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Carbon 14 dating in practice I


Alignments to Content Standards: F-IF.C.8.b F-LE.B.5

Task

A preserved plant is estimated to contain $1$ microgram (a millionth of a gram) of Carbon $14$. The amount of Carbon 14 present in the preserved plant is modeled by the equation $$ f(t) = A\left(\frac{1}{2}\right)^{\frac{t}{5730}} $$ where $t$ denotes time since the death of the plant, measured in years, and $A$ is the amount of Carbon $14$ present in the plant at death, measured in micrograms.

  1. How much Carbon $14$ was present in the living plant assuming it died $5000$ years ago?
  2. How much Carbon $14$ was present in the living plant assuming it died $10000$ years ago?
  3. The half-life of Carbon $14$ is the amount of time it takes for half of the Carbon $14$ to decay. What half-life does the expression for the function $f$ imply for Carbon $14$?

IM Commentary

In the task ''Carbon $14$ Dating'' the amount of Carbon $14$ in a preserved plant is studied as time passes after the plant has died. In practice, however, scientists wish to determine when the plant died and, as this task shows, this is not possible with a simple measurement of the amount of Carbon $14$ remaining in the preserved plant. Carbon $14$ dating requires many other hypotheses as will be addressed in ''Carbon $14$ dating in practice II.''

The equation for the amount of Carbon $14$ remaining in the preserved plant is in many ways simpler here, using $\frac{1}{2}$ as a base. This base is particularly convenient because the exponential rate of decay is determined by the half-life which, in this case, is seen in the denominator of the exponent, as shown in part (c) of the task. This should be contrasted with the equation in the task ''Carbon $14$ dating.''

All three parts of this task can be used for assessment or for instruction although the intention of this task is as a prelude to ''Carbon $14$ dating in practice II'' where the actual method used by scientists is discussed.

Solution

  1. We are given that $$ f(5000) = 1 $$ since there is one microgram of Carbon 14 remaining in the preserved plant after $5000$ years. This means, using the equation for $f(t)$, that $$ 1 = A\left(\frac{1}{2}\right)^{\frac{5000}{5730}}. $$ Solving for $A$ we find that $$ A = \left(\frac{2}{1}\right)^{\frac{5000}{5730}}. $$ Evaluating this expression on a calculator we find that there was a little more than $1.8$ micrograms of Carbon $14$ in the plant when it died.
  2. Repeating the reasoning for part (a), we find that $f(10000) = 1$ or $$ 1 = A\left(\frac{1}{2}\right)^{\frac{10000}{5730}} $$ Solving for $A$ here we find $$ A = \left(\frac{2}{1}\right)^{\frac{10000}{5730}}. $$ Evaluating this expression with a calculator, there were between $3.3$ and $3.4$ micrograms of Carbon $14$ in the plant when it died.

    Because the decrease in the amount of Carbon $14$ remaining in the preserved plant is exponential, it decreases by equal factors over equal intervals. During each of the two periods of $5000$ years it has decreased by a factor of a little over $1.8$. This is reasonable because the half-life of Carbon $14$ is about $5700$ years.

  3. The expression for $f$ is ideally suited for revealing the half-life of Carbon $14$. To see why, suppose we denote by $h$ the half-life of Carbon $14$. Then since there are $f(0) = A$ micrograms in the plant when it died, this means that $f(h) = \frac{A}{2}$. Writing out this equation we find $$ A\left(\frac{1}{2}\right)^{\frac{h}{5730}} = \frac{A}{2}. $$ Dividing both sides by $A$ gives $$ \left(\frac{1}{2}\right)^{\frac{h}{5730}} = \frac{1}{2}. $$ This equation is true when the exponent $\frac{h}{5730}$ is equal to one, in other words when $h = 5730$. So the half life of Carbon $14$ implied by the equation for $f$ is $5730$ years.