Question 18.4: Carbon-14 is a radioactive nucleus with a half-life of 5760 ...
Carbon-14 is a radioactive nucleus with a half-life of 5760 years. Living matter exchanges carbon with its surroundings (for example, through CO_{2} ) so that a constant level of ^{14}C is maintained, corresponding to 15.3 decay events per minute. Once living matter has died, carbon contained in the matter is not exchanged with the surroundings, and the amount of ^{14}C that remains in the dead material decreases with time due to radioactive decay. Consider a piece of fossilized wood that demonstrates 2.4 ^{14}C decay events per minute. How old is the wood?
The ratio of decay events yields the amount of ^{14}C present currently versus the amount that was present when the tree died:
\frac{\left[ ^{14}C\right]}{\left[ ^{14}C\right]_{0}}=\frac{2.40 \min^{-1}}{15.3\min^{-1} }=0.157The rate constant for isotope decay is related to the half-life as follows:
k=\frac{\ln 2}{t_{1/2}}=\frac{\ln 2}{5760 years}=\frac{\ln 2}{1.82 ×10^{11} s}=3.81×10^{-12} s^{-1}With the rate constant and ratio of isotope concentrations, the age of the fossilized wood is readily determined:
\frac{\left[ ^{14}C\right]}{\left[ ^{14}C\right]_{0}}=e^{-kt}\ln\left( \frac{\left[ ^{14}C\right]}{\left[ ^{14}C\right]_{0}} \right)=-kt
-\frac{1}{k}\ln\left( \frac{\left[ ^{14}C\right]}{\left[ ^{14}C\right]_{0}} \right)=-\frac{1}{3.81 ×10^{-12}s}\ln\left( 0.157 \right)=t
4.86×10^{11} s=t
This time corresponds to an age of roughly 15,400 years.