Question

Suppose you find a rock originally made of potassium-40, half of which decays into argon-40 every 1.25 billion years.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.

Answer 1

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_"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 `color(red)(1)` half-life

`1/2 * A_"K-40" = A_"K-40"/2^color(red)(1) ->` atoms of potassium-40

`A_"K-40" - A_"K-40"/2^color(red)(1) ->` atoms of argon-40

• After `color(red)(2)` half-lives

`1/2 * A_"K-40"/2^1 = A_"K-40"/2^color(red)(2) ->` atoms of potassium-40

`A_"K-40" - A_"K-40"/2^color(red)(2) ->` atoms of argon-40

• After `color(red)(3)` half-lives

`1/2 * A_"K-40"/2^2 = A_"K-40"/2^color(red)(3) ->` atoms of potassium-40

`A_"K-40" - A_"K-40"/2^color(red)(3) ->` atoms of argon-40

At this point, we can use this pattern to say that after `color(red)(n)` half-lives pass, the sample will contain

`A_"K-40"/2^color(red)(n) ->` atoms of potassium-40

`1 - A_"K-40"/2^color(red)(n) ->` 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

`(A_"K-40" - A_"K-40"/2^color(red)(n))/(A_"K-40"/(2^color(red)(n))) = 31`

This is equivalent to

`(color(blue)(cancel(color(black)(A_"K-40"))) - color(blue)(cancel(color(black)(A_"K-40")))/2^color(red)(n))/(color(blue)(cancel(color(black)(A_"K-40")))/(2^color(red)(n))) = 31`

`(2^color(red)(n) - 1)/color(blue)(cancel(color(black)(2^color(red)(n)))) * color(blue)(cancel(color(black)(2^color(red)(n))))/1 = 31`

which gives you

`2^color(red)(n) = 32`

Since

`32 = 2^5`

you can say that

`2^color(red)(n) = 2^5 implies color(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 color(red)(cancel(color(black)("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.