A wall that a factor of two makes impassable

Seven alpha emitters, all of them heavy nuclei behaving in the same way for the same reason. The alpha particles they release span a factor of 2.2 in energy — 4.08 MeV at one end and 8.95 at the other, which is the difference between a warm cup of tea and a slightly warmer one if the comparison were about temperature. Their half-lives span twenty-four orders of magnitude.

The relation was found empirically by Geiger and Nuttall in 1911, before anybody had a nucleus to attribute it to, and it sat as a curious regularity for seventeen years. What it needed was not more data but a mechanism in which an energy could appear inside an exponent, and there was no such mechanism in physics until 1928.

The problem, stated so that it is a problem #

An alpha particle inside a heavy nucleus is held by the strong interaction, which is short-ranged and enormously deep. Outside the nuclear surface there is no strong interaction left and the alpha, carrying charge $+2e$, is repelled by the daughter’s charge $+Ze$. So the potential rises from a deep well to a Coulomb peak at the nuclear radius and then falls off as $1/r$.

The energetics say the alpha should go: the daughter plus the alpha weigh less than the parent, and the difference is the 4 to 9 MeV the alpha carries away. The geometry says it cannot: the barrier peaks at five times that energy, and a classical particle at 5 MeV inside a 26 MeV barrier stays inside for ever. Both statements are correct. The nucleus is a body that is free to fall apart and unable to.

The shape of that barrier is worth being explicit about, because it is what makes the problem hard rather than merely awkward. A Coulomb repulsion falls as $1/r$, and a $1/r$ curve crosses a low energy a very long way out — so the region the alpha has to get through is not a wall but a long thin ramp, whose width is set by where the curve meets the particle’s own energy. The geometry that produces the $1/r$ is the ordinary one: the field of a point source spreads over a surface growing as $r^2$, so its potential falls as $1/r$, and the slowness of that fall is precisely the difficulty.

The one thing quantum mechanics adds #

Every object in the barrier picture is a wave, because everything with a momentum has a wavelength and an alpha at 5 MeV has one of about six femtometres — comparable with the nucleus, which is what makes the whole problem quantum rather than ballistic. Inside the barrier the local kinetic energy is negative, so the wavenumber is imaginary, and the solution of the wave equation there is not an oscillation but a real exponential. That is the whole of the rung below this one: the wavefunction does not stop at a barrier it cannot climb, it decays through it, and if the barrier ends before the decay has finished there is amplitude on the far side.

For a barrier of constant height the transmitted fraction is $e^{-2\kappa d}$ with $\kappa$ set by how far below the top the particle sits. A Coulomb barrier is not of constant height, so $\kappa$ has to be integrated along the way out:

$$2G = \frac{2}{\hbar}\int_R^b \sqrt{2\mu\left(V(r) - E\right)}\;dr ,$$

between the nuclear radius $R$ and the turning point $b = 2Ze^2/4\pi\epsilon_0 E$ where the barrier falls back to the particle’s energy. That integral is the Gamow exponent, it has a closed form for a pure Coulomb barrier, and it is the only quantity in the whole problem.

Why the sensitivity is so violent #

Now put the two together. Raising the alpha’s energy does two things at once, and both of them shrink the exponent.

It lowers the height of the region that has to be crossed, which reduces the integrand. And — this is the larger effect — it moves the turning point inward, because $b \propto 1/E$: a 9 MeV alpha turns around at half the radius a 4.5 MeV one does. The width of the forbidden region falls roughly as $1/E$ while the integrand falls as $\sqrt{V-E}$, so the exponent falls faster than the first power of the energy and, over the range where alpha decay happens, close to $1/\sqrt{E}$.

That is exactly the Geiger–Nuttall coordinate. Plotting the logarithm of the half-life against $1/\sqrt{E}$ turns the exponent into a straight line, and the straightness of the measured points is the evidence that the mechanism is a barrier integral and not anything else.

The arithmetic is worth doing once. For a 5 MeV alpha leaving a nucleus of charge 84 the exponent comes out at about 69, so the escape probability per attempt is $e^{-69}$, near $10^{-30}$. The alpha rattles inside the well at roughly a tenth the speed of light across a nucleus ten femtometres wide, which is $10^{21}$ attempts a second. Multiplying gives a rate of $10^{-9}$ per second and a half-life of a few decades — for a number obtained from three constants and a radius formula.

What is linear in all of this is the logarithm, and the eye has no intuition for it. A rate whose logarithm runs linearly in some reciprocal is the shape a chemist knows as an Arrhenius plot, where a thermal barrier’s height is read off a slope; alpha decay produces the same shape of graph from an entirely different mechanism, the exponent being a barrier integral over $\hbar$ rather than a barrier height over a temperature. The general lesson about exponentials is the same in both cases, and it is that a modest change in the exponent is not a modest change in anything.

What the model gets away with #

The calculation above is crude in several specific ways, and it is instructive that none of them matters much.

The alpha is assumed to exist inside the nucleus before it leaves. It does not, in any simple sense: the nucleus is a system of protons and neutrons, and the probability that four of them are momentarily arranged as an alpha is a genuinely hard nuclear-structure problem. That probability is a prefactor — it multiplies the attempt frequency — and a prefactor wrong by a factor of a hundred moves the predicted half-life by two decades, which on an axis spanning twenty-four is a small error.

The nuclear radius is taken from a one-line formula. $R = 1.2(A^{1/3} + 4^{1/3})$ femtometres, which is a touching-spheres estimate with a single empirical constant in it. The radius sets the inner limit of the integral, so it matters, and getting it wrong by ten per cent shifts the exponent by a few units.

The barrier is taken to be pure Coulomb right down to the surface. The strong interaction has a tail, so the real barrier is rounded rather than cusped, and the alpha’s own charge polarises the daughter.

Each of those is a factor of a few to a hundred in the answer. The exponent is twenty-four decades wide. That ratio is the reason the model works: a mechanism that produces the right exponent can be forgiven almost anything in front of it, and a mechanism that produces the wrong exponent cannot be rescued by any prefactor whatever.

The attempt frequency in that estimate deserves a word, because it looks like a fudge and is not. A particle confined to a region the size of a nucleus has a momentum of order $\hbar$ over that size — this is the same confinement energy that decides the width of any bound state, traded against the depth of the well holding it — and dividing the resulting speed by the size gives the number of times a second the particle arrives at the wall. It comes out around $10^{21}$, and it is the least important number in the calculation: the exponent covers twenty-four decades, so an attempt frequency wrong by a factor of ten moves the answer by almost nothing.

Why an alpha, and not something else #

Nothing above explains the choice of projectile. A heavy nucleus is energetically free to emit a proton, a deuteron, a carbon-14 nucleus or to split roughly in half, and all of those are barrier problems of the same kind. What decides is a competition between two numbers in the exponent.

Why an alpha, and not some other fragment, is answered by the binding-energy curve. Plot the missing mass per nucleon across the chart of nuclides and the alpha particle sits at a sharp local peak far above its neighbours — four nucleons bound by 28.3 MeV, where lithium and beryllium of similar size are bound by much less. A tightly bound fragment is one that can be emitted while leaving more energy over, and energy is exactly what shrinks the exponent. The alpha is not chosen because it is small; it is chosen because it is cheap to make and expensive to break.

Against that, the exponent contains the product of the charges, so a heavier fragment is penalised: emitting carbon-14 means a charge of 6 rather than 2 against the daughter, which multiplies the Gamow integral. The alpha wins because it is unusually tightly bound for its charge, and cluster decay — carbon-14 emission from radium, which does happen — is a billion times rarer for exactly this reason and was not observed until 1984.

Fission escapes the trade entirely by splitting the charge two ways at a separation where the Coulomb energy is already most of the released energy, which is why it is a competitive channel only in the heaviest nuclei and why its barrier is a shape rather than a distance.

The lines that are not one line #

The half-decade agreement quoted above is honest and it hides a structure worth seeing, because the departures from the straight line are not scatter.

Plot enough emitters and the points separate into families. Nuclei with even numbers of both protons and neutrons lie tightly on one line. Nuclei with an odd number of one or the other lie systematically above it — they live longer than the barrier integral says — and nuclei odd in both lie higher still, sometimes by three decades.

The barrier does not know about that. The Gamow integral contains a charge, an energy and a radius, and nothing about whether a nucleon count is even. What differs is the prefactor: the probability that four nucleons are momentarily arranged as an alpha and that removing them leaves the daughter in its ground state. In an even-even nucleus every nucleon is paired and the alpha can be assembled from a pair of pairs without disturbing anything. Where there is an unpaired nucleon, the emitted alpha has to be built around it or the odd particle has to be rearranged, and either costs amplitude.

The size of the cost is quoted as a hindrance factor — the ratio of the measured lifetime to what the even-even systematics predict — and it runs from a few to a thousand. That it can be read off directly is what makes it useful: a hindrance factor is a measurement of how much the emitted alpha’s configuration differs from the parent’s, which is nuclear structure information obtained from a lifetime.

There is a second, sharper departure at the closed shells. A nucleus with 126 neutrons is unusually tightly bound, so its daughter products are correspondingly less so, and the energy released jumps as that shell is crossed. Since the energy sits inside the exponent, a modest jump in energy is a large kink in the lifetime — which is why the lines break at the shell closures and why alpha systematics were one of the early pieces of evidence for the shell model at all.

Why anything is left to date with #

The exponent’s steepness has a consequence for what exists, and it is the reason the Earth can be dated.

Radioactive half-lives span at least forty decades, from the fraction of a microsecond of polonium-212 to well beyond the age of the universe. A nuclide surviving from the formation of the solar system needs a half-life of at least a few hundred million years; a nuclide useful as a clock needs one not enormously longer than the interval being measured, or too little of it will have decayed to measure.

So geochronology depends on there being nuclides in a narrow band — roughly $10^8$ to $10^{11}$ years — and on that band being populated. It is, barely: uranium-238 at 4.5 billion years, uranium-235 at 700 million, thorium-232 at 14 billion, potassium-40 at 1.25 billion, and a handful of others. Four decades of half-life, drawn from a distribution spanning forty.

That the band is populated at all is a consequence of the exponent’s sensitivity rather than in spite of it. A rate that varies by twenty-four decades over a factor of two in energy means that the nuclides which happen to sit at the right energy are spread thinly across a huge range of lifetimes — so any given decade of half-life has a few occupants, and no decade is crowded. A less sensitive mechanism would have bunched everything together somewhere, and the somewhere would probably not have been useful.

The three chains that survive are also, for the same reason, chains rather than single steps: uranium-238 decays through fourteen intermediate species to lead-206, and every one of those intermediates has a half-life far shorter than the first step. So the chain is rate-limited entirely by its slowest link, the daughters reach a steady state, and the whole chain can be treated as one clock — which is the assumption every uranium-lead date rests on.

Three people, one year #

The mechanism arrived in 1928, and it arrived twice.

George Gamow, then in Göttingen, worked out the barrier calculation and published it as an explanation of the Geiger–Nuttall relation. Ronald Gurney and Edward Condon, in Princeton, did the same thing independently and published within weeks. Neither knew of the other’s work.

What makes the episode worth recording is what it was the first of. Quantum mechanics had been assembled over the previous three years and applied to atoms, where it explained spectra; it had not been applied to the nucleus at all, which was a subject with almost no theory in it. Alpha decay was the first nuclear phenomenon anybody explained quantum-mechanically, and it was explained not qualitatively but to twenty-four decades.

It also settled a question about the wave picture that had been open. Tunnelling through a barrier is the sharpest possible demonstration that a particle’s wavefunction is not merely a bookkeeping device for probabilities of where it might be found: the amplitude has to be genuinely non-zero in the region the particle cannot classically occupy, or there is nothing to arrive on the far side. A theory in which the particle is somewhere and the wave describes ignorance about where cannot produce this.

And the accident of timing is worth noticing. Geiger and Nuttall had the empirical relation in 1911, from data alone, with no mechanism available and none conceivable — the concept required for it did not exist. Seventeen years later the concept arrived and the relation fell out immediately. That is an unusually clean example of a correct measurement waiting for a language, rather than a theory waiting for a measurement.

The randomness, and where it is not #

A single nucleus does not have a lifetime. It has a probability per unit time, constant from the moment it forms until the moment it goes, and no property that changes in between. That is the argument of the decay ladder’s first rung, and tunnelling is what supplies the constant. Where quantum mechanics elsewhere produces a discrete list of allowed values, here it produces a probability per second, and the difference is that the state in question is not bound.

The randomness is worth locating precisely, because it is easy to put in the wrong place. A sample of nuclei follows an exponential and the individual events do not: the curve is a property of the population and the unpredictability is a property of each nucleus, which is the ordinary relationship between a rate and a population and needs no quantum mechanics to state. What tunnelling supplies is not the randomness but the value of the rate — the single number, covering twenty-four decades across the observed alpha emitters, that comes out of a barrier integral.

The constancy is itself a consequence of the barrier picture, and worth stating. The escape probability per attempt does not depend on how many attempts have already been made, because nothing about the barrier or the wavefunction records them. A nucleus that has waited ten billion years is in exactly the state it was in at the start. There is no ageing, no accumulated damage, no gradual weakening — which is what makes the exponential exact rather than approximate, and what makes radiometric dating possible. The same memorylessness in a different setting is what makes a neutrino’s chance of interacting independent of how far it has already come.

It is worth distinguishing three exponentials that turn up in this subject and are easily run together, because they are about different variables. What survives passage through a slab falls exponentially with thickness, from a constant probability of interaction per unit length. What survives of a decaying sample falls exponentially with time waited, from a constant probability per unit time. And the tunnelling amplitude falls exponentially with distance inside a forbidden region, from the decay of a wavefunction where the kinetic energy would be negative. Only the last of the three is a wave phenomenon at all; the first two would look the same in a world with no quantum mechanics in it.

Where the model stops #

It is one-dimensional and the nucleus is not. The integral above is taken along a radius, and a real emission has an angular momentum barrier on top of the Coulomb one whenever the alpha carries any. That extra term suppresses transitions to excited states of the daughter, which is why alpha spectra have fine structure with wildly unequal branch intensities — a line spectrum whose relative brightnesses are the physics rather than a decoration on it.

The WKB approximation needs the barrier to be smooth on the scale of the wavelength, and it fails near the turning points where the local wavelength diverges. The standard connection formulae patch it, and they contribute a factor of order one — again lost in the exponent.

Deformed nuclei are not spheres. A prolate nucleus has a thinner barrier along its long axis, so it emits preferentially in that direction and faster overall than a spherical one of the same mass, and the departures from the Geiger–Nuttall line in the actinides are largely this.

And nothing here explains why alpha rather than anything else. Emission of a single proton, of a heavier cluster, and of a fission fragment are all barrier problems with the same structure, and which one dominates is decided by the balance between the energy released and the charge product in the exponent. The alpha wins in the actinides because it is unusually tightly bound for its size, and that is a fact about nuclear binding rather than about tunnelling.

What the pictures cannot show #

The barrier figure draws a potential against a radius, which is a one-dimensional summary of an object with no meaningful classical shape. The alpha is not at a place inside the well; there is an amplitude distributed over the well, and the drawing of a particle rattling between walls is a device for getting the attempt frequency, not a description.

Nor can any figure convey the size of the exponent honestly. $e^{-69}$ cannot be drawn beside $1$ on a linear axis at any scale, and on a logarithmic one it is a modest distance — which is precisely how a logarithmic axis lies about a quantity that spans thirty decades. The straight line in the hero figure looks like an ordinary correlation, and each unit along it is a factor of ten in something.

Where this ladder goes next #

The rung below establishes that a wave gets through a barrier at all; this one is about the only thing that then matters, which is how much. And the answer turns out to be an integral over the shape of the barrier — not its height, not its width alone, but the accumulated square root of the shortfall along the way — so that a problem about a particle becomes a problem about a curve.

The next rungs on this ladder are the ones where that integral is put to work rather than merely evaluated. A barrier whose shape is controlled is a scanning tunnelling microscope, in which the exponential sensitivity that made alpha lifetimes span decades becomes a way of measuring a height to a hundredth of an atomic radius. A barrier that a field tilts is field emission and the Fowler–Nordheim law, and it is why a sharp point emits electrons at a voltage a flat plate would ignore. And a barrier crossed in the other direction is fusion, where the same Coulomb integral has to be beaten by thermal energy rather than by an already-bound state, and where its steepness sets the temperature of a stellar core.

The habit worth carrying away is about where to look when a quantity varies absurdly. A range of twenty-four decades is never a sum of effects; it is a single exponent doing its job. The right response is to find the exponent, ask what is inside it, and stop trying to account for the range with anything in front.