- #1

- 43

- 4

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

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

$$Activtiy_B \sim \lambda _BN_B$$