Scattering Light from a Free Electron

Compton scattering

Teach me

Scattering of a photon by a free electronall arrows share one scale — the incoming arrow p is drawn one fixed length

added wavelength λ′ − λ against scattering angle Θ

Guided Inquiry

1 · The set-up

Here’s our set-up. A photon of wavelength λ arrives from the left and strikes a free electron at rest — the dot marked e⁻. One panel shows the incident, scattered and recoil arrows meeting that dot; the other shows the same three momenta closing into a triangle. You can turn the detector (Θ) and set λ yourself.

2 · What comes out?

Let’s start with the big question. The detector is back at Θ = 0, with λ = 0.071 nm. Once you swing it away from 0° to larger angles, how do you predict the scattered wavelength will compare to the wavelength that went in?

3 · Where is the shift largest?

Now let’s find where the effect is strongest. We’re still at λ = 0.071 nm, detector at Θ = 0. At which angle is the added wavelength largest?

4 · Change λ — what happens to the curve?

Time to change the light itself. The detector is at Θ = 90° and λ = 0.071 nm — read the added wavelength in the Readouts panel. Before moving λ up to 1 nm, what do you predict the curve in the inset will do?

5 · The same shift, seen by visible light

Now let’s stretch the wavelength a long way. The detector is at Θ = 180°: the shift is 6.8 % of λ = 0.071 nm. Before sliding λ up to 500 nm (visible light), what do you predict that percentage will read?

6 · The triangle at back-scatter

Now over to the momentum triangle. We’ve set λ to 0.005 nm, where the two photon arrows differ noticeably, and the detector to 60°. Before turning it on to 180°, what do you predict the momentum triangle will look like?

7 · Recoil from a proton

One last what-if. Stay at 180° and read the added wavelength (0.00485 nm). The target here is always an electron — so what do you predict this shift would be if the photon recoiled from a proton instead, whose mass is 1836× the electron’s?

8 · The one thing to notice

Let’s pull it all together. The added wavelength is set by the scattering angle and by the mass of whatever recoils — never by the light that went in; only its ratio to λ changes.

Listen

Controls

Scattering angle Θ90°

measured in the electron’s initial rest frame

Incident wavelength λ0.071 nm

logarithmic · 0.005 nm to 500 nm

Readouts

scattered λ′

—

added λ (shift)

—

shift as % of λ

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The Physics

Compton scattering · what this sim shows, the physics behind it, and every control

At a glance

This sim scatters one photon off one free electron at rest. You choose the wavelength λ and the detector angle Θ, and it shows the scattered wavelength and the electron's recoil. Compton measured this in 1923 as evidence that light carries momentum.

What you'll learn

How energy and momentum conservation fix the scattered wavelength λ′ and the recoil.
How the change in wavelength depends on Θ, on λ, and on the recoiling mass.
How to read the momentum triangle and the recoil angle φ.
How the change compares with λ itself, and which light shows the effect.

The sketch is not to scale. A photon hits an electron at rest; the photon leaves at angle Θ and the electron recoils at angle φ.

What's on screen

Photon arrows p and p′ (cyan, wavy): incoming and scattered photon, with Θ marked by an arc.
Electron e⁻ and recoil arrow pₑ (violet): the electron starts at the dot; pₑ is its momentum afterwards, at angle φ.
Two panels: the collision, and the momentum triangle (the same arrows head to tail), on one scale.
Inset graph (blue): added wavelength against Θ.
Readouts: scattered λ′, the added wavelength, and the added wavelength as a percentage of λ.

See itOpen the sim at its starting setting, Θ = 90° and λ = 0.071 nm. Watch for: the collision panel and the momentum triangle, each with arrows p, p′ and pₑ.

The controls

Press ⌖ on a row to see where that control sits in the sim.

Control	What it changes	Range
Scattering angle Θ	The detector angle: how far the scattered photon is turned from its original direction, measured in the electron's initial rest frame.	0°–180° opens at 90°
Incident wavelength λ	The wavelength of the photon that arrives. The slider is logarithmic.	0.005–500 nm opens at 0.071 nm
Readouts	The scattered wavelength λ′, the added wavelength (λ′ − λ), and the added wavelength as a percentage of λ, in nm and %.	panel
Inset graph	Plots the added wavelength against Θ, with a dot at your current angle.	0°–180°
↻ Reset	Returns Θ to 90° and λ to 0.071 nm, and restarts the mode you are in.	top bar

Fixed: the target is one free electron at rest, and h, c and mₑ take their SI values. The sim shows where the scattered photon's wavelength ends up, not how many photons reach each angle.

One photon, one free electron

The sim treats the light as a single photon and the target as a single electron that is free and at rest (the dot e⁻). A photon of wavelength λ carries energy E and momentum p:

A photon is massless, so its energy and momentum differ only by a factor c.

In the collision, total energy and total momentum are the same before and after:

The electron's energy is the relativistic one, √(pₑ²c² + mₑ²c⁴), so this holds at every λ the sim offers. At 0.005 nm the photon carries 248 keV, about half of mₑc².

Given λ and Θ, these are three equations (the energy, and the two momentum components in the plane) for three unknowns: λ′, the size of pₑ and its direction φ. That is why two sliders are enough to draw the whole event.

See itScatter a 0.005 nm photon through Θ = 90°. Watch for: p′ drawn shorter than p, and the second panel closing p, p′ and pₑ into a triangle.

The shift is set by the angle

Eliminating the electron's momentum and energy leaves one result for the added wavelength:

λ_C = h/mₑc is the Compton wavelength of the electron.

Never shorter. 1 − cos Θ is never negative, so λ′ ≥ λ. The electron leaves with kinetic energy it took from the photon, so the photon's energy hc/λ′ is lower.
Θ = 0°: Δλ = 0. The photon carries on undeflected and the electron is untouched, so pₑ = 0.
Θ = 90°: Δλ = λ_C = 0.002426 nm.
Θ = 180°: Δλ = 2λ_C = 0.004853 nm, the largest possible shift. The photon comes straight back.

The inset curve: Δλ rises smoothly with Θ, with no special angle in between.

See itTurn the detector to Θ = 0°. Watch for: Added λ reading 0.000000 nm, and the recoil arrow pₑ gone.

See itTurn the detector to Θ = 180°. Watch for: Added λ reaching 0.004853 nm, and the inset dot at the top of the curve.

The shift does not depend on λ

In the result above, the incident λ has cancelled: Δλ contains only Θ, h, c and the mass of the recoiling particle. Divided by λ it gives the fractional shift, and energy conservation gives the electron's kinetic energy:

The same law as a fraction of λ, and as the energy the electron leaves with.

Θ λ

These sliders move the sim too.

The nm figure follows Θ only. The % figure follows λ too, and is large only when λ is comparable to λ_C.
The mass in λ_C is the mass that recoils. For a free proton, h/m_pc = 1.32 × 10⁻⁶ nm, 1836 times smaller. The sim shows only the electron.

See itScatter through Θ = 90° at λ = 0.071 nm, then at about 7.1 nm. Watch for: Added λ reading 0.002426 nm in both runs, while its percentage of λ falls about a hundredfold.

The recoil closes the momentum triangle

Momentum conservation says p = p′ + pₑ as vectors, so the three arrows close into a triangle. The cosine rule gives the electron's momentum from the photon's:

p = h/λ comes from the incident λ; p′ = h/λ′ from the scattered one.

At Θ = 90°, cos Θ = 0 and pₑ² = p² + p′². Beyond 90° the last term adds to pₑ².
At Θ = 180° all three arrows lie on one line. p′ points backwards, so the lengths add: pₑ = p + p′.
For every Θ ≥ 90° the electron leaves with more momentum than the photon brought in.

See itBack-scatter a 0.005 nm photon, Θ = 180°. Watch for: in the triangle panel, p, p′ and pₑ on one line, with pₑ as long as p and p′ together.

X-rays make the shift visible

The Compton wavelength sets the whole size of the effect.

Because the shift is a fixed length, what matters is how large it is next to λ. At Θ = 180° it is 97% of λ at 0.005 nm, 6.8% at 0.071 nm, and 0.00097% at 500 nm, which is visible light.

The same Δλ at Θ = 180°, as a percentage of λ across the slider.

For X-rays the shifted and incident wavelengths can be told apart. For visible light they differ by about one part in 10⁵.

See itBack-scatter visible light, λ = 500 nm. Watch for: Added λ still reading 0.004853 nm, but only about 0.001% of λ.

See itBack-scatter an X-ray, λ = 0.071 nm. Watch for: the same Added λ of 0.004853 nm, now 6.8% of λ.

Where it matters

Compton's experiment (1923). X-rays of about 0.0709 nm (molybdenum Kα) scattered from graphite showed a second line at a longer wavelength, shifted by λ_C(1 − cos Θ). A classical wave would give no shift, so this supported the photon picture. Compton shared the 1927 Nobel Prize in Physics for it.
Medicine and detectors. In water and soft tissue, Compton scattering is the main way photons from about 30 keV to about 30 MeV lose energy. That range includes 140 keV nuclear imaging and radiotherapy beams.
Astrophysics. Hot electrons in galaxy clusters scatter microwave-background photons to higher energy (the Sunyaev–Zel'dovich effect).

Go deeper: what the sim leaves out

Bound electrons. Compton's data also show an unshifted line. For a tightly bound electron the whole atom recoils, so h/mc uses the atom's mass (about 2 × 10⁴ mₑ for carbon) and the shift is negligible.
How many photons. The formula gives λ′ at each Θ, not how often photons go there; that is the Klein–Nishina cross-section.
Inverse Compton. If a fast electron meets a soft photon, energy can pass to the photon. The formula above is for an electron at rest, as in the sim.

Summary

In one line. A photon scattered by a free electron at rest loses energy to it, and its wavelength grows by (h/mₑc)(1 − cos Θ), set by the angle and the recoiling mass, not by λ.

Key points

Δλ = λ_C(1 − cos Θ): 0 at 0°, λ_C = 0.002426 nm at 90°, 2λ_C = 0.004853 nm at 180°.
Δλ does not contain λ; only the percentage Δλ/λ changes with λ.
The shift is easy to see only when λ is close to λ_C: X-rays, not visible light.

Common mix-ups

"A longer wavelength is shifted more." The added length is the same; only its share of λ is smaller.
"A classical wave would be lengthened too." A classical electron re-radiates at the driving frequency, with no shift. Here the photon gives up energy.
"At 180° the electron's momentum is p − p′ in size." As vectors pₑ = p − p′, but p′ points backwards there, so the lengths add: pₑ = p + p′.

Things to try on your own

Set Θ = 180° and find the λ at which the shift is 1% of λ, then check it against 100 × 2λ_C.
At λ = 0.005 nm, find the angle where pₑ is as long as p (near 67°). The cosine rule says p′ = 2p cos Θ there.

Predict before reveal	Six gated predictions commit the student before the readouts and inset expose each result.
Target the misconception	Distractors encode the "shift scales with λ" and "binding matters" errors; feedback names each one.
Invariant made visible	One column holds still while another collapses — which is which is the thing to find.
Faithful geometry	All six arrows share one scale with \|p\| fixed; the triangle closes by the cosine rule, not by hand.
Honest limits	The degenerate figures at Θ = 0 and Θ = 180° are drawn as they stand.