# How to find the average half life of radioactive nuclide which undergoes two different decays?

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 $$A$$?

• A source for the exercise would really go a long way as this is a very strange problem. – Martin - マーチン Feb 26 '20 at 11:38

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 $$A$$?

Yes.

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.