Question

A sample of an ancient wooden boat is found to undergo 9 dpm/ g of $^{14}C$ . What is the approximate age of the boat? The rate of disintegration of wood recently cut down is 15 dpm/g of $^{14}C$ .

Answer 1

Hint: There is a relationship between the half-life, rate constant and the concentration of the reactant and the products and it is as follows.
\[t=\dfrac{2.303}{K}\log \dfrac{{{r}_{0}}}{r}\]
Here, t = age of the particular compound
K =rate constant
${{r}_{0}}$ = initial concentration of the compound
r = final concentration of the compound

Complete answer:
- In the question it is asked to find the approximate age of the boat with the given data in the question.
- There is a formula to calculate the approximate age of the boat with the given data and it is as follows.
\[\begin{align}
  & t=\dfrac{2.303}{K}\log \dfrac{{{r}_{0}}}{r} \\
 & t=\dfrac{2.303\times {{t}_{{1}/{2}\;}}}{0.693}\log \dfrac{{{r}_{0}}}{r} \\
\end{align}\]
Here, t = age of the particular compound
K = rate constant
${{r}_{0}}$ = initial concentration of the compound = 15
r = final concentration of the compound = 9
${{t}_{{1}/{2}\;}}$ = half-life of the carbon-14 = 5730 years
- Substitute all the known values in the above formula to get the age of the wood and it is as follows.
\[\begin{align}
  & t=\dfrac{2.303}{K}\log \dfrac{{{r}_{0}}}{r} \\
 & t=\dfrac{2.303\times {{t}_{{1}/{2}\;}}}{0.693}\log \dfrac{{{r}_{0}}}{r} \\
 & t=\dfrac{2.303\times 5730}{0.693}\log \dfrac{15}{9} \\
 & t=4224.47years \\
\end{align}\]
Therefore the age of the boat is 4224.47 years.

Note:
The concentration of the radioactive element is going to be measured in ‘dpm’. Dpm means disintegration per minute. The radioactive elements are going to decay continuously till the concentration of the radioactive element should be zero. We need half-life of the radioactive element to find the age of the particular object.