The half-life of decaying \(^{14}C\) is \(6000\) years approximately. How much time is needed to decaying \(80\%\) of \(^{14}C\)? [log 5 = 0.6990]

13937.6 years

The correct answer is option (3) 13937.6 years.

\(^{14}C\) is a radioactive isotope of carbon, so the reaction is of the first order.

Given,

Half-life \(t_{1/2}\) \(= 6000\, \ years\)

For a first-order reaction,

\(t_{1/2} = \frac{0.693}{k}\)

or, \(k = \frac{0.693}{6000}\)

or, \(k = 0.0001155\, \ years^{-1}\)

We know,

\(t = \frac{2.303}{k}log\frac{a}{a - x}\)

For \(80\%\) completion of the reaction

Let \(a = 100\) and \(a - x = 100 - 80 = 20\)

Applying the values in the equation above we get

\(t_{80\%} = \frac{2.303}{k}log\frac{100}{20}\)

or, \(t_{80\%} = \frac{2.303}{0.0001155}log\, \ 5\)

or, \(t_{80\%} = \frac{2.303}{0.0001155} × 0.6990\)

or, \(t_{80\%} = \frac{1.609797}{0.0001155}\)

or, \(t_{80\%} = 13937.63\, \ years\)

or, \(t_{80\%} \approx 13937.6\, \ years\)