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A piece of wood from an archaeological sample has 5.0 counts min $^{-1}$ per gram of $\mathrm{C} - 14,$ while a fresh sample of wood has a count of $15.0 \mathrm{min}^{-1} \mathrm{g}^{-1}$. If half-life of $\mathrm{C}-14$ is $5770 \mathrm{yr}$ the age of the archaeological sample is
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The correct answer is:
$9,200 \mathrm{yr}$
Age of archaeological sample.
$$
t=\frac{2.303}{k} \log \left(\frac{N_{0}}{N}\right)
$$
(Here, $N_{0}=$ initial activity. $\mathrm{N}=$ remaining activity)
$$
\begin{aligned}
k &=\frac{0.693}{t_{1 / 2}} \\
\quad \text { (where } t_{1 / 2} &=\text { half-life) } \\
&=\frac{0.693}{5770} \mathrm{yr}^{-1}
\end{aligned}
$$
$\therefore$
$$
t=\frac{2.303 \times 5770}{0.693} \log \left(\frac{15}{5}\right)
$$
$$
\begin{array}{l}
=19175.0 \log 3 \mathrm{yr} \\
=9148 \mathrm{yr} \approx 9200 \mathrm{yr}
\end{array}
$$
$$
t=\frac{2.303}{k} \log \left(\frac{N_{0}}{N}\right)
$$
(Here, $N_{0}=$ initial activity. $\mathrm{N}=$ remaining activity)
$$
\begin{aligned}
k &=\frac{0.693}{t_{1 / 2}} \\
\quad \text { (where } t_{1 / 2} &=\text { half-life) } \\
&=\frac{0.693}{5770} \mathrm{yr}^{-1}
\end{aligned}
$$
$\therefore$
$$
t=\frac{2.303 \times 5770}{0.693} \log \left(\frac{15}{5}\right)
$$
$$
\begin{array}{l}
=19175.0 \log 3 \mathrm{yr} \\
=9148 \mathrm{yr} \approx 9200 \mathrm{yr}
\end{array}
$$
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