Difference between revisions of "Solutions 6"
Line 30: | Line 30: | ||
<br>2: <br> <br> | <br>2: <br> <br> | ||
− | + | <br> | |
#<math>^{239}Pu</math> | #<math>^{239}Pu</math> |
Revision as of 14:59, 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.
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:
- Q-value: 191.42MeV
- 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
- 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:
- Failed to parse (syntax error): ^{232}Th+\eta \> ^{233}Th \> ^{233}Pa \> ^{233}U
- 133I.
- 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
- It will take 37hours of irradiation to form enough 233Th to give 100g 233U, but disintegration of 233Pa to 233U must be waited.
- 100g 233U: D=λN = 3.56*1010Bq(35.6Gbq)