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JoannaBlackwelder
 one year ago
When a sample of moon rock was analyzed by mass spectroscopy, the ratio of K40 to Ar40 was found to be 0.1295. Based on this ratio, how old is the moon?
JoannaBlackwelder
 one year ago
When a sample of moon rock was analyzed by mass spectroscopy, the ratio of K40 to Ar40 was found to be 0.1295. Based on this ratio, how old is the moon?

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aaronq
 one year ago
Best ResponseYou've already chosen the best response.1It's called "mass spectrometry" :P but knowing that the decay is a firstorder 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 halflife, \(t_{1/2}\): \(t_{1/2}=\dfrac{ln(2)}{k}\)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1wait i think the ratio is actually the reverse

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Right, but how does that relate to the ratio of K40 and Ar40?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0I found that K40 only decomposes to Ar40 part of the time.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1so it's 10.72% of the time, you'll have to account for that and the halflife is 1.251(3)×10^9 years let me think about this for a min

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1Is the ratio \(\dfrac{[Ar]}{[K]}=\dfrac{1}{0.1295}\) or \(\dfrac{[Ar]}{[K]}=\dfrac{0.1295}{1}\) ?

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0Your guess is as good as mine. The problem reads just as I have written it.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1i have to go, i'm not 100% on this, i can take a look at it later if you're not satisfied

JoannaBlackwelder
 one year ago
Best ResponseYou've already chosen the best response.0That make a lot of sense at first glance. Thanks so much!
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