Difference between revisions of "Solutions 6"
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<br>2: <br> <br> | <br>2: <br> <br> | ||
| − | #<span class="texhtml"><sup>239</sup>''P'''u'''''</span>'''+ | + | #<span class="texhtml"><sup>239</sup>''P'''u'''''</span>'''+ n <math>\rightarrow</math> <span class="texhtml"><sup>99</sup>''Y''</span> + 2n + <span class="texhtml"><sup>139</sup>''C''</span>'''''s''<br> |
#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> | #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> | <br> | ||
| − | 1.0g <sup>239</sup>Pu = 2.5 | + | #1.0g <sup>239</sup>Pu = 2.5 • 10<sup>21</sup> atomer. Number of fissions per seconds is σ • ϕ • N<sub>t</sub> = 1.89 • 10<sup>14</sup>, which will give an effect of 3.6 • 10<sup>16</sup>MeV (5811W) |
| + | #The formation of <sup>240</sup>Pu: σ • ϕ • N<sub>t</sub>= 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> | ||
'''3:''' <br> | '''3:''' <br> | ||
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#<span class="texhtml"><sup>232</sup>''T''''h'''''</span>'''+ η <math>\rightarrow</math> <span class="texhtml"><sup>233</sup>''T''</span>'''''h'' <math>\rightarrow</math> <sup>233</sup>Pa <math>\rightarrow</math> <sup>233</sup>U<br> | #<span class="texhtml"><sup>232</sup>''T''''h'''''</span>'''+ η <math>\rightarrow</math> <span class="texhtml"><sup>233</sup>''T''</span>'''''h'' <math>\rightarrow</math> <sup>233</sup>Pa <math>\rightarrow</math> <sup>233</sup>U<br> | ||
#<sup>133</sup>I. | #<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 | + | #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><span class="texhtml"><span style="font-family: sans-serif;">n</span></span> cm<sup>-2</sup>s-1 • 2.6 • 10<sup>27</sup>atomer= 1.91 • 10<sup>18</sup>atoms s<sup>-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. | #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 | + | #100g <sup>233</sup>U: D=λN = 3.56 • 10<sup>10</sup>Bq(35.6Gbq)<br> |
<br> | <br> | ||
<br> | <br> | ||
Revision as of 11:29, 19 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.
3:
- 232T'h+ η 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 • 1024cm2 • 1014n cm-2s-1 • 2.6 • 1027atomer= 1.91 • 1018atoms 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)
