# Carbon 14 Dating, Variation 2

Alignments to Content Standards: F-LE.A.1.c

## Task

Carbon 14 is a common form of carbon which decays exponentially over time. The half-life of Carbon 14, that is, the amount of time it takes for half of any amount of Carbon 14 to decay, is approximately 5730 years.

Suppose we have a preserved plant and that the plant, at the time it died, contained 10 micrograms of Carbon 14 (one microgram is equal to one millionth of a gram).

1. Using this information, make a table to calculate how much Carbon 14 remains in the preserved plant after $5730 \times n$ years for $n = 0,1,2,3,4$.
2. What can you conclude from part (a) about when there is one microgram of Carbon 14 remaining in the preserved plant?
3. How much carbon remains in the preserved plant after $2865 = \frac{5730}{2}$ years? Explain how you know.
4. Using the information from part (c), can you give a more precise response to when there is one microgram of Carbon 14 remaining in the preserved plant?

## IM Commentary

An essential characteristic of exponential functions is that their values change by equal factors over equal intervals, that is, if $f(x)$ is an exponential function and $b$ a fixed real number, then the quotient $$\frac{f(x_0+b)}{f(x_0)}$$ always takes the same value, that is, it does not depend on the real number $x_0$. This exploratory task requires the student to use this property of exponential functions in order to estimate how much Carbon 14 remains in a preserved plant after different amounts of time.

The task can be taken further although the numbers become more and more complex. In order to estimate when there is one microgram of Carbon 14 remaining in the preserved plant to the nearest $\frac{2865}{2}$ years, the method of part (d) can be employed again, this time over the interval from $17190$ years to $20055$ years. Each time this calculation is iterated, the estimated period of time for when one microgram of Carbon 14 remains is cut in half.

## Solutions

Solution: Estimation

1. We are given that each 5730 years, the amount of Carbon 14 remaining in the preserved plant is cut in half. Since there are 10 micrograms when the plant dies, the table is as follows:
Number of years since plant died Amount of Carbon 14 remaining
$0 = 0 \times 5730$ 10 micrograms
$5730= 1 \times 5730$ 5 micrograms
$11460 = 2 \times 5730$ $2 \frac{1}{2}$ micrograms
$17190 = 3 \times 5730$ $1 \frac{1}{4}$ micrograms
$22920 = 4\times 5730$ $\frac{5}{8}$ micrograms
2. According to the table it is somewhere between $17190$ years and $22920$ years that one microgram of Carbon 14 remains.
3. We know that the amount of carbon 14 remaining in the preserved plant decays in an exponential fashion. This means that over equal periods of time, the rate of decay is the same. In particular, over each period of $2865$ years the amount of carbon remaining will decrease by the same factor. If we write $c$ for the amount of carbon 14 remaining after 2865 years, then this means that $$\frac{c}{10} = \frac{5}{c}:$$ here $\frac{c}{10}$ represents the factor of decrease for the first period of $2865$ years while $\frac{5}{c}$ represents the factor of decrease for the second period of $2865$ years. Solving for $c$ gives $$c = \sqrt{50} = 5\sqrt{2}.$$

So there are about $7$ micrograms of carbon remaining in the preserved plant after $2865$ years.

See the extra solution for additional guidance for students on this part of the problem.

4. From part (c) we know that after a period of $2865$ years, the amount of Carbon 14 remaining in the preserved plant decreases by a multiplicative factor of $\frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Because the amount of Carbon 14 remaining in the preserved plant decays exponentially, after any elapsed period of 2865 years the remaining Carbon 14 is multiplied by $\frac{1}{\sqrt{2}}$. So after $20055= 17190+2865$ years the amount of Carbon 14 remaining will be about $\frac{1}{\sqrt{2}} \times \frac{5}{4}$ micrograms which is a little less than $\frac{9}{10}$ micrograms. So we can conclude that the amount of time needed to have one microgram of Carbon 14 in the preserved plant is between $17190$ and $20055 = 17190 + 2865$ years.

Solution: Additional Guidance for part (c)

Part (c) of this task is difficult and students may well require assistance getting on the right path. One way to help them, without going to the proportion indicated in the solution, would be to write the rate of decay over each period of $2865$ years as $r$: that is, if there are $x$ micrograms of Carbon 14 present in the preserved plant at time $t$ then after $2865$ years pass, there will be $rx$ micrograms of Carbon 14 remaining in the preserved plant.

Applying this to our situation, after $17190$ years there care about $5/4$ micrograms of Carbon 14 present in the preserved plant. After $22920 = 17190 + 2 \times 2865$ years, the reasoning in the previous paragraph shows that there will be $$\left(\frac{5}{4}\right) r^2$$ micrograms of Carbon 14. According to the table, this quantity is also $\frac{5}{8}$ of a microgram and so we find $$\left(\frac{5}{4}\right) r^2 = \frac{5}{8}.$$ Solving for $r$ gives $r = \frac{1}{\sqrt{2}}$. Once the rate $r$ has been determined, the amount of Carbon 14 remaining after $20055$ years is calculated as above and found to be about $\frac{5}{4\sqrt{2}}$ or a little less than $9/10$ of a microgram.