The fossil bone has a $^{14} C :^{12} C$ ratio, which is $(\frac{1}{16})$ of that in a living animal bone. If the half of $^{14} C$ is $5730$ years, then the age of the fossil (in years) is :-

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Solution

Here $\frac{N}{N_{0}} = \frac{1}{16} = \frac{1}{2^{t / T_{h}}}$ (Here $T_{h} \to$ half life)
$\Rightarrow \frac{1}{2^{4}} = \frac{1}{2^{t / 5730}}$
$\Rightarrow t = 5730 \times 4 = 22920$ years.

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Q. The fossil bone has a

^{14} C :^{12} C

ratio, which is

[\frac{1}{16}]

of that in a living animal bone. If the half-life of

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^{14} C :^{12} C

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The fossil bone has a 14C:12C ratio, which is (116) of that in a living animal bone. If the half of 14C is 5730 years, then the age of the fossil (in years) is :-

Here NN0=116=12t/Th (Here Th→ half life) ⇒124=12t/5730 ⇒t=5730×4=22920 years.

The fossil bone has a $^{14} C :^{12} C$ ratio, which is $(\frac{1}{16})$ of that in a living animal bone. If the half of $^{14} C$ is $5730$ years, then the age of the fossil (in years) is :-

Here $\frac{N}{N_{0}} = \frac{1}{16} = \frac{1}{2^{t / T_{h}}}$ (Here $T_{h} \to$ half life)
$\Rightarrow \frac{1}{2^{4}} = \frac{1}{2^{t / 5730}}$
$\Rightarrow t = 5730 \times 4 = 22920$ years.