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JEE Main 2019 (9 Jan, Shift 1) · Law of Radioactive Decay

Law of Radioactive Decay question from JEE Main 2019 (9 Jan, Shift 1)

Q83JEE Main 2019 · 9 JAN · SHIFT 1Medium

Law of Radioactive Decay

A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $\times$ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively:

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Answer: (c)

$A = \lambda N$ : $10 = \lambda_A (2N)$ and $20 = \lambda_B N$ (in mCi).
Dividing: $\frac{\lambda_B}{2\lambda_A} = 2 \Rightarrow \lambda_B = 4\lambda_A$ .
$T \propto 1/\lambda$ , so $T_A = 4T_B$ ; only 20 days and 5 days fits.
So the answer is (c).

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