Q 12-13-120JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
Activities of three radioactive substances $A$, $B$ and $C$ are represented by the curves $A$, $B$ and $C$, in the figure. Then their half-lives $T_{1/2}(A) : T_{1/2}(B) : T_{1/2}(C)$ are in the ratio:
Answer: (C) $2 : 1 : 3$
$\ln R = \ln R_0 - \lambda t$, so each decay constant is the magnitude of the slope of its line:
$\lambda_A = \dfrac{6}{10} = 0.6$, $\lambda_B = \dfrac65 = 1.2$, $\lambda_C = \dfrac25 = 0.4$ (per year).
$$T_{1/2} \propto \frac1\lambda:\quad \frac1{0.6} : \frac1{1.2} : \frac1{0.4} = 2 : 1 : 3$$
Solution by Sreeraj P, M.Sc Physics