Question

The cloth shroud from around a mummy is found to have a 14 activity of 9.2 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. Part A You may want to reference (Pages 913-916) Section 21.4 while completing this problem From the half-life for "C decay, 5715 yr, calculate the age of the shroud. Express your answer using two significant figures. EVO AXO O 2 ? 4.81 · 103 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

Answer 1

You need two equations to solve this problem.

t 1/2 = 0.693 / k and

ln (Nt / No) = -kt . . .where Nt / No is the ratio of radioactive nuclei at time t compared to the original number of nuclei. This is proportional to the number of disintegrations at time t versus the original number of disintegrations.

t1/2 = 0.693 / k

k = 0.693 / t 1/2 = 0.693 / 5715 yr = 1.21 x 10-4 yr-1

ln (Nt / No) = -kt

ln (9.2 / 16.3) = -(1.21 x 10-4) t

-0.57 = -1.21 x 10-4 t

4276.95 yr = t = age of shroud