A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238.Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half life for ²³⁸U $\rarr$ ²⁰⁶Pb is 4.5 × 10⁹ yr.)

Question

Quizplus · Accepted Answer

First, we need to calculate the number of Pb-206 atoms:

Number of Pb-206 atoms = (0.37 mg / 206.0 g/mol) x (6.022 x 10^23 atoms/mol) = 1.08 x 10^15 atoms

Next, we can use the ratio of U-238 to Pb-206 in the rock to calculate the number of U-238 atoms:

U-238/Pb-206 = (0.95 mg / 238.05 g/mol) / (0.37 mg / 206.0 g/mol) = 3.23

Taking the logarithm of both sides and solving for the number of U-238 atoms gives:

log(U-238 atoms) = log(Pb-206 atoms) + log(U-238/Pb-206) 
                = 15.03 + log(3.23) 
                = 15.51

Therefore, the number of U-238 atoms in the rock is:

U-238 atoms = 10^15.51 = 3.5 x 10^18 atoms

The answer is D.

A Rock Contains 0 →\rarr→ 206Pb Is 4

A Rock Contains 0 $\rarr$ ²⁰⁶Pb Is 4