From Anne Marie Helmenstine, Ph.D.,
Your Guide to Chemistry.
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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:

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 intaking 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 / t_1/2

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

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