University Physics IV · Photons and Matter Waves · 4.4
Recoil kinematics and the Compton edge
The shift tells you where the photon went. This lesson follows the electron: how much energy it takes at each angle, why that stops at a hard ceiling you can see in any gamma spectrum, and which electrons refuse to play along at all.
Build the model
Connect the measurement to the mechanism.
Once the shift λ′ − λ = (h/mₑc)(1 − cosθ) is in hand, the same three conservation equations that produced it still have more to say: they fix the recoil electron's energy and direction too. Four unknowns — the scattered photon's energy and angle, the electron's energy and angle — against energy conservation and two components of momentum leaves exactly one free parameter, so choosing θ determines the entire event. Sweep θ from 0 to 180° and the electron's energy sweeps continuously from zero up to a hard ceiling, Tₘₐₓ = 2E²/(mₑc² + 2E), the Compton edge, because a free electron cannot absorb a photon outright and some photon always escapes.
That continuum with its step is what a gamma spectrum actually shows you. The model costs two assumptions, and both fail visibly. It assumes the electron is free and at rest, so tightly bound electrons return an unmodified line at the incident wavelength — the whole atom recoils, mₑ becomes M, and the shift collapses below any linewidth — while the orbital momenta of loosely bound ones broaden the shifted line.
And it counts no photons: how many go where needs Klein–Nishina, which collapses to Thomson's energy-independent (1 + cos²θ) only when α = E/mₑc² is small.
- Simple definition
- The recoil kinetic energy T(θ) = Eα(1 − cosθ)/[1 + α(1 − cosθ)] is what the Compton ledger hands the electron at each photon scattering angle, and its ceiling at θ = 180° is the Compton edge.
- Example
- For the 662 keV caesium-137 line, α = E/mₑc² = 1.2955, so the edge sits at 2(662)²/(511 + 1324) = 477.7 keV and the backscattered photon leaves with the remaining 184.3 keV.
One measured angle fixes the whole event — the electron's share is not a free parameter you get to choose.
E, E′ and T in keV; α dimensionless; mₑc² = 511.0 keV. T + E′ = E exactly.
The step that ends the Compton continuum, and the photopeak-to-edge gap you can calibrate a detector against.
Both at θ = 180°. For the 1173 keV cobalt-60 line: 963.2 keV and 209.8 keV. Tₘₐₓ + E′ₘᵢₙ = E at every beam energy.
The electron always leaves into the forward hemisphere; φ reaches 90° only as θ → 0, where it carries no energy.
φ from the incident beam, coplanar with and on the opposite side from the scattered photon.
Replace mₑ by M and the shift vanishes into the linewidth: the unmodified peak is Compton scattering off a whole atom.
Carbon: M = 2.19 × 10⁴ mₑ, so the whole-atom shift can never exceed 2h/Mc = 2.2 × 10⁻⁴ pm at any angle.
Predicts the unmodified line dominating at small angles and dying at large ones — what Compton actually measured.
Carbon K shell, EB = 284 eV. Mo Kα at 17.44 keV gives T = 173 eV at 45° and 960 eV at 135°.
Marks the low-energy end: elastic, front-back symmetric, and independent of photon energy.
rₑ = 2.818 fm. Valid only for α ≪ 1; Klein–Nishina reduces to this as α → 0.
Three conservation laws, four unknowns, one free angle
After the collision there are four numbers to find: the scattered photon's energy E′ and direction θ, and the electron's kinetic energy T and direction φ. Conservation supplies three equations — energy, momentum along the beam, momentum across it — so one unknown survives, and the geometry of the event chooses it. Fix θ and everything else is forced. Energy conservation gives T = E − E′ outright, and the shift derivation already gave E′ = E/[1 + α(1 − cosθ)] with α = E/mₑc², so T = Eα(1 − cosθ)/[1 + α(1 − cosθ)]. For the 662 keV caesium line α = 1.2955; at θ = 90° the bracket is 1 + 1.2955 = 2.2955, so E′ = 662/2.2955 = 288.4 keV and the electron walks off with T = 373.6 keV, well over half the beam. Notice what the formula does at θ = 0: T = 0. There is no transfer without deflection, since a photon that keeps its direction keeps its energy, and that is why the electron continuum starts at zero rather than at some minimum.
Where the electron goes
The two momentum components carry the direction. Across the beam, pₑ sinφ = (E′/c) sinθ; along it, pₑ cosφ = E/c − (E′/c) cosθ. Divide the first by the second, substitute E′, and the half-angle identities collapse everything to one line: cot φ = (1 + α) tan(θ/2). Read its ends. As θ → 0, tan(θ/2) → 0 and φ → 90°: a barely deflected photon nudges an electron sideways with essentially no energy. As θ → 180°, cot φ → ∞ and φ → 0: the backscatter case throws the electron straight down the beam. In between φ falls monotonically, so the recoil electron is always emitted into the forward hemisphere, 0 ≤ φ < 90°, however hard it is struck — there is no backward recoil to go looking for. The 662 keV beam at θ = 90° gives cot φ = 2.2955 and φ = 23.5°: the photon has turned a right angle while the electron stays close to the beam axis. Raising the beam energy tips every recoil further forward, because the whole angular map is compressed by that single factor (1 + α).
The edge, and why a spectrum shows it as a step
T(θ) rises monotonically with θ, so the electron energies form a continuum from zero up to a ceiling reached at θ = 180°, where 1 − cosθ = 2: Tₘₐₓ = 2αE/(1 + 2α) = 2E²/(mₑc² + 2E). That is the Compton edge — 1117.6 keV for the 1332 keV cobalt line. A detector that stops the electron while the scattered photon escapes records precisely this continuum, so the spectrum shows a broad shelf ending in a step, with the full-energy photopeak stranded above it. The gap between shelf and peak is what the escaping photon removed, E′ₘᵢₙ = mₑc²E/(mₑc² + 2E), 214.4 keV for that line, climbing towards mₑc²/2 = 255.5 keV as the beam hardens but never arriving. The step is abrupt because counts pile up against it: dT/dθ = Eα sinθ/[1 + α(1 − cosθ)]² vanishes at θ = 180°, so every scattering angle beyond about 160° deposits within a few keV of Tₘₐₓ. What rounds the edge off in a real spectrum is detector resolution, not kinematics.
The electrons that refuse: the unmodified line
Every step above assumed a free electron at rest, and atomic electrons are neither. The free-electron branch opens only if the collision can pay the binding energy EB: where T(θ) < EB the electron cannot be ejected, the recoil is taken up by the whole atom, and mₑ in the shift formula is replaced by the atomic mass M — tens of thousands of times larger even for carbon, so the shift is suppressed by that same factor and the light returns at the incident wavelength. That is Compton's unmodified line. Setting T(θ) = EB and solving for the angle gives the threshold in closed form, 1 − cosθ = EB mₑc²/[E(E − EB)], which names what controls it: the harder the beam and the looser the electron, the smaller the angle beyond which that electron may as well be free. So the unmodified line is strongest near the forward direction, weakens as θ grows, and strengthens with atomic number, since a heavy atom holds a larger fraction of its electrons tightly. Treat the threshold as a sharp angle only inside this free-electron accounting; a real target hands over the coherent fraction gradually. The electrons that do come free are not at rest either, and the component of their orbital momentum along the momentum transfer Doppler-shifts each event, so the modified line arrives as a broadened Compton profile whose width measures the target's electron momentum distribution.
Three regimes, and where each one ends
The single parameter α = E/mₑc² decides which description is needed. For α ≪ 1 the largest fractional transfer is 2α/(1 + 2α) ≈ 2α — 7% for a 20 keV photon — so the scattering is elastic to that accuracy and the classical Thomson result holds: dσ/dΩ = (rₑ²/2)(1 + cos²θ), symmetric front to back, independent of energy, integrating to σT = (8π/3)rₑ² = 0.665 b per electron. Near α ≈ 1 both the kinematics and the cross-section change, and the angular distribution needs Klein–Nishina in full, dσ/dΩ = (rₑ²/2)(E′/E)²[E′/E + E/E′ − sin²θ], which returns the Thomson expression exactly when E′ = E. At 662 keV it leaves θ = 0 untouched, cuts the 90° value to 33% of its Thomson value and backscatter to 15%, and drops the total to 0.256 b, 38% of σT. For α ≫ 1 the cross-section decays roughly as (ln 2α)/α into a narrow forward cone. Two rival processes bracket the Compton window from outside: photoelectric absorption, scaling steeply as about Z⁴/E³, owns the low end, and pair production opens at 1.022 MeV. In water Compton scattering rules from 30 keV to 20 MeV; in lead the window narrows to roughly 0.6–4 MeV.
Change one variable at a time
Make the relationship visible.
Put θ at 180° and sweep E: the edge climbs from 7% of the beam at 20 keV to 89% at 2 MeV. The dashed first-order curve overstates the transfer by the factor 1 + 2α — only 8% at 20 keV but 3.6× at 662 keV, where the plot has to cut it off at T = E.
COMPTON EDGE Tₘₐₓ477.7 keV
TRANSFER T(θ)437.1 keV
SCATTERED E′(θ)224.9 keV
RECOIL ANGLE φ14.1 °
Live interpretationCOMPTON EDGE Tₘₐₓ: 477.7 keV. TRANSFER T(θ): 437.1 keV. SCATTERED E′(θ): 224.9 keV. RECOIL ANGLE φ: 14.1 °
Catch the common trap
Explain before calculating.
A NaI detector is exposed to a 1.25 MeV gamma line. Where does the Compton edge fall, and how far below the full-energy photopeak does it sit?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA caesium-137 source sends 662 keV gammas into a scintillator. Find the Compton edge, the energy the backscattered photon carries away, and check that the two account for the whole line.
- α = E/mₑc² = 662/511.0 = 1.2955.
- The transfer is largest at θ = 180°, where 1 − cosθ = 2: Tₘₐₓ = E × 2α/(1 + 2α) = 662 × 2.5910/3.5910 = 477.7 keV. The compact form 2E²/(mₑc² + 2E) = 876488/1835 returns the same 477.7 keV.
- The photon keeps the rest: E′ₘᵢₙ = 662 − 477.7 = 184.3 keV, which matches mₑc²E/(mₑc² + 2E) = 511 × 662/1835 = 184.3 keV.
- So the spectrum runs as a continuum from 0 to 478 keV, then a 184 keV gap, then the photopeak at 662 keV. A separate peak near 184 keV in the same spectrum is the backscatter peak — photons that turned around in the shielding and re-entered the detector.
AnswerTₘₐₓ = 477.7 keV and E′ₘᵢₙ = 184.3 keV. They sum to 662.0 keV, the whole line: nothing is lost, the missing 184.3 keV simply leaves as a photon.
MediumA 662 keV photon scatters off a free electron at rest and comes off at θ = 60°. Find the scattered photon's energy, the electron's kinetic energy, and the electron's emission angle, then verify the angle from transverse momentum.
- α = 662/511.0 = 1.2955, and 1 − cos 60° = 0.500, so α(1 − cosθ) = 0.6478.
- Scattered photon: E′ = E/[1 + 0.6478] = 662/1.6478 = 401.76 keV.
- Electron: T = E − E′ = 662 − 401.76 = 260.24 keV, which is 39% of the beam and well short of the 477.7 keV edge.
- Direction: cot φ = (1 + α) tan(θ/2) = 2.2955 × tan 30° = 2.2955 × 0.5774 = 1.3253, so tan φ = 0.7545 and φ = 37.0°.
- Check it from transverse momentum: pec = √(T² + 2Tmₑc²) = √(260.24² + 2 × 260.24 × 511.0) = 577.7 keV, and E′ sin θ = 401.76 × 0.8660 = 347.9 keV, so sin φ = 347.9/577.7 = 0.6023 and φ = 37.0° again — the same answer from a different pair of the three equations.
AnswerE′ = 401.8 keV, T = 260.2 keV, and the electron leaves at φ = 37.0° to the beam, into the forward hemisphere the cotangent rule promised.
HardMolybdenum Kα X-rays of wavelength 71.1 pm fall on graphite. Carbon's K electrons are bound by 284 eV and its valence electrons by about 11 eV. Find the largest energy the photon can give a free electron, the angle at which the transfer first exceeds the K-shell binding, and the shift the unmodified line really carries.
- E = hc/λ = 1239.84 eV nm / 0.0711 nm = 17.44 keV, so α = 17.44/511.0 = 0.0341 — deep in the Thomson regime.
- Head-on: Tₘₐₓ = 2E²/(mₑc² + 2E) = 2(17.44)²/(511.0 + 34.88) = 608.3/545.9 = 1.114 keV. Even at 180° the photon can shed only 6.4% of its energy.
- Set T(θ) equal to the K binding. With w = 1 − cosθ, T = E²w/(mₑc² + Ew) = 0.284 keV gives 304.15w = 0.284(511.0 + 17.44w), so 299.2w = 145.12, w = 0.4850, cosθ = 0.5150 and θ = 59.0°.
- Below 59° the two K electrons cannot be freed, so the whole atom recoils: mₑ → M = 2.19 × 10⁴ mₑ and the shift becomes (h/Mc)(1 − cosθ) = (2.426 pm / 2.19 × 10⁴) × 0.485 = 5.4 × 10⁻⁵ pm.
- Against the 71.1 pm line that is 7.6 × 10⁻⁷ of the wavelength, some three orders of magnitude below anything a Bragg spectrometer of the period could separate, so those electrons return light at the incident wavelength.
- The valence electrons, bound by only 11 eV, clear their own threshold at θ = 11.0°, so past 11° four of carbon's six electrons already act free, and past 59° all six do — beyond that angle the unmodified line has nothing left to scatter from, which is what Compton reported.
AnswerTₘₐₓ = 1.11 keV; the K-shell threshold sits at θ = 59.0°; below it the coherent shift is 5.4 × 10⁻⁵ pm, under one part per million of the line — an unmodified peak at the incident wavelength.