Question

The cloth shroud from around a mummy is found to have a 14C activity of 8.0 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for 14C decay, 5715 yr , calculate the age of the shroud. Express your answer using two significant figures.

Answer 1

Given:

Half life = 5715 yr

use relation between rate constant and half life of 2nd order reaction

k = 1/([14C]o*half life)

= 1/(16.3*5715)

= 1.073*10^-5 disintegration-1.yr-1

we have:

[14C]o = 16.3 disintegration

[14C] = 8.0 disintegration

k = 1.073*10^-5 disintegration-1.yr-1

use integrated rate law for 2nd order reaction

1/[14C] = 1/[14C]o + k*t

1/(8) = 1/(16.3) + 1.073*10^-5*t

0.125 = 0.06135 +1.073*10^-5*t

1.073*10^-5*t = 0.06365

t = 5929 yr

Answer: 5929 yr