The law of radioactive decay reads

$$ N(t)=N_0e^{-\lambda t}$$

Is it valid when there is less than 1 nucleus or particle to decay? Obviously, it is nonsense to consider that we have 1/2 of nucleus or a number lower than 1 in N(t). Has someone defined the time when 1 nucleus or particle are left, that is, the quantity

$$T(1)=\dfrac{\ln N_0}{\lambda}$$?

Obviously, T(1) should be related to half-life T(1/2) and mean lifetime $\tau$ , why is it not common that measure? After all, it seems that when less than 1 nucleus is left, there is no "fair" to continue with the exponential model (or yes?). I read somewhere a sentence: "Humans are not able to understand the exponential law"...Of course, that is quite non exact but the exponential laws are tricky...


2 Answers 2


To answer your question, let's look at how this equation can be derived. Say, at some time $t$, there are $N(t)$ nuclei. Let $p_t(Δt)$ be the probability that any one nucleus has not decayed (this probability is assumed to be the same for all nuclei) after an additional time $Δt$.

If we also assume that there are a lot of nuclei (this is important), we can say that at time $t+Δt$, there are $N(t+Δt) = p_t(Δt)·N(t)$ left (cf. the law of large numbers).

Rearranging this equation, we get $\frac{N(t+Δt)-N(t)}{Δt} = \frac{p_t(Δt)-1}{Δt}N(t)$.

It is reasonable to assume that the probability of a single particle not decaying is continuous and time independent (that is, if we change our time interval $Δt$, $p_t(Δt)$ will not "jump", and the probability for a fixed time interval will not change with time, which means $p_{t_1}(Δt) = p_{t_2}(Δt)$). Also $P_t(0) = 1$ obviously.

Now let's treat N as a continuous function (this is important as well) and let $Δt→0$. Then:

$\frac{dN(t)}{dt} = \lim_{Δt→0} \frac{N(t+Δt)-N(t)}{Δt} = \lim_{Δt→0} \frac{p_t(Δt) - 1}{Δt} N(t) = \frac{dP_t}{dt} N(t) =: -λN(t)$.

(We also assumed that $N$ and $p_t$ is differentiable.)

This equation is a differential equation for N(t). It has the solution $N(t) = N(0) e^{-λt}$.

Now, let's review this derivation:

To get this equation, we assumed N to be large so that the law of large numbers applies. We also assumed N to be a continuous function instead of a discrete one. This is obviously not true, because there can be no, say $\sqrt{2}$, nuclei. However, this is not a serious problem, because we assumed N to be large: If N is large, we cannot count the number of nuclei exactly anyway, so a small error is acceptable, because the result still gives a very good approximation. So no, this equation does not hold for small N, especially not for N(0) = 1.

  • $\begingroup$ However, even in particle physics we apply the exponential law to events with low statistics from a Poissonian law, so, I guess there is another reason why we could in principle apply the exponential law to "low number of nuclei" (N being discrete!). I think this question is very interesting...And I was preparing some exams for my students when I realized all this...Of course, I have guess my own answer, but I want to see what people have to say... $\endgroup$
    – riemannium
    Jun 29, 2014 at 20:33
  • $\begingroup$ @riemannium Well, dmckee's answer provides an equation that is valid for any number of nuclei, though it's not quite the same as $N(t) = e^{-λt}$. $\endgroup$ Jun 30, 2014 at 18:02

A more general way to state the physics here is something like

The probability that any given nucleus will have already decayed after a time $t$ has elapsed is $P_{decayed} = (1 - e^{-t/\tau})$ and is independent of the fate of all other nuclei.

Which remains valid for any number of nuclei and from which you can easily extract the large numbers law that you quote above.

Perhaps some people would prefer to turn the above statement around to something like the next statement

The probability that a given nucleus will have failed to decay after a time $t$ has passed is $P_{undecayed} = e^{-t/\tau}$ and is independent of the fate of all other nuclei.

but the cases are trivially related to one another.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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