JoannaBlackwelder
  • JoannaBlackwelder
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?
Chemistry
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SOLVED
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jamiebookeater
  • jamiebookeater
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aaronq
  • aaronq
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, \(t_{1/2}\): \(t_{1/2}=\dfrac{ln(2)}{k}\)
aaronq
  • aaronq
The ratio they're talking about is the rearragement of \(A_t\) and \(A_o\) \(\sf \large A_t=A_o*e^{-kt}\rightarrow \dfrac{A_t}{A_o}=e^{-kt}\)
aaronq
  • aaronq
wait i think the ratio is actually the reverse

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JoannaBlackwelder
  • JoannaBlackwelder
Right, but how does that relate to the ratio of K-40 and Ar-40?
JoannaBlackwelder
  • JoannaBlackwelder
I found that K-40 only decomposes to Ar-40 part of the time.
JoannaBlackwelder
  • JoannaBlackwelder
https://en.wikipedia.org/wiki/Potassium-40
aaronq
  • aaronq
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
JoannaBlackwelder
  • JoannaBlackwelder
Sure :-)
aaronq
  • aaronq
Is the ratio \(\dfrac{[Ar]}{[K]}=\dfrac{1}{0.1295}\) or \(\dfrac{[Ar]}{[K]}=\dfrac{0.1295}{1}\) ?
JoannaBlackwelder
  • JoannaBlackwelder
Your guess is as good as mine. The problem reads just as I have written it.
aaronq
  • aaronq
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
aaronq
  • aaronq
i have to go, i'm not 100% on this, i can take a look at it later if you're not satisfied
JoannaBlackwelder
  • JoannaBlackwelder
That make a lot of sense at first glance. Thanks so much!
aaronq
  • aaronq
no problem !

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