Question

A sample of wood from an ancient table was found to contain 17.6 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. Determine the decay constant for 14C and the age of the wood.

Answer 1

1)
we have:
Half life = 5720 yr
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(5720)
= 1.212*10^-4 yr-1
Answer: 1.212*10^-4 yr-1

2)

we have:
[14C]o = 100 M (Let initial concentration)
[14C] = 17.6 M (This is remaining)
k = 1.212*10^-4 yr-1

use integrated rate law for 1st order reaction
ln[14C] = ln[14C]o - k*t
ln(17.6) = ln(100) - 1.212*10^-4*t
2.8679 = 4.6052 - 1.212*10^-4*t
1.212*10^-4*t = 1.7373
t = 1.43*10^4 yr
Answer: 1.43*10^4 yr