Difference between revisions of "Solutions 6"
From mn/safe/nukwik
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<br>2: <br> | <br>2: <br> | ||
| − | # | + | #239Pu+n->99Y+2n+139Cs |
| − | #Q-value: 191.42MeV | + | #Q-value: 191.42MeV |
| − | #The energy which is released by disintegration after stability is reached: | + | #The energy which is released by disintegration after stability is reached: |
| − | # | + | #<sup>99</sup>Y: M(<sup>99</sup>Y)-M(<sup>99</sup>Ru)=17.4MeV <sup>139</sup>Cs: M(<sup>139</sup>Cs)-M(<sup>139</sup>La)=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.<br> | |
| − | + | ||
| + | <br> | ||
| + | |||
| + | 1.0g <sup>239</sup>Pu = 2.5 *10<sup>21</sup> atomer. Number of fissions per seconds is σ*ϕ*Nt = 1.89*10<sup>14</sup>, which will give an effect of 3.6*10<sup>16</sup>meV->5811W<br>The formation of <sup>240</sup>Pu: σ*ϕ*Nt= 6.8*10<sup>13</sup>s<sup>-1</sup>. After 100 days of irradiation 4*10<sup>-6</sup> g Pu will be made.<br><span''' | ||
| + | ''' | ||
| + | </span> '''3:''' <br> | ||
| + | |||
| + | #232Th+n->233Th->233Pa->233U | ||
| + | #<sup>133</sup>I | ||
| + | #One ton <sup>232</sup>Th equals to 2.6*10<sup>27</sup> atoms. The rate of formation for neutron capture (<sup>233</sup>Th): σ*ϕ*N<sub>t</sub> = 7.37*10^<sup>24</sup>cm<sup>2</sup>*10<sup>14</sup><math>\eta</math> cm-2s-1*2.6*10<sup>27</sup>atomer= 1.91*10<sup>18</sup>atomer <sup>s-1</sup> | ||
| + | #It will take 37hours of irradiation to form enough <sup>233</sup>Th to give 100g <sup>233</sup>U, but disintegration of <sup>233</sup>Pa to <sup>233</sup>U must be waited. | ||
| + | #100g <sup>233</sup>U: D=λN = 3.56*10<sup>10</sup>Bq(35.6Gbq)<br> | ||
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Revision as of 14:18, 18 June 2012
Nuclear reactions and nuclear reactors
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 reached:
- 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.
1.0g 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*ϕ*Nt= 6.8*1013s-1. After 100 days of irradiation 4*10-6 g Pu will be made.
<span
</span> 3:
- 232Th+n->233Th->233Pa->233U
- 133I
- One ton 232Th equals to 2.6*1027 atoms. The rate of formation for neutron capture (233Th): σ*ϕ*Nt = 7.37*10^24cm2*1014 cm-2s-1*2.6*1027atomer= 1.91*1018atomer s-1
- It will take 37hours of irradiation to form enough 233Th to give 100g 233U, but disintegration of 233Pa to 233U must be waited.
- 100g 233U: D=λN = 3.56*1010Bq(35.6Gbq)
