Difference between revisions of "Solutions 6"
From mn/safe/nukwik
(→Nuclear Reactions and Nuclear Reactors) |
(→Nuclear Reactions and Nuclear Reactors) |
||
Line 1: | Line 1: | ||
= Nuclear Reactions and Nuclear Reactors<br> = | = Nuclear Reactions and Nuclear Reactors<br> = | ||
− | + | Return to [[Problem Solving Sets]] | |
− | |||
− | Return to [[Problem Solving Sets]] | ||
<br> | <br> | ||
Line 17: | Line 15: | ||
|- | |- | ||
| <sup>235</sup>U/<sup>236</sup>U<br> | | <sup>235</sup>U/<sup>236</sup>U<br> | ||
− | | 6.55<br> | + | | align="right" | 6.55<br> |
− | | -0.004<br> | + | | align="right" | -0.004<br> |
|- | |- | ||
| <sup>238</sup>U/<sup>239</sup>U<br> | | <sup>238</sup>U/<sup>239</sup>U<br> | ||
− | | 4.81<br> | + | | align="right" | 4.81<br> |
− | | -0.012<br> | + | | align="right" | -0.012<br> |
|- | |- | ||
| <sup>239</sup>Pu/<sup>240</sup>Pu<br> | | <sup>239</sup>Pu/<sup>240</sup>Pu<br> | ||
− | | 6.53<br> | + | | align="right" | 6.53<br> |
− | | -0.003<br> | + | | align="right" | -0.003<br> |
|} | |} | ||
Line 40: | Line 38: | ||
<br> | <br> | ||
− | #1.0 g <sup>239</sup>Pu = 2.5<math>\cdot</math>10<sup>21</sup> atomer. Number of fissions per seconds is σ<math>\cdot</math>ϕ<math>\cdot</math>N<sub>t</sub> = 1.89<math>\cdot</math>10<sup>14</sup>, which will give an effect of 3.6<math>\cdot</math>10<sup>16</sup>MeV (5811W) | + | #1.0 g <sup>239</sup>Pu = 2.5<math>\cdot</math>10<sup>21</sup> atomer. Number of fissions per seconds is σ<math>\cdot</math>ϕ<math>\cdot</math>N<sub>t</sub> = 1.89<math>\cdot</math>10<sup>14</sup>, which will give an effect of 3.6<math>\cdot</math>10<sup>16</sup>MeV (5811W) |
#The formation of <sup>240</sup>Pu: σ<math>\cdot</math>ϕ<math>\cdot</math>N<sub>t</sub>= 6.8<math>\cdot</math>10<sup>13</sup>s<sup>-1</sup>. After 100 days of irradiation 4<math>\cdot</math>10<sup>-6</sup> g Pu will be made.<br> | #The formation of <sup>240</sup>Pu: σ<math>\cdot</math>ϕ<math>\cdot</math>N<sub>t</sub>= 6.8<math>\cdot</math>10<sup>13</sup>s<sup>-1</sup>. After 100 days of irradiation 4<math>\cdot</math>10<sup>-6</sup> g Pu will be made.<br> | ||
Revision as of 15:41, 22 June 2012
Nuclear Reactions and Nuclear Reactors
Return to Problem Solving Sets
1: It is noteworthy to notice the Q-value for the neutron capture and the change in binding energy per nucleon for each of the isotope pairs, see table 6.2.
Pair of nuclide |
Q-value for neutron capture (MeV) |
Change in EB/A (MeV) |
235U/236U |
6.55 |
-0.004 |
238U/239U |
4.81 |
-0.012 |
239Pu/240Pu |
6.53 |
-0.003 |
The nuclide pair 238U/239U have a significantly lower Q-value and a significantly bigger fall in EB/A than the other pairs. This can be explained by the pair-pair configuration in the 238U nucleus, which makes it less favorable to bind another neutron. On the other hand, for pair-odd nuclides it is much more favorable to bind another neutron to achieve a pair-pair configuration. This is shown from the cross sections for interaction with thermal neutrons (σ and σf).
2:
- 239Pu+ n 99Y + 2n + 139Cs
- Q-value: 191.42MeV
- The energy which is released by disintegration after stability is
99Y: M(99Y)-M(99Ru)=17.4MeV
139Cs: M(139Cs)-M(139La)=6.5MeV - 2/3 of this energy will disappear with neutrinos. Some of the disintegrations have too long half-lives to have an effect on the reactor safety.
- 1.0 g 239Pu = 2.5 1021 atomer. Number of fissions per seconds is σ ϕ Nt = 1.89 1014, which will give an effect of 3.6 1016MeV (5811W)
- The formation of 240Pu: σ
3:
- 232Th+ n 233Th 233Pa 233U
- 133I.
- One ton 232Th equals to 2.6n cm-2s-1 2.6 1027atomer= 1.91 1018atoms s-1 1027 atoms. The rate of formation for neutron capture (233Th): σ ϕ Nt = 7.37 1024cm2 1014
- It will take 37hours of irradiation to form enough 233Th to give 100 g 233U, but disintegration of 233Pa to 233U must be waited.
- 100 g 233U: D=λN = 3.56