Question

A fossil was analyzed and determined to have a carbon-14 level that is 20 % that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?

Answer 1

we have:

Half life = 5730 year

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 year-1

we have:

[C-14]o = 100

[C-14] = 20

k = 1.209*10^-4 year-1

use integrated rate law for 1st order reaction

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

ln(20) = ln(100) - 1.209*10^-4*t

2.9957 = 4.6052 - 1.209*10^-4*t

1.209*10^-4*t = 1.6094

t = 13307 year

Answer: 1.33*10^4 year