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Q.
For a certain radioactive process the graph between In $R$ and $t(s e c)$ is obtained as shown in the figure.
Then the value of half life for the unknown radioactive material is approximately :
$R=R_{0} e^{-\lambda t}$
$l n R=l n R_{0}-\lambda t$
$-\lambda$ is slope of straight line
$\lambda=\frac{3}{20}$
$t_{1 / 2}=\frac{l n 2}{\lambda}=4.62$
