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
1
half-life
1
2
⋅
A
K-40
=
A
K-40
2
1
→
atoms of potassium-40
A
K-40
−
A
K-40
2
1
→
atoms of argon-40
After
2
half-lives
1
2
⋅
A
K-40
2
1
=
A
K-40
2
2
→
atoms of potassium-40
A
K-40
−
A
K-40
2
2
→
atoms of argon-40
After
3
half-lives
1
2
⋅
A
K-40
2
2
=
A
K-40
2
3
→
atoms of potassium-40
A
K-40
−
A
K-40
2
3
→
atoms of argon-40
At this point, we can use this pattern to say that after
n
half-lives pass, the sample will contain
A
K-40
2
n
→
atoms of potassium-40
1
−
A
K-40
2
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
n
A
K-40
2
n
=
31
This is equivalent to
A
K-40
−
A
K-40
2
n
A
K-40
2
n
=
31
2
n
−
1
2
n
⋅
2
n
1
=
31
which gives you
2
n
=
32
Since
32
=
2
5
you can say that
2
n
=
2
5
⇒
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
half-lives
⋅
1.25 billion years
1
half-life
=
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.
