Question

When a sample of moon rock was analyzed by mass spectroscopy, the ratio of K-40 to Ar-40 was found to be 0.1295. Based on this ratio, how old is the moon?

Answer 1

It's called "mass spectrometry" :P

but knowing that the decay is a first-order process, we can use:

\(\sf \large A_t=A_o*e^{-kt}\)

Where \(A_t\) is the amount at time \(t\)
\(A_o\) is the initial amount
k is the decay constant
t is time

You'll need to find k first using the half-life, \(t_{1/2}\):

\(t_{1/2}=\dfrac{ln(2)}{k}\)

Answer 2

The ratio they're talking about is the rearragement of \(A_t\) and \(A_o\)

\(\sf \large A_t=A_o*e^{-kt}\rightarrow \dfrac{A_t}{A_o}=e^{-kt}\)

Answer 3

wait i think the ratio is actually the reverse

Answer 4

Right, but how does that relate to the ratio of K-40 and Ar-40?

Answer 5

I found that K-40 only decomposes to Ar-40 part of the time.

Answer 6

https://en.wikipedia.org/wiki/Potassium-40

Answer 7

so it's 10.72% of the time, you'll have to account for that

and the half-life is 1.251(3)×10^9 years

let me think about this for a min

Answer 8

Sure :-)

Answer 9

Is the ratio \(\dfrac{[Ar]}{[K]}=\dfrac{1}{0.1295}\) or \(\dfrac{[Ar]}{[K]}=\dfrac{0.1295}{1}\) ?

Answer 10

Your guess is as good as mine. The problem reads just as I have written it.

Answer 11

I tried it both ways, this way gives me a positive value the other gives me a negative, so

\(\dfrac{[Ar]}{[K]}=\dfrac{1}{0.1295}=e^{-kt}\)

need to take account of all the products:

\(\dfrac{[Ar]}{0.1072}=\dfrac{[All~products ]}{1}\rightarrow [Ar]=\dfrac{[All]*0.1072}{1}\)

Back into original equation:

\(\dfrac{[Ar]}{[K]}=\dfrac{[All]*0.1072}{[K]}=\dfrac{0.1072 }{0.1295}\)

So \(\huge \dfrac{0.1072 }{0.1295}=e^{-0.0000000001846915t}\)

t=1,023,244,884 years

1 billion years sounds plausible

Answer 12

i have to go, i'm not 100% on this, i can take a look at it later if you're not satisfied

Answer 13

That make a lot of sense at first glance. Thanks so much!

Answer 14

no problem !