
Carbon dating is the technique of comparing the concentrations of carbon12 and carbon14 to determine the age of a biological material. While the most common isotopes of carbon, ${}_{6}{}^{13}\text{C}$ and ${}_{6}{}^{12}\text{C}$, have stable nuclei, ${}_{6}{}^{14}\text{C}$ is radioactive. Carbon14 is produced steadily in Earth’s upper atmosphere, where ${}_{7}{}^{14}\text{N}$ nuclei interact with neutrons produced by cosmic rays entering the atmosphere. Living organisms constantly exchange carbon with the environment maintaining a constant ratio of about one atom of carbon14 for every 10^{13} atoms of carbon12. When an organism dies, active carbon exchange with the environment stops. Carbon14 already within the organism decays radioactively with a halflife of 5,730 years. This decay provides a natural clock which starts ticking the moment an organism dies. For example, carbon dating can determine the age of ancient Egyptian papyrus samples and the charcoal from prehistoric campfires. Carbon dating works reliably up to about 10 times the halflife, or 57,300 years. After 10 halflives there is not enough carbon14 remaining to measure accurately.

The rate of radioactive decay—how many nuclei decay per unit time—is related to the halflife t_{½}. If we have a radioactive sample that initially has N_{0} nuclei, then after time t the number of radioactive nuclei remaining in the sample is given by equation (27.2).

(27.2)  $$N={N}_{0}{\left(\frac{1}{2}\right)}^{t/{t}_{{}_{\text{\xbd}}}}$$
 N  =  amount of sample after time t  N_{0}  =  initial amount of sample  t  =  time elapsed (s)  t_{½}  =  halflife (s) 
 Decay rate equation


How can we calculate the amount of time that has elapsed to produce a particular concentration of a radioactive isotope? If we know t_{½}, N_{0}, and N, then we can solve for time t. The equation above relates the current concentration to initial concentration and halflife. Since time is in the exponent, we start by taking the logarithm of both sides of the equation: $$\begin{array}{ccc}\frac{N}{{N}_{0}}={\left(\frac{1}{2}\right)}^{t/{t}_{1/2}}& \Rightarrow & \mathrm{log}\left(\frac{N}{{N}_{0}}\right)=\mathrm{log}{\left(\frac{1}{2}\right)}^{t/{t}_{1/2}}=\frac{t}{{t}_{1/2}}\mathrm{log}\left(\frac{1}{2}\right)\end{array}=\frac{t}{{t}_{1/2}}\mathrm{log}2$$
 Now solve for the time t: $$t=\frac{{t}_{1/2}\mathrm{log}\left(N/{N}_{0}\right)}{\mathrm{log}2}$$


The general technique for dating matter using radioactivity is called radiometric dating. Besides carbon14, a number of other radioactive isotopes are used for dating. For example, climate scientists measure the concentration of oxygen18 and oxygen16 in ice core samples to understand the composition of the atmosphere over time. Geologists use this same technique to date rock samples by measuring the concentration ratio of uranium238 to plutonium239. This technique allows geologists to explore geologic times because the halflife of uranium238 is 4.5 billion years.

An ancient fern specimen has oneeighth the amount of carbon14 per gram as compared to a living creature. How old is the specimen?
Asked: 
the number of years ago the fern died 
Given: 
The mass of carbon14 has decayed to oneeighth its original value. 
Relationships: 
For the time period of one halflife, half of the radioactive atoms will decay radioactively. The halflife of carbon14 is 5,730 years. 
Solution: 
After one halflife, half of the carbon14 atoms will be left. After the second halflife, onehalf of the remaining onehalf carbon14 atoms will remain, or onefourth. After the third halflife, oneeighth will remain. The fern therefore died three halflives of carbon14 ago: 3 × 5,730 years is 17,190 years. 
Answer: 
The fern died 17,190 years ago. 

 
