In CO₂, obtained from atmospheric air, small fraction of 1.3 x 10^-12 is the radioactive isotopes ${ }_6^{14}\mathrm{C}\left(\mathrm{\lambda }=3.83\times {10}^{-12}\right)$. If 200g of carbon fragment is found in an animal’s bone in an archaeological site shows an activity of 16 decays in 1 second, then the age of bone will be (Take ,in (25/8=1.1)

${ }_{12}^6\mathrm{C}12\mathrm{g}{\text{ of }}_6^{12}\mathrm{C}=6.02\times {10}^{23}\text{ atoms }$

200gm of carbon nearly contains $1\times {10}^{25}$ atoms 

When animal is alive ratio of ${}_6{}^{14}\mathrm{C}{\text{ to }}_6^{12}\mathrm{C}\text{ is }1.3\times {10}^{-12}$ in the bone 

Number of ${}_6{}^{14}C$ nuclear at the time $=1\times {10}^{25}\times 1.3\times {10}^{-12}=1.3\times {10}^{13}\text{ atoms }$

As $\mathrm{\lambda }=\frac{\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}{{\mathrm{N}}_0}\mathrm{t}=0$

So, ${\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{t}=0}=3.83\times {10}^{-12}\times 1.3\times {10}^{13}=50{\mathrm{s}}^{-1}$

So, ${\mathrm{e}}^{\mathrm{\lambda t}}=\frac{{\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{t}=0}}{{\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}_{\mathrm{t}=\mathrm{t}}}=\frac{50{\mathrm{s}}^{-1}}{16{\mathrm{s}}^{-1}}\Rightarrow \mathrm{t}=\frac{1}{\mathrm{\lambda }}\ln \left(\frac{50}{16}\right)=9500\mathrm{yr}$