At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio (RB)/(RA) of their activities after time t itself decays with time t as e ⁻³ t. If the half-life of A is ln2, the half-life of B is: (a) 4ln2 (b) (ln 2)/(2) (c) (ln 2)/(4) (d) 2ln2

Answer: (c). RA = R₀ e-λA t, RB = R₀ e-λB t, so (RB)/(RA) = e-(λB - λA)t = e-3t. λB - λA = 3; λA = (ln 2)/(ln 2) = 1, so λB = 4. TB = (ln 2)/(λB) = (ln 2)/(4) So the answer is (c) (ln 2)/(4).

At a given instant, say t = 0, two radioactive substance A and B have… | JEE Main 2019 (9 Jan, Shift 2)

Q84JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Law of Radioactive Decay

At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio $\frac{R_B}{R_A}$ of their activities after time t itself decays with time t as e ${}^{-3}$ t. If the half-life of A is ln2, the half-life of B is:

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Answer: (c)

$R_A = R_0 e^{-\lambda_A t}$ , $R_B = R_0 e^{-\lambda_B t}$ , so $\dfrac{R_B}{R_A} = e^{-(\lambda_B - \lambda_A)t} = e^{-3t}$ .
$\lambda_B - \lambda_A = 3$ ; $\lambda_A = \dfrac{\ln 2}{\ln 2} = 1$ , so $\lambda_B = 4$ .
$T_B = \dfrac{\ln 2}{\lambda_B} = \dfrac{\ln 2}{4}$
So the answer is (c) $\dfrac{\ln 2}{4}$ .

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Law of Radioactive Decay question from JEE Main 2019 (9 Jan, Shift 2)

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