Revision as of 14:42, 18 June 2012

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.

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]^{239}Pu+\eta \; ^{99}Y+2\eta ^{139}Cs[/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*N_t= 6.8*10¹³s^-1. After 100 days of irradiation 4*10^-6 g Pu will be made.

3:

1. [math]^{232}Th+\eta \; ^{233}Th \; ^{233}Pa \; ^{233}U[/math]
2. ¹³³I.
3. One ton ²³²Th equals to 2.6*10²⁷ atoms. The rate of formation for neutron capture (²³³Th): σ*ϕ*N_t = 7.37*10^²⁴cm²*10¹⁴η cm-2s-1*2.6*10²⁷atomer= 1.91*10¹⁸atomer ^s-1
4. It will take 37hours of irradiation to form enough ²³³Th to give 100g ²³³U, but disintegration of ²³³Pa to ²³³U must be waited.
5. 100g ²³³U: D=λN = 3.56*10¹⁰Bq(35.6Gbq)