observations as "the optical analogue of the Compton effect". Read more in our post on C.V. Raman and light scattering.
What can a student take from Compton's way of working?
Apply familiar laws to an unfamiliar object. Compton used the same conservation laws you use for colliding blocks, applied to a photon and an electron.
Let the data decide. The angle dependence of the shift ruled out the classical picture.
Check the size of an effect. Knowing that 2.4 pm matters for X-rays but not for visible light explains why it was found when it was.
If your trouble is setting up the conservation equations rather than the arithmetic, use worked solutions to check your steps. The Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: repeat the worked example for θ = 180° and check that the electron now gets about twice as much energy.
