# What does the term "half life" mean for a single radioactive particle? [duplicate]

I was introduced to the term half life as the time it takes for the number of radioactive nuclei to become half of its initial value in a radioactive sample.

But there is a question in "Concept of Physics by HC Verma}" which says that a free neutron decays with a "half life" of 14 minutes. Now this is really confusing. Here it is : What does the term half life even mean for a single radioactive nucleus or for a free neutron ? Does it mean that the neutron is only "half transformed" (into a proton and the beta particle) by the given time ?

• From en.wikipedia.org/wiki/Free_neutron_decay free neutrons are unstable and have a mean lifetime of 879.6±0.8 s (about 14 minutes, 39.6 seconds). Therefore, the half-life for this process (which differs from the mean lifetime by a factor of ln(2) ≈ 0.693) is 611±1 s (about 10 minutes, 11 seconds). Nov 6, 2021 at 14:01
• Does this answer your question? What happens to a radioactive material's atom when it disintegrates? Nov 6, 2021 at 21:21
• You are not likely to see a half-decayed neutron. Once started, the process completes on a timescale of 10e-25 s. Nov 6, 2021 at 22:59
• All you have here is an example of slightly imprecise writing. Obviously, what's meant is the half life of a large number of neutrons. Nov 7, 2021 at 2:55

For a single free neutron that exists a half-life of 14 minutes would mean that, over a timespan of 14 minutes, measured in the neutron's rest frame, there is a 50% chance that it will decay into a proton, an electron (beta particle), and an electron antineutrino.

(As @PM 2Ring notes in their comment on the original question, the half-life of a free neutron in reality is about 10 minutes, and the book's question mistakenly substituted in the value of the free neutron mean lifetime.)

• i.e. it is the probability of having decayed after the half time interval Nov 6, 2021 at 9:57
• For any finite length of time, yes it is mathematically possible for a free neutron to exist for that long, though it gets incredibly unlikely very quickly. For a free neutron to have existed as a free neutron in its rest frame for 13.8 billion years, you're looking at cumulative probability of approximately $1$ in $2^{500\,000\,000\,000\,000}$ Nov 6, 2021 at 10:32
• @anna v okay I got it.. but I think you need to edit your comment.. It should be a coin and not a dice.. thanks btw.. Nov 6, 2021 at 10:50
• @JuanPerez: It does not particularly matter in this case. The two viewpoints have a different interpretation of the word "probability" (average over all possible realizations of the experiment vs. personal belief in likelihood, to oversimplify a bit), but the math works out the same either way (and, as far as we can experimentally determine, appears to be an accurate model of both coin flipping and spontaneous particle decay). Nov 6, 2021 at 18:35
• @Ankit correcting: that is a different probability estimate. take a coin, throw it once, there is 50 % probability of being "heads" or "tails" . What is the probability of being continually "tails"/ after an infinite number of throws? – Nov 6, 2021 at 19:28

It means that for a large number of neutrons, half of them would decay every 14 minutes.

• The question is specifically how to interpret that for single particle. Nov 6, 2021 at 21:35
• Even for a large (but finite) number of neutrons there is still a finite (but little) probability that none of them decays in 14' - or, e.g. all of them decay in 14'. In a very good approximation, they don't feel each other's decay. Nov 6, 2021 at 22:55
• The part highlighted in blue is not about a single particle, even though the questioner thought it was. It means that after observing a great many neutrons half of them have been found to decay after 14 minutes. Nov 7, 2021 at 0:13

It means every half-life, the atom flips a coin. Heads it remains as it is. Tails it decays.

It's more random and continuous than that, but that's the gist.

It's conceivable that it could land ten "heads" in a row and thus survive 10 times its half-life. In fact, ALL the U-235 still on earth did something along those lines.

Both @John Hunter and @notovny answered your question. The following is a little more discussion of the good comment by @llmari Karonen on the @notovny response.

We have sufficient information from observing the decay of a very large number of identical radionuclides to claim we know the decay rate, hence the probability of decay, with no uncertainty. This probability is the same using either a classical (objective or frequency) approach, or a Bayesian (subjective) approach.

For the classical approach, the probability of event $$E$$ is $$P_O = lim_{N \to \infty} {N(E) \over N}$$ where $$N$$ is the number of independent trials and $$N(E)$$ is the number of times event $$E$$ occurs. For a very large $$N$$, observing $$N(E)$$, we know $$P_O$$ with essentially no uncertainty.

For the Bayesian approach, we assume a prior value for the event, $$P_S$$, and update it to a more accurate estimate for $$P_S$$, called the posterior, as we gather more information. For a large body of information we know the updated $$P_S$$ with no essentially uncertainty.

With sufficient information, the classical objective probability $$P_O$$ and the Bayesian subjective probability $$P_S$$ for the event are the same: one value with no uncertainty.

See the text Bayesian Reliability Analysis by Martz and Waller for information on the Bayesian approach.

For the more general case where we have limited information (trials for the classical case or state of knowledge for the Bayesian case) we have uncertainty in the probability. Using classical statistical inference this uncertainty can be expressed as a confidence interval for $$P_O$$. Using the Bayesian approach we can treat the classical $$P_O$$ as a random variable and express the uncertainty in $$P_O$$ as a subjective probability distribution for $$P_O$$ based on our imperfect state of knowledge. Our uncertainty is reduced as we perform more trials or improve our state of knowledge; the confidence interval is reduced and the subjective probability distribution is "narrowed". (More trials contribute to our improved state of knowledge, but in general other factors also contribute.)

For cases with significant state of knowledge (epistemic) uncertainty, we have insufficient information to use the probability measure of uncertainty, even in a Bayesian sense. For example, using a Bayesian approach, if we have a poor prior, and little information to update to a posterior, the poor prior cannot be modified accurately to provide a good posterior and the Bayesian estimate can be way off. For such situations, we can use a broader measure of uncertainty, such as evidence theory, and estimate belief/plausibility where belief and plausibility are, respectively, lower and upper bounds on probability.

For a single particle, you need to consider the wavefunction of the system. You start out with a neutron and this will then evolves into a superposition of a neutron, and the system comprising of a proton, electron and an anti-neutrino. And because the emitted electron will then interact with atoms and molecules in the environment, this superposition will involve more and more particles. This then leads to decoherence, the environment can then be said to have effectively measured whether or not the neutron has decayed.

• This answer is missing the word "half-life". Nov 7, 2021 at 20:06