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454 Chapter 20 Radioactivity and Nuclear Chemistry Energetics of Nuclear Reactions, Mass Defect, and Nuclear Binding Energy 20.65 Given: 1.0 g of matter converted to energy Find: energy Conceptual Plan: g kg E 1000 Solution: 1.0 X 1000 1kg = 0.0010 kg then E = = (0.0010 2.9979 X = 9.0 X 10¹³ J Check: The units (J) are correct. The magnitude of the answer makes physical sense because we are converting a large quantity of amus to energy. 20.67 Given: (a) O-16 = 15.9949145 amu, (b) Ni-58 = 57.935346 amu, and (c) Xe-129 = 128.904780 amu Find: mass defect and nuclear binding energy per nucleon Conceptual Plan: isotope mass mass defect nuclear binding energy per nucleon 931.5 MeV mass defect = Z(mass + (A - (mass on) - mass of isotope (1 amu)(A nucleons) Solution: mass defect = Z(mass + (A - (mass - mass of isotope (a) O-16 mass defect = 8(1.00783 amu) + (16 - (1.00866 amu) - 15.9949145 amu = 931.5 MeV MeV 0.1370055 amu = 0.13701 amu and 0.1370055 X = 7.976 nucleons) nucleon (b) Ni-58 mass defect = 28(1.00783 amu) + (58 (1.00866 amu) 57.935346 amu = 931.5 MeV MeV 0.543694 amu = 0.54369 amu and 0.543694 X II 8.732 (1 (58 nucleons) nucleon (c) Xe-129 mass defect = 54(1.00783 amu) + (129 - 54) amu) - 128.904780 amu = 1.16754 amu and 1.16754 amu (1 931.5 MeV nucleons) = 8.431 nucleon MeV Check: The units (amu and MeV/nucleon) are correct. The mass defect increases with an increasing number of nucle- ons, but the MeV/nucleon does not change by as much (on a relative basis). 20.69 Given: + + + U-235 = 235.043922 amu, Xe-144 = 143.9385 amu, and Sr-90 = 89.907738 amu Find: energy per g of U-235 Conceptual Plan: mass of products and reactants mass defect mass defect/g of U-235 then mass defect mass defect = Σmass of reactants - Σmass of products 235.043922 g U-235 g kg = 1000 g Solution: mass defect = Σmass of reactants - Σmass of products; notice that we can cancel a neutron from each side to get 92 + + and mass defect = 235.043922 g (143.9385 g + 89.907738 g + 1.00866 g) = 0.189024 g then 235.043922 0.189024 g U-235 X 1000 1 kg = 8.04207 X g U-235 kg then E = = 8.04207 X U-235 kg 2.9979 X = 7.228 X g U-235 J Check: The units (J) are correct. A large amount of energy is expected per gram of fuel in a nuclear reactor. 20.71 Given: + H-2 = 2.014102 amu, and He-3 = 3.016029 amu Find: energy per g reactant Conceptual Plan: mass of products and reactants mass defect mass defect/g of H-2 then mass defect mass defect = Σmass of reactants - Σmass of products 2(2.014102 g H-2) Copyright © 2017 Pearson Education, Inc.