The Physics of Energy sources Nuclear Fission

The Physics of Energy sources
Nuclear Fission
B. Maffei
Bruno.maffei@manchester.ac.uk
Nuclear Fission
1
Introduction
!   We saw previously from the Binding energy vs A curve that heavy
nuclei (above A~120) will gain stability by splitting into 2
fragments of A~60 each
!   This reaction will release energy (Q) due to difference in binding
energy between the parent nucleus and the products
!   The reaction will also produce neutrons
!   This has been observed in early fission experiments in the early 40s
!   The reaction will quite probably produce γ rays
A simple energy release model
B/A energy for heavy nuclei (A~200) is about 7.6MeV
B/A energy for more stable nuclei (fragments) ~ 8.5MeV
The change of B/A is about 0.9MeV per nucleon
If we have 235U fission we have a release of energy
Q=235x0.9MeV= 211.5MeV
A
typical
Nuclear
Fission nuclear fission releases an energy of the order of 200MeV
2
Mass partition
An typical example of fission is
235
U + n→236U *→147La +87Br + 2n
Ref 2
It is not unique and many different mass
partitions are possible as shown in figure.
235U
For
fragment mass numbers vary between
70 and 160 with most probable values at about
96 and 135
Distribution of fission fragment
masses from the fission of 235U
The energy released in process will be transmitted to the fragments as Kinetic energy
Nuclear Fission
3
Neutron production
Ref 2
!   The line of stability of the nuclei
is not totally a straight line
!   There is a slight curvature leaning
towards neutron rich nuclei when A
increases
!   Ratio N/Z for A>200 is ~ 1.5
•  Uranium, Plutonium …..
!   Ratio N/Z for 70<A<160 is 1.3-1.4
•  These are the fragments
When the fission of the heavy nucleus occurs, the fragments end up having the same neutron to
proton ratio (~1.5) than the original nucleus. So compared to the line of stability for nuclei of
their mass, these fragments are neutron rich.
They will decay towards the line of stability mainly by β- emissions (long) but also by rapid
shedding of several neutrons.
These will be crucial to initiate other fission reactions  chain reaction
Several fission reactions for the same original element  Average number of neutrons produced
Nuclear Fission
4
Induced fission reaction
Imagine a typical fission reaction
239
94
124
Pu →112
Pd
+
46
48 Cd + 3n
The calculation of the Q factor will give 188MeV (in agreement with the rough
estimate of 200MeV we saw earlier).
From this figure we might think that fission is quite a probable process due to the
energy released  so could it be spontaneous ?
Actually no: do not forget the Coulomb interaction Coulomb barrier
Let s assume that the 2 fragments are uniformly charged spheres of radius rpd and rcd
We have the energy due to the Coulomb potential between the 2 fragments
E coulomb
e 2 Z Pd ZCd
=
4 "# 0 r
r being the distance between them
Let s try to bring these 2 nuclei as close as possible to each other
!
1/ 3
r = rCd + rPd = r0 ( ACd
+ A1/Pd3 )
Nuclear Fission
Reminder: nucleus radius
r = r0 A1/ 3 , where r0 = 1.2fm
5
Induced fission reaction (2)
Replacing all the parameters by their value we find
(strongly recommended to do that as an exercise)
Bc = Ecoulomb = 270MeV
Bc: Coulomb barrier
Coulomb Potential
energy (1/r variation)
c
distance
Attractive strong
nuclear interaction
Based on Ref 2
Nuclear Fission
We see that in order to be able to glue the 2
fragments together, we need to cross this
energy barrier due to the Coulomb repulsion
Only when the 2 nuclei will be close enough
(basically coming in contact) the strong force
will be able to overcome the repulsion and
keep them together as one nucleus
So working the other way round, in order to have a high
probability of spontaneous fission of the parent nucleus,
the energy Q that can be released during the fission
should be comparable to the Coulomb barrier.
It is not the case here
6
Induced fission reaction (3)
Note: Spontaneous fission does exist
•  Even with a low probability, it will occur statistically
•  Occurs quantum mechanically by tunnelling effect
•  Some very heavy isotopes are so instable that fission overcomes the Coulomb barrier
We would need A large (~300) and roughly Z2/A>47
However for some isotopes spontaneous fission is a non negligible decay mode
Example: 252Cf is used as a radioisotope (spontaneous fission branch ratio~3%)
More generally for fission reaction to happen, it needs to be induced (triggered)
We need to excite the nucleus to above the Coulomb barrier
That will happen through absorption of a neutron
Having no charge, the neutron will not have to go through a Coulomb barrier in
order to be absorbed by the nucleus
The addition of the neutron will leave the resulting nucleus in an excited state
235
U + n→236U *
Nuclear Fission
7
Activation energy
!   The previous crude model was mainly to explain why fission needs to be induced.
!   However, in reality the necessary extra energy amount to make fission happen is
not that large. This is what we call activation energy.
Energy plot from our simple model of Pu fission
Coulomb barrier
270 MeV
188 MeV
Fission energy released
Equivalent to
Reference: Energy of
2 fragments far apart
Energy state of Pu nucleus
Graphs based on Ref 3
So to make sure that the reaction will occur (probability=1), the neutron needs to
provide an energy at least equal to the activation energy
Nuclear Fission
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Activation energy: crude model to more realistic one
!   The actual activation energy is not as high as we did calculate
!   Mainly for heavy nuclei
!   Part of that can be explained with our simple liquid drop model
n+
235U
236U*
+n(s)
Due to nucleus stretching, the overall binding energy of the nucleus has changed
Consider the semi-empirical
mass formula from the LDM
2
ac .Z(Z "1) asym (A " 2Z)
2/3
B(Z, A) = av .A " as .A "
"
+ # (A,Z)
A1/ 3
A
When the nucleus is stretched, the volume is constant but the surface increases:
!  Volume term is
! the same
!  Surface and Coulomb terms will be modified
!  Binding energy decreased (for more details see ref 3)
Nuclear Fission
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Variation of activation energy with mass number
!   In order to get a better value of the activation energies more
detailed models are used taking into account more
sophisticated effects than the LDM (i.e. shell model).
Z
Around Uranium
5MeV typically
Dark curve: LDM
Thin curve: shell structure model
Nuclear Fission
Figs based on Ref 3
A~280
Spontaneous
fission
Activation energy for heavy nuclei.
A and Z dependence
10
Application to uranium
235
U + n→236U *
When 235U captures a neutron to form a compound state 236U* the excitation energy is:
Qex = [ m( 236U * ) " m( 236U)] # c 2
Assuming the kinetic energy of the neutron small, the energy of the compound state 236U* can
be found directly by the mass energies of 235U and the neutron:
!236
m( U * ) = m( 235U ) + mn = (235 .043924u + 1.008665u ) = 236 .052589u
Qex = (236.052589u − 236.045563) × 931.5MeV / u = 6.5MeV
The activation energy of 236U is 6.2MeV  235U can be fissioned with zero energy neutron.
Similar calculation for
238
U + n→239U *
gives Qex=4.8MeV
Activation energy for 239U is 6.6MeV. We need a neutron of at least ~2MeV to get fission.
The difference between the excitation energies (6.5 and 4.8MeV) is one of the explanations
for the extreme difference in the fissionability of 235U and 238U
Nuclear Fission
11
Fission energy budget
Example on thermal-neutron induced fission of 235U (all figures from ref 2)
For this reaction, there is an
average of 2.4 neutrons per fission
!   About 87% of the total energy is released
promptly
Mean energy of each
released neutron ~ 2MeV
!   90% of this in fragments KE.
!   ~13% in radioactive decay
!   Electrons, γ rays energy converted into heat
!   Neutrino energy not recoverable
Nuclear Fission
Energy spectrum of emitted neutrons
12
Neutrons released
We have seen that there is a prompt release of a few neutrons per reaction
These prompt neutrons have a range of kinetic energy
By convention we have the following designation:
!  Thermal:
E ~ 0.025 eV
!  Epithermal:
E ~ 1 eV
Delayed neutrons
!  Slow:
E ~ 1 keV
Timescale ~ 6sec
!  Fast:
E ~ 100 keV to 10 MeV
About 1% of
released neutrons
These are released within about
10-14 sec after fission
Other neutrons are emitted
during some of the fission decay
chains (from heavy fragments)
Ex:
93
37
About 99% of
released neutrons
sec
93
92
Rb56 ⎯6⎯→
⎯
Sr
→
38
55
38 Sr54 + n
Decay branch β- followed by neutron em. 1.4% probability
Nuclear Fission
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Cross section – A brief explanation
For a more precise development see references at the end
Intensity I=φA
Flux φ
Consider an incident flux of particles on a surface A of width dx
(Number of particles/m2/sec)
Intensity (number of incident particles per second) = φ × A
A
If N target nuclei are exposed to the beam
(number of atoms in volume A.dx)
N = ρ N × dx × A
ρN being the atoms volume density (Number/m3)
dx
Let s suppose that we have a reaction (i.e. fission) between the incident particles and the
atoms in the volume A.dx. If the nuclei in the target act independently, the reaction rate will
be proportional to the number of incident particles ( φ or I), and the number of atoms N.
Rate R ∝ N × φ
R ∝ ρ N × dx × A × φ
event rate per nucleus
σ=
incident flux
Nuclear Fission
The constant of proportionality is called the cross section σ
σ is the reaction rate per target atom per unit of flux.
It is equivalent to a probability of the process to occur.
and
R = σ × N ×φ
Note: σ as units of area (cross section). R number of reaction per unit of time
14
A classical approach of the cross-section
(but wrong!)
Consider the reaction
R1
R
R2
135
Xe + n→136Xe*
The neutron (radius R1) can be captured by the nucleus
of Xeon (radius R2) when the strong interaction act 
distance between their centres is r ≤ R = R + R
1
2
Neglecting R1 in comparison with R2, R=R2
Then a simple picture of the cross section would give:
σ = π × R 2 = π × r02 × A2 / 3 ≈ 120 fm 2
However, the cross section for this specific reaction is actually σ~106 fm2
The classic approach is not the right one. In order to get the real values of
the cross-sections a proper quantum mechanical treatment has to be
perform and other physical processes have to be taken into account.
Units: we use a more appropriate one for cross-section – 1 barn=100 fm2= 10-28 m2
Nuclear Fission
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Example
Given 10g of natural uranium (mass 238.029u) and a neutron flux of 1013cm-2s-1, a
thermal fission cross section of 235U of 584b, find the fission rate.
Fission rate = Nσφ with N being the number of 235U atoms.
Abundance of 235U in natural uranium= 0.72%
10 g × 6.023 ×10 23
N=
× 0.0072 = 1.82 ×10 20 atoms
238.029
Fission rate = 1.82 ×10 20 ×1013 × 584 ×10 −24 = 1.06 ×1012 reactions / sec
Note: if several reactions are possible, each with a cross section σi, total cross section:
σ total = ∑σ i
i
Nuclear Fission
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Cross section data
Thermal neutrons
Fast neutrons
Cross sections of neutron induced
fission of 235U and 238U as a function
of the incident neutron KE
Note the poor cross
section of 238U
Nuclear Fission
Figures from Ref 3
17
What happens to the emitted neutron?
!   There is an average of η neutrons emitted per reaction
!   Not all emitted neutrons will be able to trigger another fission.
!   There are other reactions that enter in competition
!   Not all neutrons are useful in a chain reaction
!   Each competitive reaction has a probability to occur  Reaction rate
4 possible reactions: 3 competitors to fission
scattering
U + n→236U * ⎯elastic
⎯ ⎯
⎯⎯→235U + n
235
Back to square one
Scattering
U * ⎯inelastic
⎯ ⎯scattering
⎯ ⎯
⎯→235U * + n
Re-emitted neutron as lower KE
leaving 235U in an excited state
U * ⎯radiative
⎯ ⎯capture
⎯⎯→236U + γ
Heavy nucleus + low energy incident
neutron: most probable decay
236
236
Nuclear Fission
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Fission vs capture
!   Each of the previous reactions has a corresponding cross section.
!   The cross sections are normalised: we can sum them
σ fission + σ capture = σ f + σ c = σ absorption = σ a
σ elastic + σ inelastic = σ e + σ i = σ scattering = σ s
with
σ total = σ a + σ s
In order to get a good fission rate, we want σfission to be the dominant cross section
From previous plot we saw that the fission cross section of 235U is strongly dependent of the
incident neutron KE. σsis relatively independent of the energy ~ 10b
σfission= 584b for thermal neutrons (~0.025eV)
σfission~ 1.5b for fast neutrons (~2MeV), ~1/6 of σs
Fast neutrons (mean of 2MeV) are released during fission
Nuclear Fission
We need to slow them down to thermal energies to sustain chain reaction
à Moderation
19
Number of useful neutrons
!   Each 235U fission reaction, initiated by a thermal neutron, produces an average of
η=2.4 neutrons.
!   Some neutrons are lost through radiative capture both with 235U and 238U.
Suppose that we have pure 235U
Let s call the number of released neutrons available
η
to induce fission ηa (also called
)
σf
σc
235U
584 96
238U
-
σa
σs
680
10
2.72 2.72 8.3
σf
ηa = η
= 2.06
σ f +σc
Suppose that we have natural uranium which contains 0.72% 235U
⎡
⎤
0.72 × σ f ( 235U )
ηa = η ⎢
⎥ = 1.328
235
238
⎢⎣ 0.72 × σ a ( U ) + 99.28 × σ c ( U ) ⎥⎦
We can vary ηa by changing the enrichment (proportion of 235U in natural uranium)
Note: an enrichment of 1.6% leads to ηa = 1.654
Nuclear Fission
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Reactor kinetics
!   We introduce the neutron reproduction (or multiplication) factor k.
!   This is the ratio in number of neutron from one generation to the next.
!   Neutrons are characterised by a time constant τ, mean time before absorption occurs
!   Includes the time necessary to moderate the neutron (10-6 s)
!   Includes a diffusion time at thermal energies before absorption (10-3 s)
If there are N neutrons at time t, then there will be on average kN neutrons at
time t+τ, k2N at time t+2τ and so on.
Then in a short time interval dt, the increase will be:
dt
dN = (kN " N)
#
Giving
Nuclear Fission
dN(t)
dt
= (k "1)
N(t)
#!
$
N(t) = N 0e
t
(k "1)
#
21
Energy rate
The rate of energy released during fission will be:
Q value per fission x number of absorbed neutrons leading to fission in time dt
% (k $1)t (
N(t)
dE = Q "
dt = QN 0 exp'
*dt
& # )
#
The integration gives
making
!
!
We then have 3 cases
Nuclear Fission
$ (k #1)t '
N 0"
E =Q
exp&
) + C st
% " (
(k #1)
% (k #1)t (
E " exp'
*
& $ )
!   k<1 the number of neutrons decreases with time. The energy
produced also decreases. Sub-critical: the chain reaction will stop
after a while
!   k~1 The number of neutron!remains constant, so is the produced
energy. Critical: the chain reaction will be steadily sustained.
!   k>1 The number of neutrons and the energy will increase
exponentially with time. Super-critical: we get a bomb
22
How to keep k~1 ?
!   An ideal reactor would have k=ηa
!   We have seen that each fission gives an average of ηa fast neutrons
!   However other processes impact the neutron production
!   Before moderation, a small fraction of fast neutrons will create
fission with 238U (even with a small cross section ~ 1b). We have
εηa with ε ~ 1.03 (fast fission factor)
!   During moderation the neutrons slowing down will encounter some
resonances in the 238U capture cross section. We now have εpηa
with p ~ 0.75 (resonance escape probability)
!   There is a probability of neutron capture by the moderator  εpfηa
with f ~ 0.83 (thermal utilisation factor)
!   There will be neutron leakage in the reactor ~ 3%
All these factors will depend on the reactor design. These are typical values
k
ηa =
≈ 1.656 if we want k ~ 1
0.97εpf
Nuclear Fission
This can be achieved with an uranium
enrichment of 1.5% typically
23
Moderation of fast neutrons
!   Neutrons need to be slowed down as rapidly as possible
!   Neutrons will be slowed down through successive scattering with moderator nuclei
Let s have a look at the kinetic energy loss of one neutron after one scattering
Consider fast neutrons (mass mn) with a speed v.
Originally the moderator is static (nucleus mass M).
After scattering neutrons will have a speed v1, the moderator nuclei a speed v2
2
Suppose that the neutron is scattered through 180deg
#
&
2
mn 2
m
2
n
v
"
v
=
v
+
v
%
(
(
)
(
1)
1
$M'
M
Conservation of
Conservation of energy
momentum
# mn &
2
1
1
1
(v " v1)(v + v1) = %$ M ('(v + v1)
mn v = "mn v1 + Mv 2
mn v 2 = mn v12 + Mv 22
2
2
2
+
!
v1 ( mn + M ) = v( M " mn )
2
" mn %
2
2
m
v
=
v
+
v
$
'
( 1)
2
v1 ( M " mn )
v 22 = n ( v 2 " v12 )
#M&
=
!
M
!
v (mn + M )
!
For the neutron:
When M>>mn
neutron does not lose much energy
!
Moderator should have a small atomic mass number
Nuclear Fission
KE
! after
KE before
1
mn v12 # M " m & 2
n
=2
=%
(
1
m
+
M
'
mn v 2 $ n
24
2
Summary
!   Fission reaction key points:
!   Why/how can we have a fission reaction?
•  Explanation from B/A curve
•  Activation energy
•  Products
!   Energy produced  How to calculate Q
!   Neutron production
!   How to trigger efficiently a fusion reaction
!   What is a cross-section?
!   What are the mean free path and mean free time associated to cross-section
•  See example 6
!   Take into account the various cross-sections
!   Neutron moderation
!   Reactor kinetics - why and how to keep k~1
Nuclear Fission
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References
Ref 2: Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006)
Ref 3: Kenneth S. Krane, Introductory Nuclear Physics (Wiley 1988)
Nuclear Fission
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