InterviewSolution
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InterviewSolution
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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. 45840 yearsD. 22921 years |
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Answer» Correct Answer - D (d) After `n` half-lives (`Le`., at `t = nT`) the number of nuclides left undecayed, `N = N_0 ((1)/(2))^n` Given, `(N)/(N_0) = (1)/(16)` `:. (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 = 4 T` or `t = 4 xx 5730 = 22920 years` `(because T = 5730 years)`. |
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