Suppose I have a sample which can decay via two processes and/or decay to two separate products, $$A \to B, A \to C$$

The total half life can be calculated $t= (t_A^{-1} + t_B^{-1})^{-1}. $

I am interested in something else though. Suppose $A$ goes through the following process: $$A\to B \to C$$ where the half life from $A \to B$ is different than the half life from $B \to C.$

What would be the total half-life $A \to C?$ Does it suffice simply to add the two half lives?

  • 12
    $\begingroup$ The problem is that the concatenation of these two processes leads to a process that is no longer Poissonian (in fact it should follow an Erlang-2 distribution), and hence the time dependence is no longer exponential. Typically, if we quote a half life, we are assuming the process is Poisonnian. $\endgroup$
    – march
    Commented Oct 25, 2022 at 23:57

3 Answers 3


You have another answer which states correctly that the concentrations as a function of time must be found numerically. However, in the common case where the lifetimes are very different, the concentrations will reach "secular equilibrium."

We can find the rate of change of the intermediate population $N_B$ as

$$ \frac{\mathrm d}{\mathrm dt} N_B = +\frac{N_A}{\tau_{AB}} - \frac{N_B}{\tau_{BC}} $$

That is, all of the $B$ are coming from the exponential decay of $A$, but if you were to remove all of the $A$, then the remaining $B$ would obey the classic exponential decay rule. This has an equilibrium when the derivative is zero, which happens at a population ratio of

$$ \frac{N_B}{N_A} = \frac{\tau_{BC}}{\tau_{AB}} $$

If the first decay $A\to B$ is fast, this ratio is large, because all of the $A$ disappear before the $B$ have had much of a chance to decay. However, if the second decay $B\to C$ is fast, then the equilibrium population of $B$ will be a constant fraction of the population of $A$. If you are running your experiment over a period much briefer than the slow $\tau_{AB}$, you can approximate $N_A$ as a constant and convince yourself that $N_B$ will approach this secular-equilibrium population exponentially, from above or from below, with time constant $\tau_{BC}$.

Two examples:

First, a common lab experiment is the cesium-barium generator. Cesium-137 has a half-life of about thirty years; it has a substantial branching fraction to an excited state in barium-137, which emits a gamma ray with half-life about two minutes. Because cesium and barium have different chemistries, you can wash the cesium source with a weak acid, and the acid will carry away barium ions which you can drip into a bottlecap and place under a student's Geiger counter. Your students' goal is to measure that the barium activity fades away with half-life of two minutes; your more challenging experiment would be to measure whether the barium solution "recharges" at the same rate.

Second, a common general-public radiation issue is radon gas accumulation in basements. This happens when a house's concrete foundation, or possibly shallow bedrock, contains a nontrivial concentration of uranium and its decay products. The longest half-life is the uranium, at a few billion years. All of the other decay products will tend towards their secular equilibrium populations. Radon, however, has noble-gas chemistry rather than actinide-metal chemistry, so the radon can transport itself out of bedrock and out of concrete. If you have a radon source in a room where the mixing with the atmosphere is slow, the radon can accumulate in that room. The most common radon isotope lives for about a week, and this is also the measurement time on high-quality radon test kits.

  • $\begingroup$ Ah yes, thats the exact lab experiment I am teaching at the moment $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 4:27
  • 5
    $\begingroup$ In my advanced lab in college, we used air blowers and filters to collect radon-saturated dust from the basement of the physics building, and then ran the filters up to the lab, cut them up, and put them into our NaI detector to track the decay chain. Quite a unique experiment, actually seeing the signals decay to nothing rather than the standard commercial sources with essentially constant activity. $\endgroup$
    – J. Murray
    Commented Oct 26, 2022 at 6:01
  • $\begingroup$ @J.Murray That was a fun experiment! $\endgroup$ Commented Oct 26, 2022 at 19:31
  • $\begingroup$ In accelerator experiments, there is usually some detector which picks up some beam-induced radioactive activation which was not intended in the experiment design. A sufficiently annoyed graduate student can watch this induced activity build up when the beam is on, then dissipate when the beam goes off. $\endgroup$
    – rob
    Commented Oct 27, 2022 at 13:00

Concentrations of substances $c_A(t),c_B(t),c_C(t)$ satisfy equations $$ c'_A=-k_1c_A $$ $$ c'_B=k_1c_A-k_2c_B $$ $$ c'_C=k_2c_B, $$ where $k_1=\frac{\ln{(2)}}{t_1}$ and $k_2=\frac{\ln{(2)}}{t_2}$.

The solution for initial conditions $c_A(0)=c_0$, $c_B(0)=c_C(0)=0$ is: $$ c_A=c_0e^{-k_1 t} $$ $$ c_B=\frac{k_1}{k_2-k_1}c_0(e^{-k_1 t}-e^{-k_2t}) $$ $$ c_C=\frac{1}{k_2-k_1}c_0\left(k_2(1-e^{-k_1 t})-k_1(1-e^{-k_2t})\right) $$

If half-life time $t$ is defined by: $c_C(t)=c_0/2$, then it is given by equation: $$ 2k_2(1-e^{-k_1 t})-2k_1(1-e^{-k_2t})=k_2-k_1 $$ $$ 2k_1e^{-k_2 t}-2k_2e^{-k_1t}=k_1-k_2 $$ $$ 2t_2e^{-\ln{(2)} t/t_2}-2t_1e^{-\ln{(2)} t/t_1}=t_2-t_1 $$ This equation has no analytic solutions.

  • $\begingroup$ Why is $t$ in the final equation? Surely the half lives do not depend on time $\endgroup$ Commented Oct 25, 2022 at 23:37
  • $\begingroup$ $t$ is the half-life $\endgroup$
    – atarasenko
    Commented Oct 25, 2022 at 23:39
  • 3
    $\begingroup$ Interestingly, it does not seem the half-life is independent of initial conditions. In other words, the idea of a “half-life” is not useful here because the solution is not proportional under time translation. $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 4:40

For an elementary decay reaction, the average decay time $\langle t \rangle$, decay half-life $t_{1/2}$, and decay rate $\lambda$ are all directly related and often used interchangeably in the sense that they "describe the same thing"

$\langle t \rangle=\frac{1}{\lambda}$ and $t_{1/2} = \frac{\ln2}{\lambda}$.

If you have two sequential decay processes you must be careful to not mix the different concepts. Say, the decay rates of the processes are $\lambda_1$ and $\lambda_2$ the average time for the complete decay process is indeed given by

$\langle t_{tot} \rangle = \langle t_1 \rangle+\langle t_2 \rangle$

where $t_i = \frac{1}{\lambda_i}$ as makes intuitive sense.

However, the time until half of your initial number of atoms have decayed is not simply given by the sum of the two half-lives. While we could calculate it, we do not usually do that because the whole point of the half-life is to describe an exponentially decaying process that only has one time-scale as already mentioned in another answer. If you tell someone that a certain process has a half-life of X, you are telling them not only when half the initial stuff has decayed but also that the process is a single exponential with a fixed rate. That's how the terminology is commonly used.


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