(a) Show that for a specific value (n, l), there exists a large degeneracy relative to the energy characterized by the quantum…

(a) For a 3D isotropic harmonic oscillator, H = −(ħ²/2m)∇² + ½mω²r². In spherical coordinates the energy eigenvalues are E_n,l = (2n + l + 3/2) ħω, n = 0,1,2,...; l = 0,1,2,... Define N = 2n + l. Then E_N = (N + 3/2) ħω. Thus the energy depends only on N, not separately on n and l. For a fixed N, l can be N, N−2, N−4,... with the same parity as N, and n = (N−l)/2. Hence many (n,l) states are degenerate.

For a given l, the magnetic degeneracy is (2l+1). Therefore the orbital degeneracy of level N is ∑_l=N,N−2,... (2l+1) = (N+1)(N+2)/2. Including spin s = 1/2 multiplies by 2, so D_N = (N+1)(N+2). The cumulative occupancy up to level N is S_N = ∑_N'=0^N (N'+1)(N'+2) = (N+1)(N+2)(N+3)/3. Evaluating:

• N = 0: S = 2
• N = 1: S = 8
• N = 2: S = 20
• N = 3: S = 40
• N = 4: S = 70
• N = 5: S = 112
• N = 6: S = 168

Thus the pure harmonic-oscillator shell closures are 2, 8, 20, 40, 70, 112, 168. With the spin-orbit interaction included, the empirical magic numbers become 2, 8, 20, 28, 50, 82, 126.

(b) Let atomic radius be r and assume hard uniform spheres touch along nearest-neighbour directions.

Simple cubic: Atoms per unit cell = 8 corners × 1/8 = 1. Touching along cube edge: a = 2r. Packing fraction f = [1 × (4/3)πr³]/a³ = [(4/3)πr³]/(8r³) = π/6 ≈ 0.524.

BCC: Atoms per unit cell = 8 corners × 1/8 + 1 body centre = 2. Touching along body diagonal: √3 a = 4r, so a = 4r/√3. Packing fraction f = [2 × (4/3)πr³]/[64r³/(3√3)] = π√3/8 ≈ 0.680.

FCC: Atoms per unit cell = 8 corners × 1/8 + 6 face centres × 1/2 = 4. Touching along face diagonal: √2 a = 4r, so a = 2√2 r. Packing fraction f = [4 × (4/3)πr³]/[(2√2 r)³] = π/(3√2) = π√2/6 ≈ 0.740.

Thus numbers per unit cell are 1, 2, 4 and packing fractions are π/6, π√3/8, π√2/6 for SC, BCC, FCC respectively.

(c)(i) The Weizsäcker semi-empirical mass formula is B(A,Z) = a_V A − a_S A^(2/3) − a_C Z(Z−1)/A^(1/3) − a_A (A−2Z)²/A + δ(A,Z), where δ = +a_P A^(−3/4) for even-even, δ = 0 for odd-A, δ = −a_P A^(−3/4) for odd-odd. Typical coefficients are a_V ≈ 15.75 MeV, a_S ≈ 17.8 MeV, a_C ≈ 0.711 MeV, a_A ≈ 23.7 MeV, a_P ≈ 34 MeV.

• Volume term a_V A: short-range nuclear attraction saturates; binding is proportional to A.
• Surface term −a_S A^(2/3): surface nucleons are less bound; area ∝ A^(2/3).
• Coulomb term −a_C Z(Z−1)/A^(1/3): proton-proton repulsion in a sphere of radius ∝ A^(1/3).
• Asymmetry term −a_A (A−2Z)²/A: Pauli principle favours N ≈ Z; neutron excess costs energy.
• Pairing term δ: even-even nuclei are extra stable, odd-odd less stable, odd-A intermediate.

(c)(ii) For α-decay: ₉₂²³⁸U → ₉₀²³⁴Th + ₂⁴He. Qα = B(234,90) + B(4,2) − B(238,92). Using the above SEMF coefficients: B(238,92) ≈ 1814.6 MeV, B(234,90) ≈ 1790.6 MeV, B(4,2) = 28.3 MeV (helium binding energy). Hence Qα ≈ 1790.6 + 28.3 − 1814.6 = +4.3 MeV > 0.

For β⁻-decay: ₉₂²³⁸U → ₉₃²³⁸Np + e⁻ + anti-ν. Qβ⁻ = 0.782 MeV + B(238,93) − B(238,92). Using SEMF, B(238,93) ≈ 1813.4 MeV, so Qβ⁻ ≈ 0.782 + 1813.4 − 1814.6 = −0.4 MeV < 0.

Thus β⁻-decay is energetically forbidden. Alpha decay is allowed and favoured because Z = 92 is very large: reducing Z by 2 lowers Coulomb repulsion strongly, and the daughter ₉₀²³⁴Th is even-even, gaining pairing stability. β⁻-decay would increase Z to 93, increase Coulomb energy, and produce odd-odd ₉₃²³⁸Np, losing pairing stability. Therefore ₉₂²³⁸U is an α-emitter and not a β⁻-emitter.