(a) ¹⁴C has 6 protons and 8 neutrons. Is it neutron-rich, and how does it decay? (b) ²³⁸U (Z=92, N=146, N/Z≈1.59) is unstable. Why don't extra neutrons stabilize it? Solution: (a) N/Z = 8/6 ≈ 1.33 — high for such a light nucleus (stable ¹²C has N/Z = 1). It is neutron-rich and undergoes β⁻ decay: n → p + e⁻ + ν̄, becoming ¹⁴N (7p, 7n, N/Z = 1) — stable. (b) Beyond Z ≈ 82, Coulomb repulsion (∝ Z²) overwhelms even the extra neutrons' added attraction; no N/Z ratio can bind the nucleus, so it alpha-decays instead.