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

A stable terminal product obeys ${\dot{N}}_c={\lambda }_b{N}_b$. Conservation of the number of chain members gives $N_a+N_b+N_c=N_{a0}+N_{c0}$ when $N_{b0}=0$. Therefore, for distinct nuclear decay constants,Set $N_{c0}=0$ for an initially absent terminal product. For equal nuclear decay constants,Both expressions have ${\dot{N}}_c(0)=0$ because the intermediate population initially vanishes; their leading growth is $\frac{1}{2}{N}_{a0}{\lambda }_a{\lambda }_b{t}^2$. For positive nuclear decay constants, $N_c$ increases monotonically to $N_{c0}+N_{a0}$.