Radioactive secular equilibrium and the relationship between the half lives of the parent and the daughter nucleus

Question

It's known that

In nuclear physics, secular equilibrium is a situation in which the quantity of a radioactive isotope remains constant because its production rate (e.g., due to decay of a parent isotope) is equal to its decay rate. Secular equilibrium can occur in a radioactive decay chain only if the half-life of the daughter radionuclide B is much shorter than the half-life of the parent radionuclide A.

My problem:why can secular equilibrium occur in a radioactive decay chain only if the half-life of the daughter radionuclide B is much shorter than the half-life of the parent radionuclide A.

My understanding Say I've got two radioactive substances $A$ and $B$ such that $A$ decays to $B$ and $B$ to some other stable substance $C$ $$ A \rightarrow B \rightarrow C $$ From Radioactive decay law we know that$$ -\frac{d N}{d t}=\lambda N $$ where$N$ is the number of undecayed particles. Let $N_a,$$ \lambda_{a} $ be the number of undecayed particles and the decay constant of substance $A$ similarly let $N_b,$$ \lambda_{b} $ be the number of undecayed particles and the decay constant of substance $B$. Now for the above defined secular equilibrium to exist for substance $B$ one can write using the above decay law $$ \lambda_{a} N_{a}=\lambda_{b} N_{b}$$

Using $$\lambda=\frac{\ln 2}{t \frac{1}{2}} $$ we can rewrite the above equilibrium equality as$$ \frac{N a}{t_{x \frac{1}{2}}}=\frac{N_{b}}{t_{y \frac{1}{2}}}$$ where$$ t_{x \frac{1}{2}}, t_{y \frac{1}{2}} $$ are the half lives of substances $A$ and $B$ respectively . From here onwards how can one deduce that secular equilibrium can occur in a radioactive decay chain only if the half-life of the daughter radionuclide B is much shorter than the half-life of the parent radionuclide A. Thank you a lot.

Answer 1

Your math is completely correct. The key missing piece is conceptual.

Specifically: if the lifetime of the parent is long compared to the half-life of the daughter ($\tau_p \gg \tau_d$), then there is a time-scale in between them which is long compared to $\tau_d$ but short compared to the half-life of the parent $\tau_p$. Over this intermediate time-scale, the number of atoms of the parent is roughly constant (since $e^{-t/\tau_p} \approx 1$ when $t \ll \tau_p$), and the number of daughter atoms is given by the ratio you found above.

In the graph below, I have plotted the number of parent and daughter atoms for $\tau_p = 10^3 \tau_d$ (in units where $\tau_d = 1$.) You can see the "secular equilibrium" period between about $t = 10$ and $t = 500$: the times that are significantly greater than $\tau_d$ but still less than $\tau_p$. Over this period, both $N_p$ and $N_d$ are roughly constant. Eventually, of course, the decay of the parent becomes significant, and there's not really an equilibrium any more.

On the other hand, if the half-life of the parent is much smaller than the half-life of the daughter, then there's no intermediate time-scale over which the number of parent atoms can be said to be constant. The parent decays into the daughter very quickly, and then the daughter decays away slowly. There's never really an "equilibrium" between parent and daughter, because the parent is basically gone by the time the daughter's decay becomes significant.