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Every radioactive-decay chapter ends with a carbon-dating question, and most students solve it by pattern: halve, halve again, multiply by the half-life. The person behind the method did something more interesting. Willard Libby predicted how much carbon-14 living things should contain, built a counter sensitive enough to measure it, and then tested the whole idea on objects whose ages were already known. This post covers what he did, two worked dating calculations, and the mistakes to avoid.
Who was Willard Libby?
The Nobel Prize facts page records that Libby was born on 17 December 1908 in Grand Valley, Colorado, and died on 8 September 1980 in Los Angeles. He received the 1960 Nobel Prize in Chemistry alone, "for his method to use carbon-14 for age determination in archaeology, geology, geophysics, and other branches of science". The same page notes that he developed the method in 1949.
Where does carbon-14 come from?
In his Nobel lecture of 12 December 1960, Libby explains the chain. Cosmic rays striking the upper atmosphere produce neutrons, about two for every square centimetre of Earth's surface each second. Nitrogen-14 captures these neutrons and becomes carbon-14, releasing a proton. The carbon-14 joins the planet's carbon pool: carbon dioxide in the air, living things and carbon dissolved in the ocean.
Libby then did a back-of-the-envelope calculation that is worth repeating. That pool holds about 8.5 grams of carbon per square centimetre of surface. If production and decay have balanced, 2.0 decays per second are shared among 8.5 g, which is 2.0 ÷ 8.5 = 0.235 decays per second per gram, or about 14 per minute. He reported that measured living carbon agreed with this prediction to within about 10 percent.
How did Libby test the method before trusting it?
A living organism keeps topping up its carbon-14 by breathing or eating. At death the top-up stops and the activity falls. Libby wrote that a sample 5,600 years old should be half as radioactive as living matter.
Measuring that was the hard part. His counter, described in the lecture, sat inside eight inches of iron, and was surrounded by a ring of Geiger counters wired to switch the central counter off whenever a cosmic-ray particle passed through. That cut the background from about 500 counts per minute to between 1 and 6. With 5 grams of carbon inside, living carbon gave about 75 counts per minute, 5,600-year-old carbon 37.5, and 56,000-year-old carbon only 0.07.
Then came the test. Libby says historians told him that reliable history reached back only about 5,000 years, so he and James Arnold asked for samples of known age, mainly from Egypt. The oldest were wood from the First Dynasty tombs of the Vizier Hemaka and King Zet at Saqqara, agreed to be about 4,900 ± 200 years old, followed by cedar from the pyramid of Sneferu at Dahshur. Arnold and Libby published the comparison as Age Determinations by Radiocarbon Content: Checks with Samples of Known Age (Science, 1949). In the lecture Libby calls the plot the "Curve of Knowns", drawn with no adjustable constants, and judges the agreement "satisfactory".
How do you calculate a radiocarbon age?
Radiocarbon labs still follow the convention set out by Stuiver and Polach (1977): a "conventional radiocarbon age" uses the Libby half-life of 5,568 years, which gives a mean life of 8,033 years, and t = −8033 ln(A/A₀). The same paper notes that the better half-life is 5,730 years and that the convention assumes the atmospheric carbon-14 level was constant, which is why archaeologists later convert these ages to calendar years.
Worked example 1. Living carbon shows 14 decays per minute per gram. A piece of charcoal shows 3.5. The ratio is 3.5/14 = 1/4, which is two half-lives. Age = 2 × 5,568 = 11,136 years. The formula agrees: 8,033 × ln 4 = 8,033 × 1.386 ≈ 11,136 years.
Worked example 2. A wooden bowl shows 10.5 decays per minute per gram. The ratio is 0.75, not a neat power of a half. Use the formula: t = 8,033 × ln(14/10.5) = 8,033 × ln(1.333) = 8,033 × 0.2877 ≈ 2,311 years.
Our post on radioactive decay and half-life problems covers the exam-style versions of these sums.
What mistakes do students make?
Treating decay as linear. In example 2 a quarter of the activity is gone, so it is tempting to say "half of a half-life", 2,784 years. The true answer is 2,311 years, because the fastest loss happens first.
Mixing half-lives. Using 5,730 years in example 1 gives 11,460 years. Neither is wrong, but say which you used. In school problems, use the value given.
Forgetting the limit. At 56,000 years Libby's counter saw 0.07 counts per minute against a background of 1 to 6. Very old samples are hard to date this way.
Confusing activity with amount. A ratio of activities per gram of carbon equals the ratio of carbon-14 atoms, since activity = λN. Check that both numbers are per gram before you divide.
When the jump from N = N₀e^(−λt) to a date feels shaky, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
What can a student learn from Libby's way of working?
Libby's lecture shows three habits. He estimated before he measured, so he knew roughly what number to expect. He removed noise deliberately, with iron shielding and a ring of counters, instead of hoping the signal would show through. And he checked his method against cases with known answers before using it on unknown ones.
You can copy all three. Estimate the answer to a numerical before you solve it. Identify the "background" in your errors, such as unit slips, and remove it on purpose. And test any new formula on a problem whose answer you already know.
Try it now: estimate an age for a sample at 7 decays per minute per gram, then check it with t = 8,033 ln(A₀/A).
