Fossil bone has a {}^{14}C:{}^{12}C ratio, which is left[ {frac{1}{{16}}} right] &

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13.Nuclei

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The fossil bone has a ${}^{14}C:{}^{12}C$ ratio, which is $\left[ {\frac{1}{{16}}} \right]$ of that in a living animal bone. If the halflife of ${}^{14}C$ is $5730\, years$, then the age of the fossil bone is ..........$years$

A

$11460$

B

$17190$

C

$22920$

D

$45840$

(AIIMS-2006)

Solution

$\frac{{^{14}{\text{C}}}}{{^{12}{\text{C}}}} = \frac{1}{{16}} = \frac{{\text{N}}}{{{{\text{N}}_0}}}$

$\because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$

$\Rightarrow \quad \frac{1}{16}=\left(\frac{1}{2}\right)^{n} \Rightarrow\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{n}$

or, $\quad \mathrm{n}=4$

or $\quad \frac{t}{\mathrm{T}}=4$

or $\quad \mathrm{t}=4 \times \mathrm{T}=4 \times 5730=22920$ $years$

Standard 12

Physics

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