Suppose you find a rock originally made of potassium-40, half of which decays into argon-40 every 1.25 billion years. You open the rock and find 31 atoms of argon-40 for every atom of potassium-40. How long ago did the rock form?

It's for my astronomy class.

Oct 7, 2017

Here's what I got.

Explanation:

The idea here is that the ratio that exists between the number of atoms of argon-40 and the number of atoms of potassium-40 will give you the number of half-lives that passed.

As you know, the half-life of a radioactive nuclide tells you the time needed for half of the atoms of said nuclide to undergo radioactive decay.

In your case, you know that potassium-40 has a half-life of $1.25$ billion years because that's how long it takes for half of the number of atoms present in the sample to decay to argon-40.

Now, let's say that your sample started with ${A}_{\text{K-40}}$ atoms of potassium-40 and $0$ atoms of argon-40.

You can thus say that the sample will contain--keep in mind that the atoms of potassium that decay form argon-40!

• After $\textcolor{red}{1}$ half-life

$\frac{1}{2} \cdot {A}_{\text{K-40" = A_"K-40}} / {2}^{\textcolor{red}{1}} \to$ atoms of potassium-40

${A}_{\text{K-40" - A_"K-40}} / {2}^{\textcolor{red}{1}} \to$ atoms of argon-40

• After $\textcolor{red}{2}$ half-lives

$\frac{1}{2} \cdot {A}_{\text{K-40"/2^1 = A_"K-40}} / {2}^{\textcolor{red}{2}} \to$ atoms of potassium-40

${A}_{\text{K-40" - A_"K-40}} / {2}^{\textcolor{red}{2}} \to$ atoms of argon-40

• After $\textcolor{red}{3}$ half-lives

$\frac{1}{2} \cdot {A}_{\text{K-40"/2^2 = A_"K-40}} / {2}^{\textcolor{red}{3}} \to$ atoms of potassium-40

${A}_{\text{K-40" - A_"K-40}} / {2}^{\textcolor{red}{3}} \to$ atoms of argon-40

At this point, we can use this pattern to say that after $\textcolor{red}{n}$ half-lives pass, the sample will contain

${A}_{\text{K-40}} / {2}^{\textcolor{red}{n}} \to$ atoms of potassium-40

$1 - {A}_{\text{K-40}} / {2}^{\textcolor{red}{n}} \to$ atoms of argon-40

Now, you know that sample contains $31$ atoms of argon-40 for every $1$ atom of potassium-40, which means that you have

$\left({A}_{\text{K-40" - A_"K-40"/2^color(red)(n))/(A_"K-40}} / \left({2}^{\textcolor{red}{n}}\right)\right) = 31$

This is equivalent to

$\left(\frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{A}_{\text{K-40"))) - color(blue)(cancel(color(black)(A_"K-40")))/2^color(red)(n))/(color(blue)(cancel(color(black)(A_"K-40}}}}}}{{2}^{\textcolor{red}{n}}}\right) = 31$

$\frac{{2}^{\textcolor{red}{n}} - 1}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{2}^{\textcolor{red}{n}}}}}} \cdot \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{2}^{\textcolor{red}{n}}}}}}{1} = 31$

which gives you

${2}^{\textcolor{red}{n}} = 32$

Since

$32 = {2}^{5}$

you can say that

${2}^{\textcolor{red}{n}} = {2}^{5} \implies \textcolor{red}{n} = 5$

This means that $5$ half lives must pass in order for the sample to contain $31$ atoms of argon-40 for every $1$ atom of potassium-40.

Consequently, you can say that the age of the rock is

$5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{half-lives"))) * "1.25 billion years"/(1color(red)(cancel(color(black)("half-life")))) = color(darkgreen)(ul(color(black)("6.25 billion years}}}}$

I'll leave the answer rounded to three sig figs, but keep in mind that you have two significant figures for the number of atoms of argon-40 present per atom of potassium-40.