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MHT CET 2026 (19 May, Shift 1) · Law of Radioactive Decay

Law of Radioactive Decay question from MHT CET 2026 (19 May, Shift 1)

Q6MHT CET 2026 · 19 MAY · SHIFT 1Medium

Law of Radioactive Decay

Initially $8 \times 10^{10}$ atoms are present in a radioactive sample. The half-life of the sample is 1 hour. The number of atoms that will decay in the time interval $t = 2$ hour to $t = 5$ hour, is

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Answer: (c)

$N = N_0/2^{t/T}$ with $T = 1$ h.
At $t = 2$ h: $N = \frac{8\times10^{10}}{4} = 2\times10^{10}$ .
At $t = 5$ h: $N = \frac{8\times10^{10}}{32} = 0.25\times10^{10}$ .
Decayed $= 2 - 0.25 = 1.75\times10^{10}$ , so the answer is (c).

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