(a) The total binding energies of ¹⁵₈O, ¹⁶₈O and ¹⁷₈O are 111·96 MeV, 127·62 MeV and 131·76 MeV respectively. Determine the…

(a)

Let B(¹⁵O)=111.96 MeV, B(¹⁶O)=127.62 MeV and B(¹⁷O)=131.76 MeV. The neutron separation energy is Sₙ(ᴬZ) = B(ᴬZ) − B(ᴬ⁻¹Z).

For ¹⁶O: Sₙ(¹⁶O) = 127.62 − 111.96 = 15.66 MeV. In ¹⁶O the 8th neutron occupies the 1p₁/₂ shell.

For ¹⁷O: Sₙ(¹⁷O) = 131.76 − 127.62 = 4.14 MeV. In ¹⁷O the extra neutron occupies the 1d₅/₂ shell.

Therefore the energy gap between the 1p₁/₂ and 1d₅/₂ neutron shells is ΔE = Sₙ(¹⁶O) − Sₙ(¹⁷O) = 15.66 − 4.14 = 11.52 MeV. Equivalently, ΔE = 2B(¹⁶O) − B(¹⁵O) − B(¹⁷O) = 11.52 MeV. This assumes single-particle shell energies dominate and residual interactions are small. Final answer: 11.52 MeV.

(b)

The basic assumption of the single-particle shell model is the independent-particle approximation. Each nucleon moves independently in an average central potential produced by all the other nucleons. The short-range nucleon-nucleon interaction is replaced by a smooth mean field. Nucleons occupy single-particle states labelled by n, l, j, mⱼ, obey the Pauli exclusion principle, and fill levels from the bottom upward. Closed shells occur at magic numbers, and residual interactions are treated as perturbations.

A three-dimensional spherical harmonic oscillator has potential V(r) = ½ mω²r² and energies E_N = (N + 3/2)ℏω, where N = 2n + l. For fixed N, states with different l are degenerate, e.g. 2s and 1d. The centrifugal term ℏ²l(l+1)/(2mr²) is already present in the radial equation, but in the pure oscillator its effect is compensated by the radial node number, so alone it does not lift the N degeneracy. In the nuclear shell model an additional l-dependent centrifugal/orbit term, often D l², is introduced. This makes the effective potential depend on l and shifts different l states within the same oscillator shell, thereby removing the l-degeneracy. With D < 0, higher-l states are pushed down (for example 1d below 2s), mimicking the flatter bottom of a realistic Woods–Saxon well.

The spin-orbit term is ΔV_ls = −V_ls(r) L·S. For a single nucleon, s = 1/2, so L·S = ½ℏ²[j(j+1) − l(l+1) − s(s+1)]. For j = l + 1/2: L·S = +lℏ²/2. For j = l − 1/2: L·S = −(l+1)ℏ²/2. Because the nuclear spin-orbit interaction is attractive, the j = l + 1/2 member is lowered relative to j = l − 1/2. Thus each l level splits into two j levels, removing the j-degeneracy; the splitting is proportional to (2l + 1)ℏ²/2, so it grows with l. For large l the lowered j = l + 1/2 level (1f₇/₂, 1g₉/₂, 1h₁₁/₂, ...) drops into the shell below, producing the magic numbers 28, 50, 82 and 126 in addition to 2, 8 and 20. Together, the l-dependent and spin-orbit terms remove the degeneracy of the spherical harmonic oscillator and produce the observed shell structure and magic numbers.

(c)

Leptons are spin-1/2 fermions that do not feel the strong interaction. They occur in three families: the first contains e⁻ and νₑ; the second contains μ⁻ and ν_μ; the third contains τ⁻ and ν_τ. Their antiparticles are e⁺, ν̄ₑ; μ⁺, ν̄_μ; τ⁺, ν̄_τ. Charged leptons have electric charge −1, and neutrinos are neutral.

Lepton number is assigned as L = +1 for leptons, L = −1 for antileptons, and L = 0 for non-leptons. Total lepton number is conserved. In the Standard Model, the separate family lepton numbers Lₑ, L_μ and L_τ are also conserved in ordinary weak interactions, except for small neutrino-oscillation effects.

(i) π⁻ → μ⁻ + ν̄_τ Initial state: L = 0, L_μ = 0, L_τ = 0. Final state: μ⁻ has L_μ = +1, while ν̄_τ has L_τ = −1. Total lepton number is +1 − 1 = 0, so total L is conserved. However, muon lepton number changes from 0 to +1 and tau lepton number changes from 0 to −1. Therefore this reaction violates separate muon and tau lepton-number conservation. Under the usual family lepton-number conservation, it is not possible as a normal weak decay. The allowed decay is π⁻ → μ⁻ + ν̄_μ. If only total lepton number were imposed, it would be allowed, but separate family conservation forbids it.

(ii) n → p⁺ + e⁻ + ν̄ₑ Initial state: neutron is not a lepton, so L = 0 and Lₑ = 0. Final state: e⁻ has Lₑ = +1, and ν̄ₑ has Lₑ = −1. Thus Lₑ = (+1) + (−1) = 0, and total L = 0. Both total and electron lepton number are conserved. This is the usual β⁻ decay of the neutron. Therefore this reaction is possible.