6
$\begingroup$

My first question on Stackexchange (if it is formatted wrong or something please tell me so I know in future) - here it is:

Given an unstable nucleus (exactly which nucleus is not particularly pertinent) - what decides precisely when it will decay?

I am somewhat familiar with the concept- a nucleus becomes 'unstable' when the mutual repulsion between the protons exceeds the binding force of the strong nuclear force. That being said, if we were to theoretically isolate a single unstable atom, there are a few possibilities (correct me if I am wrong):

  1. atom immediately decays
  2. atom decays x seconds later
  3. atom doesn't decay

There is an unpredictable nature (which I assume arises from something to do with quantum uncertainty). But getting back to my question, what is it that suddenly makes an 'unstable nucleus' decay?

Any further reading would be appreciated, although nothing too complicated (university level is my limit I think).

UPDATE

Thanks all, my first time using this forum and I was not let down :) With regards to the question itself, I'm a bit disappointed that there is no precise mechanism, but I guess Einstein was wrong when he quoted: "God does not play dice"

$\endgroup$
5
$\begingroup$

As far as we know, nuclear decay is truly random, that is, random in the quantum mechanical sense. That is, when you observe the system, there is a probability that you will see the decay products rather than the original nucleus, because the wave function of the system is a superposition of the parent nucleus state and the daughter nucleus state (+alpha particles or whatever). As time goes on, the coefficients of the superposition evolve so that the probability of observing the parent nucleus approaches zero while the probability of observing the daughter nucleus (or further decay products) approaches one.

Thus, there is nothing immediately happening that causes the nucleus to decay; rather, as the parent nucleus is unstable, time evolution gradually eliminates it from the system in a continuous manner! In the Copenhagen interpretation, observing the nucleus causes it to collapse into a state where you can definitely ascertain whether it has decayed or not. The longer you wait, the more likely.

$\endgroup$
  • $\begingroup$ One should add, that the actual decoherence time of a nucleus because of the continuous measurement process by the electromagnetic radiation field is probably short. If one adds the hohlraum radiation to the (theoretical) experiment, the decoherence of the system is assured by the third law of thermodynamics, which doesn't allow us to reduce the temperature to 0K. $\endgroup$ – CuriousOne Sep 10 '14 at 7:16
2
$\begingroup$

Your question addresses a general principle in quantum mechanics. If we have an initial state $i$ and a final state $f$ then we can calculate the probability of a transition from $i$ to $f$, but this is only a probability - we cannot say when the transition will happen, only the probability that it will happen in some time interval. This isn't because we don't know enough about what's going on, but rather it's a fundamental principle in QM.

In your example of a radioactive nucleus, we could in principle write down Schrodinger's equation for the nucleus and solve it to calculate the eigenfunctions. These eigenfunctions are the wavefunctions that describe the ground state, the first excited state, the second excited state and so on. Let's call the first excited state $\psi_i$ and the ground state $\psi_f$, then the decay corresponds to transition $\psi_i \rightarrow \psi_f$.

In order for the decay to happen there must be some physical process that acts on the initial state $\psi_i$ and changes it to something else. Typically the process will change $\psi_i$ to a mixture of $\psi_i$ and $\psi_f$, in other words it changes our initial excited state to a superposition of the initial excited state and the final ground state. The physical process will be some complicated differential equation, but let's represent it by the symbol $\hat{V}$, so the action of the operator can be written as:

$$ \hat{V}\psi_i = c_i\psi_i + c_f\psi_f \tag{1} $$

So the operator produces a superposition that is some fraction $c_i$ of the initial state and some fraction $c_f$ of the final state. Over time $c_i$ will decrease and $c_f$ will increase, so over time the superposition looks less and less like the initial state and more and more like the final state but there is no sharp cutoff between the two.

If you want to calculate the transition probability then you use an equation called Fermi's Golden Rule. I'll write this down, though don't bother about the details because they get involved:

$$ P_{i\rightarrow f} = \frac{2\pi}{\hbar}\langle\psi_f|\hat{V}|\psi_i\rangle\rho \tag{2} $$

where $P_{i\rightarrow f}$ is the probability that the transition will occur per unit time.

The only important bit of this is the $\langle\psi_f|\hat{V}|\psi_i\rangle$ because this picks out the value of $c_f$ from equation (1) above. If $c_f$ is small, i.e. if the superposition is mostly composed of the initial state, then $\langle\psi_f|\hat{V}|\psi_i\rangle$ will be small and the transition probability will be small. Conversely if the superposition is mostly the final state $\langle\psi_f|\hat{V}|\psi_i\rangle$ will be high and the transition probability will be high.

To make this a bit more concrete, in gamma decay the operator $\hat{V}$ is the function that creates a photon so it describes the process:

$$ \text{nucleus} \rightarrow \text{nucleus} + \text{photon} $$

Beta decay is more complex because the beta decay 1. destroys a neutron, 2. creates an electron, 3. creates an antinueutrino and 4. creates a proton, so in this case $\hat{V}$ describes the process:

$$ n \rightarrow p + e + \bar{\nu} $$

Still, in both cases the decay probability per unit time is (in principle) still obtained by plugging $\hat{V}$ into equation (2). I say in principle because in practice the calculations are usually too hard to do except as approximations.

$\endgroup$
0
$\begingroup$

What mechanism decides when an unstable nucleus decays? [...]

Let me rephrase that question so:

"Given some initial number of (otherwise equal) objects, and having measured the sequence of their subsequent decays (if any),
what can we conclude about the mechanism, or "barrier", which had prevented them from each having decayed/disintegrated right away?"

To consider the simplest possibilities first:

1: If they had indeed all decayed/disintegrated right away, then (we say that) there was no "barrier" against that to speak of.

2: If none had decayed/disintegrated within some (non-zero) duration of a trial, then, obviously, there had been some "barrier" (or perhaps rather: "prohibition") against such decays, which had been impenetrable (or "rigorous") so far, in the trial under consideration.

3: If we're given and looking at only precisely one object and if it had been seen to decay in the trial under consideration (and, as a distinction to 1:, if that decay had not occured "right away", but after some finite duration of that object having "lived") then:
we can conclude that there had been some barrier; but we cannot draw further conclusions (such as will be described below), with any confidence.

4: If we're given and considering some "suitably large number" $N_0$ of objects, and if it is found that the measured sequence of their decays followed the "usual exponential law" (including "incidental statistical deviations"), $$\frac{1}{N[~T~]}~\frac{\Delta N[~T~]}{\Delta T} \approx \text{constant},$$ then it can be concluded (with confidence increasing as $N_0$ increases, and as $\frac{N_0 - N_{\text{end of trial}}}{N_0}$ increases, and with the "likelihood of incidental statistical deviations") that the "(potential) barrier" had been equal for all objects and constant throughout the trial, and that the "mechanism" is well described as quantum tunneling. In case of $\alpha$-decay especially: through a "barrier" due to strong force binding nucleons together; according to the analyses by Gamow, Gurney, Condon ...

So: the conclusion, from a trial as described, about the "(potential) barrier" having been (as good as) equal for all given unstable objects/nuclei, and constant (with respect to external "conditions" or "internal parameters"), is derived just due to the decays having been independent of each other, and "statistically random", and the "decay mechanism" having been perfectly universal instead of determining the specific "precise" life duration for each individual unstable nucleus.

To illustrate still more possibilities:
5: If we're given and considering some "suitably large number" $N_0$ of objects, and if it is found that, after having lived for a while, they all decayed together at once, then:
the "mechanism responsible" for the "sudden drop of the barrier" might be called a "trigger", and even a "perfect trigger", with confidence increasing with $N_0$.

6: If there were found some (more or less "systematic") deviations from "usual exponential law" as described in 4, then different "mechanisms" might come into consideration, and "(potential) barriers" other than constant might prove more probable; attributable to "conditions having varied" in the course of the trial.

$\endgroup$
0
$\begingroup$

At least some radioactive decay is non-random. The measured modulation of decay rates on a 6 month or seasonal interval strongly suggests that the flux of solar neutrinos has a significant effect on radioactive decay. The degree of the coupling is SO pronounced that I wonder if ANY decay is truly spontaneous.

$\endgroup$
  • 2
    $\begingroup$ This seems to be busted: phys.org/news/… $\endgroup$ – Sebastian Riese Oct 18 '15 at 21:11
  • $\begingroup$ This has been discussed in several questions, and the overall feeling it not optimistic that this represent a real effect. In principle neutrino rates should matter for some beta decays, but getting a measurement will be difficult. Nor would I term those events "decay". $\endgroup$ – dmckee Oct 18 '15 at 22:29

protected by Qmechanic Oct 18 '15 at 20:45

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.