Question

If a log contains 60% of the 14C present in a living tree, how long has the log been dead? The half-life of 14C is 5730 years.

Answer 1

Given:

Half life = 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 initially is be 100)

[14C] = 60 (since 60 % is remaining)

k = 1.209*10^-4 yr-1

use integrated rate law for 1st order reaction

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

ln(60) = ln(100) - 1.209*10^-4*t

4.094345 = 4.60517 - 1.209*10^-4*t

1.209*10^-4*t = 0.510826

t = 4224 yr

Answer: 4224 yr