University Physics V · Finite Potential Wells and Quantum Tunneling · 6.8

Alpha Decay & the Geiger–Nuttall Law

An alpha sits 17 MeV below the top of its own barrier and gets out anyway. This is where barrier algebra stops being a toy: one WKB integral over a Coulomb tail, one attempt frequency, and the model turns a factor of two in decay energy into twenty-four decades of half-life — then tells you which part of that number to distrust.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Alpha decay is the one place a tunnelling exponent is measured across twenty-four decades and survives. The model is Gamow's, from 1928. Assume a four-nucleon cluster is already assembled inside the daughter, rattling in a well whose wall stands at a contact radius R ≈ 1.2(4^⅓ + A_d^⅓) fm; outside that radius the alpha sees nothing but the daughter's Coulomb tail 2Z_d e²/4πε₀r, which for ²¹²Po reaches 26.2 MeV at contact while the alpha carries Q = 8.954 MeV. Classically it is caged.

Quantum mechanically the WKB exponent between the turning points, 2G = (2/ħ)∫ from R to b of √(2μ(V − Q)) dr, is finite, and for a bare Coulomb barrier it integrates in closed form to an arccos of √(R/b). Attach an attempt frequency ν = v/2R ≈ 10²¹ s⁻¹ and a fitted preformation factor P, and λ = P ν e^(−2G). What that buys is the Geiger–Nuttall law: the point-Coulomb part of the exponent is 3.92 Z_d/√Q, the finite-radius correction is almost independent of Q, so ln λ falls linearly in Z_d/√Q and a factor of 2.2 in decay energy becomes 24 orders of magnitude in half-life. What it costs is precision.

Everything outside the exponent — cluster preformation, the well depth, the real diffuse surface — is swept into a prefactor nobody derives, and the exponent's grip on R is so tight that half a femtometre moves the predicted half-life by a factor of five. The model predicts slopes, not half-lives.

Simple definition: Alpha decay is the tunnelling of a preformed alpha cluster through the Coulomb barrier of its daughter, so the decay constant is an attempt frequency times exp(−2G), where 2G is the WKB action across the forbidden gap between the nuclear surface and the outer turning point.
Example: For ²¹²Po → ²⁰⁸Pb + α, Q = 8.954 MeV faces a 26.2 MeV barrier reaching out to b = 26.4 fm: 2G = 32.3, so e^(−2G) = 9.5 × 10⁻¹⁵, and 1.2 × 10²¹ attempts a second give t½ ≈ 63 ns against a measured 0.30 µs.

Decay energy and the alpha's share of itQ = [M_P − M_D − M_α]c², T_α = Q(A − 4)/A

Q is what the barrier is measured against, but a spectrometer reads T_α. Convert before touching the exponent.

Atomic masses: the Z electrons cancel. ²¹²Po: Q = 8.954 MeV, T_α = 8.785 MeV, recoil 0.169 MeV.

Coulomb barrier and outer turning pointV(r) = 2Z_d e²/4πε₀r, b = 2Z_d e²/4πε₀Q

Fixes both limits of the forbidden region. Everything after this is one integral from R to b.

e²/4πε₀ = 1.440 MeV fm with r in fm. ²¹²Po: V(R) = 26.2 MeV at R = 9.02 fm, and b = 26.4 fm.

The Gamow exponent, in closed form2G = (2b/ħ)√(2μQ) [arccos√x − √(x(1 − x))], x = R/b

Substituting r = b sin²θ does the whole integral, so a bare Coulomb barrier never needs quadrature.

μ is the reduced mass, μc² = 3657 MeV for α on ²⁰⁸Pb; 2G is dimensionless. ²¹²Po: x = 0.342, 2G = 32.3.

Point-Coulomb limit: the Sommerfeld term2G → 2πη = 4πZ_d e²/(4πε₀ħv) = 3.92 Z_d/√Q

The only piece carrying Q, and therefore the piece that makes a Geiger–Nuttall plot straight.

The R → 0 limit, with Q in MeV and v = √(2Q/μ). ²¹²Po: 107.5, cut to 32.3 by the finite radius.

From an exponent to a rateλ = P ν e^(−2G), ν = v/2R, t½ = ln2/λ

Everything sharp is in the exponent and everything sloppy is in the prefactor — so quote ratios, not rates.

ν = 1.16 × 10²¹ s⁻¹ for ²¹²Po. P is the dimensionless preformation factor, fitted, typically 1 to 10⁻³.

The Geiger–Nuttall lawlog₁₀(t½/s) = a Z_d/√Q + c, a ≈ 1.50, c ≈ −47.5

Two nuclides fix the line, and it then places four more members of the ²³⁸U chain within a fifth of a decade.

Q in MeV, Z_d the daughter's charge. Fitted here to ²³²Th and ²¹²Po alone; leading theory gives a = 1.70.

Set the barrier up before doing anything quantum

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) = 9.02 fm and V(R) = 2 × 82 × 1.440/9.02 = 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, nearly two nuclear diameters. 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.

The Coulomb WKB integral has a closed form

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 32.29 to 32.60 and cuts the predicted rate by a third, 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 0.9464 − 0.4743 = 0.4721, so 2G = 2 × 26.37 × 1.297 × 0.4721 = 32.29 and e^(−2G) = 9.5 × 10⁻¹⁵.

An exponent is not a rate

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) = 1.10 × 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 1.45 × 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.

Why the law is straight in Z_d/√Q

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) = 1.50 Z_d/√Q − 47.5, 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.

The error budget is R and Q, not the integral

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 and the rate by e^0.54 = 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 9.02 to 9.50 fm — 5%, comfortably inside the spread between radius parametrisations — drops 2G from 32.29 to 30.58 and multiplies λ by 5.5. 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.

What is dropped, and what each omission costs

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 230 by which ²³⁵U lags the even-even systematics 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Decay energy Q8.95 MeV

Daughter charge Z_d82

Contact radius R9.0 fm

Hold Z_d at 82 and slide Q from 8.95 down to 4.00 MeV: b marches from 26 fm to 59 fm, 2G climbs from 32 to 83, and the predicted half-life crosses 22 decades. Then move R alone from 9.0 to 9.5 fm and watch the same half-life fall fivefold — that is the model's real error bar.

OUTER TURNING POINT b26.4 fm

BARRIER AT CONTACT V(R)26.2 MeV

GAMOW EXPONENT 2G32.4

PREDICTED log₁₀(t½/s)-7.2

Live interpretationOUTER TURNING POINT b: 26.4 fm. BARRIER AT CONTACT V(R): 26.2 MeV. GAMOW EXPONENT 2G: 32.4. PREDICTED log₁₀(t½/s): −7.2

Catch the common trap

Explain before calculating.

Misconception check

Students leave believing the alpha borrows energy from the uncertainty principle, ΔE Δt ≈ ħ, climbs the 26 MeV barrier on the loan, and repays it outside. Nothing in the calculation does that. The energy is fixed at E = Q throughout: the same eigenvalue appears inside the well, under the barrier and in the outgoing wave. Between R and b the wavefunction does not oscillate at a higher energy — it is a real decaying exponential, and κ² = 2μ(V − Q)/ħ² > 0 says precisely that no kinetic energy is available there. Nor is there a time to borrow one for: 2G is an integral over position, and the 10²¹ s⁻¹ that turns it into a rate comes from the interior motion, not the barrier. And the negative kinetic energy is unobservable: localising the alpha within 1/κ ≈ 1 fm gives it a momentum spread ħκ whose energy is exactly V − Q, enough to lift it over the barrier you meant to catch it under.

Two even-even alpha emitters share a daughter charge Z_d = 86 and a contact radius R = 9.13 fm, but their decay energies are 5.00 and 6.00 MeV. Using λ = P ν e^(−2G) with the same preformation factor, by roughly what factor do their half-lives differ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFor ²¹²Po → ²⁰⁸Pb + α the atomic masses are 211.988868 u, 207.976652 u and 4.002603 u. Find Q, the alpha's kinetic energy, the outer turning point b, and the barrier height at the contact radius R = 9.02 fm. Take e²/4πε₀ = 1.440 MeV fm and 1 u = 931.494 MeV/c².

Atomic masses carry their electrons, and 84 = 82 + 2, so the electron masses cancel: Δm = 211.988868 − 207.976652 − 4.002603 = 0.009613 u.
Q = 0.009613 × 931.494 = 8.954 MeV. This is the energy the barrier is measured against, not what a detector records.
Momentum conservation splits Q by mass: T_α = Q(A − 4)/A = 8.954 × 208/212 = 8.785 MeV, leaving 0.169 MeV to the recoiling ²⁰⁸Pb.
Outer turning point, from V(b) = Q: b = 2Z_d e²/(4πε₀Q) = 2 × 82 × 1.440/8.954 = 236.2/8.954 = 26.4 fm.
Barrier at contact: V(R) = 236.2/9.02 = 26.2 MeV, which is 2.9 times Q. The alpha is classically forbidden across 26.4 − 9.0 = 17.4 fm of vacuum.

AnswerQ = 8.954 MeV, T_α = 8.785 MeV, b = 26.4 fm, V(R) = 26.2 MeV — a barrier 2.9 times the available energy, with a 17.4 fm forbidden gap to cross.

MediumContinue with ²¹²Po, taking R = 9.02 fm, b = 26.37 fm, Q = 8.954 MeV, μc² = 3657 MeV and ħc = 197.33 MeV fm. Evaluate the Gamow exponent in closed form, attach an attempt frequency, and predict the half-life. Compare with the measured 0.30 µs.

Wavenumber scale: √(2μQ)/ħ = √(2 × 3657 × 8.954)/197.33 = 255.9/197.33 = 1.297 fm⁻¹, so the prefactor 2b√(2μQ)/ħ = 2 × 26.37 × 1.297 = 68.4.
Thickness parameter: x = R/b = 9.02/26.37 = 0.342, so √x = 0.5846, arccos(0.5846) = 0.9464 rad, and √(x(1 − x)) = √0.2250 = 0.4743.
Bracket = 0.9464 − 0.4743 = 0.4721, so 2G = 68.4 × 0.4721 = 32.3 and e^(−2G) = 9.5 × 10⁻¹⁵: one escape per 1.1 × 10¹⁴ attempts.
Attempt frequency: v = c√(2Q/μc²) = c√(17.91/3657) = 0.0700c = 2.10 × 10²² fm s⁻¹, and ν = v/2R = 2.10 × 10²²/18.03 = 1.16 × 10²¹ s⁻¹.
λ = ν e^(−2G) = 1.16 × 10²¹ × 9.5 × 10⁻¹⁵ = 1.10 × 10⁷ s⁻¹, so t½ = ln2/λ = 6.3 × 10⁻⁸ s = 63 ns.
The measured 0.30 µs is 4.7 times longer, so the model overpredicts the rate by 4.7. Read that as a preformation factor P ≈ 0.21, not as a failure of the exponent — a 2% shift in R would have covered it.

Answer2G = 32.3, e^(−2G) = 9.5 × 10⁻¹⁵, ν = 1.16 × 10²¹ s⁻¹ and t½ = 63 ns, against 0.30 µs measured: a preformation factor P ≈ 0.21.

HardRun the same pipeline for ²³²Th → ²²⁸Ra + α with Q = 4.082 MeV, Z_d = 88, R = 9.24 fm and μc² = 3663 MeV. Then fit a Geiger–Nuttall line through ²³²Th (t½ = 1.405 × 10¹⁰ yr) and ²¹²Po (t½ = 0.30 µs), and test it on ²²⁶Ra (Z_d = 86, Q = 4.871 MeV, t½ = 1600 yr).

b = 2 × 88 × 1.440/4.082 = 253.4/4.082 = 62.09 fm, so x = 9.24/62.09 = 0.1488: the barrier is now nearly seven contact radii thick, against 2.9 for ²¹²Po.
√(2μQ)/ħ = √(2 × 3663 × 4.082)/197.33 = 172.9/197.33 = 0.8763 fm⁻¹, and the bracket is arccos(0.3857) − √(0.1488 × 0.8512) = 1.1748 − 0.3559 = 0.8190.
2G = 2 × 62.09 × 0.8763 × 0.8190 = 89.1, so e^(−2G) = 2.0 × 10⁻³⁹. With v = 0.0472c, ν = 7.7 × 10²⁰ s⁻¹, λ = 1.5 × 10⁻¹⁸ s⁻¹ and t½ = 4.6 × 10¹⁷ s = 1.45 × 10¹⁰ yr, against 1.405 × 10¹⁰ measured.
Now the empirical line. Z_d/√Q is 88/2.020 = 43.56 for ²³²Th and 82/2.992 = 27.40 for ²¹²Po, while log₁₀(t½/s) is 17.65 and −6.52. Slope = 24.17/16.15 = 1.50, intercept = −6.52 − 1.50 × 27.40 = −47.5.
Test on ²²⁶Ra: Z_d/√Q = 86/2.207 = 38.97, so log₁₀(t½/s) = 1.50 × 38.97 − 47.5 = 10.78, i.e. 6.0 × 10¹⁰ s = 1900 yr against 1600 yr measured — 20% out, from a two-point fit spanning 24 decades.
The same line puts ²¹⁴Po at 10^(−3.69) s against 1.6 × 10⁻⁴ s measured. What it cannot do is ²³⁵U, 230 times slower than these even-even systematics: an odd nucleon hinders preformation, and hindrance moves the intercept, never the slope.

Answer2G = 89.1 gives t½ = 1.45 × 10¹⁰ yr against 1.405 × 10¹⁰ measured. The two-point line log₁₀(t½/s) = 1.50 Z_d/√Q − 47.5 puts ²²⁶Ra at 1900 yr against 1600 yr.