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
Concept. Activity halves every half-life: $\dfrac{A}{A_0} = \left(\dfrac12\right)^{t/T_{1/2}}$.
Step 1. Number of half-lives. $\dfrac{1}{16} = \left(\dfrac12\right)^4$, so $4$ half-lives.
Step 2. Time. $t = 4 \times 20 = 80$ minutes.
Answer. $80$ minutes (Option D).
Solution by Sreeraj P, M.Sc Physics