Simulation · Physics · Class 12
Alpha-particle scattering
From the lesson Alpha-particle trajectory in Atoms. Change the values and watch what happens.
The idea behind it
NCERT §12.2, §12.2.1
- The foil is so thin that an alpha-particle is taken to suffer at most one scattering, so it is enough to follow one alpha-particle past one nucleus.
- An alpha-particle is a helium nucleus, with charge 2e and the mass of a helium atom. A gold nucleus has charge Ze with Z = 79, and it is about 50 times heavier than the alpha-particle, so it is treated as stationary.
- The path follows from Newton's second law and Coulomb's law. The repulsive force at separation r is F = (1/4πε₀)(2e)(Ze)/r² (Eq. 12.1), directed along the line joining them; its size and direction change all the time as the particle approaches and recedes.
- The impact parameter b is the perpendicular distance of the alpha-particle's initial velocity from the centre of the nucleus. In a beam, the particles have nearly the same kinetic energy but a spread of impact parameters, so they scatter in different directions with different probabilities.
- Small b means a close pass and a large scattering angle θ. A head-on collision has the smallest b and the particle rebounds (θ ≈ π). A large b leaves it nearly undeviated (θ ≈ 0).
- Only a small fraction of alpha-particles rebound, so few collisions are head-on. That means the mass and positive charge are concentrated in a small volume, and Rutherford scattering gives an upper limit to the size of the nucleus.
- Distance of closest approach (head-on): all the kinetic energy becomes electric potential energy, K = (1/4πε₀)(2e)(Ze)/d, so d = 2Ze²/(4πε₀K).
- Example 12.2: for a 7.7 MeV alpha-particle (1.2 × 10⁻¹² J, the most energetic of natural origin), d = 3.84 × 10⁻¹⁶ Z m, and for gold (Z = 79) d = 3.0 × 10⁻¹⁴ m = 30 fm (1 fm = 10⁻¹⁵ m). The gold nucleus is therefore smaller than 30 fm; its actual radius is about 6 fm, so the alpha-particle turns back without touching it.
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