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

(N/A) $(i)$ In the present case, the graph of $R$ versus $t$ is an exponential decay curve as shown in the figure.
At time $t = 0$, $R_0 = 100 \text{ MBq}$.
The half-life $\tau_{1/2}$ is the time at which the activity becomes half of its initial value, i.e., $R = \frac{R_0}{2} = 50 \text{ MBq}$.
From the graph, at $R = 50 \text{ MBq}$, the corresponding time is $t = 0.66 \text{ h}$.
Therefore, the half-life $\tau_{1/2} = 0.66 \text{ h} = 0.66 \times 60 \text{ min} = 39.6 \text{ min} \approx 40 \text{ min}$.
$(ii)$ According to the law of radioactive decay, $R = R_0 e^{-\lambda t}$.
Taking the natural logarithm on both sides:
$\ln \left( \frac{R}{R_0} \right) = -\lambda t$.
This is an equation of a straight line $y = mx + c$ where $y = \ln \left( \frac{R}{R_0} \right)$, $x = t$, slope $m = -\lambda$, and intercept $c = 0$.
The slope of this graph is $-\lambda$. By calculating the slope from the data points, we find $\lambda \approx 1.05 \text{ h}^{-1}$.
Using the relation $\tau_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{1.05} \approx 0.66 \text{ h}$.