(a) Does the nucleus possess magnetic moment ? Justify your answer. Define nuclear magneton (μN) and Bohr magneton (μB)…

(a) The nucleus possesses a magnetic moment because its charged and spinning constituents generate magnetic dipoles. Protons contribute through spin and orbital motion; neutrons, though neutral, have intrinsic magnetic moments due to their internal charged quark structure. In even-even nuclei paired nucleons cancel, but odd-A nuclei have an unpaired nucleon, giving a net moment that lies between the Schmidt limits. Experimentally, nuclear moments are revealed by the hyperfine structure of spectral lines and measured directly by Rabi's molecular-beam and NMR methods; the proton and neutron spin g-factors are 5.586 and −3.826. The nuclear magneton is μN = eℏ/(2mp), the natural unit for nuclear moments; the Bohr magneton is μB = eℏ/(2me), the natural atomic unit. Using e=1.602×10⁻¹⁹ C, ℏ=1.055×10⁻³⁴ J s, mp=1.673×10⁻²⁷ kg, me=9.109×10⁻³¹ kg, μN≈5.05×10⁻²⁷ J T⁻¹ and μB≈9.27×10⁻²⁴ J T⁻¹; μB/μN=mp/me≈1836. Nuclear moments are therefore of order μN, much smaller than atomic moments.

(b) (i) The semi-empirical mass formula for binding energy is B(A,Z)=a₁A−a₂A²ᐟ³−a₃Z²/A¹ᐟ³−a₄(A−2Z)²/A+δ, where the terms are volume, surface, Coulomb, asymmetry and pairing. The volume term reflects saturation, the surface term reduces binding of surface nucleons, Coulomb repulsion penalises protons, asymmetry penalises neutron-proton imbalance, and pairing favours even-even nuclei. Equivalently, M(A,Z)=Zm_H+(A−Z)m_n−B/c². For fixed A, δ is independent of Z, so the most stable Z maximises B. Differentiating the Z-dependent terms, −2a₃Z/A¹ᐟ³+4a₄(A−2Z)/A=0, so 2a₃Z/A¹ᐟ³=4a₄(A−2Z)/A. Solving gives Zmin=A/[2+a₃A²ᐟ³/(2a₄)]. With a₃=0.711 MeV and a₄=23.702 MeV, a₃/(2a₄)=0.0150, hence Zmin=A/(2+0.0150A²ᐟ³).

(ii) For ⁹Be+⁴He→¹²C+n, use neutral atomic masses. The left atoms contain 4+2=6 electrons and the right carbon atom contains 6, so electron masses cancel exactly; the neutron mass is used as given. Q=[9.015060+4.003874−12.003815−1.008986]u×931.5 MeV u⁻¹ =0.006133×931.5=5.71 MeV. The positive Q-value means the reaction is exothermic. This is the reaction in which Chadwick identified the neutron in 1932.

(c) The conservation laws to be checked are energy-momentum, electric charge, baryon number, lepton number (electron, muon and tau separately), angular momentum, and for strong/EM processes strangeness and isospin; weak interactions may change strangeness and violate parity. Energy-momentum can usually be satisfied by kinetic energies, so the decisive checks are the additive quantum numbers.

For π⁺ + n → K⁰ + K⁺: charge is +1 on both sides. Baryon number is not conserved as written: left B=0+1=1, right B=0+0=0. Strangeness also changes from 0 to +1+1=+2. Hence the reaction is forbidden; it cannot proceed by strong or EM interaction, and baryon-number conservation forbids it as written even for weak interaction.

For ν̄e + p → n + e⁻: charge is +1 on the left but −1 on the right, so electric charge is not conserved. Electron lepton number is −1 on the left and +1 on the right, giving ΔLe=+2. Baryon number is conserved, but the failure of charge and lepton number makes the reaction impossible. The allowed inverse beta process is ν̄e + p → n + e⁺.