Find the average life of a radio nuclide which decays by parallel paths, \begin{align} A &\rightarrow B\\ 2A &\rightarrow B, \end{align} where the decay constants are $\lambda_1 = \pu{0.018 s-1}$ and $\lambda_2 = \pu{0.001 s-1}$, respectively.

I used the formula $$\frac{1}{t} = \frac{t_1 t_2}{t_1 + t_2,}$$ where $t$ represents the mean life. The equation can be written as: $$\frac{1}{t} = \lambda_1 + \lambda_2$$ But I am getting the correct answer only if I take $\lambda_2$ as 2 times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

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    $\begingroup$ A source for the exercise would really go a long way as this is a very strange problem. $\endgroup$ Commented Feb 26, 2020 at 11:38
  • $\begingroup$ The first answer and question is misleading, second order reactions do not occur during radioactive decay. I think that you need to consider branching decays such as that of bismuth-212. This will give you a better way of dealing with a real world problem of multiple reactions from the same nuclide. $\endgroup$ Commented Dec 7, 2022 at 11:53

2 Answers 2


But I am getting the correct answer only if I take $\lambda_2$ as $2$ times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?


The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations. Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is $$ \left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A], $$ and that of reaction 2 is $$ \left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A]. $$

Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of $2$ in the differential equation for reaction 2 is because two $A$ nuclei are involved per "unit" of reaction progression.

The total rate of loss of $A$ is the sum of the above: $$ \frac{\mathrm{d}[A]}{\mathrm{d}t} = -\lambda_1[A] -2\lambda_2[A] = -\left(\lambda_1 + 2\lambda_2\right)[A]. $$

This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.

While it seems quite unusual to me to have a second-order reaction ($2A\longrightarrow B$) with a first-order rate constant (units of $\mathrm{s}^{-1}$), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.

As well, simultaneously having $A \longrightarrow B$ and $2A \longrightarrow B$ processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. $2A \longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.


I think that both the person asking the question and hBy2Py have both got things rather wrong. Have no fear a nuclear worker is here.

There are some radionuclides such as neptunium-238 which can decay in multiple reactions.

For this neptunium isotope there are three different modes, alpha, beta- and beta+. We have three rate constants.

Lalpha, Lbeta+ and Lbeta-

The rate constant for the disappearance of the radionuclide is given by summing the rate constants for the three routes to get the total rate constant for radioactive decay.

Lphysical = Lalpha + Lbeta+ + Lbeta-

Things get more complex if the neptunium is inside the body of an animal. Then we have to consider the biological half life. If Tbio is the biological half life for neptunium in the animal then we can write

Kbio = ln(2)/Tbio

Then Ktotal for the rate of loss of the neptunium from the animal (first order reaction) is given by

Ktotal = Kbio + Lalpha + Lbeta- + Lbeta+

If we have a radionuclide whose decay does not branch then the algebra is a bit more simple. Consider Cs137

Then we can write

Teff (effective half life) = 1/((1/Tbio)+(1/Tphys))

Tphy is the phsyical half life which is the radioactive half life.

  • $\begingroup$ I am very curious about the particular example of such a real case, as I cannot remember any. It is common to have parallel decays ( E.g. potassium-40 decays by all 3 beta decay modes -- e-, e+, e-capture). But A -> B vs 2A -> B rather smells by mass/energy conservation law breach of a not well thought through made up task. $\endgroup$
    – Poutnik
    Commented Dec 7, 2022 at 11:56
  • $\begingroup$ If we consider a nuclide which has two decay branches, say for arguments sake that 90 % of decays are alpha and 10 % are beta-. Then 1 in 10 of the atoms will undergo beta decay while 90 % undergo alpha. Then the rate constant for the beta decay will be ten times lower than the rate constant for the disappearance of the nuclide $\endgroup$ Commented Dec 7, 2022 at 12:07
  • $\begingroup$ That would be clear - for usual branching. But these branches will have different products, not like for the OP, not speaking about different reagent /product ratio. Unless those 2 reactions are some error. // If the 2 at 2A is a typo and should be just A, I can imagine $\ce{^{40}K -> ^{40}Ar + e+}$ and $\ce{^{40}K + e- -> ^{40}Ar}$ $\endgroup$
    – Poutnik
    Commented Dec 7, 2022 at 12:10
  • $\begingroup$ I think that the OP dreamed up something which is possible to write maths for but is impossible to happen in real life $\endgroup$ Commented Dec 7, 2022 at 12:15
  • $\begingroup$ Rather, she(?) had some funny (or sad) teacher or textbook. $\endgroup$
    – Poutnik
    Commented Dec 7, 2022 at 12:19

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