A rock contains 0.688 mg 206Pb for every 1.000 mg 238U present. Assuming that no lead was originally present, that all the 206Pb formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between 238U and 206Pb is negligible, calculate the age of the rock. (For 238U, t1/2 = 4.5 X 10 9 years.)
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