Lesson 9 of 15

Rutherford Scattering

Rutherford Scattering

In 1911, Ernest Rutherford directed alpha particles at a thin gold foil and found that some scattered at very large angles — impossible if the atom were a diffuse "plum pudding." He deduced that the atom has a tiny, dense, positively charged nucleus.

Coulomb Differential Cross Section

For a particle of charge Z1eZ_1 e scattering off a nucleus of charge Z2eZ_2 e via the Coulomb force, the differential cross section is:

dσdΩ=(Z1Z2αc4Ekin)21sin4(θ/2)\frac{d\sigma}{d\Omega} = \left(\frac{Z_1 Z_2 \alpha \hbar c}{4 E_{\text{kin}}}\right)^2 \frac{1}{\sin^4(\theta/2)}

The combination αc=1.44×103\alpha \hbar c = 1.44 \times 10^{-3} GeV·fm is the Coulomb strength parameter. With energies in GeV, the result is in fm²/sr.

Key features:

  • Diverges as θ0\theta \to 0 (forward scattering, long-range Coulomb force)
  • Falls steeply with increasing angle (sin4\sin^{-4})
  • Scales as Z12Z22Z_1^2 Z_2^2 — heavy nuclei scatter much more strongly

Distance of Closest Approach

For a head-on collision, all kinetic energy converts to Coulomb potential energy at the turning point:

Ekin=Z1Z2e24πε0rminrmin=Z1Z2αcEkinE_{\text{kin}} = \frac{Z_1 Z_2 e^2}{4\pi\varepsilon_0 r_{\text{min}}} \quad \Rightarrow \quad r_{\text{min}} = \frac{Z_1 Z_2 \alpha \hbar c}{E_{\text{kin}}}

For 5 MeV alpha particles on gold: rmin45.5r_{\text{min}} \approx 45.5 fm — safely outside the nuclear radius (~7 fm), confirming pure Coulomb scattering.

Your Task

Implement two functions. All constants must be defined inside each function body.

  • rutherford_dsigma_dOmega(Z1, Z2, E_kin_GeV, theta_rad) — result in fm²/sr
  • closest_approach_fm(Z1, Z2, E_kin_GeV) — result in fm
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