Compton scattering — wavelength shift derivation: Worked examples

Before anything, pin down the numbers we reuse everywhere:

The scenario matrix

Every Compton problem is one of these cells. The worked examples below are tagged with the cell they hit.

Cell	Case class	What makes it different	Example
A	$θ = 0^{\circ}$ (degenerate, no deflection)	$1 - cos θ = 0 \Rightarrow Δ λ = 0$	Ex 1
B	$θ = 9 0^{\circ}$ (clean midpoint)	$cos 9 0^{\circ} = 0 \Rightarrow Δ λ = λ_{C}$	Ex 2
C	$θ = 18 0^{\circ}$ (backscatter, limiting max)	$1 - cos θ = 2 \Rightarrow Δ λ = 2 λ_{C}$	Ex 3
D	General angle $0 < θ < 9 0^{\circ}$	must actually compute $cos θ$	Ex 4
E	General angle $9 0^{\circ} < θ < 18 0^{\circ}$	$cos θ < 0$ , shift exceeds $λ_{C}$	Ex 5
F	Energy given to electron (recoil KE)	use energy conservation, not the shift alone	Ex 6
G	Fractional shift $Δ λ / λ$ (why X-rays not light)	same $Δ λ$ , wildly different fraction	Ex 7
H	Wrong-mass trap / unshifted peak	proton or whole atom → tiny shift	Ex 8
I	Inverse problem / exam twist (find $θ$ from $Δ λ$ )	solve $cos θ$ backwards, check range	Ex 9

Worked examples

Worked example Ex 1 — Cell A: the no-deflection degenerate case (

$θ = 0^{\circ}$ ) A 20 pm X-ray photon grazes an electron and continues straight ahead ( $θ = 0^{\circ}$ ). What is its new wavelength $λ^{'}$ ?

Forecast: Guess before reading — does the wavelength grow, shrink, or stay the same?

Write the formula: $Δ λ = λ_{C} (1 - cos θ)$ . Why this step? It is the only relation between angle and wavelength change; everything starts here.
Put $θ = 0^{\circ}$ , so $cos 0^{\circ} = 1$ : $Δ λ = λ_{C} (1 - 1) = 0$ . Why this step? At $θ = 0^{\circ}$ the photon didn't change direction, so it transferred no momentum — no momentum transfer means no energy loss, so no shift.
Therefore $λ^{'} = λ + 0 = 20 pm$ . Why this step? $λ^{'} = λ + Δ λ$ by definition of the shift.

Verify: A photon that misses entirely should be unchanged — $λ^{'} = λ$ is the physically correct "nothing happened" answer. Units: pm + pm = pm. ✓

Figure — Compton scattering — wavelength shift derivation

Worked example Ex 6 — Cell F: energy delivered to the recoil electron

The backscatter case from Ex 3 had $λ = 10.0$ pm $\to λ^{'} = 14.86$ pm. How much kinetic energy did the electron pick up?

Forecast: Guess — will it be a few eV (like the photoelectric effect) or many keV?

Energy conservation: the electron's KE equals the photon's energy loss, $K E = E_{γ} - E_{γ}^{'} = h c (\frac{1}{λ} - \frac{1}{λ ^{'}})$ . Why this step? The shift formula only gives wavelengths; to get energy we must return to Conservation of energy — the photon's lost energy goes entirely to the electron (it started at rest).
Use $h c = 1240$ eV·nm and convert pm to nm ( $10.0$ pm $= 0.0100$ nm, $14.86$ pm $= 0.01486$ nm): $K E = 1240 (\frac{1}{0.0100} - \frac{1}{0.01486}) eV = 1240 (100 - 67.294) eV .$ Why this step? $h c$ in eV·nm with wavelengths in nm makes the answer come out directly in eV — a units trick worth memorising.
$K E = 1240 \times 32.706 \approx 4.055 \times 1 0^{4} eV \approx 40.6 keV$ .

Verify: This is comparable to the photon's own $\sim 124$ keV energy, which makes sense: backscatter is the maximum-energy-transfer geometry. Far above photoelectric-scale eV energies — as expected for X-rays. ✓ (Compare with Photoelectric effect scales.)

Worked example Ex 7 — Cell G: why X-rays and not visible light

Compute the fractional shift $Δ λ / λ$ at $θ = 9 0^{\circ}$ for (a) an X-ray of $λ = 24.3$ pm and (b) green light of $λ = 500$ nm.

Forecast: The absolute shift $Δ λ$ is the same for both (2.43 pm). So why is Compton scattering an X-ray phenomenon?

Both have $Δ λ = λ_{C} = 2.43$ pm (Cell B, $θ = 9 0^{\circ}$ ). Why this step? The formula is blind to the incoming wavelength — the shift is identical.
X-ray fraction: $\frac{Δ λ}{λ} = \frac{2.43}{24.3} = 0.100 = 10%$ . Why this step? A 10% change is easy to resolve in a spectrometer.
Green-light fraction: $500 nm = 500000$ pm, so $\frac{2.43}{500000} = 4.86 \times 1 0^{- 6} \approx 0.000486%$ . Why this step? The same 2.43 pm is a vanishing sliver of a 500 nm wave — undetectable.

Verify: The absolute shifts are equal but the fractions differ by a factor $500000/24.3 \approx 20576$ . This is precisely why Compton used X-rays. ✓

Worked example Ex 8 — Cell H: the wrong-mass trap (proton / bound atom)

Repeat the $θ = 9 0^{\circ}$ case but let the photon scatter off a proton ( $m_{p} \approx 1836 m_{e}$ ) instead of a free electron. What is the shift?

Forecast: Will it still be 2.43 pm?

The Compton wavelength uses the recoiling particle's mass: $λ_{C, p} = \frac{h}{m _{p} c} = \frac{λ _{C}}{1836}$ . Why this step? Re-read the derivation — $m$ in $h / (m c)$ is the mass of whatever recoils. Using the wrong mass is the classic error the parent note warns about.
$Δ λ = λ_{C, p} (1 - cos 9 0^{\circ}) = \frac{2.43}{1836} pm \approx 1.32 \times 1 0^{- 3} pm .$ Why this step? Heavier target → smaller Compton wavelength → tinier shift.

Verify: The shift is about 1836 times smaller — essentially zero on an X-ray scale. This is exactly the origin of the unshifted peak: photons that hit tightly-bound electrons make the whole heavy atom recoil, so $m ≫ m_{e}$ and $Δ λ \approx 0$ . ✓

Worked example Ex 9 — Cell I: inverse problem / exam twist (find the angle)

An experiment measures a Compton shift of $Δ λ = 3.645$ pm. At what angle $θ$ did the photon scatter? Is such a shift even possible?

Forecast: Can you get $θ$ back out, and is there more than one answer?

Rearrange the formula for $cos θ$ : $Δ λ = λ_{C} (1 - cos θ) \Rightarrow cos θ = 1 - \frac{Δ λ}{λ _{C}} .$ Why this step? We invert the relation to solve for the unknown angle, isolating $cos θ$ first because that's the only place $θ$ appears.
$cos θ = 1 - \frac{3.645}{2.43} = 1 - 1.5 = - 0.5$ . Why this step? Plug in the measured shift and the known constant.
$θ = arccos (- 0.5) = 12 0^{\circ}$ . Why this step? $arccos$ answers "which angle has this cosine?" On $[0^{\circ}, 18 0^{\circ}]$ — the only physical range for a scattering angle — there is exactly one answer, so no ambiguity.

Verify: First, feasibility: $Δ λ / λ_{C} = 1.5 \leq 2$ , so it's within the allowed $[0, 2 λ_{C}]$ band — a valid measurement. Second, this is the reverse of Ex 5, which sent $12 0^{\circ} \to 3.645$ pm. Round trip closes. ✓

Recall Quick self-test across the matrix

Shift at $θ = 0^{\circ}$ ? ::: $0$ (Cell A — no deflection, no momentum transfer). Shift at $θ = 9 0^{\circ}$ ? ::: $λ_{C} = 2.43$ pm (Cell B). Maximum possible shift and where? ::: $2 λ_{C} = 4.86$ pm at $θ = 18 0^{\circ}$ (Cell C). Shift at $θ = 12 0^{\circ}$ for any $λ$ ? ::: $1.5 λ_{C} = 3.645$ pm (Cell E — obtuse, exceeds $λ_{C}$ ). If $Δ λ = 1.215$ pm, what angle? ::: $cos θ = 0.5 \Rightarrow θ = 6 0^{\circ}$ (Cell D/I). Why doesn't visible light show Compton scattering? ::: Same 2.43 pm shift is a negligible fraction of a 500 nm wave (Cell G). Photon off a proton at $9 0^{\circ}$ ? ::: $\approx 1.32 \times 1 0^{- 3}$ pm, i.e. ~1836× smaller (Cell H). Largest valid $Δ λ / λ_{C}$ ? ::: $2$ ; anything above is unphysical (Cell I).

Connections

Parent: Compton scattering derivation
Photon momentum and energy — supplies $p = h / λ$ and $E = h c / λ$ used in every example.
Conservation of energy and Conservation of momentum — the two laws behind the shift and the recoil-energy example.
Relativistic energy-momentum relation — why the electron recoil in Ex 6 is treated relativistically.
Photoelectric effect — contrast the eV energy scale there with the keV recoil here.
X-ray production and Bremsstrahlung — where the incoming X-rays come from.
de Broglie wavelength — the same $h / (m c)$ -style length scale in a different guise.