Rutherford Scattering

Equivalent forms

\frac{d\sigma}{d\Omega}\propto \csc^4(\theta/2)

A purely classical Coulomb calculation gives the exact quantum answer here — a rare coincidence that let Rutherford discover the nucleus without quantum mechanics.

Symbol	Name	SI unit	Dimension	Range
$d\sigma/d\Omega$	Differential cross-sectionoutput Scattered intensity per unit solid angle	m^2/sr	L^2	—
$\theta$	Scattering angle Deflection angle of the projectile	rad	1	0.1 – 3
$E$	Kinetic energy Incident particle kinetic energy	J	ML^2T^-2	1e-13 – 1e-12
$Z_1$	Projectile charge number Atomic number of the projectile	1	1	—
$Z_2$	Target charge number Atomic number of the target nucleus	1	1	—
$\varepsilon_0$	Vacuum permittivity Electric constant	F/m	M^-1L^-3T^4*I^2	—

Where it holds

Non-relativistic point charges, single Coulomb scattering. Breaks down at very small angles (screening by electrons) and very large energies where the alpha probes the nuclear force (deviations revealed nuclear size).

Dimensional analysis

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

per steradian

(Z^{2}e^{2}/\varepsilon _{0}E)^{2}

→ (energy

\cdot

length/energy

)^{2} = [L^2]

e^{2}/(4\pi \varepsilon _{0})

has units

J\cdot m

, so the bracket is a length; squared gives an area.

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Discovery

Ernest Rutherford (with Geiger & Marsden) · 1911

Geiger and Marsden's 1909 experiment counted rare large-angle deflections of alpha particles off gold foil. Rutherford spent 18 months puzzling over it before realizing in 1911 that only a concentrated central charge could do it. He derived the scattering law classically and estimated the nucleus is <10⁻¹⁴ m — ten thousand times smaller than the atom.

Research status: stable

Real-world applications

Rutherford Backscattering Spectrometry (RBS) for thin-film composition
Ion-beam analysis of materials and artworks
The conceptual basis of every particle-collider scattering experiment

Common misconceptions

Most alphas barely deflect — large-angle events are rare, roughly 1 in 8000
The formula is classical, yet agrees with the full quantum (Born) result for a Coulomb potential
Backscattering requires a near head-on approach, not a glancing hit

Experimental verification

Geiger–Marsden confirmed the angular, energy

(1/E^{2})

, and charge

(Z^{2})

dependence. Deviations at high energy later measured the finite nuclear radius

(\sim 1.2

A^{1/3} fm).

Derivation

Treat the alpha as a classical particle in the repulsive Coulomb field of the nucleus.

Conservation of energy and angular momentum give a hyperbolic orbit; relating the impact parameter b to the deflection angle

\theta

yields

b = (Z_{1}Z_{2}e^{2}/8\pi \varepsilon _{0}E)\cdot \cot (\theta /2)

.

The cross-section

d\sigma = 2\pi b|db

|, and

d\Omega = 2\pi \sin \theta d\theta

; dividing and simplifying with the half-angle identity gives the

\csc ^{4}(\theta /2)

law.

What if…

What if you double the alpha energy?

The cross-section falls by $4\times (\propto 1/E^{2})$ — faster projectiles come closer but scatter less at a given angle.

What if the nucleus had been spread out (plum pudding)?

The maximum deflection would be a fraction of a degree — large-angle backscattering would be impossible, which is exactly why it disproved the model.

1

Angular fall-off

Given ·

θ₁:: $30^{\circ}$
θ₂:: $90^{\circ}$

Find · ratio of scattered intensities

Steps

Cross-section $\propto 1/\sin ^{4}(\theta /2)$
$\sin ^{4}(15^{\circ})=0.00448$ , $\sin ^{4}(45^{\circ})=0.25$
Intensity at $30^{\circ} is \sim 55\times$ that at $90^{\circ}$

Result · ratio

= \sin ^{4}(45^{\circ})/\sin ^{4}(15^{\circ}) \approx 0.25/0.0045 \approx 55

coulomb lawbohr modelimpact parameter