Radioactive Decay question?

Question

A rock contains 0.688 mg of Pb-206 for every 1.000 mg of U-238 present. Assuming tha tno lead was originally present, that all the Pb-206 formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between U-238 and Pb-206 is negligible, calculate the age of the rock. (For U-238, half life = 4.5 * 10^9 years).

The answer is 3.8 * 10^9 years but I would like an explanation why. Thank you.

I know! And that's the same answer I got but I was told that the U and Pb are in 1:1 molar ratio not in a 1:1 gram ratio and that I have to change it to moles but I got lost along the way because I didn't understand what I was doing and didn't know what I was doing.

So I guess my real question is how do I obtain the answer of 3.8 * 10^9 years if I have to use the molar ratio?

Answer 1

Well, it all goes to law of radioactive decay:
N = N_0 Exp(-k t)
with N the number of nuclei at time t, N_0 the initial number of nuclei, t the time, and k the constant of radioactive decay. At half life, the number of nuclei is N = N_0 / 2, so you can obtain k.

To figure the time (age of the rock), you need to figure the current number of uranium nuclei (from its mass), and the initial number of nuclei. Obviously, this is formed by adding the number of nuclei still present with that of nuclei which decayed and now are Pb nuclei (again, you can figure this from its mass). Well, now you just have to plug in these in the decay equation and obtain the age...

You can take it from here.

-Added after comments -
I guess your problem is finding the number of nuclei from the mass. Well, just divide the total mass by the mass of one atom (same as the mass of the nucleus, since the electrons have negligible mass). For uranium, this is 1.000mg/(238*u), where u is the unit of atomic mass; you get similar relation for Pb. Since the decay equation involves, at the end, ratio of numbers, the unit of atomic mass cancels out and you don't need to know its value.

Exact masses for the nuclei were measured; they are different from the simple A*u used here (due to binding energy), but the differences are negligible.

Answer 2

I don't like half-life very much, because it forces student to use math procedures that are only used for that purpose, while rate constants are in the mainstream of math use. That said, we can convert the half life to a rate constant.
0.693 = k (4.5x10^9 years)
k = 1.45x10-10/year.
Going back to the problem, the 0.688 of Pb-206 was originally U-238, so we adjust for mass lost from time=0. 0.688x238/206 = 0.800 appx
Based on the problem statement, at time=0, there was 1.8 mg of U-238, and now there is only 1 mg. So
C(time) = Co exp(-kt), t=time in years
1.0 = 1.8 exp(- 1.45x10-10 t)
0.55 = exp (-1.45x10-10 t)
0.72 = 1.45x10-10 t
5x10+9 years = t All calcs are approximate, I don't use a calculator.