Question 29.7: Should We Report This to Homicide? Goal Apply the technique ...
Should We Report This to Homicide?
Goal Apply the technique of carbon-14 dating.
Problem A 50.0-g sample of carbon is taken from the pelvis bone of a skeleton and is found to have a carbon-14 decay rate of 200.0 decays / \mathrm{min}. It is known that carbon from a living organism has a decay rate of 15.0 decays / \mathrm{min} \cdot \mathrm{g} and that { }^{14} \mathrm{C} has a half-life of 5730 \mathrm{yr}=3.01 \times 10^{9} \mathrm{~min}. Find the age of the skeleton.
Strategy Calculate the original activity and the decay constant, and then substitute those numbers and the current activity into Equation 29.19.
t=-{\frac{\ln\left({\frac{R}{R_{0}}}\right)}{\lambda}} (29.19)
Calculate the original activity R_{0} from the decay rate and the mass of the sample:
R_{0}=\left(15.0 \frac{\text { decays }}{\min \cdot \mathrm{g}}\right)(50.0 \mathrm{~g})=7.50 \times 10^{2} \frac{\text { decays }}{\min }
Find the decay constant from Equation 29.5:
T_{1/2}={\frac{\ln2}{\lambda}}={\frac{0.693}{\lambda}} (29.5)
\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{3.01 \times 10^{9} \mathrm{~min}}=2.30 \times 10^{-10} \mathrm{~min}^{-1}
R is given, so now we substitute all values into Equation 29.19 to find the age of the skeleton:
\begin{aligned} t & =-\frac{\ln \left(\frac{R}{R_{0}}\right)}{\lambda}=-\frac{\ln \left(\frac{200.0 \text { decays } / \mathrm{min}}{7.50 \times 10^{2} \text { decays } / \mathrm{min}}\right)}{2.30 \times 10^{-10} \mathrm{~min}^{-1}} \\ & =\frac{1.32}{2.30 \times 10^{-10} \mathrm{~min}^{-1}} \\ & =5.74 \times 10^{9} \mathrm{~min}=1.09 \times 10^{4} \mathrm{yr} \end{aligned}
Remark For much longer periods, other radioactive substances with longer half-lives must be used to develop estimates.