Rutherford Scattering Cross Section

Equivalent forms

dσ/dΩ ∝ 1/sin⁴(θ/2)

A formula whose 1/sin⁴(θ/2) divergence at small angles is so distinctive that any deviation revealed the strong force decades later.

Symbol	Name	SI unit	Dimension	Range
$dσ/dΩ$	Differential cross sectionoutput Scattering probability per unit solid angle	m²/sr	L²	0 – 1e-25
$θ$	Scattering angle Lab-frame scattering angle	rad	1	0.01 – 3.14
$E$	Kinetic energy Incident particle kinetic energy	J	M·L²·T⁻²	1e-15 – 1e-10
$Z_1$	Projectile charge number Atomic number of incident particle	—	1	1 – 100
$Z_2$	Target charge number Atomic number of target nucleus	—	1	1 – 110

Unit systems

Symbol	SI	CGS	Natural
$dσ/dΩ$	m²/sr	cm²/sr	1/MeV²
$θ$	rad	rad	rad
$E$	J	erg	MeV
$Z_1$	—	—	—
$Z_2$	—	—	—

Where it holds

Valid for non-relativistic, spinless, point-like Coulomb scattering. Fails when projectile energy approaches nuclear radius scale

(\sim 10\,\mathrm{MeV}

for medium nuclei), where strong force takes over.

Dimensional analysis

d\sigma /d\Omega \to [L^{2}]

(e^{2}/(\varepsilon _0\,\mathrm{E}))^{2} \times

(dimensionless

) \to [I^{2}T^{2}/((M^{-1}L^{-3}T^{4}I^{2})(ML^{2}T^{-2}))]^{2} = [L^{2}]^{2} / [L^{2}]

→ after squaring and unit reduction:

[L^{2}]

Both sides are area per steradian.

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Discovery

Ernest Rutherford · 1911

Geiger and Marsden's gold-foil experiment showed alpha particles occasionally backscattering — 'as if you had fired a 15-inch shell at tissue paper and it came back to hit you.' Rutherford derived this formula and concluded the atom must have a nuclear core.

Try this

Why does an alpha particle sometimes bounce straight back from a gold foil?

Compute the differential cross section for 5 MeV alpha particles scattered by gold (Z=79) at 60°.

Research status: stable

Real-world applications

Rutherford backscattering spectrometry (RBS) for thin-film analysis
Ion-beam analysis in materials science
Historical evidence of the nuclear atom (the experiment itself)
Calibration of nuclear physics detectors

Common misconceptions

The formula breaks down at very small angles (screening by electrons) and very high energies (nuclear structure).
It assumes a spinless point projectile; for electrons, Mott scattering adds a $1 - \beta ^{2}\sin ^{2}(\theta /2)$ correction.
Total cross section diverges — Coulomb scattering has no finite range.

Experimental verification

Geiger-Marsden 1909-1913 experiments scattered alpha particles off gold foil and verified the angular dependence quantitatively. Modern electron scattering experiments use deviations from Rutherford to map nuclear charge distributions.

Derivation

Hyperbolic orbit in Coulomb field has impact parameter

b = (Z_1\,\mathrm{Z}_2\,\mathrm{e}^{2})/(8\pi \varepsilon _0\,\mathrm{E}) \times \cot (\theta /2)

.

Differential cross section

d\sigma /d\Omega = (b/\sin \theta ) |db/d\theta

|.

Algebra gives the Rutherford formula.

The

1/\sin ^{4}(\theta /2)

is characteristic of inverse-square forces.

Limiting cases

\theta \to 0

⟶

d\sigma /d\Omega \to \infty

Coulomb force is long-range; forward scattering diverges (logarithmic divergence of total cross section).

\theta = \pi

⟶

d\sigma /d\Omega

minimumBackscattering rare but finite — Rutherford's famous backscatter signature.

E \to \infty

⟶

d\sigma /d\Omega \to 0

High-energy projectiles barely deflect — and eventually penetrate the nucleus, breaking the formula.

What if…

What if Z_2 doubled?

Cross section quadruples — heavy nuclei scatter much more strongly.

What if E doubled?

Cross section drops by factor 4 — faster particles scatter less.

1

5 MeV α on gold at 60°

Given ·

θ:: 1.047
E:: 8.01e-13
Z 1:: 2
Z 2:: 79

Find ·

d\sigma /d\Omega

Steps

Step 1: Compute $Z_1\,\mathrm{Z}_2\,\mathrm{e}^{2} = 158 \times 2.566e-38 = 4.05 \times 10^{-36} C^{2}$ .
Step 2: Denominator $16\pi \varepsilon _0\,\mathrm{E} = 16\pi \times 8.854e-12 \times 8.01e-13 = 3.566 \times 10^{-22}$ .
Step 3: Square the ratio: $(4.05e-36 / 3.566e-22)^{2} \approx 1.29 \times 10^{-28}$ .
Step 4: Divide by $\sin ^{4}(30^{\circ}) = (0.5)^{4} = 0.0625$ ; $d\sigma /d\Omega \approx 2.06 \times 10^{-27} m^{2}/sr$ .

Result · Prefactor

\approx (2\times 79\times (1.602e-19)^{2} / (16\pi \times 8.854e-12 \times 8.01e-13))^{2} \approx 1.3 \times 10^{-28} m^{2}/sr

mott scatteringcoulomb potential