Question 2 of 7: Poisson Statistics of Radioactive Decay

Question 2: Poisson Statistics of Radioactive Decay (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The normalized Poisson distribution $P(n;x)=x^n e^{-x}/n!$ and the series identity $\sum_{n=0}^{\infty}x^n/n!=e^x$.

Find. (a) the physical justification for a Poisson model of decay; (b) $E[n]=\lambda t$; (c) the survival-then-decay probability density; (d) the mean lifetime $\tau=1/\lambda$; (e) whether parent–daughter equilibrium exists when $\lambda_1\approx \lambda_2$, with a mathematical check.

Approach. Treat each atom's decay as an independent Bernoulli trial with a tiny per-interval probability, so the population obeys the Poisson (binomial) limit; use the series identity directly to evaluate the Poisson mean; differentiate the survival function for (c); integrate $t$ against the decay-time density for (d); and solve the two-nuclide Bateman system at $\lambda_1=\lambda_2$ for (e).

1. Part (a) — why Poisson fits radioactive decay. In a sample of $N$ identical, non-interacting radionuclides, each individual nucleus has the same small, constant probability $p=\lambda\,\delta t$ of decaying in a short interval $\delta t$, independently of every other nucleus and of its own past (decay has no "memory" — an old nucleus is exactly as likely to decay in the next instant as a freshly-formed one). This is precisely the binomial set-up ($N$ independent trials, probability $p$ each) in the limit of a very large number of trials ($N\to\infty$) and a very small per-trial probability ($p\to0$) with the product $Np$ held finite — the classical Poisson limit of the binomial distribution. Because $N\sim10^{20}$ or more for any macroscopic sample while $p=\lambda\,\delta t\ll1$ for any measurement interval, the number of decays observed in a fixed interval is Poisson-distributed to excellent approximation.
2. Part (b) — mean number of decays equals $\lambda t$. With $x=\lambda t$, the expected (mean) number of decays in time $t$ is $$E[n]=\sum_{n=0}^{\infty} n\,P(n;x) = \sum_{n=0}^{\infty} n\,\frac{x^n e^{-x}}{n!} = e^{-x}\sum_{n=1}^{\infty}\frac{x^n}{(n-1)!}$$ Re-indexing with $m=n-1$ and using the given identity on the resulting sum, $$e^{-x}\sum_{n=1}^{\infty}\frac{x^n}{(n-1)!} = e^{-x}\,x\sum_{m=0}^{\infty}\frac{x^m}{m!} = e^{-x}\,x\,e^{x} = x$$ $$\boxed{E[n] = x = \lambda t}$$ So the mean number of decays in an interval $t$ is exactly $\lambda t$ — which is why "activity" (decays per unit time) and the Poisson rate parameter are the same quantity.
3. Part (c) — probability of surviving to $t$ then decaying by $t+\delta t$. The probability that a single nucleus (decay constant $\lambda$) has not decayed by time $t$ is the exponential survival function $S(t)=e^{-\lambda t}$ (the $n=0$ term of the Poisson distribution with $x=\lambda t$, since "zero decays by $t$" and "survives to $t$" are the same event for one nucleus). The probability of then decaying in the next short interval $\delta t$ is $\lambda\,\delta t$ (the definition of the decay constant), independent of the nucleus's age. Multiplying the two independent-in-sequence probabilities: $$\boxed{P(\text{survive to }t,\text{ then decay in }[t,t+\delta t]) = \lambda\,e^{-\lambda t}\,\delta t}$$ This is exactly the decay-time probability density $f(t)=\lambda e^{-\lambda t}$, which is used directly in part (d).
4. Part (d) — mean lifetime. The mean lifetime is the expectation of the decay time under the density found in part (c): $$\tau = \int_0^\infty t\,f(t)\,dt = \int_0^\infty t\,\lambda\,e^{-\lambda t}\,dt$$ Integrating by parts ($u=t$, $dv=\lambda e^{-\lambda t}dt$, so $v=-e^{-\lambda t}$): $$\tau = \Big[-t\,e^{-\lambda t}\Big]_0^\infty + \int_0^\infty e^{-\lambda t}\,dt = 0 + \left[-\frac{1}{\lambda}e^{-\lambda t}\right]_0^\infty$$ $$\boxed{\tau = \frac{1}{\lambda}}$$ (Equivalently, $\tau=T_{1/2}/\ln2$, since $\lambda=\ln2/T_{1/2}$.)
5. Part (e) — equilibrium when $\lambda_1\approx\lambda_2$. Start from the Bateman solution for a parent (1) decaying to a daughter (2), $N_1(t)=N_1(0)e^{-\lambda_1 t}$ and, for the general case $\lambda_1\neq\lambda_2$, $$N_2(t) = \frac{\lambda_1 N_1(0)}{\lambda_2-\lambda_1}\big(e^{-\lambda_1 t}-e^{-\lambda_2 t}\big)$$ This expression is $0/0$ as $\lambda_2\to\lambda_1$; applying L'Hopital's rule in $\lambda_2$ at $\lambda_2=\lambda_1=\lambda$ gives the degenerate (equal-half-life) solution $$N_2(t) = \lambda\,N_1(0)\,t\,e^{-\lambda t}$$ The daughter's activity is $A_2(t)=\lambda N_2(t)=\lambda^2 N_1(0)\,t\,e^{-\lambda t}$, while the parent's activity is $A_1(t)=\lambda N_1(0)e^{-\lambda t}$, so their ratio is $$\boxed{\frac{A_2(t)}{A_1(t)} = \lambda t}$$ This ratio grows without bound as $t$ increases — it never settles at a constant value, so no equilibrium (neither secular nor transient) is reached when the parent and daughter half-lives are equal. Physically, the daughter population keeps being fed by a parent that is decaying no faster than the daughter itself can clear, so the daughter never "catches up" to a fixed activity ratio with its parent.