(a) Establish the Rutherford's scattering cross section formula for α-particle by considering the standard assumptions and…

(a) Rutherford's scattering cross section for α-particles

Assumptions: the α-particle (charge ze) and the nucleus (charge Ze) interact only by the Coulomb force F = kzZe²/r², k = 1/4πε₀; the nucleus is a point charge, heavy enough to stay at rest; motion is non-relativistic; a thin foil gives single scattering; the α-particle does not penetrate the nucleus; atomic electrons are ignored. Let the α-particle have speed v and kinetic energy E = ½mv² at large distance, and let b be the impact parameter and θ the scattering angle. Angular momentum is conserved, mvb = mr²φ̇, so dt = r²dφ/(vb). Energy conservation means the speed after scattering is again v, so the momentum change has magnitude Δp = 2mv sin(θ/2), directed along the axis of symmetry of the hyperbolic path. The path is symmetric about the axis, and the asymptotes make angle φ₀ = (π − θ)/2 with it. The impulse along the axis is Δp = ∫F cosφ dt = (kzZe²/vb) ∫₋φ₀^φ₀ cosφ dφ = (2kzZe²/vb) sinφ₀ = (2kzZe²/vb) cos(θ/2). Equating the two expressions for Δp: b = (kzZe²/mv²) cot(θ/2) = (zZe²/8πε₀E) cot(θ/2). Particles with impact parameter between b and b + db scatter into angle θ to θ + dθ, so dσ = 2πb|db| = (dσ/dΩ) 2π sinθ dθ, and dσ/dΩ = (b/sinθ)|db/dθ|. With A = zZe²/8πε₀E, b = A cot(θ/2) and |db/dθ| = (A/2) csc²(θ/2): dσ/dΩ = [A cot(θ/2)/(2 sin(θ/2)cos(θ/2))] (A/2) csc²(θ/2) = (A²/4) csc⁴(θ/2). dσ/dΩ = (zZe²/16πε₀E)² / sin⁴(θ/2), and for z = 2: (Ze²/8πε₀E)² / sin⁴(θ/2). Check: the units are m² per steradian. The count of scattered α-particles varies as 1/sin⁴(θ/2), Z² and 1/E², as verified by Geiger and Marsden.

(b) Highest-occupied nucleon level in ⁵⁶Fe, cubical box model

For a cubical box of side L = 10⁻¹² cm = 10⁻¹⁴ m: E = E₀(n₁² + n₂² + n₃²), with n = 1, 2, 3, … and E₀ = π²ħ²/(2mL²). E₀ = π²(1.055×10⁻³⁴)²/(2 × 1.67×10⁻²⁷ × 10⁻²⁸) = 3.29×10⁻¹³ J = 2.05 MeV. ⁵⁶Fe has Z = 26 protons and N = 30 neutrons. By the Pauli principle each spatial level holds two nucleons of one kind (spin up and down), and protons and neutrons fill levels independently. The levels, in order of S = n₁² + n₂² + n₃², have these states and cumulative capacity per kind: S = 3: 1 state, 2; S = 6: 3 states, 8; S = 9: 3 states, 14; S = 11: 3 states, 20; S = 12: 1 state, 22; S = 14: 6 states, 34. Both 26 protons and 30 neutrons exceed 22 and fit within 34, so the highest occupied level has S = 14. E_max = 14E₀ = 14 × 2.05 = 28.7 MeV (4.6×10⁻¹² J). This treats nucleons as independent particles in a rigid box, so it gives the order of magnitude (a few tens of MeV, comparable to the Fermi energy).

(c) Nuclear forces and the meson theory of exchange forces

Nuclear forces are the strong forces between nucleons that bind the nucleus. Their features are: short range (about 1-2 fm); much stronger than the Coulomb force at that range; charge independence (n-n, p-p and n-p forces are nearly equal); dependence on spin and a non-central (tensor) part; saturation; and a repulsive core at about 0.5 fm. Yukawa's meson theory (1935): nucleons interact by exchanging virtual mesons. A nucleon emits a meson of mass m, violating energy conservation by ΔE ≈ mc² for a time Δt ≈ ħ/mc², and another nucleon absorbs it. The meson can travel at most cΔt, so the range is R ≈ ħ/(mc). For R ≈ 1.4 fm, mc² = ħc/R ≈ 140 MeV, matching the pion discovered in 1947. Charged pions give charge exchange: p → n + π⁺, n → p + π⁻. Neutral pions give p-p and n-n forces. The exchange gives the Yukawa potential V(r) = −g² e^(−r/R)/r, which falls off faster than the Coulomb 1/r. Pion exchange also gives the spin-dependent and tensor parts (the deuteron's quadrupole moment shows the tensor force). Heavier mesons (ρ, ω) account for the short-range repulsion and saturation.