- #1
crick
- 39
- 4
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
$$Activtiy_B \sim \lambda _BN_B$$
$${\frac {\mathrm {d} N_{B}}{\mathrm {d} t}}=-\lambda _{B}N_{B}+\lambda _{A}N_{A}$$
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
$$Activtiy_B \sim \lambda _BN_B$$