Suppose you find a rock originally made of potassium-40, half of which decays into argon-40 every 1.25 billion years.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

1 Answer
Oct 7, 2017

Answer:

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.