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kmoh111
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Homework Statement
I'm working a homework problem that states:
A very old specimen of wood contained 1012 atoms of 14C in 1986.
• How many 14C atoms did it contain in the year 9474 B.C.?
• How many 14C atoms did it contain in 1986 B.C.?
Homework Equations
N(t) = N(0)exp^(-lambda*t)
The Attempt at a Solution
For C-14, lambda = ln2/T1/2 = 0693/5730 yrs = .000121 yr-1
Since we do not know N0, I took the ratio of C-14 atoms in 1986 (A.D.) to C-14 atoms in 9474 B.C.:
Let t1 reflect the time from t0 to 1986.
Let t2 reflect the time from t0 to 9474 B.C.
then the ratio of N(t1)/N(t2) is:
N0 * e-lambda*t1
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N0 * e-lambda*t2
N0 cancels and we know that N(t1) = 1012 atoms.
Solving for N(t2) gives us:
1012
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e-lambda(t2-t1)
We know that the difference t2 - t1 is equal to 1986 + 9474 = 11460 yrs. Plugging the numbers and solving for N(T2), gives us
1012
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e(-0.693(11460/5730))
N(t2) = 4 x 1012 ===> number of C-14 atoms in 9474 B.C.
Is this the correct approach? I used the approach for the second part of the question for number of C-14 atoms in 1986 B.C.. I got 1.6 x 1012 atoms.
Thanks for the help.