**Introduction**

In the 1950s W.F. Libby and others (University of Chicago) devised a method of estimating the age of organic material based on the decay rate of carbon-14. Carbon-14 dating can be used on objects ranging from a few hundred years old to 50,000 years old.

Carbon-14 is produced in the atmosphere when neutrons from cosmic radiation react with nitrogen atoms:

^{14}_{7}N + ^{1}_{0}n → ^{14}_{6}C + ^{1}_{1}H

Free carbon, including the carbon-14 produced in this reaction, can react to form carbon dioxide, a component of air. Atmospheric carbon dioxide, CO_{2}, has a steady-state concentration of about one atom of carbon-14 per every 10^{12} atoms of carbon-12. Living plants and animals that eat plants (like people) take in carbon dioxide and have the same ^{14}C/^{12}C ratio as the atmosphere.

However, when a plant or animal dies, it stops intaking carbon as food or air. The radioactive decay of the carbon that is already present starts to change the ratio of ^{14}C/^{12}C. By measuring how much the ratio is lowered, it is possible to make an estimate of how much time has passed since the plant or animal lived. The decay of carbon-14 is:

^{14}_{6}C → ^{14}_{7}N + ^{0}_{-1}e (half-life is 5720 years)

**Example Problem**

A scrap of paper taken from the Dead Sea Scrolls was found to have a ^{14}C/^{12}C ratio of 0.795 times that found in plants living today. Estimate the age of the scroll.

**Solution**

The half-life of carbon-14 is known to be 5720 years. Radioactive decay is a first order rate process, which means the reaction proceeds according to the following equation:

log_{10} X_{0}/X = kt / 2.30

where X_{0} is the quantity of radioactive material at time zero, X is the amount remaining after time t, and k is the first order rate constant, which is a characteristic of the isotope undergoing decay. Decay rates are usually expressed in terms of their half-life instead of the first order rate constant, where

k = 0.693 / t_{1/2}

so for this problem:

k = 0.693 / 5720 years = 1.21 x 10^{-4}/year

log X_{0} / X = [(1.21 x 10^{-4}/year] x t] / 2.30

X = 0.795 X_{0}, so log X_{0} / X = log 1.000/0.795 = log 1.26 = 0.100

therefore, 0.100 = [(1.21 x 10^{-4}/year) x t] / 2.30

t = 1900 years