Radioactive Dating
A geologist finds a rock which contains 3 atoms of Argon for every 1 atom of Potassium. She assumes the rock had no Argon when it solidified because Argon is a gas even at normal temperatures and while it might be trapped in a bubble in the rock, it could not get trapped as isolated atoms in solid rock. Therefore, all the Argon came from the radioactive decay of Potassium. Potassium only decays into Argon, so she knows that that 3/4 of the Potassium has decayed into Argon and only 1/4 of the original Potassium is left. In each radioactive half-life, 1/2 of the atoms decay. If only 1/4 of the original Potassium is left, then two half lives (1/2 of 1/2) of Potassium have elapsed since the rock was formed. The geologist knows that Potassium has a half life of 1.3 billion years, so the rock the geologist found is 2.6 billion years old.
Question 1. The geologist finds another rock that has seven Argon atoms for every Potassium atom, or 1/8 of the original Potassium. How old is this rock?
Potassium^40 → Argon^40 + e =1.3 billion years
Avogadro's number tells us 6 x 10^23 atoms in a mole. One mole would produce 3,000,000 decays per second.
Plate Tectonics
Measurements using radio telescopes and satellites show that Europe and North America are moving apart by about 4 cm per year because of plate tectonics.
Question 2. How old would the oldest rocks on Earth be if the entire crust was created by the rift zone in the Mid-Atlantic and was destroyed half way around the world by a subduction zone in the Pacific? (Hint: Figure out how big half the circumference of the Earth is in cm. The radius r of the Earth is 637 million cm, or 637, 000, 000 cm or 6.37x108 cm. The full circumference is C = 2 π r where π = 3.14. Then calculate how many years would it take to travel half way around the world at 4 cm per year.)
Bonus Question Is the rock in question 1 older than the age of crust in question 2? If so, how might this happen? (Hint: Where might the geologist have found the very old rock in question 1?)
Thank you.