Q. At a given instant, say t = 0, two radioactive substances 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 $e^{-3t}$ . [f the half-life of A is $m_2$, the half-life of B is :

Solution:

Half life of $A = \ell n^2$
$t_{1/2} = \frac{\ell n2}{\lambda}$
$\lambda_A = 1$
at $t = 0 \; R_A = R_B$
$N_A e^{- \lambda AT} = N_B e^{-\lambda BT}$
$N_A = N_B $ at $t = 0$
at $t = t \; \frac{R_B}{R_A} = \frac{N_0 e^{\lambda_Bt}}{N_0 e^{- \lambda_A t}}$
$e^{-(\lambda_B - \lambda_A)t} = e^{-t}$
$\lambda_B - \lambda_A = 3$
$\lambda_B = 3 + \lambda_A = 4$
$t_{1/2} = \frac{\ell n2}{\lambda_B} = \frac{\ell n 2}{4}$

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions