Α, β, and gamma decay modes

Write a nuclide as ᴬX with Z protons and N neutrons, so A = Z + N: ²³⁸U has Z = 92 and N = 146. Z fixes the element and its chemistry; isotopes share Z, isotones share N, isobars share A. The roughly 250 stable nuclides fill a narrow valley on a chart of Z against N. At fixed A the mass is close to a parabola in Z: the symmetry energy, proportional to (N − Z)²/A, is least at N = Z, while the Coulomb energy, proportional to Z²/A¹⁄³, is least at small Z. Set ∂M/∂Z = 0 at fixed A and the bottom lies at Z₀ ≈ A/(1.98 + 0.0155 A²⁄³). That gives Z₀ = 7.7 at A = 16, 50.9 at A = 120 and 82.4 at A = 208, matching ¹⁶O, ¹²⁰Sn and ²⁰⁸Pb, with N/Z climbing from 1 to 1.54. A nuclide off the valley floor is unstable, and the side it sits on decides how it decays.

A neutron-rich nuclide lowers its mass by converting a neutron, n → p + e⁻ + ν̄ₑ, taking (Z, N) to (Z + 1, N − 1) at fixed A. A proton-rich one runs the reverse, p → n + e⁺ + νₑ, which a free proton cannot do, being 1.293 MeV lighter than a neutron; inside a nucleus the gain in binding pays for it. Each step moves one place down the isobar's mass parabola, so an odd-A chain ends on a single stable isobar: ¹⁰¹Mo β⁻-decays to ¹⁰¹Tc and then to ¹⁰¹Ru, and Z₀(101) = 43.6 rounds to ruthenium's 44. β⁺ must also create a positron, which with atomic masses means M(A, Z) − M(A, Z − 1) > 2mₑ, or 1.022 MeV/c². Below that gap electron capture, p + e⁻ → n + νₑ, is the only route: ⁷Be lies 0.862 MeV above ⁷Li and decays by capture alone. Even-A isobars sit on two parabolas split by pairing, which is why odd-odd ⁴⁰K goes both ways: β⁻ in 89% of decays, capture in 11%.

Let a nucleus at rest split into two bodies and momentum conservation sends them back to back with fixed shares of Q. For ²¹⁰Bi, Q = 1.162 MeV, so every electron would carry 1.162 MeV less 6.5 eV of recoil: one sharp line. Instead the electrons emerge with every energy from zero to a sharp endpoint at Q, averaging 0.39 MeV. Calorimetry in 1927 showed the missing energy was not being lost inside the source, and angular momentum failed too: ²¹⁰Bi has spin 1 and ²¹⁰Po spin 0, and an electron's spin ½ plus any whole-number orbital angular momentum can never make up a whole number. Pauli's answer in 1930 was a third particle, neutral, spin ½ and nearly massless, that shares the energy with the electron in any proportion: Kₑ + K_ν + K_R = Q. The endpoint is where the neutrino takes almost nothing. Reines and Cowan detected reactor antineutrinos in 1956, and the shape of tritium's spectrum just below its 18.6 keV endpoint now holds the neutrino mass below 0.45 eV/c².

Emitting any cluster costs the binding the parent loses and returns the cluster's own: Q = B(daughter) + B(cluster) − B(parent), with B = 0 for a lone nucleon. For ²³⁸U, losing a neutron costs 6.15 MeV, so that is forbidden. Losing two protons and two neutrons costs 1801.69 − 1777.66 = 24.03 MeV, but helium-4 returns 28.30 MeV, so Q_α = +4.27 MeV. Put another way, Q_α ≈ 28.3 MeV − 4 ∂B/∂A along the α path, positive only once a nucleon costs less than 7.07 MeV to remove, which happens in heavy nuclei, above roughly A ≈ 150, where B/A is falling. The α and the daughter are two bodies, so α energies come as sharp lines. Q only opens the door. At the surface of ²³⁴Th the Coulomb barrier is about 28 MeV, over six times the α's 4.2 MeV, so the α must tunnel, and the rate is exquisitely sensitive to Q: ²¹²Po, with Q_α = 8.95 MeV, has a half-life of about 0.3 μs, while ²³⁸U, at 4.27 MeV, has one of 4.5 × 10⁹ years: a factor of 2 in Q, a factor near 5 × 10²³ in lifetime.

An α removes equal numbers of protons and neutrons, so N − Z is unchanged while A falls by 4. The valley wants less neutron excess at smaller A: A − 2Z₀ is 53.2 at A = 238 but only 42.5 at A = 206. So each α step leaves the daughter about two-thirds of a proton further to the neutron-rich side of its isobar's valley point, and β⁻ decays, each cutting N − Z by 2, must follow. The ledger for ²³⁸U → ²⁰⁶Pb is exact. A falls by 32, so there are 8 α decays. N − Z must fall from 54 to 42, so there are 6 β⁻ decays. Charge closes: 92 − 16 + 6 = 82. Because only α changes A, and always by 4, every heavy nuclide belongs to one of four families, A = 4n, 4n + 1, 4n + 2 or 4n + 3, ending on ²⁰⁸Pb, ²⁰⁹Bi, ²⁰⁶Pb and ²⁰⁷Pb. β⁺ essentially never appears in these chains: the α steps only ever push the nuclides the other way.

α and β decays often leave the daughter in an excited level, and it falls to the ground state by emitting photons, A and Z unchanged: ¹³⁷Cs β⁻-decays mostly to a ¹³⁷Ba level 0.662 MeV up, whose 662 keV gamma ray calibrates detectors worldwide. Why MeV, when hydrogen's lines are eV? Confine a nucleon to R ≈ 5 fm and its energy scale is (ħc)²/(2mc²R²) = (197 MeV fm)² / (2 × 939 MeV × 25 fm²) ≈ 0.8 MeV. That scale also walls the nucleus off. Chemical bonds shift energies by a few eV, parts per million of a typical Q. Heat can populate a level ΔE up only with Boltzmann factor e^(−ΔE/kT): for ΔE = 100 keV at 300 K the exponent is −3.9 × 10⁶, and even in the Sun's 1.5 × 10⁷ K core it is −77, a factor of 10⁻³⁴. So a half-life is a constant of the nuclide. The exceptions use atomic electrons, electron capture chief among them: it needs an atomic electron at the nucleus, so ⁷Be's half-life shifts by up to about 1% between chemical settings, and a bare ⁷Be ion cannot decay at all.