Question

A piece of petrified wood is examined determined that only 25% of the or life of "C is 5,730 years, how ol examined via radioactive C dating, a first-order process. It is 25% of the original amount is still present in the sample. If the half- years, how old is the petrified wood? (4 points)

Answer 1

Let initial amount of Carbon-14 is 100% i.e. A0 = 100

Then, remaining amount of C-14 is = At = 25*100/100 = 25

We need to calculate time 't'.

Half life = 5730 years

Rate constant (K) = 0.693/half life = 0.693/5730 = 0.0001209424 years^-1

Using first order reaction;

K * t = 2.303 * log (A0/At)

0.0001209424 * t = 2.303 * log(100/25)

t = 11464.5 years ...Answer