The half-life for radioactive decay of $$C-14$$  is 5730 years. An archaeological artifact containing wood had only $$80\% $$  of the $$C-14$$  found in a living tree. The age of the sample is

Solution :
Radioactive decay follows first order kinetics.
$$\eqalign{ & {\text{Therefore,}} \cr & {\text{Decay constant}}\left( \lambda \right) = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.693}}{{5730}} \cr & {\text{Given,}}\left[ {{R_0}} \right] = 100 \cr & \therefore \,\,\left[ R \right] = 80 \cr & {\text{and}}\,\,t = \frac{{2.303}}{\lambda }\log \frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & = \frac{{2.303}}{{\left( {\frac{{0.693}}{{5730}}} \right)}}\log \frac{{100}}{{80}} \cr & = \frac{{2.303 \times 5730}}{{0.693}} \times 0.0969 \cr & = 1845\,{\text{years}} \cr} $$