Q 12-13-003NEETNEET 2023Top questionEasy
The half life of a radioactive substance is $20$ minutes. In how much time, the activity of substance drops to $\left(\dfrac{1}{16}\right)^{th}$ of its initial value?
Answer: (D) $80$ minutes
$\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, so 4 half-lives are needed.
$$t = 4 \times 20 = 80\ \text{minutes}$$
Solution by Sreeraj P, M.Sc Physics