Question

A normal adult jawbone contains 200 mg of Carbon-14 in a living person. If scientists found a jawbone that only had 50mg of Carbon-14, how old is the bone? (The half-life of C-14 is 5730 years).

Answer 1

Carbon-14 is a radioisotope of carbon that decays following first-order kinetics. There are four values of interest in this problem: the "normal" (or original) amount of carbon-14 for a jawbone (
`\mathrm{N_0}`), the actual amount of carbon-14 in a jawbone (
`\mathrm{N}`), the half-life of carbon-14 (
`\mathrm{t_(1/2)}`), and the actual time elapsed (
`\mathrm{t}`) from the original time. There is an equation that ties all these values in together,

`N= N_0 e^(-kt)`

where k is the rate constant, which, for first-order decay, is related to the half-life by

`k = (\ln 2)/( t_(1/2) ).`

What you want to find here is the time elapsed (t). So, you can substitute the latter equation for k into the k in the former equation to get

`N= N_0 e^{(-\ln 2 \;t)/(t_(1/2)).`

Rearranging to solve for t, the equation becomes

`t = \left((\ln (N_0)/(N))/(\ln 2) \right) t_(1/2).`

You are given all three of the values necessary to solve for t: The normal amount of carbon-14 is 200 mg; the actual amount of carbon-14 in the sample is 50 mg; and the half-life of carbon-14 is 5730 years. Plugging them into the above equation, we get

`t = \left(\frac{\ln \frac{200 \text{ mg}}{50 \text{ mg}}}{\ln 2} \right) \left(5730 \text{ years} \right) = 11460 \text{ years}.`

So the jawbone found is 11460 years old (or 11000 if accounting for sig figs).