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

€

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

€

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

€

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.

€

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

50

100

150

200

250250

30 40 50 60 70 80 90 100

20

30

40

50

60

70

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