The half-life for radioactive decay of $14_{C}$ is 5730 years. An archaeological artifact containing wood had only 80% of the ${}^{14}C$ activity found in living tree. Estimate the age of the sample or calculate the age of the artefact.

Solution

We have given
t_1/2 = 5730 years

\frac{N}{N_{0}} = 80 % = 0.80

Decay constant,

λ = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5730} = 1.21 \times 10^{- 4} {year}^{- 1}

All radioactive nuclear decay are first order process.
Therefore, decay constant,

λ = \frac{2.03}{t} \log \frac{N_{0}}{N} λ = \frac{2.30}{t} \log \frac{10}{8}

time, t = \frac{2.303}{λ} \log 1.25 = \frac{2.303}{1.21 \times 10^{- 4}} \log 1.25 = \frac{2.303}{1.21 \times 10^{- 4}} \times 0.0969 = 1845 years . Age of sample = 1845 years .

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