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

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(C) Let the initial ratio of ${}^{14}C:{}^{12}C$ be $R_0$. In the fossil,the ratio is $R = \frac{1}{16} R_0$.Since the amount of ${}^{12}C$ remains co

Vedclass · Accepted Answer

$22920$

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(C) Let the initial ratio of ${}^{14}C:{}^{12}C$ be $R_0$. In the fossil,the ratio is $R = \frac{1}{16} R_0$.Since the amount of ${}^{12}C$ remains constant in the bone,the amount of radioactive ${}^{14}C$ must have reduced to $\frac{1}{16}$ of its initial value.The decay law is given by $N(t) = N_0 \left( \frac{1}{2} ight)^n$,where $n$ is the number of half-lives.Given $\frac{N(t)}{N_0} = \frac{1}{16} = \left( \frac{1}{2} ight)^4$.Therefore,the number of half-lives $n = 4$.The age of the fossil is $t = n 	imes T_{1/2} = 4 	imes 5730 	ext{ years} = 22920 	ext{ years}$.

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 half-life of ${}^{14}C$ is $5730 \text{ years}$,then the age of the fossil bone is ......... $\text{years}$.

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