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The fossil bone has a `.^(14)C` : `.^(12)C` ratio, which 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 :A. `11460` yearsB. `17190 `yearsC. `22920` yearsD. `45840` years |
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Answer» Correct Answer - c After `n` half-lives (i.e., at `t=nT`) the number of nuclides left undecayed `N=N_(0)((1)/(2))^(n)` Given, `(N)/(N_(0))=(1)/(16)implies(1)/(16)=((1)/(2))^(n)` or `((1)/(2))^(4)=((1)/(2))^(n)` Equating the powers, we obtain `n=4` i.e., `(t)/(T)=4` or `t= 4T` or `t=4xx5730= 22920 years` `( because T= 5730 years)` |
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