Radiometric Dating/Example/Radium in Lead/1000 Years
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Example of Radiometric Dating
Let $Q$ be a sample of lead.
Let it be established that $10 \%$ of the radium decays in $200$ years.
After $1000$ years, there will be approximately $59.05 \%$ of the original amount of radium in $Q$.
Proof
From First-Order Reaction, we have:
- $x = x_0 e^{-k t}$
where:
- $x$ is the quantity of radium at time $t$
- $x_0$ is the quantity of radium at time $t = 0$
- $k$ is a positive number.
We are given that when $t = 200$, $x = 0.9 \times x_0$.
Hence:
\(\ds 0.9 x_0\) | \(=\) | \(\ds x_0 e^{-200 k}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0.9\) | \(=\) | \(\ds e^{-200 k}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds -200 k\) | \(=\) | \(\ds \ln 0.9\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k\) | \(=\) | \(\ds -\dfrac {\ln 0.9} {200}\) |
So after $1000$ years, we have:
\(\ds \dfrac x {x_0}\) | \(=\) | \(\ds e^{-1000 \times \paren {-\ln 0.9 / 200} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{5 \times \ln 0.9}\) | ||||||||||||
\(\ds \) | \(\approx\) | \(\ds 0.59045\) |
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $1$: How Differential Equations Originate: Exercise $1.1$