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MHT CET 2026 (17 Apr, Shift 2) · Law of Radioactive Decay

Law of Radioactive Decay question from MHT CET 2026 (17 Apr, Shift 2)

Q8MHT CET 2026 · 17 APR · SHIFT 2Medium

Law of Radioactive Decay

The activity of a radioactive sample is measured as $N_0$ counts per minute at time $t = 0$ and $\frac{N_0}{e}$ counts per minute at time $t = 3$ minute. The time (in minute) in which the activity reduces to half the value, is

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Answer: (b)

$\frac{N_0}{e} = N_0e^{-3\lambda} \Rightarrow 3\lambda = 1$ , $\lambda = \frac{1}{3}\text{ min}^{-1}$ .
$T = \frac{\log_e2}{\lambda} = 3\log_e2$ min.
So the answer is (b) $3\log_e2$ .

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