Question

What is the age in years of a bone in which the ${}^{14}{\text{C}}^{12}\text{C}$ratio is measured to be $1.65\times {10}^{-13}$?

Short answer

The age of bone is 17,058.783 years.

Step-by-step solution

Step.1.

Ratio of ${}^{14}\text{C}$to ${}^{12}\text{C}$in living organisms is,

$\frac{N_0\left({}^{14}\text{C}\right)}{N_0\left({}^{12}\text{C}\right)}=1.3\times {10}^{-12}$

Ratio of ${}^{14}\text{C}$to ${}^{12}\text{C}$in bone is,

$\frac{N\left({}^{14}\text{C}\right)}{N\left({}^{12}\text{C}\right)}=1.65\times {10}^{-13}$

As $N\left({}^{12}\text{C}\right)=N_0\left({}^{12}\text{C}\right)$,

$\begin{array}{l}\frac{N\left({}^{14}\text{C}\right)}{N_0\left({}^{14}\text{C}\right)}=\frac{1.65\times {10}^{-13}}{1.3\times {10}^{-12}}\\ =1.27\times {10}^{-1}\\ =0.127\end{array}$

Therefore,

$N\left({}^{14}\text{C}\right)=0.127N_0\left({}^{14}\text{C}\right)$

Step.2.

Decay equation in terms of half-life is,

$\begin{array}{l}N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_1}}\\ {\left(\frac{1}{2}\right)}^{\frac{t}{t_1}}=\frac{N}{N_0}\end{array}$

Here,$\text{N}$is the present number of ${}^{14}\text{C}$atoms, $N_0$represent the initial number of atoms ${}^{14}\text{C}$atoms and $t_{\frac{1}{2}}$represents the half-life of ${}^{14}\text{C}$.

Substitute 5730 years for $t_{\frac{1}{2}}$and 0.127 for $\frac{N}{N_0}$,

${\left(\frac{1}{2}\right)}^{\frac{1}{5730\text{years}}}=0.127$

$\left(0.5\right)\frac{t}{5730\text{years}}=0.127$

Take natural log on both sides,

$\begin{array}{l}\frac{t}{5730\text{years}}\ln \left(0.5\right)=\ln \left(0.127\right)\\ t=\frac{\ln \left(0.127\right)}{\ln \left(0.5\right)}\left(5730\text{years}\right)\\ =\left(2.9771\right)\left(5730\text{years}\right)\\ =17,058.783\text{years}\end{array}$

Step.3.

Hence, the age of bone is 17,058.783 years.