C¹⁴ has a half-life of 5760 years. 100 mg of the sample containing ¹⁴C is reduced to 25 mg in

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17128 years

1440 year

11520 year

2880 years

answer is A.

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Detailed Solution

We know that $N = {(\frac{1}{2})}^{n} N_{0}$

$\begin{array}{l} ∴ 25 = {(\frac{1}{2})}^{n} 100 \\ \frac{25}{100} = {(\frac{1}{2})}^{n} \Rightarrow n = 2 \\ and T = n \times t_{1 / 2} \\ ∴ T = 2 \times 5760 = 11520 years \end{array}$

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