Formula for Radiocarbon Dating

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Formula_for_Radiocarbon_Dating&oldid=465182"