Difference between revisions of "Solutions 6"

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(Fission and Nuclear Reactors)
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Latest revision as of 09:02, 9 July 2012

Fission 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.

Table 6.2: Calculated Q-values and change in binding energy per nukleon
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: 

  1. 239Pu+ n [math]\rightarrow[/math] 99Y + 2n + 139Cs
  2. Q-value: 191.42MeV
  3. The energy which is released by disintegration after stability is
    99Y: M(99Y)-M(99Ru)=17.4MeV
    139Cs: M(139Cs)-M(139La)=6.5MeV
  4. 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. 1.0 g 239Pu = 2.5[math]\cdot[/math]1021 atoms. Number of fissions per seconds is σ[math]\cdot[/math]ϕ[math]\cdot[/math]Nt = 1.89[math]\cdot[/math]1014, which will give an effect of 3.6[math]\cdot[/math]1016MeV (5811W)
  2. The formation of 240Pu: σ[math]\cdot[/math]ϕ[math]\cdot[/math]Nt= 6.8[math]\cdot[/math]1013s-1. After 100 days of irradiation 0.232 g Pu will be made.


4:

  1. 232Th+ n [math]\rightarrow[/math] 233Th [math]\rightarrow[/math] 233Pa [math]\rightarrow[/math] 233U
  2. 133I.
  3. One ton 232Th equals to 2.6[math]\cdot[/math]1027 atoms. The rate of formation for neutron capture (233Th): σ[math]\cdot[/math]ϕ[math]\cdot[/math]Nt = 7.37 [math]\cdot[/math]1024cm2[math]\cdot[/math]1014n cm-2s-1[math]\cdot[/math]2.6 [math]\cdot[/math]1027atomer= 1.91 [math]\cdot[/math]1018atoms s-1
  4. It will take 37hours of irradiation to form enough 233Th to give 100 g 233U, but disintegration of 233Pa to 233U must be waited.
  5. 100 g 233U: D=λN = 3.56 [math]\cdot[/math]1010Bq(35.6Gbq)