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Dating Mars rocks

Radiation and the atom · level 8, lesson 5 of 6

A question to start

In 2013, Curiosity drilled a rock it called Cumberland and baked the powder in its oven. From the argon that came out, the team told how old the rock was: the first age ever measured on another planet. How?

The idea

Potassium is in most rock. A tiny share of it is potassium-40, which decays with a half-life of 1.25 billion years. Most of it becomes calcium-40, but about 10.5% becomes argon-40, a gas.

When rock is molten, argon escapes. Once the rock sets hard, the argon-40 made from then on is trapped. So the rock is a clock: the more argon-40 next to the potassium-40, the older it is.

Let \(R\) be the number of argon-40 atoms for each potassium-40 atom still left, and \(f = 0.105\) the share of decays that give argon. For each potassium-40 atom left, \(R/f\) have decayed, so the start was \(1 + R/f\) times what is left:

\[2^{t/T} = 1 + \frac{R}{f} \quad\Rightarrow\quad t = T \log_2\!\left(1 + \frac{R}{f}\right).\]

For Cumberland, the team found an age between 3.86 and 4.56 billion years. That matched what crater counts had suggested, which was itself a test that the method works on Mars.

rock forms4.2 billion years ongrey: calcium-40argon

Orange: potassium-40. Blue: argon-40. Grey: calcium-40. Out of 10 potassium-40 atoms, after about 4.2 billion years 1 is left, 1 has become argon-40, and 8 have become calcium-40. (Drawn as whole atoms; the true shares are 9.7%, 9.5% and 80.8%.)

Worked examples

1. Cumberland's middle age. At 4.21 billion years, the share of potassium-40 left is \((\tfrac12)^{4.21/1.25} = (\tfrac12)^{3.37} = 0.097\). Decayed: \(0.903\); argon made: \(0.903 \times 0.105 = 0.095\). So \(R = 0.095 / 0.097 \approx 1\): about one argon-40 atom for each potassium-40 atom.

2. Back to the age. With \(R = 1\): \(1 + 1/0.105 = 10.5\) and \(\log_2(10.5) = 3.39\), so \(t = 1.25 \times 3.39 = 4.24\) billion years.

3. The range. At 3.86 billion years, \(R = 0.79\); at 4.56 billion, \(R = 1.21\). The rover had to measure tiny amounts of argon well enough to tell these apart.

Try it

  1. Why does the clock start when the rock sets hard?
  2. What share of potassium-40 is left after 2.5 billion years?
  3. Find the age of a rock with \(R = 0.5\).
  4. If some argon leaked out of a rock, would its age come out too old or too young?
Answers
  1. Before then, argon escapes from the molten rock; after, it is trapped.
  2. Two half-lives: a quarter.
  3. \(1.25 \times \log_2(1 + 0.5/0.105) = 1.25 \times \log_2(5.76) = 1.25 \times 2.53 \approx 3.2\) billion years.
  4. Too young: less argon looks like less time.

Common mistake

Forgetting the share that becomes argon. Only about 1 in 10 potassium-40 decays makes argon. Treating every decay as argon gives \(t = T \log_2(1 + R)\), far too young.

Where it leads