61. (a) The mass of a 238U atom is (238 u)(1.661 ×10– 24g/u) = 3.95 ×10– 22g, so the number of uranium atoms in the rock is NU= (4.20 ×10– 3g)/(3.95 ×10– 22g) = 1.06 ×1019. (b) The mass of a 206Pb atom is (206 u)(1.661 ×10– 24g) = 3.42 ×10– 22g, so the number of lead atoms in the rock is NPb= (2.135 ×10– 3g)/(3.42 ×10– 22g) = 6.24 ×1018. (c) 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
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.