Potassium-40 decays to argon-40 with a half-life of 1.27*109 yr. What is the age of a rock in which the mass ratio of 40Ar to 40K is 4.2?

Age of the rock is 3.02x10^9 years

Explanation:

First, we need to get clear in which is the expression we need to use here:

N = No * e (-kt) (1)

Where:

N: mass, moles, concentration of one element after t time has passed (Depends of the units)

No: innitial mass, mol, concentration of one element

t: time that has passed, it could also be time of half life.

k: constant of decay.

Now, the problem already gave us the ratio between the Ar and K, which is 4.2, this means the following

rAr / rK = 4.2

or simply 4.2 : 1

This means that the innitial concentration of both of them, is just sum the value of both of them so:

No = 4.2 + 1 = 5.2

and the final N, after half life has passed, we can assume it's one (Because Argon is decaying to potassium. If the ratio is 4.2 : 1, means that Argon decays from 4.2 to 1).

With No, N and t, we can calculate the constant of decay, and then, the age of the rock. First, we'll find the constant with the expression:

k = ln (2) / t1/2 (2)

k = ln (2) / 1.27x10^9

k = 5.46x10^-10 yr^-1

With this value, we'll use equation (1) to get the age of the rock

If:

N = No * e (-kt)

then:

N/No = e (-kt)

ln (N/No) = - kt

t = - ln (N/No) / k (2)

Solving for t:

t = - ln (1/5.2) / 5.46x10^-10

t = 3.02x10^9 yr