Question:

Activities of three radioactive substances A, B and $\mathrm{C}$ are represented by the curves $\mathrm{A}, \mathrm{B}$ and C, in the figure. Then their half-lives $\mathrm{T}_{\frac{1}{2}}(\mathrm{~A}): \mathrm{T}_{\frac{1}{2}}(\mathrm{~B}): \mathrm{T}_{\frac{1}{2}}(\mathrm{C})$ are in the ratio :

1. $3: 2: 1$

2. $4: 3: 1$

3. $2: 1: 3$

4. $2: 1: 1$

Correct Option: 3,

Solution:

$\mathrm{R}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$

$\ln \mathrm{R}=\ln \mathrm{R}_{0}-\lambda \mathrm{t}$

$\lambda_{\mathrm{A}}=\frac{6}{10} \Rightarrow \mathrm{T}_{\mathrm{A}}=\frac{10}{6} \ln 2$

$\lambda_{\mathrm{B}}=\frac{6}{5} \Rightarrow \mathrm{T}_{\mathrm{B}}=\frac{5 l \mathrm{n} 2}{6}$

$\lambda_{\mathrm{C}}=\frac{2}{5} \Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{5 l \mathrm{n} 2}{2}$

$\frac{10}{6}: \frac{5}{6}: \frac{15}{6}:: 2: 1: 3$