Nuclei A decays to B with decay constant λ 1 and B decays to C with decay constant λ 2 . Initially at t = 0 , number of nuclei of A and B are 2 N 0 and N 0 respectively. At t = t 0 , number

Question

Nuclei A decays to B with decay constant λ 1 ​ and B decays to C with decay constant λ 2 ​ . Initially at t = 0 , number of nuclei of A and B are 2 N 0 ​ and N 0 ​ respectively. At t = t 0 ​ , number of nuclei of B stop changing for an instant. If at this instant number of nuclei of B are 2 3 N 0 ​ ​ .

TestHub · Accepted Answer

B stops changing when its production equals its decay:

$λ_{1} N_{A} = λ_{2} N_{B}$

$N_{A} = \frac{λ _{2}}{λ _{1}} \cdot \frac{3 N _{0}}{2}$

$2 N_{0} e^{- λ_{1} t_{0}} = \frac{3 N _{0} λ _{2}}{2 λ _{1}}$

$t_{0} = \frac{1}{λ _{1}} ln (\frac{4}{3} \frac{λ _{1}}{λ _{2}})$

Options (A), (C)

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