In part (b) of Practice Problem 16 17, why is the number of 238 92 U atoms initially present in the rock equal to the sum of the number of 238 92 U atoms and 206 82 Pb atoms present in the rock today Data from Problem 16 17 A rock from an asteroid contains 2 57 g of 238 92 U and 3 83 g of 206 82 Pb...

The Answer is in the image, click to view ...

In part (b) of Practice Problem 16.17, why is the number of 238 92 U atoms initially

Question:

In part (b) of Practice Problem 16.17, why is the number of ²³⁸₉₂U atoms initially present in the rock equal to the sum of the number of ²³⁸₉₂U atoms and ²⁰⁶₈₂Pb atoms present in the rock today?

Data from Problem 16.17

A rock from an asteroid contains 2.57 g of ²³⁸₉₂U and 3.83 g of ²⁰⁶₈₂Pb. The molar mass of ²⁰⁶₈₂Pb is 205.974 46 g/mol, the molar mass of ²³⁸₉₂U is 238.029 g/mol, and the halflife of ²³⁸₉₂U is 4.46 × 10⁹ years. Assume that all the ²⁰⁶₈₂Pb came from the radioactive decay of the ²³⁸₉₂U.