Formula for Radiocarbon Dating
Theorem
Let $Q$ be a quantity of a sample of dead organic material (usually wood) whose time of death is to be determined.
Let $t$ years be the age of $Q$ which is to be determined.
Let $r$ denote the ratio of the quantity of carbon-14 remaining in $Q$ after time $t$ to the quantity of carbon-14 in $Q$ at the time of its death.
Then the number of years that have elapsed since the death of $Q$ is given by:
- $t = -8060 \ln r$
Proof
Let $x_0$ denote the ratio of carbon-14 to carbon-12 in $Q$ at the time of its death.
Let $x$ denote the ratio of carbon-14 to carbon-12 in $Q$ after time $t$.
Thus:
- $r = \dfrac x {x_0}$
It is assumed that the rate of decay of carbon-14 is a first-order reaction.
Hence we use:
\(\ds x\) | \(=\) | \(\ds x_0 e^{-k t}\) | First-Order Reaction | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac x {x_0}\) | \(=\) | \(\ds e^{-k t}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -k t\) | \(=\) | \(\ds \map \ln {\dfrac x {x_0} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -k t\) | \(=\) | \(\ds \map \ln r\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds t\) | \(=\) | \(\ds -\dfrac {\ln r} k\) |
Chemical analysis tells us that after $10$ years, $99.876 \%$ of the carbon-14 that was present in the organic matter still remains.
Thus at time $t = 10$, we have:
\(\ds 0.99876 x_0\) | \(=\) | \(\ds x_0 e^{-10 k}\) | substituting for $t$ and $x$ in $(1)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln 0.99876\) | \(=\) | \(\ds -10 k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k\) | \(\approx\) | \(\ds 0.000124\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 k\) | \(\approx\) | \(\ds 8060\) |
Hence the result.
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate