Potassium-40 has a half life of 1.2 billion years and decays into the stable daughter isotope Argon-40. If you find a rock that contains 20 Potassium-40 atoms and 60 Argon-40 atoms, how old is the rock?

1.2 billion

2.6 billion

To determine the age of the rock, we need to use the concept of radioactive decay and the ratio of Potassium-40 to Argon-40 isotopes.

Given that Potassium-40 has a half-life of 1.2 billion years, it means that over a period of 1.2 billion years, half of the Potassium-40 atoms would have decayed into Argon-40. So, the ratio of Potassium-40 to Argon-40 in a newly formed rock would have been 1:0.

In the given rock, we have 20 Potassium-40 atoms and 60 Argon-40 atoms. To find the age of the rock, we need to determine how many half-lives have elapsed for this ratio to establish.

Let's start by dividing the number of Argon-40 atoms by the number of Potassium-40 atoms: 60 / 20 = 3.

This means that 3 half-lives have passed, because each half-life converts a 1:1 ratio of Potassium-40 to Argon-40 into a 1:2 ratio (1 Potassium-40 atom and 2 Argon-40 atoms).

Since each half-life represents 1.2 billion years, we can multiply the number of half-lives (3) by the length of each half-life (1.2 billion years).

Hence, the age of the rock would be 3.6 billion years.