Compton scattering and photon momentum
Photon momentum h/λ, made measurable by one collision.
Compton scattering turns photon momentum from a slogan into a number. Set it up like any collision, keep the electron relativistic, and the algebra returns one length, h/mₑc = 2.43 pm, that says when the shift is visible. The harder part is saying what the measurement proves, and where the free-electron premise fails.
Build the model
Light carries momentum h/λ, and one collision with an electron makes it measurable.
1The big idea
Give each photon the momentum a massless particle must carry, p = E/c = h/λ, and scattering off a free electron at rest becomes a relativistic two-body collision. Energy and the two momentum components give three equations for four unknowns after the collision, so once the detector's position fixes the photon's angle θ, everything else is determined. Squaring the momentum triangle removes the electron's recoil angle, squaring the energy balance removes its speed, and what is left is λ′ − λ = (h/mₑc)(1 − cos θ). That is an absolute shift, from zero straight ahead to 2h/mₑc = 4.85 pm in backscatter, and it contains neither the incident wavelength nor the beam's intensity nor anything about the target atom. Classical wave theory predicts no such shift, because an electron driven by a wave reradiates at the driving frequency; that is Thomson scattering, and it is exactly what the formula returns as h → 0. The shift is only noticeable when λ is within a hundred or so times 2.43 pm, which is why the experiment needs X-rays. And the model stops at its own premise: an electron bound more tightly than the energy it would receive cannot recoil alone, so the whole atom takes the kick and returns an unshifted line, while the motion of the quasi-free electrons broadens the shifted one.
2In one sentence
Compton scattering is the collision of a single photon with a free electron, in which the photon hands over part of its momentum and energy and leaves with a longer wavelength, λ′ = λ + (h/mₑc)(1 − cos θ).
Example. A 20.0 pm X-ray scattered through 90° by a free electron leaves at 22.43 pm, a 12% change. Scattered through 90°, 500 nm light shifts by the same 2.43 pm, under 5 parts in a million.
3A photon's momentum is fixed by its energy
Relativity ties energy to momentum through E² = (pc)² + (mc²)². For a massive particle the familiar p = γmv sits inside that relation; for a massless one it leaves only p = E/c. Classical electromagnetism says the same in bulk: a pulse carrying energy U carries momentum U/c, which is why light exerts radiation pressure. What the photon hypothesis adds is that the momentum comes in portions of h/λ, one per quantum, delivered to one electron at a time. A 30.0 pm X-ray photon of 41.3 keV carries 2.21 × 10⁻²³ kg m s⁻¹; a 530 nm green photon carries 1.25 × 10⁻²⁷ kg m s⁻¹. One consequence shapes the whole topic: a free electron cannot swallow a photon. Absorption would leave it with energy pc + mₑc² and momentum p, and (pc + mₑc²)² = (pc)² + (mₑc²)² forces p = 0. A free electron can only scatter the photon and send a weaker one on; outright absorption needs a third body, an atom or a lattice, to take up momentum.
4Set up the collision and count the unknowns
Put the incident photon along +x with momentum p = h/λ and the electron at rest with energy mₑc². Afterwards the photon travels at θ above the axis with p′ = h/λ′, and the electron at φ below it with momentum pₑ and energy Eₑ = √((pₑc)² + (mₑc²)²). The collision stays in one plane, so conservation gives three equations, energy and the x and y components of momentum, for four unknowns: λ′, θ, pₑ and φ. One has to be chosen, and the experiment chooses θ by where it puts the detector. Keep the electron relativistic even when it is slow. Backscatter a 1.24 MeV photon and the electron leaves at 0.94c. At X-ray energies the Newtonian K = pₑ²/2mₑ comes close but not exactly: it replaces mₑc by mₑc − (p − p′)/2 in the final result, a 0.16% error for a 30 keV photon at 90° that grows to 59% for a 500 keV photon backscattered. The relativistic energy makes the result exact.
5Eliminate the recoil angle, then the recoil speed
The electron's direction is not measured, so remove it first. Momentum conservation says the vectors form a triangle, p = p′ + pₑ, with θ the angle between p and p′. The law of cosines on the side opposite θ gives pₑ² = p² + p′² − 2pp′ cos θ, and φ is gone. Next remove the speed. Energy conservation, pc + mₑc² = p′c + Eₑ, gives Eₑ = (p − p′)c + mₑc². Square it and set it equal to (pₑc)² + (mₑc²)²: the p², p′² and (mₑc²)² terms cancel, leaving mₑc(p − p′) = pp′(1 − cos θ). Divide by mₑc pp′ to get 1/p′ − 1/p = (1 − cos θ)/(mₑc), and multiply by h, since h/p is the wavelength: λ′ − λ = (h/mₑc)(1 − cos θ). Both moves carry over to any two-body collision: squaring a vector equation removes an unmeasured angle, and squaring the energy balance removes an unmeasured speed.
6Check it: units, limits and the classical answer
Units first: h/(mₑc) is (kg m² s⁻¹)/(kg m s⁻¹) = m, a length as it must be, worth 2.426 pm. At θ = 0 the photon is undeflected and Δλ = 0; at θ = 180° the momentum transfer is largest and Δλ = 2λC = 4.85 pm. Because the shift is absolute, the experiment lives or dies on the fractional shift, at most 2λC/λ: 24% for λ = 20 pm, but 10⁻⁵ at 500 nm, finer than an ordinary grating spectrometer resolves. Now the classical correspondence. Let h → 0 at fixed λ and the shift vanishes, leaving scattered light at the incident wavelength: Thomson scattering, in which a free electron driven by the wave's electric field oscillates at the driving frequency and reradiates at it. The energy form agrees, since E′ = E/[1 + (E/mₑc²)(1 − cos θ)] tends to E when E ≪ 511 keV. A classical electron pushed along by radiation pressure would give a Doppler shift that grows with intensity and exposure time; the measured shift depends on neither.
7Measure h/mₑc, and say what the measurement settles
A crystal spectrometer reads λ′ at several angles. Plot Δλ against 1 − cos θ: the model predicts a straight line through the origin whose slope is h/mₑc. Change the incident wavelength and the line should not move; change the light-element target and it should not move either, because the slope contains the electron's mass and nothing about the atom. With mₑ and c known independently, the slope times mₑc returns h, agreeing with the photoelectric value from a completely different measurement. Be exact about what this establishes. It shows that the scatterer is a free electron, that the momentum exchanged is fixed by h/λ and the angle, and it rules out any account in which the shift depends on intensity. It does not by itself prove that light is made of particles: in 1927 Schrödinger reproduced the formula with a classical light wave scattering off an electron described as a wave. The event-by-event evidence came from coincidence counting, begun by Bothe and Geiger in 1925, which caught each scattered photon together with its own recoil electron.
8When the electron is not free: the unshifted line
The derivation assumed a free electron at rest, and a real target supplies neither. The free-electron picture holds when the energy handed over is large compared with the binding energy. A 30 keV photon scattered through 90° gives its electron K = 1.66 keV. In aluminium that is hundreds of times the few-eV binding of the conduction electrons and over ten times the 0.07–0.12 keV of the L shell, but close to the 1.56 keV of the K shell. An electron that cannot be freed does not recoil alone: the whole atom takes the momentum, mₑ in the shift becomes the atomic mass, about 49,200 mₑ for aluminium, and the line moves by 5 × 10⁻⁵ pm. That is the unshifted line that sits beside the Compton line at every angle, and it is stronger for heavier targets, which hold a larger share of tightly bound electrons. The quasi-free electrons are not at rest either: their motion Doppler-broadens the shifted line, so its width carries information about the electrons' momenta.
Key equations
What each relation says, and the units it is written in.
Photon momentum
p = E/c = h/λ
E² = (pc)² + (mc²)² with m = 0 forces p = E/c, and E = hc/λ prices it per photon: a 1.00 MeV γ-ray, λ = 1.24 pm, carries 5.34 × 10⁻²² kg m s⁻¹.
p in kg m s⁻¹, E in J, λ in m; h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m s⁻¹
Compton shift
Δλ = λ′ − λ = (h/mₑc)(1 − cos θ)
Only θ appears, so every incident wavelength shifts by the same length: 0 straight ahead, 2.43 pm at 90°, 4.85 pm at 180°.
λ, λ′ incident and scattered wavelengths in m; θ the photon's scattering angle; mₑ = 9.109 × 10⁻³¹ kg
Compton wavelength of the electron
λC = h/(mₑc) = 2.426 × 10⁻¹² m
Sets the scale: the fractional shift is at most 2λC/λ, which is 6.8% for the 71.1 pm molybdenum Kα line Compton used but under 0.01% at any wavelength beyond 50 nm, so the effect needs X-rays or γ-rays.
a length, since J s ÷ (kg m s⁻¹) = m; equivalently hc/λC = mₑc² = 511.0 keV
Scattered photon energy
E′ = E / [1 + (E/mₑc²)(1 − cos θ)]
For E ≪ 511 keV, E′ → E, the Thomson limit. For E ≫ 511 keV the backscattered photon tends to mₑc²/2 = 255.5 keV whatever E is.
E, E′ in keV with mₑc² = 511.0 keV; the recoil electron's kinetic energy is K = E − E′
Compton edge: the electron's largest energy
Kₘₐₓ = 2E² / (mₑc² + 2E)
A soft photon gives little away, a hard one most of itself: 1.45 keV of 20 keV, but 796 keV of 1.00 MeV. In a detector this is the sharp top of the recoil continuum.
reached at θ = 180°; E and Kₘₐₓ in keV
Recoil direction of the electron
cot φ = (1 + E/mₑc²) tan(θ/2)
φ is always under 90°, so the electron is always knocked forward. Measuring one angle in an event fixes the other.
φ the electron's angle below the incident direction, θ the photon's angle above it; both in degrees or radians
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold θ at 90° and drag λ from 60 pm down to 1 pm: the spectrum never moves, because Δλ does not contain λ, but p′/p falls from 0.96 to 0.29 and the electron swings from 44° to 16° below the axis. Then take θ to 180° for the largest shift, 4.85 pm.
Live results
- SHIFT Δλ = λ′ − λ
- 2.426 pm
- PHOTON ENERGY E
- 124.0 keV
- ELECTRON KINETIC ENERGY K
- 24.21 keV
- RECOIL ANGLE φ
- 38.8 °
Catch the common trap
The mistake this lesson is built to prevent.
The trap is reaching for p = mv, concluding that a massless photon carries no momentum, and picturing the electron as simply soaking up some energy from a wave. p = mv, and its relativistic form γmv, is the momentum of a massive particle; the general statement is E² = (pc)² + (mc²)², and with m = 0 it gives p = E/c, not zero. Classical electromagnetism agrees, since light exerts measurable radiation pressure. The collision itself settles it: set the photon's momentum to zero and momentum conservation keeps the electron at rest, so it can take no energy and the predicted shift is zero, against 2.43 pm measured at 90°. Inventing a photon mass E/c² and writing p = mc lands on the right number, but the reason is the massless limit of the energy-momentum relation, not a hidden mass.
Test it: what changes?
- Shift Δλ stays the same
- Fractional shift Δλ/λ grows
- Recoil electron gains more kinetic energy
- Momentum ratio p′/p falls
- Shift grows from 0 to 2h/mₑc
- Scattered photon energy E′ falls
- Electron kinetic energy K rises
- Electron recoil angle φ shrinks towards 0°
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A 40.0 pm X-ray photon scatters off a free electron at rest and is detected at θ = 75.0°. Find the scattered wavelength, the photon's momentum before and after, and the recoil electron's kinetic energy, then check that energy against the momentum the electron must carry.
View solution →
- Shift: 1 − cos 75.0° = 0.7412, so Δλ = 2.426 pm × 0.7412 = 1.798 pm and λ′ = 40.0 pm + 1.798 pm = 41.80 pm, a 4.5% change.
- Momenta: p = h/λ = 6.626 × 10⁻³⁴ J s ÷ 4.00 × 10⁻¹¹ m = 1.657 × 10⁻²³ kg m s⁻¹, and p′ = h/λ′ = 1.585 × 10⁻²³ kg m s⁻¹.
- Energies, with hc = 1239.84 keV pm: E = 1239.84/40.0 = 30.996 keV and E′ = 1239.84/41.798 = 29.663 keV, so the electron takes K = E − E′ = 1.333 keV.
- Cross-check: pₑ² = p² + p′² − 2pp′ cos 75.0° gives pₑ = 1.974 × 10⁻²³ kg m s⁻¹, only 0.072 mₑc, so the Newtonian pₑ²/2mₑ = 1.335 keV should nearly agree. It does, to 0.1%, and that 0.1% is the relativistic correction.
Answer. λ′ = 41.80 pm; p falls from 1.657 × 10⁻²³ to 1.585 × 10⁻²³ kg m s⁻¹; K = 1.33 keV.
MediumExample 2
A spectrometer records the shifted line from a light-element target lit by 56.0 pm X-rays: λ′ = 56.34, 57.19, 58.45, 59.62 and 60.55 pm at θ = 30°, 60°, 90°, 120° and 150°. An unshifted line at 56.0 pm appears at every angle. Fit the data to find h/mₑc, then use it to obtain h.
View solution →
- The model predicts Δλ = (h/mₑc)x with x = 1 − cos θ, a line through the origin, since at θ = 0 there is no collision and no shift. The x values are 0.1340, 0.5000, 1.000, 1.500 and 1.866; the shifts are 0.34, 1.19, 2.45, 3.62 and 4.55 pm.
- A least-squares line through the origin has slope Σxy/Σx². Here Σx² = 7.000 and Σxy = 17.011 pm, so the slope is 2.430 pm.
- The residuals are +0.014, −0.025, +0.020, −0.025 and +0.015 pm with no trend, and a fit with a free intercept returns −0.0005 pm, consistent with zero. The scatter gives a slope uncertainty √[Σr² / ((N − 1)Σx²)] = 0.009 pm.
- h = slope × mₑc = 2.430 × 10⁻¹² m × 9.109 × 10⁻³¹ kg × 2.998 × 10⁸ m s⁻¹ = 6.636 × 10⁻³⁴ J s, with uncertainty 0.024 × 10⁻³⁴ J s. The accepted 6.626 × 10⁻³⁴ J s lies within half a standard uncertainty.
- The unshifted line comes from electrons bound too tightly to recoil alone. It stays at 56.0 pm and is the zero from which every shift is read.
Answer. h/mₑc = 2.430 ± 0.009 pm, so h = (6.64 ± 0.02) × 10⁻³⁴ J s.
HardExample 3
In an electron-tracking gamma detector, a 250 keV photon Compton-scatters and the recoil electron's track is measured at φ = 35.0° to the incident direction. Find the photon's scattering angle θ, its scattered energy and the electron's kinetic energy, and confirm that momentum balances along and across the beam.
View solution →
- Link the angles. Across the beam p′ sin θ = pₑ sin φ, and along it p = p′ cos θ + pₑ cos φ. Dividing, tan φ = p′ sin θ / (p − p′ cos θ) = sin θ / (p/p′ − cos θ).
- The shift in momentum form, 1/p′ − 1/p = (1 − cos θ)/(mₑc), gives p/p′ = 1 + ε(1 − cos θ) with ε = E/mₑc² = 250/511.0 = 0.4892. So p/p′ − cos θ = (1 + ε)(1 − cos θ); with 1 − cos θ = 2 sin²(θ/2) and sin θ = 2 sin(θ/2) cos(θ/2) this becomes cot φ = (1 + ε) tan(θ/2).
- Invert: tan(θ/2) = cot 35.0° / (1 + ε) = 1.4281/1.4892 = 0.9590, so θ/2 = 43.80° and θ = 87.6°.
- Energies: 1 − cos 87.6° = 0.9581, so E′ = 250 keV / (1 + 0.4892 × 0.9581) = 250/1.4687 = 170.2 keV and K = 250 − 170.2 = 79.8 keV. In wavelengths, λ = 4.959 pm and λ′ = 7.284 pm: a shift of 2.325 pm, which is λC(1 − cos θ) as it must be.
- Momentum: pₑc = √(K² + 2Kmₑc²) = √(79.8² + 2 × 79.8 × 511.0) keV = 296.5 keV. Across: 296.5 × sin 35.0° = 170.1 keV and 170.2 × sin 87.6° = 170.1 keV. Along: 170.2 × 0.0419 + 296.5 × 0.8192 = 7.1 + 242.9 = 250.0 keV. The electron moves at 0.50c, so the relativistic pₑ was needed.
Answer. θ = 87.6°, E′ = 170 keV and K = 79.8 keV; the transverse momenta agree at 170.1 keV/c.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
The shift depends only on angle: Δλ = λC(1 − cos θ).
Treat the event as a relativistic collision between a photon of momentum h/λ and a free electron at rest. The shift runs from zero straight ahead to 2h/mₑc in backscatter, so it shows only for X-rays and γ-rays. Tightly bound electrons give an unshifted line.