Carbon 14 Dating of Organic Material

Carbon-14 dating can tell you how old this dinosaur skull is.
Carbon-14 dating can tell you how old this dinosaur skull is. Image Source, Getty Images
by Anne Marie Helmenstine, Ph.D.
Updated March 06, 2017

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:

147N + 10n → 146C + 11H

Free carbon, including the carbon-14 produced in this reaction, can react to form carbon dioxide, a component of air. Atmospheric carbon dioxide, CO2, has a steady-state concentration of about one atom of carbon-14 per every 1012 atoms of carbon-12. Living plants and animals that eat plants (like people) take in carbon dioxide and have the same 14C/12C ratio as the atmosphere.

However, when a plant or animal dies, it stops taking in carbon as food or air. The radioactive decay of the carbon that is already present starts to change the ratio of 14C/12C. 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:

146C → 147N + 0-1e (half-life is 5720 years)

Example Problem

A scrap of paper taken from the Dead Sea Scrolls was found to have a 14C/12C 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:

log10 X0/X = kt / 2.30

where X0 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 / t1/2

so for this problem:

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

log X0 / X = [(1.21 x 10-4/year] x t] / 2.30

X = 0.795 X0, so log X0 / 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