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A piece of bone of an animal from a ruin is found to have ${ }^{14} \mathrm{C}$ activity of 12 disintegrations per minute per gm of its carbon content. The ${ }^{14} \mathrm{C}$ activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die? (Given half life of ${ }^{14} \mathrm{C}$ is $\mathrm{t}_{1 / 2}=5760$ years)
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The correct answer is:
2391 years
2391 years
Given, for ${ }^{14} \mathrm{C}$
$$
\begin{aligned}
&A_0=16 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{~A}=12 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{t}_{1 / 2}=5760 \text { years }
\end{aligned}
$$
Now, $\lambda=\frac{0.693}{\mathrm{t}_{1 / 2}}$
$\lambda=\frac{0.693}{5760}$ per year
Then, from, $\mathrm{t}=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_0}{\mathrm{~A}}$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}
$$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} 1.333
$$
$=\frac{2.303 \times 5760 \times 0.1249}{0.693}$
$=2390.81 \approx 2391$ years
$$
\begin{aligned}
&A_0=16 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{~A}=12 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{t}_{1 / 2}=5760 \text { years }
\end{aligned}
$$
Now, $\lambda=\frac{0.693}{\mathrm{t}_{1 / 2}}$
$\lambda=\frac{0.693}{5760}$ per year
Then, from, $\mathrm{t}=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_0}{\mathrm{~A}}$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}
$$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} 1.333
$$
$=\frac{2.303 \times 5760 \times 0.1249}{0.693}$
$=2390.81 \approx 2391$ years
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