Question

A 5.00 gcharcoal sample from an ancient fire pit has an ${}^{\mathbf{14}}\mathit{C}$activity of $\mathbf{63}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{disintegrations}\mathbf{/}\mathbf{min}$. A living tree has anactivity ofrole="math" localid="1661591390811" $\mathbf{15}\mathbf{.}\mathbf{3}\mathbf{}\mathit{d}\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{g}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{/}\mathit{m}\mathit{i}\mathit{n}$per 1.00 g. The half-life of${}^{\mathbf{14}}\mathit{C}$is 5730 y. How old is the charcoal sample?

Short answer

The charcoal sample is$1.61\times {10}^{3}y$ old.

Step-by-step solution

Given data

• Mass of the charcoal sample, m = 5g
• Activity of the ${}^{14}C$nuclide of the charcoal sample, $R=(63\mathrm{disintergrations}/\min )per5.00g$
• Activity of the ${}^{14}C$nuclide of a living tree,$R_0=\left(15.3\mathrm{disintegrations}/\min \right)\mathrm{per}1.00\mathrm{g}$
• Mass of the living tree,${\mathrm{m}}_{\mathrm{tree}}=1.00\mathrm{g}$
• Half-life of ${}^{14}C$,$T_{1/2}=5730y$

Understanding the concept of activity of the sample  

The activity of the sample is the disintegrations of the sample per second. The exponential relation of the activity of the sample with the disintegration constant and time gives the required time value of the decaying charcoal.

Formulae:

The rate of undecayed sample after a given time, $R=R_0{e}^{-\lambda t}\dots \dots ..\left(1\right)$

The disintegration constant, $\lambda =\frac{\ln 2}{T_{1/2}}.........\left(2\right)$

where, $T_{1/2}$is the half-life of the substance.

Calculation of the age of the charcoal

Substituting equation (2) in equation (1) with the given data, the age of the charcoal can be given as follows:

$$\begin{gathered} t=\frac{1}{\lambda }\ln \left(\frac{R_0}{R}\right) \\ =\left(\frac{T_{1/2}}{\ln 2}\right)\ln \left(\frac{R_0}{R}\right) \\ =\left(\frac{5730y}{\ln 2}\right)\ln \left(\frac{15.3/1.00g}{63.0/5.00g}\right) \\ =1.61\times {10}^{3}y \end{gathered}$$

Hence, the charcoal is $1.61\times {10}^{3}y$old.