Question

A piece of wood from the ruins of an ancient building was found to have ${C^{14}}$ activity of $12$ disintegration per minute per gram of its carbon content. The ${C^{14}}$ activity of the living wood is 16 per minute per gram. How long ago did the tree, from which the wooden sample came? Given the half-life of ${C^{14}}$ is $5760$ years.

Answer 1

To calculate the time from which the ancient wooden sample came, we need to use the radioactive decay law which states that the number of nuclei undergoing decay per unit time is proportional to the total number of nuclei in the sample. Then we use the formula for decay constant.

Formulae used:
The radioactive decay law is given by
$R = {R_0}{e^{ - \lambda t}}$
And decay constant $\lambda = \dfrac{{0.6931}}{{{T_{\dfrac{1}{2}}}}}$
Where, $R$ - rate of disintegration after time $t$, ${R_0}$ - Initial rate of disintegration and ${T_{\dfrac{1}{2}}}$ - half life period.

Complete step by step answer:
Let us write the given data in an understanding manner. Rate of disintegration in old wood sample of ${C^{14}}$ atoms is $12$ disintegration per minute per gram and Rate of disintegration in new wood sample of ${C^{14}}$ atoms is $16$ disintegration per minute per gram. So, $R = 12$ and ${R_0} = 16$.
And half life period of ${C^{14}}$ atoms, ${T_{\dfrac{1}{2}}} = 5760$ years
We have to calculate the value of $t$ , using the radioactive decay law
$R = {R_0}{e^{ - \lambda t}}$
$\Rightarrow 12 = 16{e^{ - \lambda t}} \\\ \Rightarrow {e^{\lambda t}} = \dfrac{{16}}{{12}} \\\ \Rightarrow {e^{\lambda t}} = \dfrac{4}{3}$

Taking log on both sides, we get
${\log _e}{e^{\lambda t}} = {\log _e}\left( {\dfrac{4}{3}} \right)$
$\Rightarrow \lambda t = {\log _e}\left( {\dfrac{4}{3}} \right)$
Now, substituting the decay constant formula in above equation, we get
$t = \dfrac{{2.303 \times {{\log }_{10}}\left( {\dfrac{4}{3}} \right)}}{\lambda }$
$\Rightarrow t = \dfrac{{2.303 \times \left( {\log 4 - \log 3} \right) \times 5760}}{{0.6931}}$
$\Rightarrow t = \dfrac{{2.303 \times \left( {0.6060 - 0.4771} \right) \times 5760}}{{0.6931}}$
Solving, we get
$\therefore t = 2391.2$ years
The tree was $2391.2$ years old.

Note: We should know the formulae for radioactive decay law and decay constant.Keep in mind that to convert ${\log _e}$ to ${\log _{10}}$ , we multiply it by $2.303$. We should know how to use the logarithmic table to know the values. We can also use the formula $t = \dfrac{1}{\lambda }\log \dfrac{{{R_0}}}{R}$.