Solutions 6

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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.

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
6.55
-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: 


  1. [math]^{239}Pu[/math]
  2. Q-value: 191.42MeV
  3. The energy which is released by disintegration after stability is reached:
  4. 99Y: M(99Y)-M(99Ru)=17.4MeV                                                                             139Cs: M(139Cs)-M(139La)=6.5MeV
  5. 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:

  1. Failed to parse (syntax error): ^{232}Th+\eta \> ^{233}Th \> ^{233}Pa \> ^{233}U
  1. 133I.
  2. One ton 232Th equals to 2.6*1027 atoms. The rate of formation for neutron capture (233Th): σ*ϕ*Nt = 7.37*10^24cm2*1014η cm-2s-1*2.6*1027atomer= 1.91*1018atomer s-1
  3. It will take 37hours of irradiation to form enough 233Th to give 100g 233U, but disintegration of 233Pa to 233U must be waited.
  4. 100g 233U: D=λN = 3.56*1010Bq(35.6Gbq)