The activity $R$ of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

$t(h)$	$0$	$1$	$2$	$3$	$4$
$R(MBq)$	$100$	$35.36$	$12.51$	$4.42$	$1.56$

$(i)$ Plot the graph of $R$ versus $t$ and calculate half-life from the graph.

$(ii)$ Plot the graph of $\ln \left( {\frac{R}{{{R_0}}}} \right) \to t$ versus $t$ and obtain the value of half-life from the graph.

$(i)$ In the present case, graph of $\mathrm{R} \rightarrow t$ is obtained as hyperbola as shown below.

At time $t=0, \mathrm{R}_{0}=100 \mathrm{MB} q$

$\text { and at time } t=0.66 \mathrm{~h}, \mathrm{R}=50 \mathrm{MB} q=\frac{\mathrm{R}_{0}}{2}$

$\Rightarrow t=0.66 \mathrm{~h}=\tau_{1 / 2}$

$\Rightarrow \text { Half life } \tau_{1 / 2}=0.66 \mathrm{~h}$

$=0.66 \times 60 \mathrm{~min}$

$\therefore \tau_{1 / 2}=39.6 \mathrm{~min} \approx 40 \mathrm{~min}$

$(ii)$ According to exponential law,

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

$\therefore \ln \mathrm{R}=\ln \mathrm{R}_{0}+\ln \left(e^{-\lambda t}\right)$

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

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

$\therefore \ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=(-\lambda) t+0$$....(1)$