Question

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 $T_{\frac{1}{2}}\left(A\right):T_{\frac{1}{2}}\left(B\right):T_{\frac{1}{2}}\left(C\right)$ are in the ratio

• A. $2:1:1$
• B. $3:2:1$
• C. $2:1:3$
• D. $4:3:1$

Answer 1

The correct option is C $2:1:3$

Since, $R=R_0{e}^{-\lambda t}$
$\ln R=\ln R_0+(-\lambda \ln t)$
Slope of $\ln R\text{vs}\ln t$ is $\lambda =\frac{\ln 2}{T_{\frac{1}{2}}}$${\lambda }_A=\frac{6}{10}\Rightarrow T_A=\frac{10}{6}\ln 2$
${\lambda }_B=\frac{6}{5}\Rightarrow T_B=\frac{5\ln 2}{6}$
${\lambda }_C=\frac{2}{5}\Rightarrow T_C=\frac{5\ln 2}{2}$
$\therefore T_{\frac{1}{2}A}:T_{\frac{1}{2}B}:T_{\frac{1}{2}C}=\frac{10}{6}:\frac{5}{6}:\frac{15}{6}=2:1:3$