Question

A fossil is found to have a 14C level of 78.0% compared to living organisms. How old is the fossil?

Answer 1

Half life of 14 C is = 5730 yr

use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(5730)

= 1.209*10^-4 yr-1

we have:

[14C]o = 100 (Let living organism have activity of 100)

[14C] = 78 (activity remaining is 78.0 % of living organism)

k = 1.209*10^-4 yr-1

use integrated rate law for 1st order reaction

ln[14C] = ln[14C]o - k*t

ln(78) = ln(100) - 1.209*10^-4*t

4.3567 = 4.6052 - 1.209*10^-4*t

1.209*10^-4*t = 0.2485

t = 2054 yr

Answer: 2054 yr