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Week11,Lecture2–FissionProcess
NuclearPower,NuclearReactors‐‐Overview‐‐‐Reactortypes‐‐‐ReactorsinMichigan‐‐‐ReactorsinFrance‐‐NuclearFissionprocess‐‐‐fissionenergeBcs,fissilenuclei‐‐‐limitstoheavynuclei‐‐‐dynamicalprocess‐‐NuclearFissionOperaBons‐‐‐controlandreacBvity
7thHomeworkdueMonday
235Uvs.238UforFission
1n+235U‐>(236U144)*+QwhereQ=6.545MeV
1n+238U‐>(239U147)*+Q’whereQ’=4.406MeV
It is obvious to take nuclei with the highest atomic number for fission since they are the most unstable due to the coulomb force. But there are two isotopes of uranium available in nature but only one is ‘fissile’.
Recall that the cross section for neutron capture increases dramatically at low energies (1/v). So it is best to use as low a neutron energy as possible from the stand point of capturing the neutron ...
The excitation energy is the Q value plus a very small kinetic energy from the neutron:
The formation of the Even-Even nucleus gives about 2 MeV more energy than the formation of the Even-Odd nucleus – this makes all the difference in driving the fission process.
FissileNuclei
Those nuclei that are Even-Odd and undergo fission after low energy neutron capture are called ‘fissile’ nuclei – there are only a few practical cases.
235U … natural 239Pu … make from 238U 233U … make from 232Th
CriBcal(large)SizeforNuclei
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Sphere :BE(Z,A) = aV A − aSA2 / 3 − aC
Z 2
A1/ 3− aA
A − 2Z( )2
A± δ
Recall that the equilibrium shape of (most) nuclei is a sphere. Since fission converts a single nucleus into two separate nuclei – one needs to consider how the energy of a nucleus changes with deformation.
Small deformations of the sphere: • constant volume • no change in numbers of nucleons • Surface area grows • Coulomb energy drops
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EC ≈ ECo (1− α
2
5) =
?ES ≈ ES
o(1+2α 2
5)
EC − ECo = −
ECoα 2
5=?
ES − ESo =2ES
oα 2
5
−EC
oα 2
5=2ES
oα 2
5→
ECo
2ESo =1 Eq. 11.3
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ECo
2ESo =
0.7Z 2 /A1/ 3
2*17.2A2 / 3=1
Z 2
A=34.40.7
= 49.1
Largest Nucleus:
FissionBarrier
The fission process is an energy amplifier inthatasmallamountofexcitaBonenergycanleadtotheoutputofalargeenergyrelease.
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VCoulomb =Z1Z2e
2
R1 + R2=Z1 92 − Z1( )1.439MeV1.2[A1
1/ 3 + (238 − A1)1/ 3]
The deformation process reaches a point of no return that is called the fission barrier.
!100
!500
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30 40 50 60 70 80 90 100
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EnergyContours
Z
N
MassandChargeofFF’s
1n+235U‐>(236U144)*+6.545MeV‐>A1Z1+A2Z2+Qf
A1=A2=236/2–rareeventA1~140,A2~(236‐140)–common
UnchangedChargeRaBoZ/A=92/238=Z1/A1=Z2/A2
ModernCalculaBonofFission
Thefigureshowstheenergysurfaceofthetransuranicelement258Fmcalculatedwiththeself‐consistentnucleardensityfuncBonaltheoryasafuncBonoftwocollecBvevariables:(a)thetotalquadrupolemomentQ20represenBngtheelongaBonofnuclearshape,(b)thetotaloctupolemomentQ30represenBngtheleh‐rightshapeasymmetry
www.scidacreview.org/0704/html/unedf.html
quadrupoleshape(L=2)
octupoleshape(L=3)
258Fm158‐>50Sn+50Sn
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