Past exam of the nuclear and particle physics course of the University of Oxford / 2022 / 3 / d / ii / Solution

For $N_{b0}=0$, differentiate the distinct-nuclear decay constant result:Equating the production and loss terms gives $e^{({\lambda }_b-{\lambda }_a)t}={\lambda }_b/{\lambda }_a$, henceFor positive nuclear decay constants and $N_{a0}>0$, this is positive regardless of their ordering. Initially ${\dot{N}}_b>0$, while $N_b\to 0$ at late times; the stationary point is the unique maximum. At that time ${\lambda }_a{N}_a={\lambda }_b{N}_b$. If the nuclear decay constants are equal, differentiating $\lambda N_{a0}t{e}^{-\lambda t}$ gives $\boxed{t_{\max }=1/\lambda }$, also the continuous limit of the expression above.