Two radioactive nuclei A and B are initially in the ratio 1 : 4. Also initial activities of the nuclei are in the ratio 1 : 8. Given that, half-life of A is 2 yrs, Choose the correct alternatives.

$R=\lambda N_0 and t_{1/2}=\frac{\log 2}{\lambda }$

$\begin{array}{l}\frac{{\left(N_0\right)}_A}{{\left(N_0\right)}_B}=\frac{1}{4}\\ \frac{R_{0_A}}{R_{0_B}}=\frac{{\mathrm{\lambda }}_A{N}_{0A}}{{\mathrm{\lambda }}_B{N}_{0B}}=\frac{1}{8}\\ \Rightarrow \frac{{\mathrm{\lambda }}_A}{{\mathrm{\lambda }}_B}=\frac{1}{2}\\ {\mathrm{\lambda }}_B=2{\mathrm{\lambda }}_A\\ \Rightarrow {\left(t_{1/2}\right)}_A=2\times {\left(t_{1/2}\right)}_B\left(\because t_{1/2}=\frac{0.693}{\mathrm{\lambda }}\right)\\ Given, {\left(t_{1/2}\right)}_A=2yrs , so {\left(t_{1/2}\right)}_B=1yr \end{array}$

$\text{ At }4\text{ yr, A will be }\frac{1}{4}\text{ and }B\text{ will be }\frac{1}{16}$

$\begin{array}{l}\text{ (4 half - lives). }\\ \frac{{\left(N_t\right)}_A}{{\left(N_t\right)}_B}=\frac{\frac{1}{4}{N}_{0A}}{\frac{1}{16}{N}_{0B}}=4\times \frac{1}{4}=1\\ \text{ And }\frac{{\left(R_t\right)}_A}{{\left(R_t\right)}_B}=\frac{N_{tA}\cdot {\mathrm{\lambda }}_A}{N_{tB}\cdot {\mathrm{\lambda }}_B}\\ =1\times \frac{1}{2}=\frac{1}{2}\end{array}$

$\text{ At }6\mathrm{yr},\mathrm{A}\text{ will be }\frac{1}{8}\text{ and }\mathrm{B}\text{ will be }\frac{1}{64}$

$\begin{array}{l}\frac{{\left(N_t\right)}_A}{{\left(N_t\right)}_B}=\frac{\frac{1}{8}{N}_{0A}}{\frac{1}{64}{N}_{0B}}=2\\ \text{ Amount of nuclei decayed after one mean life time : }\\ \mathrm{\Delta }{N}_A=N_{0A}\left(1-\frac{1}{e}\right)\\ \mathrm{\Delta }{N}_B=N_{0B}\left(1-\frac{1}{e}\right)\end{array}$

$\begin{array}{l}Since, N_{0B}>N_{0A}\\ \text{}\Rightarrow f_2>f_1\end{array}$