Question

The C-14 activity of a wooden sample is 29.1 counts/min as compared to an activity of 58.2 counts/min for a standard zero age. The half-life for C-14 decay is 5715 years. Calculate the age of the sample.

a) 7755 yr

b) 2822 yr

c) 1355 yr

d) 228.8 yr

e) 5715 yr

Answer 1

Given:

Half life = 5.715*10^3 years

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(5.715*10^3)

= 1.213*10^-4 years-1

we have:

[A]o = 58.2

[A] = 29.1

k = 1.213*10^-4 years-1

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(29.1) = ln(58.2) - 1.213*10^-4*t

3.371 = 4.064 - 1.213*10^-4*t

1.213*10^-4*t = 0.6931

t = 5715 years

Answer: e