As we know.
$\left[\frac{N}{N_{0}}\right]\quad\quad\quad\quad\ldots\left(i\right)$
n = no. of half life
N - no. of atoms left
$N_{0}$- initial no. of atoms
By radioactive decay law.
k - disintegration constant
$\therefore \frac{\frac{dN}{dt}}{\frac{dN_{0}}{dt}} = \frac{N}{N_{0}}\quad\quad\quad\ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ we get
$\frac{\frac{dN}{dt}}{\frac{dN_{0}}{dt}} = \left[\frac{1}{2}\right]^{n}$
or, $\left[\frac{100}{1600}\right] = \left[\frac{1}{2}\right]^{n}\quad \Rightarrow \left[\frac{1}{2}\right]^{4} = \left[\frac{1}{2}\right]^{n}$
$\therefore n = 4$, Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts $s^{-1}$.
$\therefore$ Half life, $T_{\frac{1}{2}} = 2\,sec$
In 6 sec, 3 half life will occur
$\therefore \left[\frac{\frac{dN}{dt}}{1600}\right] = \left[\frac{1}{2}\right]^{3}$
$\Rightarrow \frac{dN}{dt} = 200$ counts $s^{-1}$