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

Question

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 $A$?

Answer 1

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.

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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.

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^† 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.

Answer 2

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.