CH-43 - Chapter 43 1. If R is the fission rate, then the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 43 1. 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. 2. 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. 3. (a) The mass of a single atom of 235 U is (235 u)(1.661 10 – 27 kg/u) = 3.90 10 – 25 kg, so the number of atoms in 1.0 kg is (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. 42-21. (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. 1661
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHAPTER 43 The conversion factor 3.156 10 7 s/y is used to obtain the last result. 4. Adapting Eq. 42-21, there are N M M NA Pu sam Pu g 239 g / mol / mol) F H G I K J 1000 602 10 25 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. 5. 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. 6. (a) We consider the process 98 49 Mo Sc Sc. 49 The disintegration energy is Q = ( m Mo 2 m Sc ) c 2 = [97.90541 u – 2(48.95002 u)](931.5 MeV/u) = +5.00 MeV. (b) The fact that it is positive does not necessarily mean we should expect to find a great deal of Molybdenum nuclei spontaneously fissioning; the energy barrier (see Fig. 43-3) is presumably higher and/or broader for Molybdenum than for Uranium. 7. (a) Using Eq. 42-20 and adapting Eq. 42-21 to this sample, the number of fission-
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 20

CH-43 - Chapter 43 1. If R is the fission rate, then the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online