§ 34. Rutherford's Gold Foil Experiment & Planetary Model of the Atom (Grade 9–11)
Hyperbolic trajectories in the Coulomb field of a heavy nucleus, impact parameter b, and scattering angle θ. The backscattering of fast alpha particles from a thin gold foil revealed that the entire positive charge and mass of the atom are concentrated in a tiny nucleus.
Theory Summary & Scientific Mechanism
Before 1911, J. J. Thomson’s “plum-pudding” model pictured positive charge spread diffusely across the entire atomic volume (Rₐtom sim 10⁻¹⁰ m), predicting that fast α-particles (q_α = +2e) would traverse thin gold foil with negligible deflection (θ 90^°!
Rutherford concluded that all positive charge +Ze and 99.97% of atomic mass reside in a tiny central atomic nucleus (Rₙuc sim 10⁻¹⁵ m). Repelled by the inverse-square Coulomb force F(r) = (1)/(4πε₀)(2Ze²)/(r²), an α-particle of kinetic energy E_α and impact parameter b follows a hyperbolic trajectory with scattering angle given by Rutherford’s formula: cot ≤ft((θ)/(2) ) = (4πε₀ E_α b)/(Z e²) = (2b)/(dₘin).
In a head-on collision (b → 0, θ → 180^°), kinetic energy converts entirely into electrostatic potential energy at the distance of closest approach dₘin = (2Ze²)/(4πε₀ E_α), establishing the planetary nuclear model of the atom.
Everyday Analogy
Imagine firing artillery shells at tissue paper and seeing one shell out of thousands bounce straight back at you! That was Rutherford’s shock when alpha particles rebounded from gold foil, proving that nearly all of an atom’s mass sits in a tiny dense nucleus!
Interactive Question & Derivation
How does the scattering angle θ of an α-particle depend on its impact parameter b (the perpendicular distance from the initial velocity line to the center of the target nucleus +Ze)?
- A: As impact parameter b decreases (b → 0), Coulomb repulsion grows stronger and scattering angle increases up to head-on backscattering (θ → 180^°)
- B: All α-particles scatter by the exact same small angle < 1^° regardless of b
- C: Scattering angle θ is larger for higher kinetic energy E_α at fixed b
Study Hint: How does the Coulomb scattering angle θ depend on the impact parameter b (how closely the alpha particle aims at the nucleus)?
Proof & Explanation: An α-particle (q_α = +2e) moving in the repulsive Coulomb field of a heavy nucleus (Q = +Ze) follows a hyperbola governed by Rutherford's relation cot (θ/2) = (2b)/(rₘin), where rₘin = (2Ze²)/(4πε₀ E_α) is the closest approach distance in a head-on collision (b = 0). When b is large, the particle passes far from the nucleus (θ ≈ 0^°); when b 90^°).
Key Equations & 3D Simulation Parameters
θ = 2arctan ≤ft((rₘin)/(2b) ), rₘin ≈ (2.88 Z)/(E_α [MeV]) fm
- Impact parameter b: 3 … 60 fm (default: 18 fm)
- α-particle kinetic energy E_α: 3 … 9 MeV (default: 5.5 MeV)
- Target nuclear charge Z: 20 … 92 e (default: 79 e)
- Beam spread width: 0.5 … 1.5 × (default: 1 ×)