question_answer

A piece of wood from the ruins of ancient building was found to have \[a{}^{14}C\] activity of 12 disintegrations per minute per gram of its carbon content. The \[{}^{14}C\] activity of the living wood is 16 disintegrations per minute per gram. How long did the tree, from which wooden sample came, die? Given, half-life \[{}^{14}C\] is 5760 yr.

Answer:

            Given,   R  = 12 dis/min/g \[{{R}_{0}}=16\] dis/min/g \[{{T}_{1/2}}=5760yr=\frac{In2}{\lambda }=\frac{0.6931}{\lambda }\] According to radioactive law of decay,             \[R={{R}_{0}}{{e}^{-\lambda \tau }}\Rightarrow \frac{R}{{{R}_{0}}}={{E}^{-\lambda \tau }}\] \[\Rightarrow \]   \[\frac{{{R}_{0}}}{R}={{e}^{\lambda \tau }}\]                                    ?(i) Taking log on both the sides of Eq. (i), we get \[\lambda t{{\log }_{e}}e={{\log }_{e}}\frac{{{R}_{0}}}{R}\] \[\Rightarrow \]   \[\lambda t=\left[ {{\log }_{10}}\left( \frac{16}{12} \right) \right]\times 2.303\] \[\Rightarrow \]   \[t=\frac{2.303(\log 4-\log 3)}{\lambda }\]             \[=\frac{2.303(0.602-0.4771)\times 5760}{0.6931}\] \[\Rightarrow \]   \[t=2391.20\,yr\]