Solutions 6
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Nuclear Reactions and Nuclear Reactors
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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 |
6.58 |
-0.007 |
238U/239U |
4.64 |
-0.015 |
239Pu/240Pu |
6.53 |
-0.007 |
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.
3:
- 1.0 g 239Pu = 2.5 1021 atoms. 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
4:
- 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