University Physics IV · Photons and Matter Waves · 4.3
The Compton shift, derived
Compton's 1923 result is the cleanest evidence that light carries momentum in lumps. Here you build it from scratch: three conservation equations, one unwanted variable to eliminate, and an answer that turns out to depend on the scattering angle and on nothing else — not the wavelength, not the target, not the beam.
Build the model
Connect the measurement to the mechanism.
Treat the X-ray as a particle with energy hc/λ and momentum h/λ, let it strike a free electron sitting at rest, and demand that relativistic energy and both components of momentum balance. That is three equations in four unknowns — the scattered wavelength, the electron's energy, its momentum and its recoil angle — and the invariant E² = (pc)² + (mc²)² supplies the fourth relation. Square and add the two momentum equations, square the energy equation, subtract, and the recoil angle disappears along with the electron's energy, leaving one line: λ′ − λ = (h/mₑ c)(1 − cos θ).
A single length survives the algebra, the Compton wavelength h/mₑ c = 2.4263 pm, and it is fixed by the mass of whatever recoils rather than by anything the experimenter chose. The cost of that clean result is the assumption that bought it. A free electron initially at rest is a fiction — real electrons are bound and moving — so the derivation is a limit, trustworthy only when the photon energy dwarfs the binding energy, and it stays silent about how often scattering happens at all.
Conservation fixes where the line lands, never how bright it is.
- Simple definition
- The Compton shift is the increase in a photon's wavelength when it scatters from a free electron at rest, equal to the Compton wavelength h/mₑ c multiplied by (1 − cos θ), with θ the photon's deflection and no dependence on the incident wavelength.
- Example
- A 71.10 pm molybdenum Kα X-ray scattered through 90° returns at 71.10 + 2.43 = 73.53 pm, a 3.4% shift; 500 nm green light scattered through the same 90° gains exactly the same amount and comes back at 500.0024 nm.
One line for the whole two-body ledger: hand it θ and it returns the shift, from 0 forward to 2h/mₑ c straight back. Δλ is never negative, so scattering can only redden the photon.
θ is the photon's deflection from the incident direction; λ and λ′ in the same unit of length.
The only length the derivation leaves behind, and it belongs to the recoiling body. Put a proton there instead and it shrinks by 1836, to 1.32 fm.
h = 6.626×10⁻³⁴ J s, mₑ = 9.109×10⁻³¹ kg, c = 2.998×10⁸ m s⁻¹
The only physics input. Everything after this point is algebra, which is why the result carries no adjustable constant.
Four unknowns λ′, Eₑ, pₑ, φ; the invariant Eₑ² = (pₑ c)² + (mₑ c²)² closes the system.
The electron's direction never reaches the answer, so the experiment needs to catch only the scattered photon.
Square and add the two momentum equations: cos²φ + sin²φ = 1 removes φ in a single step.
The form a detector that reads energies rather than wavelengths needs: a 511 keV photon deflected through 90° comes back at exactly half its energy, 255.5 keV.
mₑ c² = 511 keV. Identical content to the wavelength form, since E = hc/λ.
That ceiling is 6.8% of a 71 pm X-ray but 10 parts per million of 500 nm light, and an instrument weighs Δλ against λ, never against zero.
Ceiling reached at θ = 180°: 4.8526 pm added, whatever λ was.
What the wave picture predicts, and what Compton saw
Classical Thomson scattering has the incident electric field drive the electron into oscillation at the incident frequency, and an oscillating charge radiates at the frequency at which it oscillates. So the scattered light must return at exactly the incident wavelength, at every angle, from every target — the wave account has no free knob with which to change a colour. What Compton measured in 1923, sending 71.10 pm molybdenum Kα X-rays onto graphite, was a second line beside the unshifted one, displaced towards longer wavelengths, with the displacement growing as the detector swung from forward to backward angles and taking the same value whatever the scatterer was made of. No repair to the wave account produces a wavelength change without a moving source. A photon–electron collision does, with nothing left free to adjust.
Set up the ledger, and name the assumption
Give the photon energy E = hc/λ and momentum p = h/λ, read off the m = 0 branch of the energy–momentum invariant. Put one electron at rest at the origin and let it be free: no lattice, no binding, nothing else available to absorb momentum. Three quantities are then conserved. Energy: hc/λ + mₑ c² = hc/λ′ + Eₑ. Momentum along the beam: h/λ = (h/λ′)cos θ + pₑ cos φ. Momentum across it: 0 = (h/λ′)sin θ − pₑ sin φ. The electron has to be handled relativistically even for modest X-rays, so Eₑ and pₑ are tied together by Eₑ² = (pₑ c)² + (mₑ c²)². Four unknowns, four relations, and the physics stops here — the assumption of a free electron at rest is the entire physical content, and everything that follows is bookkeeping.
Eliminate the angle nobody measures
The recoil angle φ is in the equations but not in the question. Isolate the electron terms: pₑ cos φ = h/λ − (h/λ′)cos θ and pₑ sin φ = (h/λ′)sin θ. Square both and add. On the left cos²φ + sin²φ = 1 collapses everything to pₑ², and multiplying through by c² gives (pₑ c)² = E² + E′² − 2EE′cos θ. Now take the energy equation in the form Eₑ = E − E′ + mₑ c², square it, and use the invariant to replace Eₑ² − (mₑ c²)² by (pₑ c)²: that route gives (pₑ c)² = (E − E′)² + 2mₑ c²(E − E′). Set the two expressions for (pₑ c)² equal. The E² and E′² terms appear on both sides and cancel, and what survives is 2EE′(1 − cos θ) = 2mₑ c²(E − E′).
Divide by EE′ and the wavelength form appears
Divide 2EE′(1 − cos θ) = 2mₑ c²(E − E′) through by 2EE′ and the right-hand side becomes a difference of reciprocals: 1 − cos θ = mₑ c²(1/E′ − 1/E). Since E = hc/λ, the reciprocal 1/E is just λ/hc, so mₑ c²(λ′ − λ)/hc = 1 − cos θ, and λ′ − λ = (h/mₑ c)(1 − cos θ). Now read what has vanished. λ is gone from the right-hand side, so 71 pm and 500 nm shift by the same amount. The target material is gone, because only one mass entered and it was the electron's. The intensity never appeared at all. What remains is the angle and a single constant with the dimensions of length, h/mₑ c = 2.4263 pm, multiplied by (1 − cos θ), which runs from 0 forward, through 1 at 90°, to 2 straight back — so no Compton shift can ever exceed 4.8526 pm.
Absolute, not fractional: why it took X-rays
A spectrometer never measures Δλ on its own; it measures Δλ against the smallest separation λ/R it can resolve at that wavelength, so visibility is set by the ratio Δλ/λ. At 90° that ratio is 2.4263 pm divided by λ. For Compton's 71.10 pm line it is 3.4% — with a crystal spectrometer of resolving power 2000 that is 68 resolution elements, an unmissable displacement. For 500 nm green light the same absolute 2.4263 pm is 4.9 parts per million, so you would need R above 2 × 10⁵ merely to watch the line move, and at a photon energy of 2.5 eV no electron in the target is remotely free anyway. Every optical scattering experiment before 1923 was therefore consistent with no shift at all. The shift is universal; its detectability is not.
What the free-electron assumption costs
Nothing in the derivation is approximate — granted the assumption, the result is exact. The assumption is the weak point. A real target electron is bound by anything from a few eV to tens of keV and is moving inside the atom. Binding matters once it stops being negligible beside the photon energy: molybdenum Kα at 17.44 keV is sixty-one times carbon's 284 eV K-shell binding, so the free-electron ledger holds and the shifted line is sharp. Push the photon energy down towards the binding energy and the electron can no longer be treated as free; the recoiling body effectively becomes the whole atom, and since the shift goes as 1/m it collapses — that is the origin of the unshifted line, taken up in the next topic. The initial motion of the bound electrons smears the shifted line into a Compton profile. And conservation says nothing at all about how many photons scatter into a given angle: that needs the Klein–Nishina cross-section.
Change one variable at a time
Make the relationship visible.
Hold θ at 180°, where the shift hits its ceiling of 4.85 pm, then drag λ from 20 pm to 220 pm: the shift readout never moves, while the line below slides back towards the incident one, from 24% of a wavelength to 2%. That gap is why Compton needed X-rays.
SHIFT Δλ2.426 pm
SCATTERED λ′73.43 pm
Δλ / λ3.42 %
ENERGY LOST BY PHOTON3.30 %
Live interpretationSHIFT Δλ: 2.426 pm. SCATTERED λ′: 73.43 pm. Δλ / λ: 3.42 %. ENERGY LOST BY PHOTON: 3.30 %
Catch the common trap
Explain before calculating.
The same graphite target is illuminated first with 71.10 pm molybdenum Kα X-rays and then with 500 nm green light, and in each case the scattered light is examined at θ = 90°. What does the Compton formula predict?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA molybdenum Kα beam of wavelength 71.10 pm scatters from the loosely bound electrons in graphite, and a detector sits at θ = 60°. Find the wavelength shift, the scattered wavelength, and the shift as a percentage of the incident wavelength.
- The shift needs the angle and nothing else: Δλ = (h/mₑ c)(1 − cos θ), with h/mₑ c = 2.4263 pm.
- At θ = 60°, cos θ = 0.5000, so 1 − cos θ = 0.5000 and Δλ = 2.4263 × 0.5000 = 1.2132 pm.
- Scattering always lengthens the wavelength: λ′ = 71.10 + 1.21 = 72.31 pm.
- As a fraction, Δλ/λ = 1.2132/71.10 = 0.01706, or 1.71%. The incident wavelength was needed only for this last step — never for the shift itself.
AnswerΔλ = 1.213 pm, λ′ = 72.31 pm, a shift of 1.71% of the incident wavelength.
MediumIn a repeat of Compton's experiment the incident line is at 71.10 pm and the detector — at an angle the student forgot to record — shows the shifted line at 74.55 pm. Recover the scattering angle, then find the photon energy before and after the collision and the energy handed to the electron.
- Δλ = 74.55 − 71.10 = 3.45 pm.
- Invert the shift formula: 1 − cos θ = Δλ/(h/mₑ c) = 3.45/2.4263 = 1.4219, so cos θ = −0.4219 and θ = 115.0°. That 1 − cos θ exceeds 1 tells you at once the detector was behind the target.
- Photon energies from E = hc/λ with hc = 1239.84 eV nm: E = 1239.84/0.07110 = 17.44 keV and E′ = 1239.84/0.07455 = 16.63 keV.
- The electron takes the difference: T = 17.44 − 16.63 = 0.81 keV.
- Check against the energy form: E′ = 17.44/[1 + (17.44/511)(1.4219)] = 17.44/1.0485 = 16.63 keV, which agrees.
Answerθ = 115.0°; E = 17.44 keV and E′ = 16.63 keV, so the recoiling electron receives 0.81 keV — 4.6% of the incident energy.
HardA spectrometer separates two lines only when they differ by more than λ/R, with resolving power R = 2000. (a) What is the longest incident wavelength whose Compton shift such an instrument could still resolve? (b) Convert that wavelength to a photon energy. (c) Compton nevertheless worked at 71 pm, some 140 times shorter. Say why.
- Take the best case, backscattering at θ = 180°, where 1 − cos θ = 2 and the shift reaches its ceiling: Δλₘₐₓ = 2 × 2.4263 = 4.8526 pm.
- Resolvable means Δλ ≥ λ/R, so λ ≤ R Δλₘₐₓ = 2000 × 4.8526 pm = 9705 pm = 9.71 nm.
- That wavelength is a photon of E = 1239.84 eV nm ÷ 9.71 nm = 128 eV.
- But resolution is only one of two conditions. The derivation assumed a free electron, which requires the photon energy to dwarf the binding energy — carbon's K shell alone is 284 eV, more than the 128 eV on offer, so at 9.71 nm the ledger's central assumption has already collapsed.
- Binding, not the spectrometer, therefore sets the floor, and clearing it comfortably costs about two decades in energy above 128 eV. That is exactly the factor of 140 in wavelength between 9.71 nm and Compton's 71 pm.
Answerλ ≤ 9.7 nm, i.e. E ≥ 128 eV, satisfies the spectrometer alone; the free-electron assumption needs E ≫ 284 eV, so the measurement belongs in the hard X-ray band — where Compton put it.