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!