Question

An igneous rock contains a Pb−206/U−238 mass ratio of 0.367.

How old is the rock? (U−238 decays into Pb−206 with a half-life of 4.50×109 yr.)

I'm not sure how to get started on this :/

Answer 1

I don't know how advanced this class is but there is an easy way (and not quite so accurate) and a harder way (but a little more accurate).
Easy way.
You start with 100 atoms U and they decay over the years to form x atoms Pb which leaves 100-x atoms U. The ratio of Pb/U is 0.367 so substitute
(x/100-x) = 0.367 and solve for x and call that grams U. You should get approx 27g U but you can work it out more accurately if you wish. This is a close approximation. Then use
ln(No/N) = kt
No = 100
N = 100-27 = approx 73
k = 0.693/t_1/2 from the problem. Solve for t. If I didn't make an error the answer is 2.02E9 years for the age of the rock. This is the easy way. What's wrong with it? We have used atoms U decaying to atoms of Pb (which is ok) BUT we have used a MASS RATIO and not an atom ratio; i.e., U weighs 238 g/mol and Pb only weighs 206 g/mol so can't really equation atoms with grams.

The longer but a little more accurate way. Convert atoms U to grams and atoms Pb to grams, then we can use the mass ratio without any problems. It's a little messier math problem but here is how you do it.
U initially = 100 atoms. mass 100 atoms is (238*100/6.022E23) = ?
x atoms decay. They have a mass of (238x/6.022E23)
mass U remaining is the difference. I'll let you do that.

They decay to form Pb. Those atoms of x will have a mass of (206x/6.022E23)

Substitute mass Pb/mass U and solve for x. I obtained approx 30g but I'll let you do the math.

Then ln(No/N) = kt
No = 100
N = approx 70
k = same as the first above
t = solve and I obtained about 2.29E9 years. If you count years difference between 2.02E9 and 2.29E9 that's a bunch of years(something like 2.7E8 or 270,000,000 years) but it's small compared to the time scale. Only about 10% error. So doing it the easy way saves a lot of time and doesn't make that much error.