The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to (1/16) th of its initial value? (a) 40 minutes (b) 60 minutes (c) 80 minutes (d) 20 minutes

Answer: (c). A/A₀ = 1/16 = (1/2)⁴, so 4 half-lives are needed. t = 4 × 20 = 80 min. So the answer is (c) 80 minutes.

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Law of Radioactive Decay

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\frac{1}{16}\right)$ th of its initial value?

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Answer: (c)

$\frac{A}{A_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$ , so 4 half-lives are needed.
$t = 4 \times 20 = 80\text{ min}$ .
So the answer is (c) 80 minutes.

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