3
$\begingroup$

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 $A$?

$\endgroup$
  • 1
    $\begingroup$ A source for the exercise would really go a long way as this is a very strange problem. $\endgroup$ – Martin - マーチン Feb 26 at 11:38
8
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.