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

3 × 10³yrs

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

The decay constant (\(\lambda\)) of a radioactive substance is related to its half-life (\(T_{1/2}\)) by the following equation:

\(\lambda = \frac{\ln(2)}{T_{1/2}}\)

Given that the decay constant (\(\lambda\)) for \(^{6}C^{14}\) is \(2.31 \times 10^{-4} \, \text{year}^{-1}\), we can rearrange the equation to solve for the half-life (\(T_{1/2}\)):

\(T_{1/2} = \frac{\ln(2)}{\lambda}\)

\(T_{1/2} = \frac{\ln(2)}{2.31 \times 10^{-4} \, \text{year}^{-1}}\)

Calculating the half-life:

\(T_{1/2} \approx 3000 \, \text{years}\)

Therefore, the correct answer is \((c) \, 3 \times 10^{3} \, \text{years}\).