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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.

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  • $\begingroup$ Please read carefully our guidelines for homework and exercise questions, and particularly those for check-my-work questions. $\endgroup$ – Emilio Pisanty Aug 3 '20 at 13:50
  • $\begingroup$ Is it wrong to write a question if one gets stuck while going through any text? $\endgroup$ – Kashmiri Aug 3 '20 at 13:52
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    $\begingroup$ 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. $\endgroup$ – Michael Seifert Aug 4 '20 at 13:58
  • $\begingroup$ @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. $\endgroup$ – BioPhysicist Aug 5 '20 at 13:52
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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.

enter image description here

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.

enter image description here

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  • $\begingroup$ Thank you Michael, can you please explain it a little more.I did get how the parent number remains constant in the intermediate time, but how can you deduce that the number of daughter is constant as well? $\endgroup$ – Kashmiri Aug 7 '20 at 6:18
  • $\begingroup$ @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. $\endgroup$ – Michael Seifert Aug 7 '20 at 11:40

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