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
3. The attempt at a solution
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^-(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.
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!