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Unformatted text preview: 1635 Chapter 43 1. If M Cr is the mass of a 52 Cr nucleus and M Mg is the mass of a 26 Mg nucleus, then the disintegration energy is Q = ( M Cr – 2 M Mg ) c 2 = [51.94051 u – 2(25.98259 u)](931.5 MeV/u) = – 23.0 MeV. 2. Adapting Eq. 4221, there are N M M NA Pu sam Pu g 239 g / mol / mol) = = F H G I K J × = × 1000 6 02 10 2 5 10 23 24 ( . . plutonium nuclei in the sample. If they all fission (each releasing 180 MeV), then the total energy release is 4.54 × 10 26 MeV. 3. If R is the fission rate, then the power output is P = RQ , where Q is the energy released in each fission event. Hence, R = P / Q = (1.0 W)/(200 × 10 6 eV)(1.60 × 10 – 19 J/eV) = 3.1 × 10 10 fissions/s. 4. We note that the sum of superscripts (mass numbers A ) must balance, as well as the sum of Z values (where reference to Appendix F or G is helpful). A neutron has Z = 0 and A = 1. Uranium has Z = 92. (a) Since xenon has Z = 54, then “Y” must have Z = 92 – 54 = 38, which indicates the element Strontium. The mass number of “Y” is 235 + 1 – 140 – 1 = 95, so “Y” is 95 Sr. (b) Iodine has Z = 53, so “Y” has Z = 92 – 53 = 39, corresponding to the element Yttrium (the symbol for which, coincidentally, is Y). Since 235 + 1 – 139 – 2 = 95, then the unknown isotope is 95 Y. (c) The atomic number of Zirconium is Z = 40. Thus, 92 – 40 – 2 = 52, which means that “X” has Z = 52 (Tellurium). The mass number of “X” is 235 + 1 – 100 – 2 = 134, so we obtain 134 Te. (d) Examining the mass numbers, we find b = 235 + 1 – 141 – 92 = 3. 5. (a) The mass of a single atom of 235 U is m = (235 u)(1.661 × 10 – 27 kg/u) = 3.90 × 10 – 25 kg, so the number of atoms in m = 1.0 kg is CHAPTER 43 1636 N = m / m = (1.0 kg)/(3.90 × 10 – 25 kg) = 2.56 × 10 24 ≈ 2.6 × 10 24 . An alternate approach (but essentially the same once the connection between the “u” unit and N A is made) would be to adapt Eq. 4221. (b) The energy released by N fission events is given by E = NQ , where Q is the energy released in each event. For 1.0 kg of 235 U, E = (2.56 × 10 24 )(200 × 10 6 eV)(1.60 × 10 – 19 J/eV) = 8.19 × 10 13 J ≈ 8.2 × 10 13 J. (c) If P is the power requirement of the lamp, then t = E / P = (8.19 × 10 13 J)/(100 W) = 8.19 × 10 11 s = 2.6 × 10 4 y. The conversion factor 3.156 × 10 7 s/y is used to obtain the last result. 6. The energy released is Q m m m m m c n n = + = = ( ( . . . . U Cs Rb ) u u u u)(931.5 MeV / u) MeV. 2 23504392 100867 140 91963 92 92157 181 2 7. (a) Using Eq. 4220 and adapting Eq. 4221 to this sample, the number of fission events per second is R N T M N M T A fission sam U 23 17 fission fission g)(6.02 10 mol) ln2 g / mol)(3.0 10 y)(365 d / y) fissions / day....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Energy, Mass

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