# Activity of daugther isotope in secular equilibrium

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## Main Question or Discussion Point

Consider chain of two radioactive decays $A \to B\to C$. The equation that regulates $N_B$ is

$${\frac {\mathrm {d} N_{B}}{\mathrm {d} t}}=-\lambda _{B}N_{B}+\lambda _{A}N_{A}$$

I can't understand why the activity of $B$ is get as $\lambda_B N_B$, for example at page 20 here http://www.umich.edu/~ners311/CourseLibrary/bookchapter13.pdf. In my view that the activity should be $-\lambda _{B}N_{B}+\lambda _{A}N_{A}$ instead, i.e. $dN_B/dt$ itself.

In particular, in secular equilibrium, $\lambda_A << \lambda_B$ and it is easy to prove that, as $t\to \infty$ $$\lambda_B N_B \to \lambda_A N_A=Activity_A\sim const$$
But, as in the pdf linked, it is usually said that this should also prove that the activity of B becomes equal to the activity of A. I can't understand if in this particular case it is right to say

$$Activtiy_B \sim \lambda _BN_B$$

## Answers and Replies

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Mentor
"Activity of B" means how many decays of B are there per time. The A->B decay is irrelevant for that (although it has an indirect effect via refilling the amount of B).