Engage your students with effective distance learning resources. ACCESS RESOURCES>>

Carbon 14 dating in practice II

Alignments to Content Standards: F-LE.A.2 F-LE.A.4


In order to use Carbon $14$ for dating, scientists measure the ratio of Carbon $14$ to Carbon $12$ in the artifact or remains to be dated. When an organism dies, it ceases to absorb Carbon $14$ from the atmosphere and the Carbon $14$ within the organism decays exponentially, becoming Nitrogen $14$, with a half-life of approximately $5730$ years. Carbon $12$, however, is stable and so does not decay over time.

Scientists estimate that the ratio of Carbon $14$ to Carbon $12$ today is approximately $1$ to $1,000,000,000,000$.

  1. Assuming that this ratio has remained constant over time, write an equation for a function which models the ratio of Carbon $14$ to Carbon $12$ in a preserved plant $t$ years after plant has died.
  2. In a particular preserved plant, the ratio of Carbon $14$ to Carbon $12$ is estimated to be about $1$ to $13,000,000,000$. What can you conclude about when plant lived? Explain.
  3. Dinosaurs are estimated to have lived from about $230,000,000$ years ago until about $65,000,000$ years ago. Using this information and the given half-life of Carbon 14, explain why this method of dating is not used for dinosaur remains.

IM Commentary

This problem introduces the method used by scientists to date certain organic material. It is based not on the amount of the Carbon $14$ isotope remaining in the sample but rather on the ratio of Carbon $14$ to Carbon $12$. This ratio decreases, hypothetically, at a constant exponential rate as soon as the organic material has ceased to absorb Carbon $14$, that is, as soon as it dies.

Carbon $14$ dating is a fascinating topic and much information can be found on Wikipedia.

Many factors limit the accuracy of using Carbon 14 for dating including

  1. the hypothesis that levels of Carbon $14$ in the environment have been relatively constant. These levels can be influenced by climate, by natural processes such as volcanoes, and in recent times, by human activity.
  2. the accuracy of measurement for the amount of Carbon $14$ in a given sample. This is a serious issue because the current ratio of $1$ to $1,000,000,000$ means that extremely precise measurements will be needed to determine how much Carbon $14$ is in a specimen.
  3. the method used to estimate the amount of Carbon $14$ in a given sample. If this is done by measuring the current decaying Carbon $14$ then it is not statistically reliable with very small samples. More recent technology actually allows scientists to measure the remaining Carbon $14$ much more accurately.

This problem is intended for instructional purposes only. It provides an interesting and important example of mathematical modeling with an exponential function. If the teacher has the time and inclination, it also reveals many of the inherent difficulties with mathematical modeling, some of which are mentioned in the previous paragraph as regards this particular example.


  1. Suppose $d$ represents the amount of Carbon $14$ in the plant at its time of death while $b$ represents the total amount of Carbon $12$. As time progresses, $b$ does not change while $d$ decreases exponentially. Hence the ratio $(d:b)$ also decreases exponentially. At the time of the plants' death, we have, assuming the same ratio of Carbon $14$ to Carbon $12$ as today, that $$ \frac{d}{b} = \frac{1}{1,000,000,000,000}. $$ Let $f$ be the function which assigns to $t$, the number of years since the plant's death, the ratio of Carbon $14$ remaining in the preserved plant to Carbon 12. So $f(0) = \frac{d}{b} = \frac{1}{1,000,000,000}$. Since the ratio of Carbon $14$ to Carbon $12$ is cut in half every $5730$ years, this means that the equation $$ f(t) = \left(\frac{1}{1,000,000,000}\right) \left(\frac{1}{2}\right)^{\frac{t}{5730}}. $$ models the ratio of Carbon $14$ to Carbon $12$ if the input $t$ is in years and the output $f(t)$ is measured in micrograms. Indeed, $f$ satisfies $f(0) = \frac{1}{1,000,000,000}$ and the values of $f$ for positive $t$ decrease exponentially, being cut in half every $5730$ years.

  2. If the ratio of Carbon $12$ to Carbon $14$ is $1$ to $13,000,000,000,000$ this means that $f(t) = \frac{1}{13,000,000,000,000}$ where $f$ was the function we computed in part (a). So we have $$ \left(\frac{1}{1,000,000,000,000}\right)\left(\frac{1}{2}\right)^{\frac{t}{5730}} = \frac{1}{13,000,000,000,000}. $$ This means that $$ \left(\frac{1}{2}\right)^{\frac{t}{5730}} = \frac{10}{13}, $$ which, since $\frac{1}{2} = 2^{-1}$ can be rewritten $$ 2^{\frac{-t}{5730}} = \frac{10}{13} .$$ Rewriting this equation using the definition of logarithm gives $$ \frac{-t}{5730} = \log_2{\left(\frac{10}{13}\right)}. $$ Solving for $t$ using a calculator we find that it has been about $21,200$ years since the plant died.
  3. If a dinosaur died $65,000,000$ years ago, this represents over $11,000$ half lives for Carbon $14$. Since $2^{10} = 1024$ is a little more than a thousand, this means that $$ 2^{10,000} = \left(2^{10}\right)^{1000} > 1000^{1000}. $$ So after being halved $11,000$ times the amount of Carbon $14$ remaining in the fossil would be zero or negligible since any measurable remains of Carbon $14$ would imply that the dinosaur had more than $1000^{1000}$ times that measurable amount of Carbon $14$ when it lived and this is not possible. For this reason, together with the limitations of technology to measure very small amounts of Carbon $14$, Carbon $14$ is only used to date artifacts up to about $60,000$ years old.