= Finite-population radioactive decay
{title2=$N(t)\sim\operatorname{Binomial}(N_0,e^{-\lambda t})$}
Take $N_0$ independently decaying <atomic nuclei>, each with lifetime $T_i$ drawn from an <exponential distribution> of rate $\lambda$. With no replenishment, the survivor count is
$$
N(t)=\sum_{i=1}^{N_0}\mathbf1_{\{T_i>t\}},\qquad
p(t)=e^{-\lambda t}.
$$
Each <indicator variable> has success probability $p(t)$, so the <binomial distribution> gives
$$
\boxed{\mathbb E[N(t)]=N_0p(t),\qquad
\operatorname{Var}(N(t))=N_0p(t)(1-p(t)).}
$$
The smooth population law is the <expected value> $\overline N(t)=\mathbb E[N(t)]$, satisfying $\overline N'=-\lambda\overline N$. One realization instead loses whole <atomic nuclei> at discrete times. An observed count above the smooth curve need not contradict the <mathematical model>.
For an original chosen example, let $N_0=1000$ and <half-life> $3\,\mathrm h$. Then $\lambda=\log2/(3\,\mathrm h)$. After $9\,\mathrm h$, the survival probability is $1/8$, giving
$$
\boxed{\mathbb E[N(9\,\mathrm h)]=125,\qquad
\operatorname{SD}(N(9\,\mathrm h))=\sqrt{109.375}\simeq10.46.}
$$
The <standard deviation> describes variation between repeated populations governed by these assumptions. It is not an uncertainty interval for a noiseless deterministic curve.
\Image[/radioactive-decay-survivors-and-count-spread.png]
{title=Random survivor counts and the smooth radioactive-decay law}
{description=Original constant-rate model with $1000$ independent parent <atomic nuclei> and a $3\,\mathrm h$ <half-life>. Simulated histories are compared with their exact <expected value> and a one-<standard deviation> band. The second panel shows the exact <binomial distribution> after three half-lives. These are model calculations, not measurements from a historical sample.}
The <half-life> and the mean lifetime answer different questions:
$$
t_{1/2}=\frac{\log2}{\lambda},\qquad
\mathbb E[T_i]=\frac1\lambda.
$$
The first is the <median> of one lifetime, or the time when half the population survives on average. The second is the lifetime's <expected value>. In this example they are $3\,\mathrm h$ and approximately $4.33\,\mathrm h$.
Back to article page
