## JoannaBlackwelder one year ago 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?

1. 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, $$t_{1/2}$$: $$t_{1/2}=\dfrac{ln(2)}{k}$$

2. aaronq

The ratio they're talking about is the rearragement of $$A_t$$ and $$A_o$$ $$\sf \large A_t=A_o*e^{-kt}\rightarrow \dfrac{A_t}{A_o}=e^{-kt}$$

3. aaronq

wait i think the ratio is actually the reverse

4. JoannaBlackwelder

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

5. JoannaBlackwelder

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

6. JoannaBlackwelder
7. 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

8. JoannaBlackwelder

Sure :-)

9. aaronq

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

10. JoannaBlackwelder

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

11. 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

12. aaronq

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

13. JoannaBlackwelder

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

14. aaronq

no problem !