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## Can someone check my Radiometric Dating Problems?

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### Can someone check my Radiometric Dating Problems?

1. The half like of carbon 14 is 5730 years. Some charcoal from an ancient village has only 1/16 of its original level of carbon 14. How old is the charcoal? My answer was 45,840 yrs.

2. The half life of potassium 40 is 1.3 billion years. How much of that potassium 40 would have decayed to argon after 3.9 billion years? My answer was 3/8.
QuantumMoron
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2. 3.9 is more than 1.3 so wouldn't that mean at least half of it was decayed?
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Enjoying one moment at a time;
Accepting hardships as the pathway to peace;
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mith
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For Question #1:
I might be over simplifying it, but 1/16 is 4 half life periods (1/2 * 1/2 * 1/2 * 1/2 = 1/16)
So 4 * 5730 years = 22,920 years

For Question #2:
3.9 / 1.3 = 3 Half-life periods so: 1/2 *1/2 * 1/2 = 1/8 of the potassium would remain. 1 - 1/8 = 7/8 of the potassium would have decayed to argon.
Chroma
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You can just use exponents if you don't want to halve many times

i.e. (.5)^(n/half life) where n is number of years in question
Living one day at a time;
Enjoying one moment at a time;
Accepting hardships as the pathway to peace;
~Niebuhr

mith
Inland Taipan

Posts: 5345
Joined: Thu Jan 20, 2005 8:14 pm
Location: Nashville, TN

I knew there was a way to do it using exponents...

1/16 = (1/2)^(n/5730)
ln(1/16) = ln((1/2)^(n/5730))
ln(1/16) = (n/5730) * ln(1/2)
n/5730 = ln(1/16) / ln(1/2) = 4
n = 5730 * 4 = 22,920
Chroma
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### Re: Can someone check my Radiometric Dating Problems?

Thanks for the help everyone
QuantumMoron
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