The half life for radioactive decay of C-14 is 5730 years. A sample of tree had only 80% of C-14 left. Estimate the age of tree. $(\log 5=0.698) (\log 4=0.602)$

Approx. 1800 years

The correct answer is Option (3) → Approx. 1800 years.

Given,

Half life of the radioactive decay, \((t_{1/2})\) = 5730 years

We know, radioactive decays are of the first order, and for a first order reaction,

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

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

or, \(k = \frac{0.693}{5730}\, \ years\)

We know, for a first order reaction,

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

Let us consider, \(a = 100\); then \(a - x = 80\)

Putting the values in the above equation, we get,

\(t = \frac{2.303}{\frac{0.693}{5730}} log \frac{100}{80}\)

or, \(t = \frac{2.303}{\frac{0.693}{5730}} log (1.25)\)

or, \(t = 2.303 \times \frac{5730}{0.693} \times 0.0969\)

or, \(t = \frac{1278.84}{0.693} \approx 1845\, \ years\)

But for choosing the correct option, we are rounding off to the nearest value, i.e., 1800 years