# Finding average half life of radioactive nuclide which undergoes two different decays

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

Attempt:

I used the formula $$\frac{1}{t}=\frac{t_1t_2}{t_1+t_2}$$ 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 $\underline2A$ on the reactant side instead of $A$?

• How did you come to this attempt? Do you know the solution? – pH13 - Yet another Philipp Aug 3 '15 at 8:22
• I don't. Only the final answer is given. I can derive the formula if you want. – Aditya Dev Aug 3 '15 at 14:57

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

$$\left(\frac{d\left[A\right]}{dt}\right)_1=-\lambda_1\left[A\right] ,$$

and that of reaction 2 is$^\dagger$

$$\left(\frac{d\left[A\right]}{dt}\right)_2=-2\lambda_2\left[A\right] .$$

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:

$$\frac{d\left[A\right]}{dt}=-\lambda_1\left[A\right] -2\lambda_2\left[A\right] = -\left(\lambda_1+2\lambda_2\right)\left[A\right].$$

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

$^\dagger$ 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.

• Good answer and +1, but this is a very weird problem. Simultaneously having A -> B and 2A -> B processes would seem to violate all kinds of conservation laws. – Curt F. Aug 4 '15 at 14:43
• @CurtF Absolutely. The problem makes no physical sense. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. – hBy2Py Aug 4 '15 at 14:58
• @CurtF $2A\longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay. – hBy2Py Aug 7 '15 at 15:15