Chapter43 - 1. If MCr is the mass of a 52Cr nucleus and MMg...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1

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

View Full DocumentRight Arrow Icon
2. 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.
Background image of page 2
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.
Background image of page 3

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

View Full DocumentRight Arrow Icon
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.
Background image of page 4
N = m / m 0 = (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. The conversion factor 3.156 × 10 7 s/y is used to obtain the last result. 5. (a) The mass of a single atom of 235 U is 0 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
Background image of page 5

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

View Full DocumentRight Arrow Icon
6. The energy released is Qmmm m m c nn =+ =− = ( (. . . . UC s R b ) u u u u)(931.5 MeV / u) MeV. 2 23504392 100867 140 91963 92 92157 181 2
Background image of page 6
7. (a) Using Eq. 42-20 and adapting Eq. 42-21 to this sample, the number of fission- events per second is R N T MN MT 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. == = × × = ln ln (. / ( // 22 10 235 16 12 (b) Since 1/2 1/ R T (see Eq. 42-20), the ratio of rates is R R T α fission 1/2 fission T y y × × 17 8 8 30 10 70 10 43 10 / . . ..
Background image of page 7

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

View Full DocumentRight Arrow Icon
8. When a neutron is captured by 237 Np it gains 5.0 MeV, more than enough to offset the 4.2 MeV required for 238 Np to fission. Consequently, 237 Np is fissionable by thermal neutrons.
Background image of page 8
2 U238 U239 () (238.050782 u 1.008664 u 239.054287 u)(931.5 MeV/u) 4.8 MeV. n Qm mm c =+ = 9. The energy transferred is
Background image of page 9

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

View Full DocumentRight Arrow Icon
10. (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.
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 56

Chapter43 - 1. If MCr is the mass of a 52Cr nucleus and MMg...

This preview shows document pages 1 - 11. Sign up to view the full document.

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