The counting rate observed from a radioactive source at t=0 was...

As we know, $$\begin{aligned}& {\left[\frac{N}{N_0}\right]=\left[\frac{1}{2}\right]^n} \\ & n=\text{ no.\text{ of }\text{ half }\text{ life } }\\ & N \text{ - no.\text{ of }\text{ atoms } left }\\ & N_0 \text{ - initial no.\text{ of }\text{ atoms } }\\ & \text{ By radioactive decay law, }\\ & \frac{d N}{d \mathrm{t}}=k N \\ & k-\operatorname{disin} t e g r a t i o n \text{ constant }\\ & \therefore \frac{\frac{d N}{d t}}{\frac{d N_0}{d t}}=\frac{N}{N_0} \end{aligned}$$From (i) and (ii) we get $\frac{\frac{d N}{d t}}{\frac{d N_0}{d t}}=$\left[\frac{1}{2}\right]^$n or,$\left[\frac{100}{1600}\right]=\left[\frac{1}{2}\right]^n \Rightarrow\left[\frac{1}{2}\right]^4=\left[\frac{1}{2}\right]^n\therefore n=4$, Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts$\mathrms^{-1}$.$\therefore$Half life,$\frac{T_1}{2}=2 \mathrm{sec}$In$6 \mathrm{sec}, 3$half life will occur$$\begin{aligned}& \therefore\left[\frac{\frac{d \mathrm{~N}}{d t}}{1600}\right]=\left[\frac{1}{2}\right]^3 \\ & \Rightarrow \frac{d \mathrm{~N}}{d t}=200 \text{ counts } \mathrms^{-1} \end{aligned}$$