Word Problems – 5

Carbon Dating

This one is about carbon dating. I find it fascinating that someone figured out this ingenious method to determine the age of once living things. This person was Willard Libby, a physicist and chemist who proposed this method in 1946.

Before I get to the word problem, let’s go through an explanation on how and why carbon dating works.

Life on earth is carbon-based, which means that its chemistry is based on carbon atoms. Living things ingest, breath in, grow from molecules which mostly include carbon atoms. But carbon atoms come in several isotope forms which means that the number of neutrons in its nucleus varies. Carbon 12 (6 protons and 6 neutrons), is the primary isotope (98.89% of all carbon). Carbon 13 (6 protons and 7 neutrons) is next at 1.11%. Both of these isotopes are stable, that is, they are not radioactive and break down into other elements. A trace amount of earth’s carbon is carbon 14 (6 protons and 8 neutrons). This isotope is radioactive. It is created mainly from the cosmic ray bombardment in earth’s upper atmosphere.

Though the percentage has varied in the past (and this is taken into consideration when very accurate results are required), it is a good approximation to assume that the percentage of carbon 14 has remained constant while life has been on earth. And this percentage remains constant in a living thing while it is alive. However, once it dies, the carbon in it is no longer being refreshed and the carbon 14 in it decays away. When a bone or any other object that was once living is found, measuring the remaining carbon 14 in it can be used to estimate its age.

Exponential Decay

Radioactive materials have a half life. The half life of a radioactive material is the time it takes for half of an original amount to remain. The amount left of a radioactive material is modelled as an exponential decay:

\[A=A_0e^{kt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\]


A = the amount of material left after t units

A0 = the original amount of material at time 0

e = an irrational number used frequently in science and engineering. e is approximately 2.71828

k = is a negative number (for exponential decay problems) that relates to the half life of a material and the units used for t

t = time in specified units. In our case, the units are years.

The Word Problem

The half life of carbon 14 is 5730 years. A bone found at an excavation site is found to have 30% of the carbon 14 it would have contained when it died. Approximately, how old is the bone?

The Solution

There doesn’t appear to be a lot of information here, but there is enough. Given the exponential decay equation (1), we have a lot of unknowns here: the original amount, the amount now, the parameter k, and of course, the answer to the question, t. But the first sentence is enough to solve for k.

If there is half of the original amount after 5730 years , then from equation (1):

\[0.5A_0=A_0e^{5730k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(2\right)\]

If you divide both side by A0, then the equation becomes:

\[0.5=e^{5730k}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(3\right)\]

This can be solved using logarithms, with a CAS calculator, or our all knowing internet. Solving this, we get k = −0.00012. Isn’t it interesting that we do not need to know the original amount to get this far?

Now that we know k, we can answer the question. Using the same trick as before, if there is only 30% carbon 14 left, then:

\[0.3A_0={A_0e}^{−0.00012t}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(4\right)\]

The A0 again cancels out and solving this we get t = 9953 years.

That’s old. No bones about it!