Answer:
3.75 billion years.
Explanation:
Let the amount of potassium-40 at the beginning be K.
We can follow the decay with its half life.
In 1 half life, the amount of potassium-40 left is K/2
In 2 half lives, the amount left will be K/2²
In 3 half lives, it will be K/2³
In n half lives, it will be K/2ⁿ
So, in n half lives, the amount of Argon left will be (K - (K/2ⁿ))
In the question, it is described that the amount of Argon left = 7 × the amount of potassium-40 left
(K - (K/2ⁿ)) = 7 (K/2ⁿ)
K(1 - 2⁻ⁿ) = 7K/2ⁿ
1 - 2⁻ⁿ = 7(2⁻ⁿ)
Divide through by 2⁻ⁿ
(1/2⁻ⁿ) - 1 = 7
2ⁿ = 7+1 = 8
2ⁿ = 2³
n = 3
This means, we'll get 7 atoms of argon-40 for every atom of potassium-40 in 3 half life
1 half life = 1.25 billion years
3 half lives = 3 × 1.25 billion years = 3.75 billion years.
