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

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(C) The radioactive decay follows the formula $N_t = N_0(1/2)^n$, where $n$ is the number of half-lives.Given that the ratio of $^{14}C : ^{12}C$ in t

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$22920$

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(C) The radioactive decay follows the formula $N_t = N_0(1/2)^n$, where $n$ is the number of half-lives.Given that the ratio of $^{14}C : ^{12}C$ in the fossil is $(1/16)$ of the living bone, we have $(1/2)^n = 1/16$.Since $(1/2)^4 = 1/16$, the number of half-lives $n = 4$.The age of the fossil is calculated as $Age = n 	imes 	ext{half-life}$.$Age = 4 	imes 5730 	ext{ years} = 22920 	ext{ years}$.

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

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