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.

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

4:

1. 232Th+ n→233Th→233Pa→233U
2. 133I.
3. 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
4. It will take 37 hours 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•1010 Bq (35.6 Gbq)

Exercises with Fission and Nuclear Reactors