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]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 13937.6 years
Radioactive decay follows first order kinetics, whose half-life does not depend on the amount present:
\( \displaystyle k = \frac{0.693}{t_{1/2}} \qquad t = \frac{2.303}{k}\log\frac{[R]_0}{[R]} \)
When 80% of the 14C has decayed, 20% is left, so \( [R]_0/[R] = 100/20 = 5 \).
\( \displaystyle t = \frac{2.303 \times t_{1/2}}{0.693}\log 5 \)
\( \displaystyle t = \frac{2.303 \times 6000}{0.693} \times 0.6990 \)
\( \displaystyle t = 19939.4 \times 0.6990 = 13937.6\ \text{years} \)
Shortcut: 80% decay lies between 2 half-lives (75%, 12000 years) and 3 half-lives (87.5%, 18000 years); only 13937.6 years falls in that range.
NCERT (2026-27 edition):
First Order Reactions – “… All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.”
Half-Life of a Reaction – “The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration.”