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

    \[\begin{tabular}{|l|l|l|l|l|l|} \hline$t(\mathrm{~h})$ & 0 & 1 & 2 & 3 & 4 \\ \hline R(MBq) & 100 & $35.36$ & $12.51$ & $4.42$ & $1.56$ \\ \hline \end{tabular}\]


(i) Plot the graph of R versus t and calculate half-life from the graph.
(ii) Plot the graph of \ln \left(\mathrm{R} / \mathrm{R}_{0}\right) versus \mathrm{t} and obtain the value of half-life from the graph.
The activity \mathrm{R} of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

    \[\begin{tabular}{|l|l|l|l|l|l|} \hline$t(\mathrm{~h})$ & 0 & 1 & 2 & 3 & 4 \\ \hline R(MBq) & 100 & $35.36$ & $12.51$ & $4.42$ & $1.56$ \\ \hline \end{tabular}\]


(i) Plot the graph of R versus t and calculate half-life from the graph.
(ii) Plot the graph of \ln \left(\mathrm{R} / \mathrm{R}_{0}\right) versus \mathrm{t} and obtain the value of half-life from the graph.

(i) Graph between \mathrm{R} versus \mathrm{t} will be an exponential curve.

From the graph at slightly more than \mathrm{t}=\frac{1}{2} \mathrm{~h} the \mathrm{R} should be 50 \% so at \mathrm{R}=50 \% the \mathrm{t}(\mathrm{h})=0.7 \mathrm{~h}=0.7 \times 60=42 \mathrm{~min}

(ii) For Graph between \log \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right), we have,

at t=0, \log _{e}\left(\frac{R}{R_{0}}\right) \log \frac{100}{100}=\log 1=0

at \mathrm{t}=1 hour \log _{\mathrm{e}}\left(\frac{35.36}{100}\right)=\log _{\mathrm{e}} 0.3536=-1.04
=2.302 \log _{\mathrm{e}} 35.36=1.04

at \mathrm{t}=2 hours \log _{\mathrm{e}}\left(\frac{12.5}{100}\right)=\log _{\mathrm{e}} 0.125
=2.303 \log _{\mathrm{e}} 0.125=-2.08

at \mathrm{t}=3 hours \log _{\mathrm{e}}\left(\frac{4.42}{100}\right)=-3.11

at \mathrm{t}=4 hours \log _{\mathrm{e}}\left(\frac{1.56}{100}\right)=-4.16
t (hours) 1234
\log _{\mathrm{e}}\left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)-1.04-2.08-3.11-4.16

know that disintegration constant,

\begin{array}{l} \lambda=\frac{\log _{\mathrm{e}}\left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)}{\mathrm{t}_{1 / 2}}=-\left(\frac{4.16-3.11}{1}\right) \\ \lambda=1.05 / \mathrm{hour} \\ \mathrm{t}_{1 / 2}=\frac{0.6931}{\lambda}=\frac{0.6931}{1.05}=42 \mathrm{~min} \end{array}