The half life for the radioactive decay of potassium- 40 to argon- 40 is 1.26 x 109 years. Suppose nuclear chemical analysis shows that there is 0.771 mmol of argon-40 for every 1.000 mmol of potassium-40 in a certain sample of rock. Calculate the age of the rock. Round your answer to 2 significant digits.

Respuesta :

Given:

Half life t1/2 for K-40 = 1.26*10⁹ years

Initial amount of K-40, A₀ = 1.000 mmol

Amount of Ar-40 formed = 0.771 mmol

To determine:

The age of the rock

Explanation:

The radioactive decay can be expressed mathematically as:

A = A₀exp(-kt)------(1)

where A₀ = initial amount

A = amount after time t

k = decay constant = 0.693/t1/2 ----(2)

Here: k = 0.693/1.26*10⁹ yr = 0.55 *10⁻⁹ yr⁻¹

Now:

A₀ of K-40= 1.000 mmol

A of K-40 = 1.000 - 0.771 = 0.229 mmol

Based on eq(1)

0.229 = 1.000 exp(-0.55*10⁻⁹t)

t = 2.68*10⁹ years

Ans: Age of rock = 2.7*10⁹ yrs