Question

Nuclei A and B convert into a stable nucleus C. Nucleus A is converted into C by emitting 2 $\alpha$-particles and 3 $\beta$ - particles whereas, nucleus B is converted into C by emitting 1 $\alpha$ - particle and 5 $\beta$ - particles. At time $t=0$, nuclei of A are $4N_0$ and nuclei of B are $N_0$. Initially, the number of nuclei of C are zero. Half-life of A and B (into conversion of C) is 1 min and 2 min respectively. The time (in minutes) at which the rates of disintegration of A and B are equal is

Answer 1

We have to find the time at which ${\lambda }_A{N}_A={\lambda }_B{N}_B$
$\left(\frac{\ln \left(2\right)}{T_A}\right)\left(4N_0{e}^{-{\lambda }_{A}t}\right)=\left(N_0\right)\left(\frac{\ln \left(2\right)}{T_B}\right)\left(e^{-{\lambda }_{B}t}\right)e^{\left({\lambda }_A-{\lambda }_B\right)t}=8\left({\lambda }_A-{\lambda }_B\right)t=\ln \left(8\right)=3\ln \left(2\right)\left(\frac{\ln \left(2\right)}{1}-\frac{\ln \left(2\right)}{2}\right)t=3\ln \left(2\right)\Rightarrow t=6min$