Over a duration of 20 hours , the count rate from a radioactive...

The initial count rate is $R_0 = 4.0 \times 10^{6} \text{ s}^{-1}$.

After $t = 20 \text{ hours}$, the count rate is $R = 1.0 \times 10^{6} \text{ s}^{-1}$.

Using the law of radioactive decay, $R = R_0 \left(\dfrac{1}{2}\right)^{n}$, where $n$ is the number of half-lives.

$1.0 \times 10^{6} = 4.0 \times 10^{6} \left(\dfrac{1}{2}\right)^{n}$

\left(\dfrac{1}{2}\right)^{n} = \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^{2}$<br><br>$\Rightarrow n = 2$<br><br>Since$2$half-lives correspond to$20 \text{ hours}$, the half-life is$T_{1/2} = 10 \text{ hours}$.<br><br>For$t' = 100 \text{ hours}$, the number of half-lives is$n' = \dfrac{100}{10} = 10$.<br><br>The count rate after$100 \text{ hours}$is$R' = R_0 \left(\dfrac{1}{2}\right)^{10}$.<br><br>$R' = 4.0 \times 10^{6} \times \dfrac{1}{1024}$<br><br>$R' = \dfrac{4000 \times 10^{3}}{1024} \approx 3.9 \times 10^{3} \text{ per second}$.$