Question

A wooden article found in a cave has only $40\% $ as much $^{14}{\text{C}}$ activity as a fresh piece of wood. How old is the article? (${t_{1/2}}$for $^{14}{\text{C}} = 5760{\text{ years}}$)

Answer 1

Hint: To answer this question, you must recall the rate law and half- life formula for radioactive decay. It is important to know that radioactive decay reactions follow first order kinetics. We shall first calculate the value of decay constant. Then, we shall substitute the values given in the formula below to calculate the time.
Formula used: $\lambda {\text{t}} = 2.303\log \dfrac{{{{\text{N}}_0}}}{{{{\text{N}}_{\text{t}}}}}$
Where, $t$ represents the age of the wooden article.
${{\text{N}}_0}$ represents the initial activity of the radioactive $^{14}{\text{C}}$ in the wooden article
${{\text{N}}_{\text{t}}}$ represents the final activity of the radioactive $^{14}{\text{C}}$ in the wooden article
And $\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$ represents the decay or disintegration constant of the reaction.
Where, ${t_{1/2}}$ represents the half- life of the wooden article.

Complete step by step answer:
Taking the initial activity of the radioactive$^{14}{\text{C}}$ in the wooden article as hundred, we get the final activity the radioactive $^{14}{\text{C}}$ in the wooden article as $40\% $ of the original $ = 40$
The first order rate equation for radioactive decay is given by $\lambda {\text{t}} = 2.303\log \dfrac{{{{\text{N}}_0}}}{{{{\text{N}}_{\text{t}}}}}$
Substituting the values to find $t$, we get, ${\text{t}} = \dfrac{{2.303}}{\lambda }\log \dfrac{{{{\text{N}}_0}}}{{{{\text{N}}_{\text{t}}}}}$
$ \Rightarrow t = \dfrac{{2.303 \times 5760}}{{0.693}} \times \log \left( {\dfrac{{100}}{{40}}} \right)$
Solving this, we get:
$\therefore t = 7617.29{\text{ years}}$

Thus, the given wooden article is $7617.29{\text{ years}}$ old.

Note:
Radiocarbon $\left( {^{14}{\text{C}}} \right)$ dating of historical wooden derived objects is based on the knowledge that the cosmic ray intensity has been practically constant for thousands of years. $^{14}{\text{C}}$ is formed in the upper atmosphere by the action of cosmic radiation on \[^{^{14}}{\text{N}}\].
The $^{14}{\text{C}}$ so produced is eventually converted into carbon dioxide , which in turn, is incorporated into plants and trees by the process of photosynthesis. Because of the natural plant- animal cycle, an equilibrium is established and all living matter contains the same small proportion of $^{14}{\text{C}}$ as in the atmosphere. Once the plant or animal dies, the uptake of $^{14}{\text{C}}$ ceases and its level in the dead begins to decrease as a result of the decay reaction. Comparison of the activity of carbon in the dead matter gives us the period of isolation of the material from the living cycle.