JEE PYQ: Nuclear Physics Question 25
Question 25 - 2019 (09 Jan Shift 2)
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 $e^{-3t}$. If the half-life of A is $\ln 2$, the half-life of B is:
(1) $4\ln 2$
(2) $\frac{\ln 2}{2}$
(3) $\frac{\ln 2}{4}$
(4) $2\ln 2$
Show Answer
Answer: (3)
Solution
Half life of $A = \ell n2$ $(t_{1/2})A = \frac{\ell n 2}{\lambda_A}$ $\therefore \lambda_A = 1$ at $t = 0$, $R_A = R_B$ $N_A e^{-\lambda_A T} = N_B e^{-\lambda_B T}$ $N_A = N_B$ at $t = 0$ At $t = t$: $\frac{R_B}{R_A} = \frac{N_B e^{-\lambda_B t}}{N_A e^{-\lambda_A t}}$ $e^{-(\lambda_B - \lambda_A)t} = e^{-3t}$ $\Rightarrow \lambda_B - \lambda_A = 3$ $\lambda_B = 3 + \lambda_A = 4$ $(t{1/2})_B = \frac{\ell n2}{\lambda_B} = \frac{\ell n2}{4}$
BETA
