Compton Scattering

Equivalent forms

\lambda' - \lambda = \lambda_C (1 - \cos\theta)

\lambda_C = h/(m_e c) \approx 2.426 \times 10^{-12}\,\text{m}

The shift depends only on geometry — not on the photon's initial wavelength.

Symbol	Name	SI unit	Dimension	Range
$\Delta\lambda$	Wavelength shiftoutput Increase in photon wavelength after scattering	m	L	0 – 5e-12
$\theta$	Scattering angle Angle between incoming and outgoing photon	rad	1	0 – 3.14159265
$h$	Planck constant Quantum of action	J·s	M·L²·T⁻¹	6.62607015e-34 – 6.62607015e-34
$m_e$	Electron mass Rest mass of the electron	kg	M	9.1093837015e-31 – 9.1093837015e-31
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Imperial
$\Delta\lambda$	m	cm	in
$\theta$	rad	rad	deg
$h$	J·s	erg·s	ft·lbf·s
$m_e$	kg	g	lb
$c$	m/s	cm/s	ft/s

Where it holds

Valid for elastic scattering of photons off free or loosely bound electrons. Breaks down when binding energy is comparable to photon energy (then bound-state corrections matter). For high energies, QED corrections (Klein–Nishina formula) modify the cross section, but the kinematic shift remains exact.

Dimensional analysis

$[h/(m_e c)] = (M\cdot L^{2}\cdot T^{-1})/(M \cdot L\cdot T^{-1}) = L \checkmark$ (length)

Discovery

Arthur Holly Compton · 1923

Compton scattered X-rays off graphite and measured a wavelength shift that classical electromagnetism could not explain. Treating light as discrete photons with momentum h/λ reproduced his data exactly. He won the 1927 Nobel Prize — definitive proof of particle behavior of light.

Try this

Why does an X-ray change color when it bounces off an electron?

A high-energy photon hits a free electron and scatters at angle θ. The scattered photon has a longer wavelength than the original. By how much?

Research status: stable

Real-world applications

Medical imaging — Compton cameras for gamma-ray localization in PET and SPECT.
Gamma-ray astronomy — INTEGRAL and Fermi telescopes track photon paths via Compton kinematics.
Radiation shielding — dominant photon-attenuation process in tissue at 100 keV–10 MeV.
Material characterization — Compton profile reveals electron momentum distribution in solids.

Common misconceptions

The shift depends on incoming wavelength — it does not; only $on \theta$ .
The electron must be at rest — actually it must be effectively free; thermal motion adds Doppler broadening but the mean shift is unchanged.
Compton scattering occurs only for X-rays — it occurs at all photon energies, but the relative shift $\Delta \lambda /\lambda is$ significant only when $\lambda \sim \lambda _C$ .

Experimental verification

Compton's 1923 graphite experiment measured

\Delta \lambda at

multiple angles using a Bragg spectrometer; values matched the formula to <2%. Replicated daily in X-ray scattering labs and used in gamma-ray astronomy detectors.

Derivation

Apply 4-momentum conservation:

p_\gamma + p_e = p_\gamma ' + p_e

'.

Squaring

(p_e')^{2} = (p_\gamma + p_e - p_\gamma ')^{2} and

using massless photons

(p_\gamma ^{2} = 0)

and

m_e c^{2} for

the electron, one isolates the dot products.

Solving

for \lambda

' yields

\lambda ' - \lambda = (h/m_e c)(1 - \cos \theta )

.

Limiting cases

\theta \to 0

⟶

\Delta \lambda \to 0

Forward scattering — the photon barely interacts; no energy transfer.

\theta \to \pi

(backscatter)⟶

\Delta \lambda \to 2\lambda _C \approx 4.85\,\mathrm{pm}

Maximum shift: the photon reverses direction, dumping maximum momentum into the electron.

Target mass

m \to \infty

⟶

\Delta \lambda \to 0

Heavy targets recoil negligibly; the photon scatters elastically (Thomson limit).

What if…

What if we scattered photons off protons instead of electrons?

The Compton wavelength of a proton $is \sim 1836\times$ smaller. The shift drops $to \sim 1.3\,\mathrm{fm}$ — essentially unmeasurable for X-rays, but relevant for high-energy gamma rays.

What if h were

10\times

larger?

$\lambda _C$ would also be $10\times$ larger $(\sim 24\,\mathrm{pm})$ . Compton shifts would be huge even for visible light — every reflection from a mirror would noticeably redshift the photon.

1

X-ray backscatter shift

Given ·

\theta:: 3.14159
h:: 6.62607015e-34
m e:: 9.1093837015e-31
c:: 299792458

Find · \Delta\lambda

Steps

$\lambda _C = h/(m_e c) = 6.626e-34 / (9.109e-31 \times 3e_{8}) \approx 2.426\,\mathrm{pm}$
$1 - \cos (\pi ) = 2$
$\Delta \lambda = 2 \times 2.426\,\mathrm{pm} \approx 4.85\,\mathrm{pm}$ — the maximum possible shift.

Result ·

\Delta \lambda = (h/m_e c)(1 - \cos \pi ) = 2\cdot \lambda _C \approx 4.85 \times 10^{-12} m

2

90° scatter of a 0.1 nm X-ray

Given ·

\theta:: 1.5707963
h:: 6.62607015e-34
m e:: 9.1093837015e-31
c:: 299792458

Find · \Delta\lambda and \lambda'

Steps

$1 - \cos (\pi /2) = 1$
$\Delta \lambda = h/(m_e c) \approx 2.426\,\mathrm{pm} = 0.00243\,\mathrm{nm}$
$\lambda ' = 0.1 + 0.00243 = 0.10243\,\mathrm{nm}$ — measurable shift, $\sim 2.4$ %.

Result ·

\Delta \lambda = \lambda _C \approx 2.43\,\mathrm{pm}

;

\lambda ' \approx 0.1024\,\mathrm{nm}

Mass-Energy Equivalence→De Broglie Wavelength→Planck Radiation Law→