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

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.

• Please read carefully our guidelines for homework and exercise questions, and particularly those for check-my-work questions. Aug 3, 2020 at 13:50
• Is it wrong to write a question if one gets stuck while going through any text? Aug 3, 2020 at 13:52
• Close-voters: I think there's a valid conceptual question here about why secular equilibrium is only possible in the case where the daughter's half-life is significantly longer than the parent's. I find it ironic that this question probably wouldn't have attracted close-votes if the OP had just asked the question and not tried to present their work. Aug 4, 2020 at 13:58
• @MichaelSeifert "I think there's a valid conceptual question here..." could be used to argue for pretty much any question to remain open. Users should judge a question based on the question, not what they think the question should be. With that being said, I agree with you that this is not a check-my-work problem, even though the close voters seem to think so. Aug 5, 2020 at 13:52

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.
• @YasirSadiq: We have to assume that the number of daughters is (roughly) constant, since that's the definition of secular equilibrium. This assumption is what allows us to prove that $\lambda_p N_p \approx \lambda_d N_d$ (as you proved above.) But there are time spans over which the number of daughter nuclei varies; look at the time span before about $t = 0.1$ on my first graph. Aug 7, 2020 at 11:40