Q 12-13-104JEE MainJEE Main 2021 (20 Jul, Shift 2)Medium
For a certain radioactive process, the graph between $\ln R$ and $t$ (sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately:
Answer: (D) 4.62 sec
$R = R_0e^{-\lambda t} \Rightarrow \ln R = \ln R_0 - \lambda t$, so $\lambda$ is minus the slope. The line falls from 6 to 0 in 40 s:
$$\lambda = \frac{6}{40} = 0.15\ \text{s}^{-1},\quad T_{1/2} = \frac{0.693}{0.15} \approx 4.62\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics