University Physics IV · Limits of Classical Physics · 1.7
Compton scattering and photon momentum
Light carries momentum h/λ, and the evidence is the electron it knocks sideways. Set one photon against one free electron, conserve energy and momentum relativistically, and the scattered wavelength comes back longer by a fixed number of picometres — set by the angle and the electron's mass, and by nothing else.
Build the model
Connect the measurement to the mechanism.
Give light the particle relations E = hc/λ and p = h/λ, and its collision with a free electron at rest becomes ordinary relativistic two-body kinematics: three conservation equations, two unwanted unknowns in the electron's momentum and direction, and one angle fixed by where the detector sits. Eliminating the electron leaves a single new length, h/mₑc = 2.4263 pm, and the scattered wavelength exceeds the incident one by that length times (1 − cos θ). Two features make the result decisive.
A classical wave driving a charge re-radiates at the driving frequency, so it predicts no shift whatever — the displaced line has no classical explanation. And the shift is absolute, a few picometres regardless of what came in, which is why it hides inside visible light and stands out against a 71 pm X-ray. The formula's one letter mₑ carries its one assumption: whatever recoils must be free to do so.
When the electron is held too tightly, the atom takes the momentum instead, the shift shrinks by the mass ratio, and an unshifted line appears beside the shifted one — exactly what Compton's graphite spectra show.
- Simple definition
- Compton scattering is the collision of a photon with a free electron, in which the photon hands over energy and momentum and emerges with a longer wavelength whose increase depends only on the scattering angle.
- Example
- Molybdenum Kα X-rays of 71.1 pm scattered through 90° from graphite return at 73.53 pm — a shift of 2.43 pm, 3.4% of the wavelength, with 0.58 keV carried off by the recoiling electron.
The m = 0 branch of E² = p²c² + m²c⁴. It gives the light something to conserve.
p in kg m s⁻¹; at λ = 71.1 pm, h/λ = 9.3 × 10⁻²⁴ kg m s⁻¹.
Zero forward, 2λC straight back, and every angle between lies on the 1 − cos θ curve.
θ measured from the incident beam; Δλ in m. No λ and no target property appear.
The only length in the result, so it sets the entire scale: max shift 2λC = 4.853 pm.
h = 6.626 × 10⁻³⁴ J s, mₑ = 9.109 × 10⁻³¹ kg, c = 2.998 × 10⁸ m s⁻¹.
Backscatter saturates: as E → ∞, E′(180°) → mₑc²/2 = 255 keV, never lower.
mₑc² = 511.0 keV. α is the dimensionless ratio that decides the regime.
For 661.7 keV gammas the edge sits at 477.4 keV, 184 keV below the photopeak.
K in keV when E is. Largest at θ = 180°, where it defines the Compton edge.
Locks the electron's angle to the photon's — the coincidence test of the two-body picture.
φ is the electron's angle on the far side of the beam; always 0 ≤ φ < 90°.
A classical wave cannot shift the wavelength
A classical electromagnetic wave of frequency ν drives a free electron into oscillation at ν, and an accelerating charge radiates at its own frequency. So Thomson scattering returns light at exactly the incident wavelength, at every angle, with intensity going as 1 + cos²θ and a total cross-section σT = 6.65 × 10⁻²⁹ m² that does not depend on λ at all. In that picture Δλ = 0 is not an approximation, it is a theorem. In 1923 Compton sent 17.4 keV molybdenum Kα X-rays into graphite and measured the scattered spectrum with a calcite crystal. At each angle he measured — 45°, 90° and 135° — he found two lines: one at the incident 71.1 pm, and a second displaced to longer wavelength by an amount that grew with the angle. Nothing you can do to a classical wave produces the second line. Giving the light momentum does.
Set it up as a two-body collision
Once the light has momentum, the rest is bookkeeping. Before: a photon along +x with energy hc/λ and momentum h/λ, and an electron at rest with energy mₑc² and no momentum. After: a photon at angle θ with hc/λ′ and h/λ′, and an electron with momentum pₑ at angle φ on the other side of the beam, carrying total energy √(pₑ²c² + mₑ²c⁴). Three conservation statements follow. Energy: hc/λ + mₑc² = hc/λ′ + √(pₑ²c² + mₑ²c⁴). Momentum along x: h/λ = (h/λ′)cos θ + pₑ cos φ. Momentum along y: 0 = (h/λ′)sin θ − pₑ sin φ. That is three equations in the four unknowns λ′, pₑ, φ and θ, so fixing one of them — in practice θ, by where the detector sits — determines the other three. Beyond the photon relations themselves, exactly two physical assumptions are buried in the setup: the electron is free, and it is initially at rest.
Eliminate the electron and a single length falls out
Neither φ nor pₑ is wanted, so eliminate them. Write A = hc/λ and B = hc/λ′. Move the photon terms of the two momentum equations to one side and square: (pₑc cos φ)² = (A − B cos θ)² and (pₑc sin φ)² = (B sin θ)². Add them, and sin²φ + cos²φ = 1 gives pₑ²c² = A² + B² − 2AB cos θ. Now the energy equation: it says Eₑ = A − B + mₑc², and squaring against Eₑ² = pₑ²c² + mₑ²c⁴ gives pₑ²c² = (A − B)² + 2(A − B)mₑc². Set the two expressions equal. The A² and B² terms cancel on both sides, leaving 2(A − B)mₑc² = 2AB(1 − cos θ), which rearranges to 1/B − 1/A = (1 − cos θ)/mₑc². That step is the whole result, because 1/A and 1/B are λ/hc and λ′/hc — the answer is naturally linear in wavelength, not in frequency. Multiply through by hc and λ′ − λ = (h/mₑc)(1 − cos θ). Every h and every c has collapsed into one length.
Absolute shift, fractional visibility
Read what survived. Only θ and mₑ remain; the incident wavelength, the beam intensity and the target material have all cancelled out. The shift runs from nothing in the forward direction, through λC = 2.4263 pm at 90°, to 2λC = 4.853 pm straight back, and those are the same picometres for every incident wavelength, so what changes between experiments is not the shift but the fraction it represents. Green light at 550 nm shifts by 2.43 pm at 90°: a relative change of 4.4 × 10⁻⁶, invisible to any optical spectrometer of the day. Compton's 71.1 pm X-rays shift by the identical 2.43 pm, which is 3.4% and easily resolved on a crystal spectrometer. A 1.87 pm gamma from ¹³⁷Cs shifts by that same 2.43 pm, now 130% of its own wavelength. Energy behaves quite differently: ΔE/E = α(1 − cos θ)/[1 + α(1 − cos θ)] with α = λC/λ, so at 90° the X-ray loses 3.3% of its energy while the gamma loses 56%. The shift is universal; the damage is not.
Whose mass is mₑ? The unshifted line
mₑ entered the formula as the mass of whatever recoils. An electron bound more tightly than the collision can pay cannot recoil alone; the momentum goes to the whole atom, and mₑ becomes M. For carbon, M = 12 u = 1.99 × 10⁻²⁶ kg, so h/Mc = 1.11 × 10⁻⁴ pm — 22 000 times smaller than λC and far below any spectrometer's resolution. That light returns at the incident wavelength, and it is the unshifted line sitting beside the shifted one in every one of Compton's spectra. Which electrons count as free is settled by comparing binding energy with the recoil energy on offer. Graphite's four valence electrons per atom sit in bands no more than about 20 eV deep; carbon's two K-shell electrons are bound by 284 eV. A 17.4 keV photon scattering at 30° hands the electron only 79 eV, so the K electrons scatter without shifting while the valence electrons recoil freely. At 90° the offer rises to 576 eV, above the K-shell binding, and the unshifted line weakens. Heavier elements, holding their electrons more tightly, show a stronger unshifted line — exactly as observed.
The recoil electron: direction, energy, and the Compton edge
The same three equations fix the electron once θ is known. Divide the y-momentum equation by the x-momentum equation: cot φ = (λ′/λ − cos θ)/sin θ, and since λ′/λ = 1 + α(1 − cos θ) with α = E/mₑc², the numerator is (1 + α)(1 − cos θ), so cot φ = (1 + α) tan(θ/2). The electron never goes backward and never leaves at 90° with any momentum worth having; it is thrown forward, more sharply the harder the photon is scattered. Its energy is whatever the photon lost, K = E α(1 − cos θ)/[1 + α(1 − cos θ)], which is largest at θ = 180°: Kₘₐₓ = 2E²/(mₑc² + 2E), the Compton edge that a γ spectrometer records as a cliff — 477 keV for the 662 keV line of ¹³⁷Cs. The parameter α is the whole story of the regime: for Compton's X-rays it is 0.034 and the electron takes a few hundred electronvolts, for a 662 keV gamma it is 1.29 and the electron takes most of the energy. The lock between θ and φ is itself testable, and Compton and Simon tested it in a cloud chamber in 1925: the recoil track and the direction of the next scattering event lay where the two-body formula said they must, event by event, and Bothe and Geiger caught the scattered X-ray and its electron in time coincidence. Whether that record also forces the field itself to be quantised is a separate question, held over to the next topic.
Change one variable at a time
Make the relationship visible.
Hold θ at 180° and drag λ from 80 pm down to 1 pm: the shift stays at 4.853 pm while Δλ/λ runs from 6% to 485%, which is why Compton needed X-rays rather than visible light.
SHIFT Δλ2.426 pm
SCATTERED λ′73.53 pm
FRACTIONAL Δλ/λ3.4 %
ELECTRON K0.58 keV
Live interpretationSHIFT Δλ: 2.426 pm. SCATTERED λ′: 73.53 pm. FRACTIONAL Δλ/λ: 3.4 %. ELECTRON K: 0.58 keV
Catch the common trap
Explain before calculating.
A 17.4 keV X-ray photon (λ = 71.1 pm) and a 662 keV gamma photon (λ = 1.87 pm) each scatter through 90° from a free electron. How do their wavelength shifts and their fractional energy losses compare?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyMolybdenum Kα X-rays of wavelength 71.1 pm scatter from graphite and are detected at 90° to the beam. Find the wavelength shift, the scattered wavelength, the fractional change in wavelength, and the kinetic energy given to the electron. Take λC = h/mₑc = 2.4263 pm and hc = 1 239 840 eV pm.
- At θ = 90°, cos θ = 0, so the bracket is 1 − cos θ = 1 and the shift is exactly one electron Compton wavelength: Δλ = 2.4263 pm.
- λ′ = λ + Δλ = 71.1 + 2.4263 = 73.53 pm, and the fractional change is 2.4263/71.1 = 0.0341, or 3.4%.
- Photon energies from E = hc/λ: E = 1 239 840/71.1 = 17 438.0 eV and E′ = 1 239 840/73.5263 = 16 862.5 eV.
- The electron takes the difference: K = 17 438.0 − 16 862.5 = 575.5 eV. Note that this is the difference of two large, nearly equal numbers, so keep the extra digits until the subtraction is done.
AnswerΔλ = 2.43 pm, λ′ = 73.53 pm, a 3.4% change in wavelength; the electron leaves with K = 0.58 keV, which is 3.3% of the incident photon's energy.
MediumA ¹³⁷Cs source emits 661.7 keV gammas. One of them backscatters through 180° from a nearly free electron in a scintillator. Find its wavelength before and after, the energy of the backscattered photon, and the electron's kinetic energy — the Compton edge. Take mₑc² = 511.0 keV and hc = 1 239 840 eV pm.
- Incident wavelength: λ = hc/E = 1 239 840 eV pm ÷ 661 700 eV = 1.874 pm. That is already comparable with λC = 2.4263 pm, so a large fractional effect is coming.
- At θ = 180°, 1 − cos θ = 2, so Δλ = 2 × 2.4263 = 4.853 pm — the largest shift the formula allows — and λ′ = 1.874 + 4.853 = 6.727 pm.
- Work the energy directly as a check on the algebra: α = E/mₑc² = 661.7/511.0 = 1.2949, so E′ = E/(1 + 2α) = 661.7/3.5898 = 184.3 keV.
- Via the wavelength instead: E′ = 1 239 840/6.727 = 184 308 eV = 184.3 keV. The two routes agree.
- The electron takes the rest: Kₘₐₓ = 661.7 − 184.3 = 477.4 keV. This is the Compton edge — the sharp cliff in a measured gamma spectrum, 184 keV below the 662 keV photopeak, with the 184 keV backscatter peak appearing separately.
Answerλ = 1.874 pm shifts to 6.727 pm; the backscattered photon carries 184.3 keV and the electron 477.4 keV. No scattering angle can give the electron more than this.
HardThe same 71.1 pm molybdenum X-ray now scatters at θ = 60°. Find the electron's kinetic energy, the direction φ of its recoil, and its speed. Then decide whether a carbon K-shell electron, bound by 284 eV, could have been the scatterer. Take mₑc² = 511.0 keV and hc = 1 239 840 eV pm.
- Shift and scattered wavelength: 1 − cos 60° = 0.5, so Δλ = 2.4263 × 0.5 = 1.213 pm and λ′ = 72.313 pm.
- Energies: E = 1 239 840/71.1 = 17 438.0 eV and E′ = 1 239 840/72.313 = 17 145.4 eV, so K = 292.6 ≈ 293 eV. Checking with the closed form, K = E⋅α(1 − cos θ)/[1 + α(1 − cos θ)] with α = λC/λ = 0.034125 gives 17 438 × 0.016776 = 292.6 eV.
- Recoil direction: α = E/mₑc² = 17.438/511.0 = 0.034125, so cot φ = (1 + α)tan(θ/2) = 1.03413 × tan 30° = 1.03413 × 0.57735 = 0.59705. Hence tan φ = 1.6749 and φ = 59.2°, on the opposite side of the beam from the photon.
- Speed: K/mₑc² = 293/511 000 = 5.7 × 10⁻⁴, so the electron is barely relativistic. β = √(2K/mₑc²) = √(586/511 000) = √(1.147 × 10⁻³) = 0.0339, giving v = 0.0339 × 2.998 × 10⁸ = 1.02 × 10⁷ m s⁻¹; the exact relativistic value differs by only 0.04%.
- The free-electron test: the collision offers just 293 eV. That clears carbon's 284 eV K-shell binding, but with only 9 eV to spare, so the free-electron formula is on the edge here. At 30° the offer falls to 79 eV, the K electrons cannot recoil at all, and they contribute to the unshifted line instead. The valence electrons, bound by at most about 20 eV, are effectively free at every angle Compton measured.
AnswerK = 293 eV, φ = 59.2° from the beam on the side opposite the scattered photon, and v = 1.02 × 10⁷ m s⁻¹ (0.034c). A carbon K electron only just qualifies as free at 60°; at 30° it does not.