64. The original amount of
238
U the rock contains is given by
m
0
= me
λt
= (3.70 mg) e
(ln 2)(260
×10
6
y)/(4.47
×10
9
y)
= 3.85mg .
Thus, the amount of lead produced is
m
= (m
0
− m)
m
206
m
238
= (3.85mg
− 3.70 mg)
206
238
= 0.132 mg .