# Number of decays in a chain reaction

It is widely known that the probability of $n$ decays from one system to another $A \rightarrow B$ (e.g., electrons decaying from one atomic energy level to another or muons decaying into neutrinos and electrons, etc) in a given period of time is given by a Poisson distribution.

However, consider the following simple chain-reaction

$$A \xrightarrow{\lambda_A} B\xrightarrow{\lambda_B} C$$

where $\lambda_i$ is the decay rate. At $t=0$ (initial time) there are no particles in $B$ or $C$, all of them are in $A$. The question is how to compute or estimate the probability of a given number of decays from $A$ to $B$ plus the number of decays from $B$ to $C$ in a given period of time. In other words, one can assume that in each of the reactions one particle, say one photon, is emitted and one would like to estimate the likelihood of getting certain number of emitted photons over a period of time.

The probability of $n$ decays from $A$ to $B$ in a time $\Delta t$ must be given by a Poisson distribution with average number $\lambda_A\, \Delta t$. However, I wonder which the probability of $m$ decays between $B$ and $C$ in a period of time is. I do not think it follows a Poisson distribution given that the probability of getting $n$ decays in a given period of time should depend on time.

This must be something well-known since it has applications in nuclear (fission), atomic (spontaneous emission), and particle physics ($\pi^-$ going to $\mu^-$ (emitting $\bar\nu_{\mu}$) followed by muon decay). And also in chemistry. Seems to be something pretty common.

References are welcome.

N.B.: I am not asking about the distribution of particles in $A,\, B,$ and $C$ as a function of time.

• I believe it is a simple convolution of the populations as a function of tine with the per-atom decay probabilities. I don't know if there is a closed form solution or not, but it is easy enough to Monte Carlo a big population to get a useful sampling. – dmckee Oct 31 '13 at 19:54
• @dmckee Thanks! I know about Monte Carlo, but I'm looking for an exact or approximate analytical solution. Can you expand your comment into an answer? I don't see where the convolution comes from. Which is the "per-atom decay probability"$? – Diego Mazón Oct 31 '13 at 20:16 • @dmckee Could you please comment on my answer below? – Diego Mazón Nov 1 '13 at 0:58 ## 1 Answer The probability of getting$n$decays in a time interval$\Delta t$is given by $$p(n)=\sum_{m=0}^{n} \,p_A (m)\cdot p_B(n-m)$$ where$p_A(n)$is a Poisson distribution with average $$\lambda_A\,\int_{t_i}^{t_i+\Delta t} N_A(t') \, dt'$$ and equivalently for$p_B(n)$. And$N_A (t)$,$N_B(t)$are, respectively, the number of$A$-particles and$B$-particles. Note that these are the current number of particles, rather than the average numbers given by $$\langle N_A(t)\rangle =N_0\,e^{-\lambda_A\,t}$$ $$\langle N_B(t)\rangle =N_0\frac{\lambda_A}{\lambda_A-\lambda_B}\left(e^{-\lambda_B\,t}-e^{-\lambda_A\,t}\right)\,,$$ as pointed out by dmckee. This averages may be a reasonable approximation to the current values as long as the populations are large enough. (I have not checked this answer, feel free to criticize) • It seems wrong because for$t=0$the probability is zero for every$n$given that$N_B(0)=0$– Diego Mazón Nov 1 '13 at 1:06 • "number of number particles"? Did you mean total? – David Z Nov 1 '13 at 1:06 • This edited version appears to be fine as a starting place, but to understand the full distribution for$p(n)$you have to understand the variation of$N_A(t)$and$N_B(t)$which is the hard part of the problem and what I meant by convolving with the populations. – dmckee Nov 1 '13 at 3:34 • The expectation for the populations is given by those nice first-order equations, but in any given run the actual populations will vary which contributes to the full variation of$p(n)$. If your populations ($N_A$and$N_B\$) are large you can neglect this, but I assumed you were asking the hard question. – dmckee Nov 1 '13 at 4:09
• Yep. This is most important when the populations are very small, but at least those are the easy cases to Monte Carlo. – dmckee Nov 1 '13 at 4:19