Question

The activity of the hair of an Egyptian mummy is ${\text{7}}$disintegration ${\text{minut}}{{\text{e}}^{ - 1}}$ of ${{\text{C}}^{{\text{14}}}}$. Find the age of mummy. Given ${{\text{t}}_{{\text{0}}{\text{.5}}}}$ of ${{\text{C}}^{{\text{14}}}}$ is $5770$ year and disintegration rate of fresh sample of ${{\text{C}}^{{\text{14}}}}$ is $14$ disintegration ${\text{minut}}{{\text{e}}^{ - 1}}$.

Answer 1

Hint: To determine the answer we should know the radioactive decay constant and half-life formula. We will substitute the half-life value in the radioactive decay constant formula and we will consider the initial concentration of radioactive nuclei as hundred percent. By substituting all values we can determine the age of mummy.


Complete step-by-step solution:
The relation between radioactive disintegration constant and half-life is as follows:
\[{{\text{t}}_{{\text{1/2}}}}\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.693}}}}{\lambda }\]
Where,
\[{{\text{t}}_{{\text{1/2}}}}\] is the half life
\[{\lambda }\] is the radioactive decay constant
The formula to determine the radioactive decay constant is as follows:
\[{\lambda }\,{\text{ = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}\log \dfrac{{{{\text{N}}_{\text{o}}}}}{{\text{N}}}\]
Where,
\[{{\text{N}}_{\text{o}}}\]is the initial concentration of radioactive substance
\[{\text{N}}\]is the concentration of radioactive substance at time t.
On substituting the value of radioactive decay constant \[{\lambda }\]from half-life formula to radioactive decay constant formula we get,
\[\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{{\text{1/2}}}}}}{\text{ = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}\log \dfrac{{{{\text{N}}_{\text{o}}}}}{{\text{N}}}\]
On rearranging the above formula for t we get,
\[{\text{t = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{\text{0}}{\text{.693}}}}{{\text{t}}_{{\text{1/2}}}}\log \dfrac{{{{\text{N}}_{\text{o}}}}}{{\text{N}}}\]
The rate of disintegration is given by,
$ - \dfrac{{{\text{dN}}}}{{{\text{dt}}}}\,{ = \lambda N}$
At initially, the rate of disintegration of fresh sample of ${{\text{C}}^{{\text{14}}}}$ is $14$ disintegration ${\text{minut}}{{\text{e}}^{ - 1}}$.
For initial rate, $ - \dfrac{{{\text{dN}}}}{{{\text{dt}}}}\,{\text{ = }}\,{\lambda }{{\text{N}}_{\text{o}}}\,\,{\text{ = }}\,{\text{14}}$….$(1)$
At initially, the rate of disintegration of sample of ${{\text{C}}^{{\text{14}}}}$ is $7$ disintegration ${\text{minut}}{{\text{e}}^{ - 1}}$.
For rate at half-life, $ - \dfrac{{{\text{dN}}}}{{{\text{dt}}}}\,{\text{ = }}\,{\lambda }{{\text{N}}_{\text{t}}}\,\,{\text{ = }}\,7$…$(2)$
On dividing equation $(1)$ by$(2)$,
$\,\dfrac{{{{\text{N}}_{\text{o}}}}}{{{{\text{N}}_{\text{t}}}}}\,\,{\text{ = }}\,\dfrac{{14}}{7}$
$\,\dfrac{{{{\text{N}}_{\text{o}}}}}{{{{\text{N}}_{\text{t}}}}}\,\,{\text{ = }}\,2$
On substituting $2$for $\dfrac{{{{\text{N}}_{\text{o}}}}}{{{{\text{N}}_{\text{t}}}}}$and $5770$ for half-life in radioactive decay formula,
\[{\text{t = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\, \times \log 2\]
\[{\text{t = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\, \times \log 0.30\]
\[{\text{t = }}\,\dfrac{{0.693}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\,\]
\[{\text{t = }}\,{\text{5770}}\,{\text{year}}\]

So, the age of mummy is \[{\text{5770}}\]year.

Note: If the initial concentration was $100$%. We know at the half-life the left concentration will be $50$%. On substituting $100$for \[{{\text{N}}_{\text{o}}}\], $50$for N and $5770$ for half-life in radioactive decay formula,
\[{\text{t = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\, \times \log \dfrac{{100}}{{50}}\]
\[{\text{t = }}\,\dfrac{{{\text{2}}{\text{.303}}}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\, \times \log 0.30\]
\[{\text{t = }}\,\dfrac{{0.693}}{{{\text{0}}{\text{.693}}}} \times {\text{5770}}\,\]
\[{\text{t = }}\,{\text{5770}}\,{\text{year}}\]
The radioactive reaction is a first-order reaction. The half-life of the radioactive reaction is inversely proportional to the radioactive decay constant. The half-life of radioactive reactions does not depend upon the initial concentration of radioactive nuclei.