Compton scattering — wavelength shift derivation: Exercises

Below, $θ$ is the photon scattering angle (measured from its original direction), and $ϕ$ is the electron recoil angle. Read the figure like a map before you start: it is the single picture every problem on this page refers back to.

Intuition Read the collision figure (Figure s01)

Follow the pale-yellow arrow: it is the incoming photon, moving left-to-right along the $x$ -axis with momentum $p = h / λ$ . It slams into the blue dot — the electron, sitting still at the origin. After the hit, two new arrows leave the origin:

the pink arrow is the scattered photon, bent up by angle $θ$ , now with the longer wavelength $λ^{'}$ (so a smaller momentum $h / λ^{'}$ );
the blue arrow is the recoiling electron, kicked down by angle $ϕ$ with momentum $p_{e}$ .

The dashed grey line is the photon's original direction — the reference from which we measure $θ$ . Notice the photon bends up and the electron recoils down: their sideways ( $y$ ) kicks must cancel, because the electron started with zero $y$ -momentum. That cancellation is exactly the equation we solve in L4·Q2.

Figure — Compton scattering — wavelength shift derivation

Figure s01 — The Compton collision. Yellow = incoming photon, pink = scattered photon (angle $θ$ ), blue arrow = recoil electron (angle $ϕ$ ), blue dot = electron at rest, dashed line = photon's original axis.

Level 1 — Recognition

Recall Solution

$Δ λ = \frac{h}{m _{e} c} (1 - cos θ) .$

$Δ λ = λ^{'} - λ$ : the increase in wavelength (always $\geq 0$ ).
$h$ : Planck's constant. $m_{e}$ : electron rest mass. $c$ : speed of light.
$θ$ : the angle the scattered photon makes with its incoming direction.
$\frac{h}{m _{e} c} = λ_{C} \approx 2.43 pm = 2.43 \times 1 0^{- 12} m$ .

Recall Solution

The factor $(1 - cos θ)$ controls everything.

$θ = 0^{\circ}$ : $cos 0 = 1$ , so $1 - cos θ = 0$ → $Δ λ = 0$ (photon passes straight through, transfers no momentum).
$θ = 18 0^{\circ}$ : $cos 18 0^{\circ} = - 1$ , so $1 - cos θ = 2$ → maximum $Δ λ = 2 λ_{C} \approx 4.86 pm$ .

Level 2 — Application

Recall Solution

$cos 6 0^{\circ} = 0.5$ , so $1 - cos θ = 0.5$ . $Δ λ = λ_{C} (0.5) = 2.43 \times 0.5 = 1.215 \approx 1.22 pm .$ $λ^{'} = 20.0 + 1.215 = 21.215 \approx 21.2 pm .$

Recall Solution

$cos 12 0^{\circ} = - 0.5$ , so $1 - cos θ = 1.5$ . $Δ λ = 2.43 \times 1.5 = 3.645 \approx 3.65 pm .$ Note this is bigger than the $9 0^{\circ}$ shift ( $2.43$ pm) but smaller than the backscatter max ( $4.86$ pm), as it should be.

Recall Solution

Set $λ_{C} (1 - cos θ) = λ_{C}$ , so $1 - cos θ = 1$ , giving $cos θ = 0$ . $θ = 9 0^{\circ} .$ This is why $9 0^{\circ}$ is the textbook "clean" angle — the shift equals exactly one Compton wavelength.

Level 3 — Analysis

Recall Solution

At $9 0^{\circ}$ , $Δ λ = λ_{C} = 2.43$ pm for both (the shift is independent of $λ$ ).

X-ray: $\frac{Δ λ}{λ} = \frac{2.43}{50.0} = 0.0486 = 4.86%$ — easily measurable.
Visible: $λ = 500 nm = 5.00 \times 1 0^{5} pm$ , so $\frac{Δ λ}{λ} = \frac{2.43}{5.00 \times 1 0 ^{5}} = 4.86 \times 1 0^{- 6} = 4.86 \times 1 0^{- 4} %$ — utterly invisible. Conclusion: the shift is the same absolute size, but only matters relative to the wavelength. Short-wavelength X-rays make the fixed $2.43$ pm a big fraction, so the effect stands out.

Recall Solution

Energy conservation says the electron's kinetic energy equals whatever energy the photon lost. The photon's energy before is $E = \frac{h c}{λ}$ and after is $E^{'} = \frac{h c}{λ ^{'}}$ (each just the photon energy $h c / wavelength$ ). So the energy handed to the electron is

= hc\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right).$$ Since $\lambda' > \lambda$, the bracket is positive — the photon truly lost energy. First $\lambda' = 50.0 + 2.43 = 52.43$ pm. Then $$KE = 1.240\times10^{6}\left(\frac{1}{50.0} - \frac{1}{52.43}\right) = 1.240\times10^{6}(0.0200000 - 0.0190730)$$ $$= 1.240\times10^{6} \times 9.270\times10^{-4} \approx 1.149\times10^{3}\text{ eV} \approx 1.15\text{ keV}.$$

Level 4 — Synthesis

Recall Solution

Convert energy to wavelength: $λ = \frac{h c}{E} = \frac{1.240 \times 1 0 ^{6}}{1.00 \times 1 0 ^{5}} = 12.40$ pm. Shift at $9 0^{\circ}$ : $Δ λ = 2.43$ pm, so $λ^{'} = 12.40 + 2.43 = 14.83$ pm. Back to energy: $E^{'} = \frac{h c}{λ ^{'}} = \frac{1.240 \times 1 0 ^{6}}{14.83} \approx 8.36 \times 1 0^{4} eV \approx 83.6 keV .$ The electron therefore gained $100 - 83.6 = 16.4$ keV.

Recall Solution

First, the symbol: == $p_{e}$ is the magnitude of the recoil electron's momentum vector $p_{e}$ == — a single positive number (in the same units as $h / λ$ ), pointing along the blue arrow at angle $ϕ$ below the $x$ -axis in Figure s01.

Now build the two momentum equations by decomposing every arrow onto the axes. Momentum is a vector, so the $x$ -total and the $y$ -total are each conserved separately.

Before: only the incoming photon moves — momentum $h / λ$ purely along $+ x$ , and $0$ along $y$ (the electron is at rest).
After: the scattered photon has momentum $h / λ^{'}$ at angle $θ$ , so its components are $\frac{h}{λ ^{'}} cos θ$ (along $x$ ) and $\frac{h}{λ ^{'}} sin θ$ (along $+ y$ ). The electron has momentum $p_{e}$ at angle $- ϕ$ , so its components are $p_{e} cos ϕ$ (along $x$ ) and $- p_{e} sin ϕ$ (along $y$ , i.e. downward).

Setting before $=$ after in each direction and isolating the electron's terms:

\qquad \underbrace{p_e\sin\phi = \frac{h}{\lambda'}\sin\theta}_{y\text{-momentum}}.$$ The $y$ equation says the electron's downward kick exactly matches the photon's upward kick — the cancellation we spotted in the figure. Divide the second by the first (the common factor $h$ cancels; work in units of $1/\text{pm}$): $$\tan\phi = \frac{\dfrac{1}{\lambda'}\sin\theta}{\dfrac{1}{\lambda} - \dfrac{1}{\lambda'}\cos\theta}.$$ With $\theta=90^\circ$ ($\sin=1,\ \cos=0$): $$\tan\phi = \frac{1/\lambda'}{1/\lambda} = \frac{\lambda}{\lambda'} = \frac{12.40}{14.83} = 0.8362.$$ $$\phi = \arctan(0.8362) \approx 39.9^\circ.$$ The electron recoils forward-and-sideways, opposite the photon's sideways kick — exactly what conservation of the $y$-momentum demands.

Level 5 — Mastery

Recall Solution

The derivation never assumed the recoiling mass was the electron's — it just called it $m$ . So the same formula holds with the recoiling mass $M$ in place of $m_{e}$ : $Δ λ = \frac{h}{M c} (1 - cos θ) .$ To reuse the number we already know, factor the electron's Compton wavelength out. Since

= \frac{\lambda_C}{M/m_e},$$ a heavier mass simply *divides down* the electron shift by the mass ratio $M/m_e$. With $M/m_e = 22032$ and $(1-\cos 180^\circ)=2$: $$\Delta\lambda = \frac{2.43}{22032}\times 2 = 2.206\times10^{-4}\text{ pm} \approx 2.2\times10^{-4}\text{ pm}.$$ This is about $22000$ times smaller than the free-electron shift — experimentally indistinguishable from **zero**. So photons that bounce off tightly-bound electrons appear at the *original* wavelength: this is the ==unshifted peak== that sits alongside the shifted peak in real spectra.

Recall Solution

Fractional energy lost by the photon:

= \frac{\lambda'-\lambda}{\lambda'} = \frac{\Delta\lambda}{\lambda + \Delta\lambda}.$$ This grows with $\Delta\lambda$, which is largest at $\theta = 180^\circ$ ($\Delta\lambda = 2\lambda_C$). So backscatter gives the maximum energy transfer. For $\lambda = 2.43$ pm $= \lambda_C$: $$\Delta\lambda = 2\lambda_C = 4.86\text{ pm}, \qquad \lambda' = 2.43 + 4.86 = 7.29\text{ pm}.$$ $$\frac{\Delta E}{E} = \frac{4.86}{7.29} = 0.6667 = 66.7\%.$$ Two-thirds of this photon's energy is handed to the electron in a head-on backscatter.

Recall Solution

(a) $θ \to 0$ : $1 - cos θ \to 0$ so $Δ λ \to 0$ — a grazing (undeflected) photon transfers no momentum, keeps its wavelength. ✔ (b) $θ = 9 0^{\circ}$ : $1 - cos 9 0^{\circ} = 1$ , so $Δ λ = λ_{C} = 2.43$ pm exactly. ✔ (c) For all $θ \in [0^{\circ}, 18 0^{\circ}]$ , $cos θ \in [- 1, 1]$ , so $(1 - cos θ) \geq 0$ . Since $λ_{C} > 0$ , we get $Δ λ \geq 0$ : the scattered wavelength is never shorter than the incoming one. A photon can only lose energy to a resting electron, never gain it — that would violate energy conservation. ✔

Connections

Photon momentum and energy — every problem uses $E = h c / λ$ and $p = h / λ$ .
Conservation of energy and Conservation of momentum — the twin laws behind L4.
Relativistic energy-momentum relation — why the recoil electron obeys $E_{e}^{2} = (p_{e} c)^{2} + (m_{e} c^{2})^{2}$ .
X-ray production and Bremsstrahlung — where these X-rays come from before they scatter.
Parent: the full derivation.