Rutherford Scattering

Why it matters

Before 1911, the leading picture of the atom was J. J. Thomson's "plum-pudding" model: electrons embedded in a smooth, positively charged sphere the size of the whole atom (~10⁻¹⁰ m). Rutherford scattering demolished it. By analysing how alpha particles ricocheted off gold, Rutherford showed the positive charge is not smeared out at all — it is packed into a nucleus roughly 10,000 to 100,000 times smaller than the atom. That single inference gave us the nuclear atom, the platform on which the Bohr model, quantum mechanics, and all of nuclear and particle physics were built.

The method matters as much as the result. Rutherford established the modern experimental strategy of learning structure from a scattering pattern: shoot known projectiles at a target, measure how many come out at each angle, and back out the force law and geometry. The same logic, scaled up in energy, later revealed quarks inside the proton (deep-inelastic scattering at SLAC) and remains the working principle of every collider and of Rutherford backscattering spectrometry (RBS) used in materials science today.

How it works, step by step

1. An alpha approaches a nucleus. An alpha particle (charge +2e, a helium-4 nucleus) heads toward a much heavier nucleus (gold, +79e). Both are positive, so the force is purely repulsive Coulomb: F = Z₁Z₂e²/(4πε₀r²), an inverse-square law identical in form to gravity but pushing apart.
2. The path bends into a hyperbola. An inverse-square repulsive force with positive total energy produces an unbound hyperbolic orbit (the repulsive, E > 0 branch of the Kepler problem). The nucleus sits at the focus outside the curve. The alpha swings past and flies off along the outgoing asymptote.
3. The impact parameter sets the angle. The perpendicular offset b between the incoming line and the nucleus fixes everything. A near-central approach (small b) yields a violent deflection; a distant fly-by (large b) yields a gentle one, via b = a·cot(θ/2).
4. Energy and angular momentum are conserved. The collision is elastic — the alpha keeps its speed, only its direction changes (assuming an effectively infinite-mass target). Conservation of energy gives the distance of closest approach; conservation of angular momentum, L = m·v·b, links b to the orbit.
5. Counting hits gives the cross section. Because b maps one-to-one onto θ, the number of alphas landing in an angular ring dΩ around θ is set by the annulus of impact parameters that funnel there. That ratio is the differential cross section, dσ/dΩ = (b/sinθ)|db/dθ|.

The governing result, the Rutherford differential cross section, is:

dσ/dΩ = ( Z₁ Z₂ e² / (16 π ε₀ E) )² · 1 / sin⁴(θ/2)

Symbol by symbol:

• dσ/dΩ — differential cross section, area per unit solid angle. Units: m²/sr (often quoted in barns/sr, 1 barn = 10⁻²⁸ m²).
• Z₁, Z₂ — atomic numbers (proton counts) of projectile and target. For an alpha on gold, Z₁ = 2, Z₂ = 79 (dimensionless).
• e — elementary charge, 1.602 × 10⁻¹⁹ C.
• ε₀ — vacuum permittivity, 8.854 × 10⁻¹² F/m (C²·N⁻¹·m⁻²).
• E — kinetic energy of the incoming particle in the lab (target at rest), in joules; often stated in MeV (1 MeV = 1.602 × 10⁻¹³ J).
• θ — scattering angle between incoming and outgoing directions, in radians (dimensionless in sin⁴).

The prefactor is often written a = Z₁Z₂e²/(8πε₀E), so that b = a·cot(θ/2) and dσ/dΩ = (a/2)²/sin⁴(θ/2). Note the strong dependences: the cross section grows as Z₂² (heavier nuclei scatter far more) and falls as 1/E² (faster particles are harder to deflect).

Distance of closest approach

For a head-on collision (b = 0), the alpha decelerates, stops, and reverses. At the turning point all kinetic energy has become Coulomb potential energy:

E = Z₁ Z₂ e² / (4 π ε₀ r_min)   ⟹   r_min = Z₁ Z₂ e² / (4 π ε₀ E)

For a 7.7 MeV alpha (the energy of alphas from ²¹⁴Po that Geiger and Marsden used) on gold (Z₂ = 79):

r_min = (2)(79)(1.602e-19)² / (4π · 8.854e-12 · 7.7·1.602e-13)
      ≈ 2.95 × 10⁻¹⁴ m  ≈ 30 fm

Because the pure-Coulomb formula kept fitting the data all the way down to this ~30 fm scale, Rutherford could bound the gold nucleus to be smaller than 30 femtometres — an atom is about 3,000 times larger still. (The actual gold nuclear radius is ~7 fm; the 7.7 MeV alpha never quite touches it, which is why the Coulomb law holds so cleanly.) For any non-head-on approach the turning point is farther out than r_min.

Worked example: from impact parameter to angle

Take a 5.0 MeV alpha on gold (Z₁Z₂ = 158). First the length scale:

a = Z₁Z₂e² / (8πε₀E)
  = 158 · (1.602e-19)² / (8π · 8.854e-12 · 5.0·1.602e-13)
  ≈ 2.27 × 10⁻¹⁴ m   (≈ 22.7 fm)

Now pick impact parameters and read off the deflection with θ = 2·arccot(b/a):

Impact parameter b	b/a	Scattering angle θ	Character
227 fm (10a)	10	≈ 11.4°	Distant fly-by, tiny nudge
45.4 fm (2a)	2	≈ 53.1°	Moderate deflection
22.7 fm (a)	1	90.0°	Right-angle turn
13.1 fm (0.577a)	0.577	120°	Backward-leaning
0 fm	0	180°	Head-on, straight back

The pattern is the crux: only alphas passing within a few tens of femtometres of the nucleus — an absurdly small target — get turned through large angles. Everything else sails past nearly undeflected, which is exactly what Geiger and Marsden saw.

How the count collapses with angle

Because of the 1/sin⁴(θ/2) factor, the yield falls precipitously as you move off the beam axis. Relative to 90°:

Scattering angle θ	sin⁴(θ/2)	dσ/dΩ relative to θ = 90°
5°	3.6 × 10⁻⁶	≈ 69,000 ×
15°	2.9 × 10⁻⁴	≈ 860 ×
45°	0.0214	≈ 11.7 ×
90°	0.25	1 (reference)
135°	0.729	≈ 0.34 ×
180°	1.0	≈ 0.25 ×

Small-angle events outnumber back-scatters by orders of magnitude — yet the curve never hits zero, so a steady trickle of large-angle events must appear. Geiger and Marsden's 1913 paper confirmed the sin⁴, Z₂², 1/E², and foil-thickness dependences over an enormous dynamic range, sealing the case for the point nucleus.

A little history

In 1909, working in Rutherford's Manchester lab, Hans Geiger and undergraduate Ernest Marsden aimed alpha particles at a thin gold film and looked, at Rutherford's suggestion, for large-angle deflections nobody expected. They found them. Rutherford later described his reaction: "It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Over 1910–1911 Rutherford worked out the scattering law for a point charge and published the nuclear atom in 1911. Geiger and Marsden then spent years verifying every prediction — the angular dependence, the material (Z²) dependence, the velocity dependence, and the linear scaling with foil thickness — in their definitive 1913 paper. It is one of the cleanest examples in physics of a bold model, a hard prediction, and a decisive measurement. Rutherford had already won the 1908 Nobel Prize in Chemistry (for the disintegration of elements); the gold-foil work is arguably his greater legacy.

Common misconceptions

• "Most alphas hit the nucleus." Almost none do. The nucleus is a vanishingly small target; the overwhelming majority of alphas pass far away and are barely deflected. Backscattering is rare precisely because it requires a near-central hit.
• "The alpha touches the nucleus and bounces off." In classical Rutherford scattering the alpha never reaches the nucleus — it is turned around at r_min by the long-range Coulomb field alone. Contact with the short-range strong force only matters at higher energies, and that is exactly where deviations from the formula reveal the nuclear size.
• "The orbit is a parabola or an ellipse." Bound orbits (E < 0) are ellipses and the marginal case (E = 0) is a parabola. A scattering alpha has E > 0, so its path is always a hyperbola with the nucleus at the external focus.
• "Rutherford discovered the electron / the proton." Thomson discovered the electron (1897). Rutherford discovered the nucleus (1911) and later named and identified the proton (1917–1919). The gold-foil experiment is specifically about the nucleus.
• "You must use temperature or Celsius somewhere." Rutherford scattering is a mechanics problem — energies (E in joules or MeV) and lengths, no thermodynamics. Keep E consistent: convert MeV to joules (×1.602 × 10⁻¹³) before plugging into SI formulas.
• "Quantum mechanics changes the answer." For a pure Coulomb (1/r) potential, the first-Born quantum calculation gives exactly the same dσ/dΩ as the classical one — a famous coincidence. Quantum effects only enter through identical-particle interference, screening, or the strong force.