Question

A wooden artifact from a Chinese temple has a ¹⁴C activity of 40.7 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age.

From the half-life for ¹⁴C decay, 5715 yr, determine the age of the artifact.

Answer 1

we have:

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:

[14C]o = 58.2 counts/min

[14C] = 40.7 counts/min

k = 1.213*10^-4 yr-1

use integrated rate law for 1st order reaction

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

ln(40.7) = ln(58.2) - 1.213*10^-4*t

3.7062 = 4.0639 - 1.213*10^-4*t

1.213*10^-4*t = 0.3577

t = 2950 yr

Answer: 2950 yr