Question

A piece of wood from the ruins of an 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 ago did the tree, from which the wooden sample came, die? Given half-life of ${}^{14}C$ is $5760$ years.

Answer 1

Given,

Rate of disintegration in old wood sample of $C-14$ radio-active atoms is $12$ atoms per min gm.

Initially rate of disintegration of $C-14$ when the tree was live $=16$ atoms per min per gm.

$T_{1/2}$ of $C-14$ nuclei $=5,760$ years

According to radio-active decay law,

$N=N_0{e}^{-\lambda t}$

$12=16e^{-\lambda t}$

$e^{\lambda t}=\frac{16}{12}$

$lo{g}_e{e}^{\lambda t}=lo{g}_e\frac{4}{3}$



$\lambda t=lo{g}_e\frac{4}{3}$

$t=\frac{2.303\times lo{g}_{10}\left(\frac{4}{3}\right)}{\lambda }$ $(\because \lambda =\frac{0.6931}{T_{1/2}})$


$=\frac{2.303\times (log4-log3)\times 5760}{0.6931}$ year




$\therefore t=\frac{2.303(0.6020-0.4771)5760}{0.6931}$


$=2391.20years$


Final Answer: $2391.20$ years