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have been in the ground for 100 million years, what fraction of the initial C-14 atoms is still there? (t1/2 = 5,370 yrs)

a. 0
b. 1 x 10^[-]10
c. 5.7 x 10^[-]5
d. 1.0 x 10^[-]3
e. 5.7 x 10^[-]1

Not sure how to answer this. Could you please show me the steps in your result so that I'll have a greater understanding of the problem?? (TIA)

2007-12-09 19:08:47 · 4 answers · asked by xxxladyxxx 1 in Science & Mathematics Chemistry

4 answers

If n0 atoms of C-14 decays over a period of t with a rate constant of k, and after time t, n atoms of C-14 are still there, then, we know that: (it's a first order decay after all!)
n/n0=exp(-kt) ------(1)
=> k= [ln(n0/n)]/t --(2)
First solve for k by plugging in n/n0 = 1/2, and t = 5370 yr in Eq(2).
Then, plug in the value of k (that you calculated) and t = 100x10^6yr (100 million yrs) in Eq.1, and calculate n/n0 (the fraction).

2007-12-09 19:20:03 · answer #1 · answered by Dr. K 3 · 0 0

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2016-05-22 10:50:54 · answer #2 · answered by garnet 3 · 0 0

Since the numbers are pretty spread out, you can cheat a little by finding number of half-lives which would be 10^8 divided by 5370. which is about 1.9x10^4. Very roughly, 20 half-lives corresponds to a fraction of 1x10^-6 (one in one million). Since you have so many more here, the answer would be essentially zero.

2007-12-09 19:24:19 · answer #3 · answered by cattbarf 7 · 1 0

Use the exponential function. You know the half life. So you know the time constant tau after which the remaining number of atoms is 1/e of the original number. Now you calculate the natural logarithm (logarithm to base e) of the fraction. That is the number you have to multiply the tau with to the correct number of years. Easy.

2007-12-09 19:13:39 · answer #4 · answered by Anonymous · 0 0

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