The decay constant of ₆C¹⁴ is 2.31 × 10^–4 year^–1. Its half-life is:

3 × l0³yrs

The correct answer is option 3. 3 × 10³yrs.

Given:
Decay constant (\(\lambda\)) = 2.31 × 10\(^{-4}\) year\(^{-1}\)

The half-life (\(t_{1/2}\)) is calculated using the formula:

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

Substituting the value of \(\lambda\) into the formula:

\(t_{1/2} = \frac{0.693}{2.31 \times 10^{-4} \text{ year}^{-1}}\)

Performing the calculation:

\(t_{1/2} \approx \frac{0.693}{2.31 \times 10^{-4}} \approx 3.0 \times 10^3\) years

Therefore, the correct answer is option (3) \(3.0 \times 10^3\) years.