Alpha decay and the Geiger–Nuttall law

The potential has two pieces and one seam. Inside the contact radius R = 1.2(4^⅓ + A_d^⅓) fm the alpha feels the nuclear well; outside it feels only the daughter's charge, V(r) = 2Z_d e²/4πε₀r, with e²/4πε₀ = 1.440 MeV fm. For ²¹²Po → ²⁰⁸Pb + α that gives R = 1.2(1.587 + 5.925) = 9.01 fm and V(R) = 2 × 82 × 1.440/9.01 = 26.2 MeV. The alpha's budget is the mass difference, Q = 8.954 MeV, of which the recoiling daughter takes 4/212, leaving the measured T_α = 8.785 MeV. So the alpha meets a wall 2.9 times its energy and stays classically forbidden all the way out to b = 2Z_d e²/4πε₀Q = 26.4 fm — a 17.4 fm gap of vacuum, wider than the lead nucleus is across. Two assumptions are already in place: that a four-nucleon cluster exists as an entity before it leaves, and that the nuclear surface is a step. Both are wrong in detail, and both are paid for later.

Write V(r) − Q = Q(b/r − 1), which is exactly what defining b as the turning point makes true. Then 2G = (2√(2μQ)/ħ) ∫ from R to b of √(b/r − 1) dr, and r = b sin²θ turns the integrand into 2b cos²θ dθ, giving 2G = (2b√(2μQ)/ħ)[arccos√x − √(x(1 − x))] with x = R/b. Two cautions before you put numbers in. The mass is the reduced mass of alpha and daughter, μc² = 3657 MeV for ²⁰⁸Pb: using m_α instead raises 2G from 32.30 to 32.61 and cuts the predicted rate by about a quarter, from a 1.9% error in a mass. And ħc = 197.33 MeV fm keeps every quantity in MeV and fm. For ²¹²Po, √(2μQ)/ħ = √(2 × 3657 × 8.954)/197.33 = 1.297 fm⁻¹, x = 0.342, and the bracket is 0.9465 − 0.4743 = 0.4722, so 2G = 2 × 26.37 × 1.297 × 0.4722 = 32.30 and e^(−2G) = 9.4 × 10⁻¹⁵.

A transmission factor is a probability per attempt, so it needs an attempt rate. Take the alpha's speed from Q alone, v = c√(2Q/μc²) = 0.0700c = 2.10 × 10⁷ m s⁻¹, and let it cross the well once each way: ν = v/2R = 1.16 × 10²¹ s⁻¹. Then λ = ν e^(−2G) = 1.094 × 10⁷ s⁻¹ and t½ = ln2/λ = 63 ns, against a measured 0.30 µs — the alpha needs 1.1 × 10¹⁴ attempts, and gets through in a fraction of a microsecond because it makes them so fast. The factor of 4.7 by which the prediction is too quick is the preformation probability P ≈ 0.21: four nucleons are not sitting there ready-made, and P counts the fraction of the time they are correlated into a cluster at the surface. Run the same recipe on ²³²Th and it returns 1.43 × 10¹⁰ yr against 1.405 × 10¹⁰ measured, so P ≈ 1 there. That is luck, not accuracy. P scatters over two or three decades; the exponent spans twenty-four.

Expand the bracket for small x: arccos√x − √(x(1 − x)) ≈ π/2 − 2√x. Then 2G ≈ π√(2μQ)b/ħ − 4√(2μQb)√R/ħ. The first term is the point-Coulomb exponent 2πη = 3.92 Z_d/√Q, with η the Sommerfeld parameter of the alpha–daughter pair. The second carries √(Qb) = √(2Z_d e²/4πε₀), which contains no Q at all: the finite-radius correction depends on Z_d and R, not on the decay energy. So log₁₀ t½ is linear in Z_d/√Q with a leading slope 3.92/ln10 = 1.70, and everything else lands in the intercept. Fit two real nuclides — ²³²Th at Z_d/√Q = 43.56 and ²¹²Po at 27.40 — and you get log₁₀(t½/s) = 1.496 Z_d/√Q − 47.53, the slope a little shallower than 1.70 because Z_d rises along the chain and the correction rises with it. That line places ²²⁶Ra, ²²²Rn, ²¹⁸Po and ²¹⁴Po within 0.2 of a decade. Use the expansion for the shape only: at x = 0.34 it is 15% low, so numbers need the exact bracket.

Differentiate before you trust anything. The leading exponent goes as Q^(−1/2), so a 1% error in Q shifts 2G by about ½ × 107 × 0.01 = 0.54, or 0.48 with the exact bracket, and the rate by e^0.48 = 1.6. That is why alpha energies are worth measuring to a keV, and mass spectrometry obliges. R is the opposite story: moving it from 9.01 to 9.50 fm — 5%, comfortably inside the spread between radius parametrisations — drops 2G from 32.30 to 30.57 and multiplies λ by 5.3. Nothing in the model pins R that well, because the sharp-edged well it labels does not exist. Two working rules follow. Never quote a predicted half-life to better than an order of magnitude. And prefer ratios: predict t½(A)/t½(B) for two neighbouring emitters and the shared errors in R, ν and P largely cancel, which is precisely why the Geiger–Nuttall slope is reproducible while its intercept is fitted.

Three omissions, in increasing severity. Angular momentum: an alpha carrying l adds l(l+1)ħ²/2μr² to the barrier, worth 0.39 MeV at contact for l = 2 in ²¹²Po, which integrates to a hindrance of only 1.6 at l = 2 and about 11 at l = 5 — so the factor of about 40 by which ²³⁵U lags the even-even line is mostly structure, not the centrifugal term. Deformation: actinides are prolate, so the barrier depends on emission angle and the alpha leaves preferentially along the long axis, where it is thinnest. And the state itself: a decaying nucleus is not a stationary state. It is a pole of the transmission amplitude at Eᵣ − iΓ/2, a Gamow–Siegert vector that is purely outgoing, grows at infinity and cannot be normalised. λ = Γ/ħ is the exact statement; λ = P ν e^(−2G) is the semiclassical estimate of it, with the attempt frequency standing in for the imaginary part of a pole.