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

$k = \dfrac{\ln 2}{ t_{1/2} }.$

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

$N= N_0 e^{\frac{-\ln 2 \;t}{t_{1/2}}.$

Rearranging to solve for t, the equation becomes

$t = \left(\dfrac{\ln \dfrac{N_0}{N}}{\ln 2} \right) t_{1/2}.$

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

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

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