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A piece of bone of an animal from a ruin is found to have $^{14}C$ activity of $12$ disintegrations per minute per gm of its carbon content. The $^{14}C$ activity of a living animal is $16$ disintegrations per minute per gm. How long ago nearly did the animal die? ............$years$ (Given halflife of $^{14}C$ is $t_{1/2} = 5760\,years$ )
$1672$
$2391$
$3291$
$4453$
Solution
Given, for $^{14} \mathrm{C}$
$A_{0}=16$ dis $\min ^{-1} g^{-1}$
$A=12$ dis $\min ^{-1} g^{-1}$
$\mathrm{t}_{1 / 2}=5760$ years
Now, $\lambda = \frac{{0.693}}{{{t_{1/2}}}}$
$\lambda=\frac{0.693}{5760}$ per year
Then, from, $t=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_{0}}{\mathrm{A}}$
$=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}$
$ = \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
