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]^{239}Pu + \eta \; ^{99}Y + 2\eta + ^(139)Cs\lt /span\gt #Q-value: 191.42MeV #The energy which is released by disintegration after stability is reached:                      \lt sup\gt 99\lt /sup\gt Y: M(\lt sup\gt 99\lt /sup\gt Y)-M(\lt sup\gt 99\lt /sup\gt Ru)=17.4MeV\lt br\gt \lt sup\gt 139\lt /sup\gt Cs: M(\lt sup\gt 1\lt /sup\gt \lt sup\gt 39\lt /sup\gt Cs)-M(\lt sup\gt 139\lt /sup\gt La)=6.5MeV\lt br\gt \lt br\gt #\lt sup\gt \lt /sup\gt 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 \lt sup\gt 239\lt /sup\gt Pu = 2.5 *10\lt sup\gt 21\lt /sup\gt atomer. Number of fissions per seconds is σ*ϕ*Nt = 1.89*10\lt sup\gt 14\lt /sup\gt , which will give an effect of \lt span class="texhtml"\gt 3.6 * 10\lt sup\gt 16\lt /sup\gt ''M''''e''''V''  (5811''W)''\lt /span\gt #The formation of \lt sup\gt 240\lt /sup\gt Pu: σ*ϕ*Nt= 6.8*10\lt sup\gt 13\lt /sup\gt s\lt sup\gt -1\lt /sup\gt . After 100 days of irradiation 4*10\lt sup\gt -6\lt /sup\gt g \lt sup\gt 240\lt /sup\gt Pu will be made. \lt br\gt 3: \lt br\gt #\lt sup\gt \lt /sup\gt \lt math\gt ^{232}Th+\eta\;^{233Th\;^{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*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)