JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Modern Physics

Activity decays exponentially, $A=A_0{e}^{-\lambda t}$, so a plot of $\ln A$ against $t$ is a straight line of slope $-\lambda$; the steeper the line, the larger the decay constant and, since $T_{1/2}=\frac{\ln 2}{\lambda }$, the shorter the half-life. Reading the three slopes from the graph gives decay constants in the ratio $\frac{6}{10}:\frac{6}{5}:\frac{2}{5}$. Because half-life varies as $\frac{1}{\lambda }$, the half-lives follow the reciprocal ratio $\frac{10}{6}:\frac{5}{6}:\frac{5}{2}$, which reduces to $T_{1/2}(A):T_{1/2}(B):T_{1/2}(C)=2:1:3$.