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21 Nuclear Chemistry Solutions to Exercises 0.002388 amu X 6.022142 1g 10²³ amu 1000g 1kg X 8.987551 s² 10¹⁶ = = 3.564490 X 10⁻¹³ = 3.564 Am = 2(1.0072765) + 2(1.0086649) - 4.001505 = 0.030378 amu 0.030378 amu 6.022142 1g X amu 1kg X 8.987551 X 10¹⁶ m² = = 4.533636 10⁻¹² = 4.5336 J ⁶Li: Am = 3(1.0072765) + 3(1.0086649) - 6.0134771 = 0.0343471 amu E 0.0343471 amu X 6.022142 1g amu 1000 1kg 8.987551 s² 10¹⁶ = = 5.126021 10⁻¹² = 5.12602 X (c) Binding energy per nucleon 3.564490 nucleons = 1.782 4.533636 X nucleons = 1.1334 ⁶Li: 5.126021 X nucleons = 8.54337 This trend in binding energy/nucleon agrees with the curve in Figure 21.12, which shows an irregular increase in binding energy/nucleon up to atomic number 56. The anomalously high value for calculated above is also apparent on the figure. 21.48 Nuclear mass is atomic mass minus mass of electrons. Nuclear binding energy is nuclear mass minus mass of the separate nucleons, converted to energy using Equation [21.22]. Divide by the total number of nucleons to find binding energy per nucleon. (a) Nuclear mass ¹⁴N: 13.999234 - 7(5.485799 10⁻⁴ amu) = 13.995394 47.935878 - 22(5.485799 10⁻⁴ amu) = 47.923809 ¹²⁹Xe: 128.904779 - 54(5.485799 X 10⁻⁴ amu) = 128.875156 amu (b) Nuclear binding energy Am = 7(1.0072765) + 7(1.0086649) - 13.995394 = 0.1161959 = 0.116196 amu E 0.1161959 amu 6.0221421 1g amu 1000g 1kg 8.987551 s² X 10¹⁶ 2 = = 1.734127 = 1.73413 J Am = 22(1.0072765) + 26(1.0086649) - 47.923809 = 0.4615614 = 0.461561 amu 0.4615614 amu 6.0221421 1g 10²³ amu 1000 1kg 8.987551 s² 10¹⁶ m² = = 6.888428 = 6.88843 J 657