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21 Nuclear Chemistry Solutions to Exercises (a) 1 Ci = 3.7 X 10¹⁰ disintegrations 1 Bq = 1 dis/s 14.3 mCi 1000 m 10¹⁰ Ci dis/s = 5.29 10⁸ = 5.3 10⁸ dis/s = Bq (b) 1 rad = 1 X J/kg; 1 Gy = 1 J/kg = 100 rad. From part (a), the activity of the source is 5.3 dis/s. 5.29 10⁸ dis/s dis X 1 = 6.1 X 10⁻³ rad X 1000 rad mrad (c) rem = rad (RBE); Sv = Gy (RBE) where 1 Sv = 100 rem mrem = 6.14 10² mrad (9.5) = 5.83 X 10³ = 5.8 10³ mrem (or 5.8 rem) 21.68 (a) 15 mCi 1000 mCi 1Ci 10¹⁰ dis/s = 5.55 = 5.6 10⁸ dis/s = 5.6 10⁸ Bq (b) 5.55x10° dis/sx 240 sx0.075x dis 65 1 = = 1.3x10⁻⁵ J/kg 1Gy Gy 100 1Gy rad = 1.3 10⁻³ rad (c) 1.3x10⁻³ rad (1.0) = rem 1000 mrem = 1.3 mrem 1 rem 1.3x10⁻⁵ Gy (1.0) = 1.3 X Sv (d) The mammogram dose of 300 mrem is ~230 times as much radiation as the dose absorbed by the 65 kg person described above. Additional Exercises 21.69 This corresponds to a reduction in mass number of (3 4 =) 12 and a reduction in atomic number of (3 2 - 2) = 4. The stable nucleus is (This is part of the sequence in Figure 21.5.) 21.70 Use nuclear masses calculated in Solution 21.50. Am = 4.00150 + 2(1.00727) 2(3.10493) = -0.19382 amu 662