1. The problem statement, all variables and given/known data
The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the original 6 12 C atoms remains?

2. Relevant equations
(N1/No)=e^-λt

3. The attempt at a solution
Variables:
t = 41,000 yrs
T1/2 of Carbon = 5730 yrs
ln2 = .693

I basically plugged in the numbers and solved because you are given all the variables.

(N1/No)=e^-λt
(N1/No) = e^-(ln2/T1/2)t
(N1/No) = e^-(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(.693/5730 yrs)(41,000 yrs)
ln(N1/No) = -(1.2094E-4)(41,000 yrs)
ln(N1/No) = -4.95863
(N1/No) = e^(-4.95863)
(N1/No) = .0102548

I then calculated for the % because the answer .012548 is a fraction.

.0102548/100 %
=1.025%

However that answer is incorrect and I’m not exactly sure why. I made sure that I was using the (ln)-function instead of the (log)-function. I don’t know if it is my math or if I’m putting it into the website incorrectly.

Any help would be greatly appreciated!

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