Worked exercise — from activity to the age of an artefact

Problem

An archaeological find shows a quantity of residual $^{14}\text{C}$ equal to $34\%$ of that of living tissue. Estimate the age of the find, knowing that the half-life of carbon-14 is $T_{1/2} = 5730$ years.

Solution

Setup. While alive, an organism exchanges carbon with the environment and keeps the fraction of $^{14}\text{C}$ constant. At death the exchange stops and $^{14}\text{C}$ starts to decay. The ratio of residual to initial atoms follows:

$\frac{N}{N_0} = 0{,}34 = 2^{-t/T_{1/2}}$

Taking logarithms. Taking the base-$2$ logarithm of both sides:

$\frac{t}{T_{1/2}} = -\log_2(0{,}34) = \log_2\!\left(\frac{1}{0{,}34}\right) \approx 1{,}556$

Age of the find.

$t = 1{,}556\cdot T_{1/2} = 1{,}556\cdot 5730\ \text{years} \approx \ev{t \approx 8920\ \text{years}}$

An age consistent with the Mesolithic-Neolithic period.

Limits of the method $\sim 50\,000$ years: beyond that, the residual $^{14}\text{C}$ is too little to be measured precisely.

Radiocarbon dating is reliable up to about ten half-lives, i.e.

Connections

Topics: Nuclear physics Concepts: Radioactive decay Skills: Using logarithmic scales