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Is there any way to predict the half-lives of radioactive isotopes from theory (that is, using only theoretical considerations, without using data about the decay)? For example, could we predict that the half life of Carbon-14 is roughly 5700 years?

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Answering this question is one of the major successes of 20th-century physics.

For strong decays, Gamow's alpha-tunneling model is quite successful. It relates the lifetime of an alpha emitter to the energy released in the decay using the approximately-valid assumption that nuclear density is constant and that the nucleus has a relatively sharp edge.

For beta decays there is quantity "$ft$" which convolves the half-life of the decay with the electrical interaction between the emitted electron and the positively-charged daughter nucleus. The $ft$-values are related in a relatively simple way to the matrix element for the decay, and for a given class of decay ("allowed", "superallowed", "first forbidden", etc., which are determined by the quantum numbers of the parent and daughter nucleus) the $ft$ values for most nuclei fall into a pretty narrow range.

Any nuclear physics textbook should have a chapter on characterizing decays. (I happen to be looking at Wong.)

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There is a large class of weak decays that can be predicted fairly well from theory (using the notion of weak universality). However, I have in mind the weak decay of heavy leptons and individual hadrons, so that is not quite what you were asking for as it occurs outside the nuclear context.

You can tackle the nuclear problem too, but the phase space computation in complicated by the nuclear structure.

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In theory it could be done. The problem is that we're dealing with hundreds of entangled nucleons. A particle can be modeled as a wave in three-dimensional space. Two particles that aren't entangled can be modeled as two waves. But if they are entangled, you have to use one wave in six-dimensional space. In order to model an atomic nucleus, you'd need 300-dimensional space. Solving that numerically would require a computer vastly larger than the universe.

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  • $\begingroup$ Sorry, I chose a bad example. I changed the example to Carbon-14. Is this doable? Also I think you are talking about estimating the half-life from computer simulation. But I think I heard once where theory directly predicts the half life of some isotope just using pen and paper calculations. $\endgroup$ – Joshua Benabou Apr 19 '15 at 0:04
  • $\begingroup$ Carbon 14 still would require 42 dimensions. If you want to look at ten positions, which I doubt is enough, you'd need 10^42 nodes. That's not going to work. Are you talking about just using the number of protons and neutrons, or using the mass? If you know the mass, you might be able to do it easily, but then you don't know why it has exactly that much binding energy, so I don't know if it really counts. $\endgroup$ – DanielLC Apr 20 '15 at 1:40
  • $\begingroup$ You can use the mass. $\endgroup$ – Joshua Benabou Apr 20 '15 at 1:43
  • $\begingroup$ As an outside observer, that's disastrous. If we can't calculate properties of stinking atoms using theory, why do we expect a solvable TOE at all? $\endgroup$ – Zach466920 Apr 21 '15 at 21:27
  • $\begingroup$ Well apparently it can be calculated perfectly a sufficiently powerful computer. $\endgroup$ – Joshua Benabou Apr 21 '15 at 23:12
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Here is a a link to a course textbook addition from the University of Michigan for a nuclear physics course. This theory about the mean lifetime seems pretty straight-forward, except you have to come up with a perturbation potential to do the quantum mechanical calculation, i.e., there needs to be some kind of reasonable potential barrier model. So, looks possible to calculate from quantum mechanics using Fermi's Golden Rule #2.

Calculating mean lifetimes, and thus half-lives

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    $\begingroup$ Welcome to Physics! Please do not post links without at least a basic summary of what they contain and how that content answers the question, since link-only answers become useless if the link rots away. Link-only answers are not considered answers here and will be deleted. $\endgroup$ – ACuriousMind Jan 22 '18 at 19:03

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