Potassium-40 decays to argon-40 with a half-life of 1.27 x 10^9 years?

Question

Potassium-40 decays to argon-40 with a half-life of 1.27 x 10^9 years.

Question: What is the age of a rock in which the mass ratio of Argon-40 to Potassium-40 is 3.3? (answer should be in years)

Answer 1

We need two equations to solve this problem.

t 1/2 = 0.693 / k  and ln (Nt / No) = -kt

t 1/2 = 0.693 / k

k = 0.693 / t 1/2 = 0.693 / 1.27 x 10^9 yr = 5.46 x 10^-10 yr^-1

ln (Nt / No) = -kt . . .Nt/ No is the fraction of K-40 that remains at time t. If the ratio of Ar-40 to K-40 is 3.3 to 1, then the fraction of K-40 remaining is 1 / (1 + 3.3) = 0.232 (23.2%).

ln (0.232) = -(5.46 x 10^-10) t

-1.46 = -5.46 x 10^-10 t

2.68 x 10^9 yr = t

Does this answer make sense? When a radioactive isotope decays, after two half-lives, there is 25% of the original amount left. Two half-lives of K-40 is 2 x 1.27 x 10^9 yr = 2.54 x 10^9 yr. In our problem we solve for time at 23.3 % of the K-40 left, which is slightly more than two half-lives. Note that our answer, 2.68 x 10^9 yr, is indeed slightly more than 2.54 x 10^9 yr.