Question

Question: The specific activity of ${}^{\mathbf{14}}\mathbf{C}$in the biosphere is $\mathbf{0}\mathbf{.}\mathbf{255}\mathbf{}\mathbf{Bq}\mathbf{}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$. What is the age of the piece of papyrus from an Egyptian tomb if its beta counting rate is $\mathbf{0}\mathbf{.}\mathbf{153}\mathbf{}\mathbf{Bq}\mathbf{}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$? The half-life of is $\mathbf{5730}$years.

Short answer

The age of the papyrus from the Egyptian tomb is $4222$years.

Step-by-step solution

Decay constant

The decay constant can be calculated from the half-life ${}^{14}\mathrm{C}$of with the help of the following expression:

$\begin{array}{l}{\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}\\ \mathrm{k}=\frac{0.693}{5730}{\mathrm{yr}}^{-1}\\ \mathrm{k}=1.21\times {10}^{-4}{\mathrm{yr}}^{-1}\end{array}$

Calculating the age of papyrus

Initial activity of ${}^{14}\mathrm{C}=0.255\mathrm{Bq}{\mathrm{g}}^{-1}$

Present activity of ${}^{14}\mathrm{C}=0.153\mathrm{Bq}{\mathrm{g}}^{-1}$

Radioactive decay follows first-order kinetics. Therefore,

$$\begin{gathered} {\mathrm{A}}_{\mathrm{t}}={\mathrm{A}}_0{\mathrm{e}}^{-\mathrm{kt}} \\ \ln \frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{t}}}=\mathrm{kt} \\ \mathrm{t}=\frac{2.303}{\mathrm{k}}\log \frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{t}}} \\ \mathrm{t}=\frac{2.303}{1.21\times {10}^{-4}{\mathrm{yr}}^{-1}}\log \frac{0.255}{0.153} \\ \mathrm{t}=4222\mathrm{years} \end{gathered}$$

The age of the papyrus found in the Egyptian tomb is$4222$years.