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 :
Correct Option: 3,
$\mathrm{R}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$
$\ln \mathrm{R}=\ln \mathrm{R}_{0}-\lambda \mathrm{t}$
$\lambda_{\mathrm{A}}=\frac{6}{10} \Rightarrow \mathrm{T}_{\mathrm{A}}=\frac{10}{6} \ln 2$
$\lambda_{\mathrm{B}}=\frac{6}{5} \Rightarrow \mathrm{T}_{\mathrm{B}}=\frac{5 l \mathrm{n} 2}{6}$
$\lambda_{\mathrm{C}}=\frac{2}{5} \Rightarrow \mathrm{T}_{\mathrm{C}}=\frac{5 l \mathrm{n} 2}{2}$
$\frac{10}{6}: \frac{5}{6}: \frac{15}{6}:: 2: 1: 3$
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