Radiocarbon dating
Tonight's idea
A tiny fraction of the carbon in the air is carbon-14, a radioactive isotope continually made by cosmic rays. Plants take it in and animals eat the plants, so carbon being built into living tissue tracks the atmosphere's carbon-14 ratio. Once that tissue stops exchanging carbon, its carbon-14 decays with nothing to replace it.
Decay is the equation of a11, rate proportional to amount: \(N(t) = N_0 e^{-kt}\), with \(N_0\) the amount when exchange stops and \(k\) the decay constant. The half-life \(T = 5{,}730\) years fixes \(k\): \(N(T) = N_0/2\) means \(e^{-kT} = \tfrac12\), so \(k = \ln 2 / T \approx 1.21 \times 10^{-4}\) per year.
A lab measures the fraction remaining \(f = N/N_0\) against modern carbon. Take logs of \(f = e^{-kt}\) to get the age:
\(t = -\dfrac{\ln f}{k} = \dfrac{T}{\ln 2}\,\ln\dfrac1f \approx 8{,}267 \ln\dfrac1f\) years.
Three things make the answer uncertain. The atmosphere's ratio has drifted over time, so raw ages are calibrated against samples of known age. Intruding younger or older carbon (contamination) shifts \(f\). And \(f\) is a measurement with an error bar, which the logarithm turns into a range of years.
Watch one: 60% remaining
Charcoal from a hearth has a carbon-14 ratio 0.60 times the modern value. Then
\[t = 8{,}267 \ln\frac{1}{0.60} = 8{,}267 \times 0.511 \approx 4{,}220 \text{ years}.\]
Check it in half-lives: \(4{,}220 / 5{,}730 = 0.737\), and \(2^{-0.737} = 0.600\). Two cautions. The date is roughly when that wood's carbon was fixed from the atmosphere (when the tree ring grew), not when the fire burned. Inner wood can therefore be centuries older than the fire. And 4,220 is a raw age; the calibration curve would move it.
Your turn
- What fraction of the original carbon-14 remains after 2,000 years? After 10,000 years?
- A bone has 12.5% of the modern ratio. Find its age twice: by counting half-lives, and with the formula.
- A lab reports \(f = 0.50 \pm 0.01\). (a) Put \(0.49\) and \(0.51\) into the formula and give the age as a range. (b) Differentiate \(t = -\tfrac1k \ln f\) to show a small measurement error \(|\Delta f|\) gives an age uncertainty \(|\Delta t| \approx |\Delta f|/(kf)\), and check against (a). (c) Evaluate \(|\Delta t|\) for the same \(\pm 0.01\) when \(f = 0.05\). Why does the method struggle with very old samples?
- A sample is really 20,000 years old. In handling, 1% of its carbon is replaced by modern carbon (\(f = 1\)). (a) Find the true \(f\), the measured \(f\), and the age the lab reports. (b) Repeat for a 40,000-year-old sample. Which way does modern contamination push a date, and when does it matter most?
Stretch: how long does one atom last?
The fraction of atoms that decay between \(t\) and \(t + dt\) is \(k e^{-kt}\,dt\). The average lifetime of an atom is therefore \(\displaystyle\int_0^\infty t\,k e^{-kt}\,dt\). Evaluate it (integration by parts, a9). Is the mean lifetime longer or shorter than the half-life, and why?
Answers
- \(e^{-k \cdot 2000} = e^{-0.242} = 0.785\), so 78.5%. \(e^{-k \cdot 10000} = e^{-1.21} = 0.298\), about 30%.
- \(12.5\% = \tfrac18 = \left(\tfrac12\right)^3\): three half-lives, \(3 \times 5{,}730 = 17{,}190\) years. Formula: \(8{,}267 \ln 8 = 8{,}267 \times 2.079 = 17{,}190\) years.
- (a) \(f = 0.51\) gives 5,566 years and \(f = 0.49\) gives 5,897: the age is 5,730 give or take about 165 years. (b) \(\dfrac{dt}{df} = -\dfrac{1}{kf}\), so \(|\Delta t| \approx \dfrac{|\Delta f|}{kf} = \dfrac{0.01}{1.21 \times 10^{-4} \times 0.5} = 165\) years, matching (a). (c) At \(f = 0.05\) (about 24,800 years) the same \(\pm 0.01\) gives a linear estimate of about \(\pm 1{,}650\) years, ten times worse. The error grows like \(1/f\): the less carbon-14 is left, the more every tiny measurement error is magnified, until past about 50,000 years there is too little left to measure.
- (a) True \(f = 2^{-20000/5730} = 0.0890\). Measured \(0.99 \times 0.0890 + 0.01 \times 1 = 0.0981\). Reported age \(8{,}267 \ln(1/0.0981) = 19{,}190\) years, about 800 years too young. (b) True \(f = 0.0079\); measured \(0.0178\); reported age 33,300 years, nearly 7,000 years too young. Modern carbon always makes a sample look younger, and the older the sample, the less of its own carbon-14 it has to dilute the intruder.
Stretch. By parts with \(u = t\), \(dv = k e^{-kt}\,dt\): \(\big[-t e^{-kt}\big]_0^\infty + \displaystyle\int_0^\infty e^{-kt}\,dt = 0 + \dfrac1k\). The mean lifetime is \(1/k = T/\ln 2 \approx 8{,}267\) years, longer than the half-life. Half the atoms are gone by 5,730 years, but the survivors include some that last for a very long time, and that long tail pulls the mean up. (It is also the 8,267 in the age formula: age equals mean lifetime times \(\ln(1/f)\).)