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- From: davidlai@unixg.ubc.ca (David TW Lai)
- Newsgroups: sci.chem
- Subject: Nuclear Chemistry (K-40 decay)
- Date: 22 Jan 93 05:55:16 GMT
- Organization: The University of British Columbia
- Lines: 37
- Message-ID: <davidlai.727682116@unixg.ubc.ca>
- NNTP-Posting-Host: unixg.ubc.ca
-
- Hi,
-
- According to my book, K-40 decays into two products. About 11%
- decays into Ar-40, and 89% decays into Ca-40. My book says that since
- K-40 follows two decay paths, TWO decay constants must be kept in doing
- K-40 decay problems. However, if I follow the following equations from
- p.69 of Principles of Isotope Geology:
-
- Le = .581 x 10^-10 (Decay constant of K-40 to Ar-40)
-
- Lb = 4.962 x 10^-10 (Decay constant of K-40 to Ca-40)
-
- L = 5.543 x 10^-10 (Total decay constant of K-40)
-
- t = (1/L) ln( (Ar-40/K-40) * (L/Le) +1 )
-
- = 1.804 x 10^9 ln (10*9.540 + 1)
-
- = 1.804 x 10^9 ln (96.40)
-
- = 1.804 x 10^9 * 4.569
-
- = 8.242 x 10^9 OR 8.242 billion years.
-
- The modern estimate of the Earth's age is 4.55 billion years only!
-
- The original question is that there is a rock with 1 atom of K-40
- and 10 atoms of Ar-40 (relatively). And determine the age of this rock.
- If this rock is really 8.242 billion years, it may have come from the
- outer space! ;-)
-
- Would someone comment on this?
-
- Regards,
-
- David.
-
-