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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.
CBSE | Class 12 | ncert exemplar | Nuclei | Physics
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,

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

i

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