At time t = 0 , the activity of a radioactive substance is 1600 Bq....

The activity of a radioactive substance at time $t$ is given by $A = A_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}$.

Given $A_0 = 1600$ Bq at $t = 0$.

At $t = 8$ s, $A = 100$ Bq.

$100 = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$

$\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$

\left(\dfrac{1}{2}\right)^4 = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$<br><br>$\dfrac{8}{T_{1/2}} = 4 \Rightarrow T_{1/2} = 2$s.<br><br>The activity at$t = 2$s is:<br><br>$A = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{2}{2}} = 1600 \times \dfrac{1}{2} = 800$Bq.<br><br>Answer:$800$ Bq