ISEGORIA / MATH ENCYCLOPEDIA
Nuclear physics: binding, decay and tunnelling
A nucleus is a drop of protons and neutrons held by a short-range force and pushed apart by the protons’ charge. Five terms of a liquid-drop formula account for almost all of its binding energy, explain why iron sits at the bottom of the energy valley and mark where nuclei stop existing. Unstable nuclei decay at random, yet chains of them settle into predictable equilibria. And alpha decay, whose half-lives range from microseconds to billions of years, is quantum tunnelling through a Coulomb barrier.
Before you begin: Quantum mechanics, exponential decay and electrostatics
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The liquid drop and the valley of stability
The binding energy B of a nucleus is the energy needed to pull it apart into free protons and neutrons. Treat the nucleus as a drop of incompressible nuclear fluid and B follows from five terms: a volume term, because each nucleon is bound only to its neighbours; a surface term, because nucleons on the surface have fewer neighbours; the Coulomb repulsion of the protons; an asymmetry term, because the Pauli principle makes unequal numbers of protons and neutrons cost energy; and a small pairing term. Together they draw the whole chart of nuclides: a narrow valley of stable nuclei, bending towards extra neutrons as the protons’ repulsion grows, bounded by drip lines. Drag across the chart to see any nucleus’s terms.
Worked example. For uranium‑238 the volume term gives 3748.5 MeV, the surface takes 683.6, the Coulomb repulsion of 92 protons 960.5, the excess of 54 neutrons 290.4, and pairing adds 0.7: \(B=1814.7\) MeV, against the measured 1801.7, a difference of 0.7%. Per nucleon that is 7.62 MeV, well below iron‑56’s 8.85 (measured 8.79), so splitting uranium into two medium nuclei releases about 1 MeV per nucleon, roughly 200 MeV per fission. The same formula gives uranium‑238 an alpha-decay energy \(Q_\alpha=4.34\) MeV; the measured value is 4.27.
Watch out. The liquid drop knows nothing of shells: nuclei with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons are bound more tightly than it predicts, by up to about 10 MeV, and those magic numbers explain why lead‑208 is so stable. It also fails for the lightest nuclei, where a few nucleons hardly form a drop (helium‑4 is bound far more strongly than the formula says). The coefficients are one common fit; others differ by a few per cent. The drip lines drawn here are where the formula’s last neutron or proton becomes unbound, and the real ones differ in detail.
Why do heavy stable nuclei have more neutrons than protons, when the asymmetry term favours equal numbers?
Because the Coulomb energy grows as \(Z^2/A^{1/3}\) while the asymmetry penalty grows only as \((N-Z)^2/A\). Minimising the sum over \(Z\) at fixed \(A\) gives \(Z_0=A/(2+0.0150A^{2/3})\) approximately, close to \(A/2\) for light nuclei but only about 0.40\(A\) at \(A=208\). Each proton swapped for a neutron saves more repulsion than the asymmetry costs, until the two rates balance.
2. Decay chains and radioactive equilibrium
Each unstable nucleus has a fixed probability per unit time of decaying, independent of its age, so a large sample decays exponentially even though no single atom’s moment can be predicted. When the product is itself radioactive, the amounts follow Bateman’s equations: the daughter is fed by the parent and drained by its own decay. Its amount rises, peaks, and then, if it is shorter-lived than its parent, settles into equilibrium, decaying in step with the parent at a fixed ratio of activities. Set the two half-lives, press Play to watch 900 atoms decay one at a time, and compare their counts with the exact curves.
Worked example. Molybdenum‑99 (half-life 65.94 h) decays to technetium‑99m (6.007 h), the most widely used isotope in medical imaging. Starting from pure molybdenum, the technetium peaks after \(\ln(\lambda_B/\lambda_A)/(\lambda_B-\lambda_A)=22.8\) h, when it is 7.2% of the original number of atoms, so hospitals milk their generators about once a day. From then on the technetium’s activity is \(\lambda_B/(\lambda_B-\lambda_A)=1.100\) times the molybdenum’s: transient equilibrium.
Watch out. The chain here has two radioactive members and a stable end; real chains such as uranium‑238’s have fourteen steps and branch, and Bateman’s solution extends to any length. In the molybdenum example about 12% of decays bypass technetium‑99m, which lowers its amount by that fraction. The 900 atoms are simulated with exact random decay times, so their counts scatter about the curves by roughly the square root of the number.
Why does secular equilibrium make the daughter’s activity equal to the parent’s?
In equilibrium the daughter’s amount is steady, so it is produced as fast as it decays: \(\lambda_AN_A=\lambda_BN_B\), which says the two activities are equal. This only works when the parent changes negligibly over the daughter’s lifetime. In transient equilibrium the parent is visibly declining, the daughter lags behind it, and the ratio becomes \(\lambda_B/(\lambda_B-\lambda_A)\), slightly above one.
3. Alpha decay by tunnelling
An alpha particle inside a heavy nucleus has positive energy Q, yet it is trapped behind a Coulomb barrier six or seven times higher. Classically it could never leave. Quantum mechanically its wavefunction leaks through, and the probability of getting out on each attempt is the Gamow factor e^(−G), where G is an integral of the momentum it would need under the barrier. Multiply by the number of times per second it strikes the wall and you have the decay rate. Because Q enters G under a square root in an exponent, a factor of two in Q changes the half-life by more than twenty orders of magnitude, which is the Geiger–Nuttall law of 1911. Drag Q on either figure.
Worked example. For uranium‑238, \(Q=4.270\) MeV and the daughter thorium has \(Z_d=90\). With \(R=1.2(A_d^{1/3}+4^{1/3})=9.30\) fm the barrier top is 27.9 MeV, and the alpha must tunnel out to \(b=60.7\) fm, giving \(G=88.0\) and a probability \(e^{-G}=6.1\times10^{-39}\) per attempt. Bouncing \(7.8\times10^{20}\) times a second, it escapes after \(4.6\times10^9\) years on average; the measured half-life is \(4.47\times10^9\) years. Polonium‑212, with \(Q=8.954\) MeV, has \(G=32.3\) and a predicted half-life of 0.063 µs, against 0.30 µs measured.
Watch out. This is Gamow’s original model: a sharp nuclear surface, a pure Coulomb barrier, an alpha particle that already exists inside the nucleus, and an attempt frequency \(v/2R\) from its speed at energy \(Q\). It ignores the probability of forming the alpha cluster, angular momentum and nuclear deformation, so its half-lives are only good to within about a factor of ten. The largest misses are at closed shells: polonium‑210, with 126 neutrons, lives forty times longer than the model says. The radius parameter 1.2 fm is a fit to these nuclei.
Why does a slightly larger Q shorten the half-life so enormously?
Because Q acts on the barrier twice: the outer turning point \(b\propto1/Q\) moves inwards, so the barrier is thinner, and the height \(V-Q\) above the alpha’s energy shrinks everywhere under it. To leading order \(G\approx\pi\sqrt{2\mu}\,b\sqrt{Q}/\hbar\propto Z_d/\sqrt Q\), and the probability is \(e^{-G}\) with \(G\) between 30 and 90, so a ten per cent change in \(G\) shifts the half-life by several factors of ten. That is the straight line of the Geiger–Nuttall plot, \(\log t_{1/2}=a+b\,Z_d/\sqrt Q\).