Solutions 6

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 E_B/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 σ_f).

2:

1. [math]239Pu+n/\gt 99Y+2n+139Cs[/math]
2. Q-value: 191.42MeV
3. 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
   
   
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.
5. 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 [math]3.6*1016MeV/\gt 5811W[/math]
6. The formation of ²⁴⁰Pu: σ*ϕ*Nt= 6.8*10¹³s^-1. After 100 days of irradiation 4*10^-6 g ²⁴⁰Pu will be made.

3:

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*10^-24cm^2*10^14n cm-2s-1*2.6*10^27atomer= 1.91*10¹⁸atomer 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.
4. 100g ²³³U: D=λN = 3.56*10¹⁰Bq(35.6Gbq)