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21 Nuclear Chemistry Solutions to Exercises (a) Am = 4.0015028 + 1.0086649 - 3.1055014 - 2.0135514 = -0.1088851 = -0.10889 amu 0.1088851 amu 1g 6.022 10²³ atomic reaction' = X 'atomic reaction' 6.022 amu mol of reaction X 10³ 1kg (2.99792458 = -9.7861 J/mol g (b) Am = 3.1049328 + 1.0086649 - 2(2.0135514) = -0.0864949 = -0.08649 amu = -7.774 J/mol (c) Am = 4.0015028 + 1.0072714 - 2.0135514 - 3.1049328 = -0.1097100 = -0.10971 amu = -9.8602 J/mol 21.51 We can use Figure 21.12 to see that the binding energy per nucleon (which gives rise to the mass defect) is greatest for nuclei of mass numbers around 50. Thus (a) 27 Co should possess the greatest mass defect per nucleon. 21.52 Nuclear mass = 61.928345 amu - 28(5.485799 X 10⁻⁴ amu) = 61.912985 Binding energy = 28(1.0072765) + 34(1.0086649) - 61.912985 = 0.585363 amu 2 E 0.585363 amu 6.0221421 1g 10²³ amu 1000 1kg 8.987551 s² X 10¹⁶ = = 8.73606 J Binding energy/nucleon = 8.73606 J / 62 = 1.40904 The value given for iron-56 in Table 21.7 is 1.41 These values are the same, to three significant figures. Effects and Uses of Radioisotopes (sections 21.7-21.9) 21.53 (a) Nal is a good source of iodine, because it is a strong electrolyte and completely dissociated into ions in aqueous solution. The I-(aq) are mobile and immediately available for bio-uptake. They do not need to be digested or processed in the body before uptake can occur. Also, iodine is a large percentage of the total mass of Nal. (b) After ingestion, (-(aq) must enter the bloodstream, travel to the thyroid and then be absorbed. This requires a finite amount of time. A Geiger counter placed near the thyroid immediately after ingestion will register background, then gradually increase in signal until the concentration of I-(aq) in the thyroid reaches a maximum. Then, over time, iodine-131 decays, and the signal decreases. (c) Analyze/Plan. The half-life of iodine-131 is 8.02 days. Use t1/2 to calculate the decay rate constant, k. Then solve Equation [21.19] for t. No = 0.12 (12% of ingested iodine absorbed); = 0.0001 (0.01% of the original ingested amount). Solve. k = 0.693 = 0.693/8.02 d = 0.086409 = 0.0864 = -kt; t = 659