Alpha Decay: Tunneling with a Half-life
An alpha particle rattles against a wall it cannot climb ten billion billion times a second. Once in 10³⁹ tries it is simply on the other side — and that one number is uranium's four-billion-year half-life.
Alpha Decay: Tunneling with a Half-life visualization
Amber is the potential the alpha sees: a well 35 MeV deep out to the nuclear radius R, then the daughter’s Coulomb repulsion falling as 1/r. The dashed sky line is Q; it meets the barrier again at b. Violet is the wavefunction, drawn about the Q line: oscillating inside, dying through the barrier, small and oscillating again beyond b. It is the exact stationary solution of the same Schrödinger engine as the tunneling and bound-state pages — for a barrier of this shape with its exponent divided by 14. At full strength the outside amplitude would be e^−G/2 ≈ 1.46 × 10^-20 of the inside, which is not a line a screen can draw.
The Geiger–Nuttall line
Add nuclides with the button in the controls. Each one drops two markers at the same horizontal position: the measured half-life (filled) and what the Gamow model predicts (hollow). From the second point on, a least-squares line is drawn through the measured ones.
| Nuclide | Q (MeV) | Z_d/√Q | G | Measured t½ | Predicted t½ | Pred ÷ meas |
|---|---|---|---|---|---|---|
| U-238 | 4.270 | 43.55 | 91.3 | 4.47 billion years | 39.9 billion years | 8.9 |
The same exponential, at the other end of the scale
A scanning tunnelling microscope is a metal tip held a few ångström above a surface. No electron has the energy to cross the vacuum gap — the work function φ is the wall — so the current is pure tunnelling, I ∝ e^−2κd. Drag the tip, or use the sliders, and watch the meter: it is logarithmic, and it has to be.
Raise the tip by exactly 1 Å from where it is now and the current falls from 722 pA to 82.1 pA — a factor of e^2κ ≈ 8.8. That is the whole trick of the instrument: a feedback loop holds the current constant, so the tip height traces the surface to a hundredth of an ångström, and the atoms appear as bumps in the scan strip at the bottom of the picture.
What you are looking at
George Gamow’s 1928 picture, which is still the picture. Inside the nucleus an alpha particle — two protons, two neutrons, already assembled — moves in a well about 35 MeV deep. At the surface the strong force lets go and the only thing left is electrostatics: the daughter’s 90 protons pushing on the alpha’s two, a barrier that is 30 MeV high at the surface and falls as 1/r.
The alpha comes out with 4.27 MeV. That is far below the top of the barrier, so classically it is trapped forever. Quantum mechanically the wavefunction does not stop at the wall; it decays through it, and by the time the barrier has fallen back to Q — at b = 61 fm, several nuclear radii out — a little of it is left. Everything past b is a free alpha leaving.
Three numbers make a half-life
The exponent. Transmission through a slowly varying barrier is T = e^−G with G = (2/ħ)∫√(2μ(V − Q)) dr from R to b — the WKB integral, the same one the tunneling page uses for its rectangular wall, except that for a 1/r barrier it can be done in closed form. For U-238 at this Q it is G = 91.3, so T = 2.14 × 10^-40.
The attempt rate. Inside the well the alpha has Q + 35 MeV of kinetic energy and crosses a nucleus 17 fm across in about 10⁻²¹ s, so it hits the wall f ≈ 2.57 × 10^21 times a second.
The product. λ = f·T is the chance per second of escape, and t½ = ln 2/λ = 39.9 billion years. Nothing in the chain is adjusted to hit the answer; the only free number is r₀, and it is the same for every nucleus on the page.
Why the plot is a straight line
In 1911 Hans Geiger and John Nuttall noticed that plotting the logarithm of the decay rate against the alpha’s range in air gave a straight line — and had no idea why. Gamow’s integral is the why. Because b ∝ Z/Q, the exponent is very nearly G ∝ Z/√Q with a correction from the nuclear radius, so log t½ against Z/√Q is close to linear with a slope the model predicts.
Plot all nine and look at the spread: 14 billion years at the top, 300 nanoseconds at the bottom — 24 orders of magnitude — from Q values that differ by barely a factor of two. Then drag the Q slider on any nucleus and watch T fly. Half a percent more energy takes a third off radium’s 1,600 years.
The hollow markers are the model. They sit close to the filled ones and they sit on a line of the same slope, which is the real test: the model does not merely fit, it explains the form of the law.
Where it misses, and why that is interesting
Select Po-210. The model predicts a half-life about six times shorter than the measured 138 days — the worst miss on the page, and it is not a numerical accident. Polonium-210 has 126 neutrons, a closed shell, and an alpha particle is much less likely to be preformed inside a nucleus that has just filled one. The real decay rate is the Gamow rate times a “preformation factor” that the simple model sets to 1; near shell closures it can be 1/30.
The other misses are the radius. Slide r₀ and every prediction moves together: one femtometre on R is nearly two orders of magnitude on t½, because R sits at the steepest part of the integral. That is why nuclear physicists quote alpha-decay half-lives as a way of measuring nuclear radii, rather than the other way round.
What is and isn't real here
The Q values and half-lives are the measured ones (AME2020 and Nubase 2020). The barrier on the canvas is drawn in real MeV and femtometres from the daughter’s charge. G, T, f and the predicted half-lives are the Gamow model with no extra factors, andlib/quantum/alphaDecay.test.tsholds them against the measured values (within 10×, order preserved), holds the WKB integral against the exact rectangular-barrier result, and holds the STM decay constant against κ worked out by hand from the electron mass.
The wavefunction is the one honest simplification: it is the true solution of the Schrödinger equation for a barrier of this shape, but with the exponent divided by 14. A real alpha wavefunction is 10⁻¹⁹ of its interior size once it gets out, and a curve like that would draw as a flat line. The well depth of 35 MeV is a typical value, not a measurement; it only enters the assault frequency, where a factor of two is invisible next to an exponent of 90.