Question

What percentage of 146C (t1/2
= 5715 years) remains in a sample estimated to be 18730 years old?

Answer 1

Given:
Half life = 5715 yr
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(5715)
= 1.213*10^-4 yr-1
we have:
[C-14]o = 100 (Let initial concentration be 100)
t = 18730.0 yr
k = 1.213*10^-4 yr-1

use integrated rate law for 1st order reaction
ln[C-14] = ln[C-14]o - k*t
ln[C-14] = ln(100) - 1.213*10^-4*18730
ln[C-14] = 4.6052 - 1.213*10^-4*18730
ln[C-14] = 2.334
[C-14] = 10.3

10.3 remains out of 100 which is 10.3 %
Answer: 10.3 %