Question

An ancient wooden club is found that contains74g of carbon and has an activity of 7.7 decays per second.

Determine its age assuming that in living trees the ratio of 14C/12Catoms is about 1.3

Answer 1

Since decay is exponential
(N/No) = e^(-t / tau)
N - current number of C14 atoms in the sample
No - initial number of C14 atoms in the sample
tau - mean lifetime of C14 atom.
t - age of the sample.
ln(N/No) = -t / tau
t = - tau x ln(N/No)
Now we have the formula. Let's calculate tau, N and No.
Half life th is the time taken for the activity of a given amount of a radioactive substance to decay to half of its initial value. It's constant characteristic to isotope. For C14
th = 5730 years.
tau = th / ln(2) = 5730 / ln(2) = 8266.64 years
Atomic weight for carbon is 12.011 g/mol. In 74g of carbon there are n = 74 / 12.011 =6.16101 mol of carbon, so there are totally Ns = 6.16101 x 6.0221 x 10^23 = 3.7102 x 10^24 atoms in the sample. (both kinds C12 and C14 together)
Originally there were No = 1.3 x 10^-12 x 3.7102 x 10^24 = 4.823288 x 10^12 atoms of C14. I calculate with total number of atoms as it was number of atoms C12 alone. This is correct enough, since there are much less C14 than C12. Total activity A is number of decays an object undergoes per second
A = N / tau
N = A x tau = 7.7 x 8266.64 x 3.15576 x 10^7 = 2.0087 x 10^12
Finally,
t = - 8266.64 x ln (2.0087 / 4.823288) = 7241 years.