The following problem shows how the radioactive decay of potassium-40 explains the presence of argon in the Earth's crust and atmosphere. The follow-up explanation shows how to calculate the age of a rock using potassium-argon dating.

- What is the crustal abundance of potassium-40 in parts per million?

- What was the crustal abundance of potassium-40 just after the Earth formed, 4.5 billion years ago?
- How much potassium-40 has decayed away in the last 4.5 billion years, in kg?

- How much argon-40 was produced by the decay of potassium-40 in the Earth's crust?

- How much argon now exists in the atmosphere?

Here are the solutions with detailed calculations. Your answers might be slightly different due to rounding.

0.012 x 0.00012 = 1.4 x 10

If we take the formation of the Earth as time 0 and the present time as 4.5 billion years, then we have:

1.4 =

where 1.4 is the present abundance of potassium-40 in parts per million,

(2.4 x 10

4.2 x 10

The total mass of the argon in the Earth's atmosphere is

(total mass of atmosphere) x 1.3% =

(5.1 x 10

Thus, the amount of argon in the atmosphere is about 1.4% of the amount of argon-40 produced in the Earth's crust over the past 4.5 billion years. Most of the argon produced in the crust remains locked in the rock. Only about 1.4% has escaped and entered the atmosphere.

The amount of argon still trapped in the Earth's crust is the amount produced minus the amount that has escaped into the atmosphere:

(4.6 x 10

The crustal abundance of argon is the mass of trapped argon-40 divided by the mass of Earth's crust:

(4.5 x 10

By measuring the amount of potassium and argon in a rock sample, the age of the rock since it solidified can be determined. For example, a rock taken from a fresh lava flow will have no argon, whereas a rock that is 1.2 billion years old (one potassium-40 half-life) will have an amount of argon equal to 11% of the amount of potassium-40 remaining in the rock. One half-life ago, there was twice as much potassium-40 in the rock, an amount that has decayed away into calcium-40 (89%) and argon-40 (11%).

Here is a typical rock dating problem:

A rock sample is found to contain 1.00 gram of potassium and 6.0 micrograms of argon. How old is the rock?

At the present time, all natural potassium is 0.012% potassium-40. Thus, the quantity of potassium-40 currently in the rock sample is:

1.00 gram x 0.00012 = 0.00012 grams =

The 6.0 micrograms of argon came from the decay of potassium-40. Potassium-40 decays to calcium-40 89% of the time and to argon-40 11% of the time. Therefore, the quantity of potassium-40 that decayed to produced the argon is:

(X micrograms potassium-40) x 0.11 = (6.0 micrograms argon-40)

X = 6.0/0.11 =

55 micrograms of potassium-40 decayed away and 120 micrograms remain in the rock. Therefore, when the rock solidified from a molten form, it contained 120 + 55 =

Now we have the starting amount, the ending amount, and the half-life of potassium-40, so we can use the decay formula to find the elapsed time

Before we plug in the numbers, let's solve the equation for

Now take the base-10 logarithm of both sides:

= (1.2 billion years)

©2013 Gray Chang