238
63. (a) The mass of a U atom is (238 u)(1.661 × 10-24 g/u) = 3.95 × 10-22 g, so the number of uranium
206
atoms in the rock is NU =(4.20 × 10-3 g)/(3.95 × 10-22 g) = 1.06 × 1019. The mass of a Pb
atom is (206 u)(1.661 × 10-24 g) = 3.42 × 10-22 g, so the number of lead atoms in the rock is
NPb =(2.135 × 10-3 g)/(3.42 × 10-22 g) = 6.24 × 1018.
(b) If no lead was lost, there was originally one uranium atom for each lead atom formed by decay,
in addition to the uranium atoms that did not yet decay. Thus, the original number of uranium
atoms was NU0 = NU + NPb =1.06 × 1019 +6.24 × 1018 =1.68 × 1019.
(c) We use
NU = NU0 e-t
where is the disintegration constant for the decay. It is related to the half-life T1/2 by =
(ln 2)/T1/2. Thus
T1/2
1 NU NU 4.47 × 109 y 1.06 × 1019
t = - ln = - ln = - ln =2.97 × 109 y .
NU0 ln 2 NU0 ln 2 1.68 × 1019
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