Question 19.PS.6: Carbon-14 Dating Charcoal fragments found in a prehistoric c......

15,300 years old

Strategy and Explanation   We will use Equation 19.2 to solve the problem

\ln (\frac{A}{A_0}) = -kt

where A is proportional to the known activity of the charcoal (2.40  min^{-1}  g^{-1}) and A_0 is proportional to the activity of the carbon-14 in living material (15.3  min^{-1}  g^{-1}). We first need to calculate k, the rate constant, using the half-life of carbon-14, 5.73 × 10^3  yr.

k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5.73 × 10^3  yr} = 1.21 × 10^{-4}  yr^{-1}

Now we are ready to calculate the time, t.

\ln (\frac{2.40  min^{-1}  g^{-1}}{15.3  min^{-1}  g^{-1}}) = -kt

\ln(0.1569) = – (1.21 × 10^{-4}  yr^{-1} )t

t = \frac{1.852}{1.21 × 10^{-4}  yr^{-1}} = 1.53 × 10^{-4}  yr

Thus, the charcoal is approximately 15,300 years old.

Reasonable Answer Check   The disintegration rate has fallen a factor of six from the rate for living material, so more than two but less than three half-lives have elapsed. This agrees with our calculated result.