1 (664)

Prévia do material em texto

21 Nuclear Chemistry Solutions to Exercises Am = 54(1.0072765) + 75(1.0086649) - 128.875156 = 1.1676425 = 1.167643 amu = 1.1676425 amu X 6.0221421 1g X 10²³ amu X 1kg 8.987551 X 10¹⁶ m² = 1.742610 X J (c) Binding energy per nucleon ¹⁴N: 1.73413 J / 14 nucleons = 1.23866 6.888428 J / 48 nucleons = 1.43509 X ¹²⁹Xe: 1.742610 X nucleons = 1.350860 X 21.49 Analyze/Plan. Use Equation [21.22] to calculate the mass equivalence of the solar radiation. Solve. (a) 1 min 60 = d = J/d Am = 1.541 = 1.714 = 10⁵kg/d (b) Analyze/Plan. Calculate the mass change in the given nuclear reaction, then a conversion factor for g 235U to mass equivalent. Solve. Am = 140.8833 + 91.9021 + 2(1.0086649) - 234.9935 = -0.19077 = -0.1908 amu Converting from atoms to moles and amu to grams, it requires 1.000 mol or 235.0 g 235U to produce energy equivalent to a change in mass of 0.1908 g. 0.10% of 1.714 10⁵ kg is 1.714 10² kg = 1.714 10⁵ g (This is about 230 tons of per day.) 21.50 First, calculate nuclear masses from atomic masses. Then, the calculated Am is for one group of single nuclides involved in a reaction, labeled reaction'. Multiplying by Avogadro's number changes the quantity to 'mol of reaction'. Since energy is released, the sign of is negative. ¹H: 1.00782 amu - 1(5.485799 10⁻⁴ amu) = 1.0072714 = 1.00727 amu 2.01410 amu 1(5.485799 10⁻⁴ amu) = 2.0135514 = 2.01355 amu ³H: 3.10605 amu 1(5.485799 X 10⁻⁴ amu) = 3.1055014 = 3.10550 amu ³He: 3.10603 2(5.485799 10⁻⁴ amu) = 3.1049328 = 3.10493 amu 4.00260 2(5.485799 X 10⁻⁴ amu) = 4.0015028 = 4.00150 amu 658