# What happens before a radioactive element decays?

What happens to a radioactive element just before it decays?

In school, I've been told that the decay process of an element is absolutely random, and it is impossible to determine which unstable element decays next. Clearly, there needs to be a triggering event. What is this event?

• "Clearly, there needs to be a triggering event." - Nope. It just happens. Quantum mechanics is weird. – OrangeDog Feb 12 '15 at 10:28
• schrodinger's cat decided it wanted to explode – ratchet freak Feb 12 '15 at 11:14
• The atom takes a trip to Monte Carlo. – Hot Licks Feb 12 '15 at 16:43
• @OrangeDog Does this also rule out hidden state? For instance, maybe we could tell when a radioactive element was going to decay were it not for the Heisenberg uncertainty principle. – Michael Feb 12 '15 at 18:40
• Nope. See others answers for the in-depth reasons. – OrangeDog Feb 12 '15 at 19:14

Nothing happens! It's random! The nucleus is in an unstable state, and unstable states have a certain small probability to decay within a given amount of time (how small depends on the nucleus). There's not much else to it! Sometimes decay can be stimulated but the type of decay you're talking about is truly random.

• I wish there was a more scientific answer than it's random. – René Nyffenegger Feb 12 '15 at 8:12
• You could calculate a wave function and a probability distribution to feel better, but that's the way quantum mechanics is--it's not deterministic. The only thing that can be predicted are probabilities. – Surgical Commander Feb 12 '15 at 8:14
• @RenéNyffenegger You're in good company. A certain physicist said pretty much the same thing nearly 90 years ago. – Phylogenesis Feb 12 '15 at 11:57
• @RenéNyffenegger your best bet is to get used to the idea that "random" is scientific. It's a very precise kind of "random". Randomness can be highly scientific, and an entire branch of mathematics is dedicated to drawing accurate conclusions about "random" things. – Roman Starkov Feb 12 '15 at 18:22
• @ReneNyffenegger Of course we'll never be able to conclusively prove that quantum processes have no patterns at all beyond the Born Rule - there could always be correlations too subtle for our tests to capture - but empirical analyses of random(?) numbers generated by quantum processes show that they're more random than those from any of our best pseudorandom number generators: arxiv.org/abs/1004.1521v1 – tparker May 23 '16 at 0:37

As the other answers state, the individual nuclei have a probability of decay and this happens randomly, as they sit there.

You are correct though in wondering about a trigger, because at the atomic level that is exactly what happens with lasing, induced-emission = induced-decay.

Spontaneous decay is random, controlled by the quantum mechanical individual atom probabilities. A trigger is not necessary for spontaneous decay, but can induce decays, enhancing the probabilities greatly; that is how the lasing phenomenon is observed.

So a further question could be: "Is it possible to induce/trigger nuclear decays so that the probabilities of decay become large?". And the answer is yes; that is how we have nuclear reactors and bombs. An induced fission reaction. A neutron is absorbed by a uranium-235 nucleus, turning it briefly into an excited uranium-236 nucleus, with the excitation energy provided by the kinetic energy of the neutron plus the forces that bind the neutron. The uranium-236, in turn, splits into fast-moving lighter elements (fission products) and releases three free neutrons. At the same time, one or more "prompt gamma rays" (not shown) are produced, as well.

It is called a chain reaction and depends on the density of neutrons available in the material, and thus is not random.

Spontaneous decays are random. Induced ones are not; they can be triggered.

• This means that in the event one nucleus decays, one Planck time before this event there is no difference in the nucleus, compared to the scenario where it doesn't decay? – vsz Feb 13 '15 at 7:15
• @vsz Yes, there is no difference in the probability distribution ( the square of the wavefunction) that would announce/be-measured-as a decay imminent. – anna v Feb 13 '15 at 12:02
• Also, in the case of electron capture where an orbit electron is captured by the nucleus, the probability can be changed by altering the electron cloud and its "density" inside the nucleus. In particular, if all electrons are stripped away from the atom (ionizing it completely), the probability of capturing an electron should become zero. But the chemical surroundings of the atom can also matter. Wikipedia has a link to the paper Change of the Be-7 electron capture half-life in metallic environments. – Jeppe Stig Nielsen Feb 13 '15 at 13:48

There really is none.

Unstable elements (and unstable elementary particles) can decay into a less energetic state. However, each kind of decay depends on a quantum mechanical process, this is tunneling for $\alpha$, a virtual $W^\pm$ for $\beta$ or a transition from one nuclear shell to another for $\gamma$. Now these underlying processes can be strongly supressed, meaning that they are incredibly unlikely. Then it will take a "long" time until the process happens, and we have an unstable material.

The lifetime of a nucleus depends on how probable the decay process is. This can be computed to some accuracy for most decays. I think one of the most easy approaches to understand this is Gamow's theory of alpha decay, where the $\alpha$ particle is trapped inside the unstable core and keeps bouncing off potential walls. At every bounce however, the $\alpha$ particle has a tiny chance of tunneling through the potential barrier to the outside, where it is free. The higher and thicker the barrier, the less frequent the decay, the longer the half-life of the nucleus.

From this example it should be clear how nuclear decay can take a long time, but it still happens spontaneousely without an external trigger event.

• Does Gamow's theory describe what really happens or just a metaphor? – René Nyffenegger Feb 12 '15 at 8:18
• Well, the bouncing about part is a simplification to help understand the process without knowing QM. In reality, the $\alpha$ particle does not bounce about, but has still only has a finite chance per second to pass through the barrier. The more crucial point that the decay happens through the quantum mechanical tunneling process, which is very rare, is perfectly correct, though. – Neuneck Feb 12 '15 at 8:20
• Shouldn't the end of the first paragraph say 'stable' ? – Spine Feast Feb 12 '15 at 8:28
• No. Even if it takes a long time to decay, such a material is considered unstable. See uranium for example which has a half-life of more than 1000yrs. – Neuneck Feb 12 '15 at 8:32
• @RenéNyffenegger yes, just a metaphor – OrangeDog Feb 12 '15 at 10:29

An " intuitive" approach is to consider that in QM, the exact location of particles doesn't exist. They're all probability waves, and you never have a 100% chance to find a particle in exactly one place.

So for unstable nuclear atoms, the probability function of the protons and neutrons are smeared out even further. There's a significant non-zero probability of finding one of the particles away from the others.

But like Schrodinger's cat, a probability is not yet a fact until there's an observation. The nuclear decay has and has not happened at the same time. Once you observe that decayed atom or the particle it emitted, the probability wave collapses. The probability of finding the original radioactive becomes zero.

All the other answers that "no, there is no triggering event, it just happens, quantum mechanics is like that" are perfectly right.

But let's look at the experimental evidence for these answers. Yes indeed, there is considerable experimental evidence that heavily falsifies the idea that there is a triggering event.

This evidence is the statistical probability density for the decay lifetime. It is found in countless experiments that the decay times are exponentially distributed. The exponential distribution is the unique continuous pdf with the following property: if we take a particle which we know has not decayed after some time, any positive time (even if it is a million times longer than the mean decay lifetime), then the probability distribution, conditioned on this knowledge, of the particle's lifetime after that point is exactly the same as the unconditional distribution for the particle's lifetime. The discrete analogue of this dense statement is the geometric distribution and the discrete version of the idea can be summed up as "a coin has no memory". That is, if you toss a coin known to be fair and get a long run of heads, the probability that the next throw will give a tail (or head) is still $1/2$.

If there were internal "clockwork" to the particle that meant there were several stages, separated by "trigger events", to the decay, then we would not see the exponential distribution. Suppose, like a fluorophore, that the particle rises into a radiative state, decays to a metastable state and then fluoresces. If the radiative state's lifetime is significant compared with the metastable state's, then the distribution of fluorescence lifetimes would be the convolution of two exponential distributions, not the exponential distribution. This is indeed what we do see if we examine fluorescence lifetimes carefully. The fluorophore has a "memory" and is like a three-state finite state machine.

I explain these ideas more in my answer to the Physics SE question, "Are There Old Aged Particles".

Consider a particle in a box, but where the box has a thin wall, and the energy level outside the box is lower than that inside. (This is, e.g., a neutron in an unstable heavy nucleus.) Follow the development of this wave in time. It will tunnel out eventually to the lower energy state and propogate away. From this, you can see that the decay is always "happening", with the usual ontological caveats regarding wavefunctions.

An observation of the particle outside the nucleus in a particular place will invalidate the wavefunction view of the system; the decay has then happened in the normal sense. And because the wavefunction before this "collapse" is extended in time, and decaying exponentially, the time of the detection will be random, though distrubuted, over several experiments, with the exponential decay.