A piece of bone of an animal from a ruin is found to have ^ 14 C...

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$