The fossil bone has a {}^{14}C:{}^{12}C& | vedclass

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Question

$\frac{{^{14}{	ext{C}}}}{{^{12}{	ext{C}}}} = \frac{1}{{16}} = \frac{{	ext{N}}}{{{{	ext{N}}_0}}}$

$\because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\l

Vedclass · Accepted Answer

$22920$

Vedclass · Answer

$\frac{{^{14}{	ext{C}}}}{{^{12}{	ext{C}}}} = \frac{1}{{16}} = \frac{{	ext{N}}}{{{{	ext{N}}_0}}}$

$\because \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}ight)^{\mathrm{n}}$

$\Rightarrow \quad \frac{1}{16}=\left(\frac{1}{2}ight)^{n} \Rightarrow\left(\frac{1}{2}ight)^{4}=\left(\frac{1}{2}ight)^{n}$

or, $\quad \mathrm{n}=4$

or $\quad \frac{t}{\mathrm{T}}=4$

or $\quad \mathrm{t}=4 	imes \mathrm{T}=4 	imes 5730=22920$ $years$

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 ..........$years$

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