Q 12-13-134JEE MainJEE Main 2017 (2 Apr)Medium
A radioactive nucleus $A$ with a half-life $T$ decays into a nucleus $B$. At $t = 0$, there is no nucleus $B$. At some time $t$, the ratio of the number of $B$ to that of $A$ is $0.3$. Then, $t$ is given by (take $\log x$ to mean $\log_e x$):
Answer: (C) $t = T\dfrac{\log 1.3}{\log 2}$
With $N_0$ nuclei of $A$ initially, $N_A = N_0e^{-\lambda t}$ and $N_B = N_0 - N_A$:
$$\frac{N_B}{N_A} = e^{\lambda t} - 1 = 0.3 \;\Rightarrow\; \lambda t = \log 1.3$$
With $\lambda = \dfrac{\log 2}{T}$:
$$t = T\frac{\log 1.3}{\log 2}$$
Solution by Sreeraj P, M.Sc Physics