First, we need to calculate how much lead (Pb206) would have been formed from the decay of the original amount of uranium (U238) in the rock, given the half-life of U238.
The decay process of U238 eventually leads to the formation of Pb206 through a series of alpha and beta decays. The decay process is as follows:
U238 → Th234 → Pa234 → U234 → Th230 → Ra226 → Rn222 → Po218 → Pb214 → Bi214 → Po214 → Pb210 → Bi210 → Po210 → Pb206
Each decay step involves the emission of alpha or beta particles, and the resulting nucleus has a different atomic number and mass number. The final product of the decay chain is Pb206, which is a stable isotope.
We can use the fact that the half-life of U238 is 4.47×10^9 years to calculate how much U238 would have decayed to Pb206 over the estimated 4.40 billion years since the rock was formed.
The half-life of U238 means that half of the original amount of U238 would have decayed to Pb206 after 4.47×10^9 years. Therefore, the fraction of U238 that has decayed to Pb206 after 4.40 billion years is:
(1/2)^(4.40/4.47) = 0.0184
This means that only 1.84% of the original amount of U238 remains, and the rest has decayed to Pb206. We can calculate the current mass of Pb206 in the rock by multiplying the current mass of U238 by this fraction:
mass of Pb206 = 1.130g × 0.0184 = 0.0208g
Therefore, the current mass of Pb206 in the moon rock is approximately 0.0208 grams.