Compton Effect Calculator

The wavelength shift Δλ = (h/mₑc)(1 − cos θ), plus the scattered photon’s energy, the electron’s recoil energy and the angle it leaves at — and the point that catches everybody: the shift in nanometres does not depend on the incident wavelength at all, only on the angle.

Last updated 11 Nov 2026

Scatter a photon off a free electron at angle θ

θ → Δλ, λ′, E′, the electron’s recoil energy and its angle

The incident photon is given asThe two are the same statement: E = hc/λ, with hc = 1239.8419843 eV nm. A hard X-ray of 70.8 pm is 17.51 keV. Whichever you pick, the wavelength SHIFT comes out the same — that is the surprising part of this effect, and the second table below is about nothing else.

…and its value isThe default, 70.8 pm, is the molybdenum Kα line Compton himself used in 1923 — he wrote it as 0.708 ångströms. Try 2.5×10⁻³ keV in energy mode for green light and watch the fractional shift vanish into nothing, which is why the effect was never seen with visible light.

Scattering angle θ of the photonIn degrees, measured from the direction the photon was already going. 0° is no deflection and gives no shift at all; 90° is the classic right-angle measurement and gives a shift of exactly one Compton wavelength; 180° is backscatter and gives the largest shift there is, two Compton wavelengths.

Which value of h/mₑc to useAll three are the same quantity to the precision each was quoted at. The CODATA figure is 2.426 310 235 38(76)×10⁻¹² m, and it carries an uncertainty only because the electron mass does — h and c are exact. Compton printed 2h/mc = 0.0484 Å in his paper; the modern value of that is 0.048 526 Å, so he had it to three figures in 1923.

The collision, drawn at your own angle. The photon comes in from the left, leaves at θ, and the electron recoils below the axis at φ — all three at their true angles. The red arrow’s length grows with the share of the energy the electron takes, so you can watch it lengthen as you turn θ towards backscatter.

2.426310pmExample

Compton’s own 1923 case: the molybdenum Kα line at 0.708 Å (70.8 pm), scattered through 90° by graphite

One formula for the photon, two more for the electron it hit

Δλ = λ′ − λ = (h/m_ec)(1 − cos θ) · h/m_ec = 2.426 310 pm · E′ = E / [1 + α(1 − cos θ)], α = E/m_ec² · cot φ = (1 + α) tan(θ/2)

h/mₑc: the Compton wavelength of the electron, 2.426 310 235 38(76) pm. It is a property of the electron and of nothing else — not of the photon, not of the target material. This is the single fact that makes the formula so persuasive: a length built from h, m_e and c predicts an X-ray measurement
(1 − cos θ): runs from 0 at θ = 0 to 2 at θ = 180°, so the shift runs from zero to twice the Compton wavelength and never further. Often written 2sin²(θ/2), which is the same thing and is the form Compton used
Δλ: an ABSOLUTE shift in length, the same 2.43 pm at 90° whether the photon arrived with 1 keV or 10 MeV. What changes with energy is how big that shift is compared with the wavelength: 0.2 per cent for a 1 keV photon, 1957 per cent for a 10 MeV one
α = E/mₑc²: the photon energy measured in units of the electron rest energy, 510.999 keV. When α is small the electron barely recoils and E′ ≈ E; when α is large the photon loses most of its energy in one hit
φ: the angle the electron leaves at, on the other side of the original direction. It runs the other way from θ: forward scattering of the photon (small θ) throws the electron out sideways at nearly 90°, and backscatter (θ = 180°) drives it straight forward at φ = 0
free and at rest: the assumption underneath all three formulas, and false on both counts. Both approximations are excellent for an outer electron hit by an X-ray and poor for a tightly bound inner one, and the difference shows up in a real spectrum as an unshifted line beside the shifted one

Worked example

Compton’s own 1923 case: the molybdenum Kα line at 0.708 Å (70.8 pm), scattered through 90° by graphite
THE CONSTANT. h/m_ec = 6.626 070 15×10⁻³⁴ ÷ (9.1093837139×10⁻³¹ × 2.99792458×10⁸) = 2.426 310 pm = 0.024 263 Å. Compton calculated 0.0242 Å from the constants he had; the agreement to three figures is the measurement of h, m and c in disguise.
THE ANGLE FACTOR. cos 90° = 0, so (1 − cos θ) = 1 exactly. At this one angle the formula has no trigonometry left: Δλ = h/m_ec = 2.426 310 pm. That is the headline, and it is why 90° is the angle to measure at.
THE SCATTERED WAVELENGTH. λ′ = 70.8 + 2.4263 = 73.2263 pm, which is 0.732 26 Å. The shift is 3.427 per cent of the incident wavelength — small, but a 1923 crystal spectrometer could resolve it, and that is the entire reason the experiment was done with X-rays rather than light.
THE ENERGIES. E = 1239 841.98/70.8 = 17.5119 keV and E′ = 1239 841.98/73.2263 = 16.9316 keV, so the electron carries away 580.2 eV — 3.313 per cent of the photon’s energy. Notice that the energy share and the wavelength share are not the same number (3.313 against 3.427), because E goes as 1/λ.
THE ELECTRON’S DIRECTION. α = 17 511.9/510 999 = 0.034 27, so cot φ = 1.034 27 × tan 45° = 1.034 27 and φ = 44.03°. The photon went up at 90° and the electron went down at 44°, and momentum balances in both directions — which is the part of the theory that Compton could not test and Bothe and Geiger later could.
WHAT COMPTON ACTUALLY MEASURED, honestly. His paper gives the predicted shift at this geometry as 0.024 Å and the measured shift as 0.022 Å — eight per cent low. He reports it as agreement, and for 1923 it was: the classical theory predicted no shift at all, so a measured 0.022 against a predicted 0.024 settles the question even though the figures do not match to three places. The scanned copy read for this page prints the scattered wavelength itself with an unreadable digit, so that number is deliberately not reproduced here.
AND THE CHECK THAT COSTS MARKS. If a question gives you the incident wavelength in ångströms and you add 2.4263 to it, you are out by a factor of 100. The shift is 0.024 263 Å, not 2.43 Å. This page works in picometres throughout for that reason and prints the ångström value separately.

A 100 keV photon, swept through every angle

Computed on this page’s own expressions. Two things to read off it. Δλ is not proportional to θ: it goes as (1 − cos θ), so the first 15° buys almost nothing and the shift is already half its maximum by 90°. And φ runs backwards — the less the photon is deflected the more sideways the electron goes, and φ never exceeds 90°, because a photon coming at an electron cannot throw it backwards.
θ	Δλ	λ′	E′	electron’s energy	share taken	φ
0°	0 pm	12.3984 pm	100.000 keV	0 keV	0%	90° (undefined in practice)
15°	0.0827 pm	12.4811 pm	99.338 keV	0.662 keV	0.66%	81.05°
30°	0.3251 pm	12.7235 pm	97.445 keV	2.555 keV	2.56%	72.24°
45°	0.7107 pm	13.1091 pm	94.579 keV	5.421 keV	5.42%	63.65°
60°	1.2132 pm	13.6116 pm	91.087 keV	8.913 keV	8.91%	55.38°
90°	2.4263 pm	14.8247 pm	83.633 keV	16.367 keV	16.37%	39.91°
120°	3.6395 pm	16.0379 pm	77.307 keV	22.693 keV	22.69%	25.77°
180°	4.8526 pm	17.2510 pm	71.871 keV	28.129 keV	28.13%	0°

The same 90° scatter at eight different energies — the point of the whole page

The Δλ column is constant all the way down and every other column changes enormously. That is the whole content of the formula, and the thing students most often get backwards: the shift is a fixed length, and whether it matters depends on what you compare it with. For green light 2.43 pm is five parts in a million of the wavelength, which is why nobody found the effect with light; for a 10 MeV photon it is twenty times the wavelength, which is why megavoltage beams are dominated by Compton scattering. Note too that the fractional shift and the energy share differ, because E ∝ 1/λ.
Incident photon	λ	Δλ	Δλ/λ	E′	share the electron takes	where you meet it
2.5 eV	495,937 pm	2.4263 pm	0.0005%	2.500 eV	0.0005%	green light — the shift is 5 parts per million, undetectable
1 keV	1239.84 pm	2.4263 pm	0.196%	0.998 keV	0.19%	a soft X-ray; just about measurable
10 keV	123.984 pm	2.4263 pm	1.957%	9.808 keV	1.92%	a dental X-ray tube
17.51 keV	70.800 pm	2.4263 pm	3.427%	16.932 keV	3.31%	molybdenum Kα — Compton’s own line
100 keV	12.398 pm	2.4263 pm	19.570%	83.633 keV	16.37%	diagnostic radiography
662 keV	1.8729 pm	2.4263 pm	129.550%	288.390 keV	56.44%	the caesium-137 gamma line
1 MeV	1.2398 pm	2.4263 pm	195.695%	338.186 keV	66.18%	a radiotherapy beam
10 MeV	0.1240 pm	2.4263 pm	1956.951%	486.156 keV	95.14%	a linac; the photon loses almost everything in one hit

The shift is a length, not a percentage — and that is the whole lesson

What the experiment showed. Shine X-rays on a light element — Compton used graphite — and look at the scattered beam through a crystal spectrometer. Classical electromagnetism says the scattered radiation must have the same wavelength as the incident radiation: an electron driven at one frequency radiates at that frequency. Compton found a second line at a longer wavelength, and the extra length depended on the angle and on nothing else. Treat the X-ray as a particle with energy hν and momentum h/λ, let it collide elastically with a free electron, conserve energy and momentum relativistically, and Δλ = (h/m_ec)(1 − cos θ) falls out in four lines. That is why this is the experiment usually credited with establishing that light carries momentum in quanta.

Why there is no incident wavelength in the formula. The collision does not scale with the photon’s wavelength; it scales with the electron’s mass, through the length h/m_ec. A 1 keV photon and a 10 MeV photon scattered at the same angle both gain 2.43 pm. What differs is what that is worth: 0.2 per cent of the wavelength for the first, nearly 2000 per cent for the second. Nobody detected the effect with visible light not because light behaves differently but because 2.43 pm out of 500 nm is five parts in a million — and that is also the answer to why you cannot see it.

Where this goes wrong. Three places, in order of how often. The first is units: Δλ is 2.43 picometres, which is 0.0243 ångströms and 2.43×10⁻³ nm, and adding it to a wavelength quoted in ångströms without converting is out by a factor of 100. The second is degrees and radians: cos θ needs θ in radians in any calculator set to radian mode, and cos 90 in radian mode is −0.448, which gives a plausible-looking wrong answer rather than an obvious one. The third is computing the electron’s energy as hc/Δλ; it is E − E′, and those are two large numbers whose difference is small, so keep the figures.

What this page assumes, and what that costs. The electron is taken to be free and at rest, and neither is true. The consequence is visible in Compton’s own data and in every spectrum since: beside the shifted line there is an unshifted one, from photons that scattered off electrons too tightly bound to be knocked free, where the recoiling mass is effectively the whole atom. Put a carbon nucleus into h/mc instead of the electron and the shift becomes about 0.0001 pm — unmeasurable, which is what “unshifted” means. This page does not compute the relative intensity of the two lines; that needs the Klein–Nishina cross-section, which is not in scope here.

How it sits beside the other quantum experiments. The photoelectric effect gives the photon its energy, hν. Compton gives it its momentum, h/λ, and shows that the two obey relativistic collision mechanics. The de Broglie wavelength then turns the statement round and gives momentum to matter. Read in that order the three fit together as one argument, and the Compton shift is the step that makes the photon a particle with a direction rather than just a packet of energy.

Frequently asked questions

Why does a real spectrum show two lines?

Because some electrons are too tightly bound to be treated as free. When the photon scatters off a loosely bound outer electron, the electron is knocked clear, takes the recoil momentum, and the photon comes out at the shifted wavelength this page computes. When it scatters off a tightly bound inner electron, the recoil is taken by the whole atom, whose mass is tens of thousands of times larger. Since the shift goes as 1/m, the shift for an atom is some twenty thousand times smaller — of order 10⁻⁴ pm — and the line appears unmoved. Compton saw both lines in graphite in 1923, and the ratio of their intensities depends on the element, which is how you can tell light elements from heavy ones this way.

Can the wavelength ever get shorter?

Not by this formula, which always gives Δλ ≥ 0 because (1 − cos θ) is never negative. The reason is the assumption: the electron starts at rest, so it can only take energy away from the photon. If the electron is already moving fast towards the photon then energy can flow the other way and the photon gains energy — that is inverse Compton scattering, and it is how some astrophysical X-rays are made. This page does not model it, and a question that asks for a decrease in wavelength is asking about a different situation from the one this formula describes.

How much energy can the electron take at most?

At θ = 180°, where the photon is sent straight back. That maximum is called the Compton edge and it is E·2α/(1 + 2α), with α = E/m_ec². For a 100 keV photon it is 28.13 keV, or 28 per cent; for the 662 keV caesium-137 line it is 478 keV, or 72 per cent; for a 10 MeV photon it is 95 per cent. Set the angle to 180 on this page and the electron row gives it. Notice that it is never 100 per cent: the photon cannot be absorbed outright in a Compton collision, because momentum would not balance. That is the difference from the photoelectric effect, where the atom takes the recoil and the photon does vanish.

Is the Compton effect proof that light is a particle?

It is strong evidence that light carries momentum in discrete amounts of h/λ, which classical wave theory does not provide, and it was taken that way in 1923. Be careful about the word “proof”, though. The derivation treats the photon as a relativistic particle and gets the right answer, and the alternative classical calculation gets a definite wrong answer (no shift at all), so the experiment discriminates between them cleanly. What it does not do is make light stop being a wave: the same crystal spectrometer that measured the shift worked by diffraction, which is wave behaviour, in the same apparatus on the same day.

Related calculators

References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt. Every constant on this page comes from that file. It marks h = 6.626 070 15×10⁻³⁴ J Hz⁻¹, c = 299 792 458 m/s and e = 1.602 176 634×10⁻¹⁹ C as (exact), because the 2019 SI defines the kilogram, the metre and the ampere through them; the masses are measured, the electron being 9.109 383 7139(28)×10⁻³¹ kg.
A. H. Compton, “A Quantum Theory of the Scattering of X-Rays by Light Elements”, The Physical Review, Second Series, Washington University, Saint Louis; submitted December 1922. A scanned copy was read on 11 November 2026 at people.isy.liu.se/jalar/kurser/QF/references/Compton1923.pdf. Verified from it: the paper gives the shift as λ_θ − λ₀ = (2h/mc)sin²(½θ) = 0.0484 sin²(½θ) ångströms, states h/mc = 0.0242 Å, uses the molybdenum Kα line at 0.708 Å, scatters it from graphite at right angles, and reports 0.022 Å (experiment) against 0.024 Å (theory). What could not be read: the scan’s OCR renders the volume and page numbers and one digit of the scattered wavelength illegibly, so this page quotes the volume, page range and that wavelength nowhere. The shift figures above are from legible text.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019 (the copy read on 11 November 2026 is marked V4.01, June 2026). §2.2 fixes the numerical values of h, c and e, which is why those three carry no uncertainty here while every mass does. A school question that says “take h = 6.63×10⁻³⁴ J s” is now rounding an exact number rather than quoting a measured one.
The relative intensity of the shifted and unshifted lines, and the angular distribution of Compton scattering, are governed by the Klein–Nishina cross-section. No source for it was read for this build, so this page computes neither, and nothing here should be used to predict how much scattering happens at a given angle — only what the wavelength and energies are when it does. That is a narrower claim than a radiation-physics page would make, and it is deliberately the narrower one.