Basic Physics of Nuclear Reactors
Prof. Paddy Regan,
Dept. of Physics
University of Surrey
Guildford, UK
p.regan@surrey.ac.uk
&
Radioactivity Group
National Physical Laboratory
Teddington, TW11 0LW
paddy.regan@surrey.ac.uk
Feb, 2015

Several components are important for a controlled nuclear
reactor:
 Fissionable fuel
 Moderator to slow down neutrons
 Control rods for safety and reactor criticality control
 Reflector to surround moderator and fuel in order to contain
neutrons and thereby improve efficiency
 Reactor vessel and radiation shield
 Energy transfer systems if commercial power is desired

Two main effects can “poison” reactors:
(1) neutrons may be absorbed without producing fission
[for example, by neutron radiative capture];
(2) neutrons may escape from the fuel zone.
Outline
•
1: Basic Nuclear Energetics
– Nuclear masses; binding energy; binding energy per nucleon; nucleon
separation energies.
•
2: Neutron Interactions
– Definition of cross-section; elastic collisions and neutron moderation;
neutron attenuation; neutron mean free path; definition of neutron flux;
definitions of neutron mean time/collision probability per unit time; reactor
power; elastic and inelastic scattering.
•
3: Basics of Nuclear Fission
– Fissile materials 235,238U; energetics of fission; beta-delayed neutrons; oddeven effects; cross-sections for neutron interactions on 235,238U; Fission
fragment mass distributions; neutrons per fission; fission neutron energy
spectra; moderator materials and kinetics.
•
4: Criticality
– Concept of the chain reaction; 4 and 6 factor formulae; Definition of
criticality and reactivity; reasons for Uranium enrichment; moderator
materials; control rod materials; neutron cycle; delayed neutrons; Simple
criticality calculations; reactor poisons; fast fission
•
5: Breeder Reactors
– Fissile and Fertile Materials; Formation of 239Pu; Use of 232Th; Breeding
ratio.
Useful web-sites for more notes, examples etc. etc:
http://canteach.candu.org/aecl.html
http://www.neutron.kth.se/courses/reactor_physics/lecturenotes.shtml
http://www.if.uidaho.edu/~gunner/Nuclear/LectureNotes/Lecture18_Reactor_Physics.pdf
http://physics.ukzn.ac.za/~moodleym/ReactorTheory/
http://www.atomeromu.hu/mukodes/lancreakcio-e.htm
Part 1: Basic Nuclear Energetics
Nomenclature
136
54
Xe
=
82
Ionic Charge
Chemical symbol
Z
N = A-Z
A
Where A = Mass Number,
Z = Proton Number
N = Neutron Number = A – Z
Ionic Charge = charge on the ion. If neutral it is left blank
Reactions :-
136
54
Xe
( d,p )
82
137
54
Xe
83
2
Where d = H
represents the deuteron and p the proton
1 1
Mass and Energy
 From Special Relativity:-
m = m0
 1-(v/c)2
when v = 0 then E = m0c 2
For an electron E0 = m0c2
= (9.11 x 10-31)(3 x 108)2
= 8.2 x 10-14 J
= 0.511 MeV
For 1 unit of atomic mass = 1/12 of the mass of a carbon atom
= 1.66 x 10-27 kg
= 931.5 MeV
 Matter can be converted into energy and energy can be converted into matter.
Super Heavies
Fewer than 300 nuclei
Proton Drip Line
Neutron Drip Line
 We can find some 283 stable or long-lived nuclei on the Earth’s surface.
 We define the proton (Sp = 0) and neutron (Sn = 0) drip-lines?. About 6-7000 species
live long enough to be studied in principle.
Binding Energy
 Z protons and N = A-Z neutrons can be brought together to form a nucleus which
has a total mass M(A,Z) which is LESS THAN THE SUM OF ZmP + (A-Z)mN
Some of the mass has been “used up” in binding the particles together.
This difference in mass is equal to B or energy B/c2 is the Binding Energy.
 Variations in atomic masses due to variations in binding energy are very small
compared with u  931 MeV
The atomic mass is often given in the form (M(A,Z) – A) called the Mass Excess
The Binding Energy increases with mass and proton number. This is reflected in
the Chart of the Nuclides, the plot of the stable nuclides as a function of N and Z.
Separation Energies.
 If we want to remove one neutron or proton from the nucleus it “costs” energy.
 This leads to the definition of
the neutron separation energy SN = B(A,Z) –B(A-1,Z)
= [ M(A-1,Z) –M(A,Z) + MN]c2
Similarly the proton separation energy is defined as
SP =B(A,Z) –B(A-1,Z-1)
= [ M(A-1,Z-1) –M(A,Z) + MH]c2
where we are working with atomic masses.
These quantities are analogous to the Ionisation energies in atoms. Effectively
they give us the amount of energy required to remove the last neutron or proton,
i.e. how well bound the last particle is.
 Plotted as a function of A they reveal evidence of Shell structure.
The variation of Binding Energy per nucleon with A
Binding Energy
8 MeV/A
Slope=
8 MeV/A
A
B (Z, A) 8 ´ A MeV
 binding is via a short range force
Deviations from the “constant” 8 MeV/A  fission and fusion can release energy
A = 56
Nuclear binding
Mnucleus < Sum of the constituent nucleon masses.
B(Z,A)/c2 = ZMp + NMn – M(Z,A) ( > 0)
Coefficients (values obtained from fits to experimental data)
av = 15.56 MeV ; as = 17.23 MeV ; ac = 0.697 MeV ; aA = 23.285 MeV ; aP = 12.0 MeV

In fission a nucleus separates into two fission fragments. One fragment is
typically somewhat larger than the other.

Fission occurs for heavy nuclei because of the increased Coulomb forces
between the protons.

We can understand fission by using the semi-empirical mass formula based on
the liquid drop model.
 For a spherical nucleus of with mass number A ~ 240, the attractive shortrange nuclear forces offset the Coulomb repulsive term.
 As a nucleus becomes nonspherical, the surface energy is increased, and
the effect of the short-range nuclear interactions is reduced.

Nucleons on the surface are not surrounded by other nucleons, and the
unsaturated nuclear force reduces the overall nuclear attraction. For a certain
deformation, a critical energy is reached, and the fission barrier is overcome.
EXAMPLE:
SOLUTION:
Part 2: Neutron Interactions
Examples of Neutron-Induced Reactions:
Standard unit for cross-section is the ‘barn’ where 1 barn = 10-28m2 = 10-24cm2
8
(see later re criticality and
discussion on effects of
delayed neutrons on
reactor control).
Reactor Power
The Reactor Power (= energy released per unit time in the reactor volume)
can be given by the product of the
Energy per fission reaction x the fission reaction rate per unit volume x reactor volume
Note that since the (fission) reaction rate per unit volume ( = Rf )
is the equal to the product of the collision probability per unit path length times the
neutron flux (in neutrons/ per unit area per unit time), then we can write
(recall s.N=S)
Power (in units of J/s = Watts) = [Joules ] x [ 1/length] x [1/length2 . time] x [length3]
Example:
A research reactor has a cubic shape, with a typical neutron flux of
1013 n/cm2s through its volume and sides of length 0.8m.
If the probability of fission per unit length, Sf=0.1cm-1, what is the
power of the reactor?
Solution
Assuming typical energy release per fission = 200 MeV
(i.e., neutron capture followed by gamma
de-excitation of the compound nucleus to
Its ground state; or followed by fission.)
i.e., neutron can interact either by
scattering (both elastically and inelastically) or can be
absorbed (leading to either gamma-emission; fission; p, a, & n emission)
The total neutron interaction cross-section (st) is the sum of all the individual components.
k
Question:
Answer:
Neutron Moderation
 There are many possible sources of neutrons.
In the list most of the neutrons produced had energies of 2-5 MeV.
There are many other reactions which can produce neutrons of much higher energy.
We cannot easily control their energies as we do charged particles.
However we can slow neutrons down to the energy we want.
This is done in collisions with atoms of various materials. This process is
called “moderating” the neutron energy and the material used is called a moderator.
The resulting neutrons are classified in broad energy ranges
Thermal
Epithermal
Slow
Fast
E  0.025eV
E ~ 1 eV
E ~ 1 keV
E ~ 100 keV – 10 MeV
It turns out that moderation of neutron energies is a very important part of reactor
design and operation.
Why do we need to moderate the neutrons (slow them down?)
The fission cross-section increases dramatically for thermal (~0.025 eV)
neutrons on 235U than for higher energies (1/v dependence).
Neutron Scattering / Moderation
A
m = neutron
mass
u = initial neutron
velocity in lab
frame.
O
v1
v’1 = neutron velocity in centre of
momentum frame
q
vcom
B
v’2 = moderator nucleus velocity in
centre of momentum frame.
M = moderator nucleus
mass
v = 0 = initial moderator
velocity in lab frame.
v2
Substituting into the equation for kinetic energy gives
Substituting into the cosine rule equation,
M = A = mass of moderator nucleus (in AMU) ;
m = 1 = mass of neutron in AMU.
Initial kinetic energy of the neutron is T0=1/2 mu2 ;
Scattered neutron kinetic energy is T1=1/2mv21
For a 1H moderator (A=1)
For a 12C moderator (A=12)
Example:
Head-on Collision
mu = -mv + MV
mu2 = mv2 + MV2
Momentum conservation
Energy conservation
mv = MV – mu
Therefore mv2 = ( MV – mu )2
m
Substitute
mu2 = ( MV – mu )2 +MV2
m
Expand and rearrange and we get
2mMu = ( Mm + M2 )V
Therefore V = 2mu
(M + m)
Neutron Interactions
Elastic scattering:- effectively a “billiard ball” collision in which there is no internal
excitation of the nucleus. Energy and momentum conserved in the collision.
Inelastic scattering:- neutron is scattered but nucleus is excited internally in the
collision.
Neutron capture:- neutron is absorbed to form a compound nucleus. The excited
state formed has no memory of its formation other than very general
conservation laws.
AX
N
Primary γ-rays
SN
Secondary
γ-rays
If target is stable nucleus
A+1X
N+1
formed is radioactive in
majority of cases.
Neutron Interactions
At low neutron energies neutron capture dominates.
 At higher energies other reactions such as (n,p), (n,2n) or (n,α) etc become more
important.
In all of these reactions you should remember that quantities such as momentum,
energy, angular momentum etc are conserved.
In the decay of the capture state in neutron capture the primary gamma –rays
populate states with J determined by selection rules in gamma decay. As a result
most of the states populated directly have spins close to that of the capture state.
The states populated by primary gamma rays are populated in a non-selective way.
their intensities are distributed in a statistical way.
In general the nuclei created in neutron capture are radioactive. Since they are on
the neutron –rich side of stability they decay to the nucleus with Z + 1. Note:-This Is
important in discussing reactors since it is the reason that heavier nuclei are bred
during reactor operation.
Part 3: Basics of Nuclear Fission
Discovery of Fission
Discovery of the neutron by Chadwick (1932 )
In following years Fermi and his co-workers bombarded many elements with
neutrons and studied the induced radioactivity.
They discovered many of the nuclei produced in neutron capture decay by βemission. In other words the excess neutron is converted to a proton.
This gave them the idea that they could make transuranic elements, not found in
in Nature, by n capture followed by beta decay.
They bombarded Uranium with neutrons and found new activities but when they
tried to separate these activities chemically they showed a chemical behaviour
similar to Ba (Barium)
In 1939 Hahn and Strassmann showed, using careful radiochemical techniques,
that the activity was due to isotopes of Ba (Z = 56)
With this information Meitner and Frisch (1939) proposed that this was due to the
neutron inducing the heavy nucleus to fission – to split into two heavy fragments.
[Note:-the name is borrowed from the description of cell division in biology]
Why does Fission occur?
If we start with a spherical nucleus and it starts
to vibrate, either naturally or because it absorbs a
nucleus then the shape is distorted and the surface
area term increases and Coulomb term decreases.
250
Energy MeV
200
B = aVA – aSA2/3 – aCZ(Z-1) – aA(A – 2Z) +/- 
A1/3
A
Separation distance
This means that as we stretch the nucleus the total
binding energy may decrease.
Once we go past the maximum point in
this curve energy is gained by continuing
and so we proceed to fission.
EB
EF
EF – energy released in fission
EB – energy of barrier
EA – Activation energy
Why does Fission occur?
 Consider the case of 238U splitting into two equal fragments
each being 119Pd with Z = 46.
R1 = R2 = 1.25(119)1/3 = 6.1F (From R = R oA1/3 )
V = 1 . Z1Z2e2 = 250 MeV
4εo
- Coulomb Barrier
R
From the S.E.M.F, energy released in this case is 214 MeV
The fission process is hindered by the barrier. Inside the barrier the two Pd nuclei
can exist but the barrier inhibits their release.
Note:- This is too crude – such estimates can easily be modified . For example if we
split the nucleus into masses 79 and 159 then the Coulomb Barrier falls to 221 MeV.
Again releasing a few neutrons means mass numbers are nearer stability and hence
more energy is released. Also the Coulomb barrier calculation assumes a sharp edge
to the nucleus which is unrealistic.
Main conclusion – Height of Coulomb Barrier  Energy released in fission.
Why does Fission occur?
The Semi-empirical mass formula
B = aVA – aSA2/3 – aCZ(Z-1) – aA(A – 2Z) +/- 
A1/3
A
The primary cause of fission is competition between the nuclear and Coulomb forces.
As nuclei get larger/heavier the total nuclear binding energy increases with A and
the Coulomb repulsion increases faster as Z2.
If now we think of the process of fission as being the emission of a heavy fragment
like the emission of an alpha particle then we can regard the heavy nuclei as sitting
near the top of the nuclear potential well.
The Coulomb Barrier is very thin here and so can be
penetrated easily. Fission can thus occur spontaneously
or it can be induced through the absorption of a relatively
low energy neutron or photon.
Energy MeV
250
200
Separation distance
Any nucleus can fission if it is given enough excitation energy.
In terms of practical applications it is only important for heavy
nuclei
Why does Fission occur?
Main conclusion – Height of Coulomb Barrier  Energy released in fission.
For some nuclei the energy release places them just below the barrier and they will
undergo Spontaneous fission since they have a reasonable chance of tunnelling
through the barrier.
Others may exist where the barrier would be zero. They would fission instantaneously.
Naturally we do not find such nuclei in Nature. The barrier would be
zero around mass 300.
Nuclei may also exist which are so far below the barrier that spontaneous fission
does not occur. However extra energy brought in by a photon or neutron may form
an intermediate state at or above the barrier so that induced fission occurs readily.
Note:-If the intermediate state is below the barrier still, then fission is inhibited and
other decay modes may dominate.
Depending on the energy of the intermediate state absorption of thermal
neutrons will push them over the barrier whereas for others several MeV of
energy may be required.
Energy versus Separation as a function of mass.
3) At A =250 one should get spontaneous fission
2) At A = 235 it requires an input of energy
1) At A = 100 it would need a lot of energy.
The Energy released in Fission
Roughly speaking we can see from the binding energy curve below that we gain
about 0.9 MeV per nucleon in the fission of Uranium. That is  200 MeV in total.
If we take a specific pair of fragments with masses 93 and 141 the Q-value
is 181 MeV. Other combinations have different Q-values but overall the assumption
of 200 MeV released is a reasonable one.
0.9 MeV
The two fragments are so large
compared to everything else
released that we can assume
that they are emitted
back-to-back to conserve
momentum
Fission in terms of the Liquid Drop model.
Basic idea:- We stretch an initially spherical nucleus. What is the effect on the B.E?
Assume stretched nucleus is an ellipsoid of revolution with volume = 4/3.ab2,
where a is the semi-major and b the semi-minor axis.
We introduce a distortion parameter ε and the distortion can be written
a = R( 1 + ε )
b = R( 1 + ε )-1/2
Note:- R3 = ab2 which means volume is constant as we stretch the spheroid.
As we stretch the spheroid the surface area S = 4R2 ( 1 + 2/5. ε2 + ---) increases.
At the same time the Coulomb energy term in the S.E.M.F is modified by the factor
(( 1 - 1/5. ε2 + ---))
Thus the decrease in the binding energy between spheroid and ellipsoid is given by
the S.E.M.F as
E = -aSA2/3 ( 1 + 2/5. ε2 + ---) – aCZ2A-1/3 ( 1 - 1/5. ε2 + ---)
+ aSA2/3 + aCZ2A-1/3
b
= ( -2/5 aSA2/3 + 1/5 aCZ2A-1/3)ε2
a
Fission in terms of the Liquid Drop model.
If second term is larger then energy difference is positive – we gain energy by
stretching the nucleus. The more it stretches the larger the gain. If so the system
is unstable against stretching and will fission.
Condition for this spontaneous fission to occur is
1/5 aCZ2A-1/3 > 2/5 aSA2/3
If we put in the values for the coefficients we find
Z2
> 47
A
 BUT remember….
a) Quantum mechanical tunnelling modifies this expression
b) Nuclei around Uranium have a permanent (‘quadrupole’) deformation.
c) As this parameter increases the spontaneous fission half life decreases
and when it is 10-20 seconds we find Z2/A  47
Note:- no such nuclei are known in Nature.
Lifetimes for Spontaneous fission
Expect the half-life for spontaneous fission
gets rapidly shorter with increasing Z2/A.
Remember this is not the actual half-life of the
radionuclei (just a ‘partial half-life) since they
can also decay by alpha or beta emission.
λT = λF + λAlpha + λBeta
V.M.Strutinsky and H.C.Pauli, Physics and Chemistry of Fission(IAEA 1969) p155
Comparison – Liquid Drop and Observed Activation Energies
Fission Cross-sections
Here we see the cross-sections for neutron
-induced fission and neutron capture on
235U and 238U
Thermal
In the thermal (low energy) part of the
spectrum the cross-section falls with energy
as 1/ v.
For 235U the cross-section for fission =
584 b, much larger than for scattering 9 b
or for capture 97 b. Overall the cross-section
falls by 3 orders-of-magnitude from thermal
to fast energies.
For 238U no fission occurs until the neutron
reaches an energy of  1 MeV.
Differences between Even and Odd Nuclei
Previous slide showed there is a significant difference in behaviour for 235 and 238.
Why?
235U + thermal neutron
E (excitation energy) = [m( 236U*) – m( 236Ugs)] assuming EN = 0
Now m( 236U*) = m(235U) + mN = 236.052589 u
So E (excitation energy) = (236.052589 u – 236.045563 u)931.502 MeV/u
= 6.5 MeV
The Activation Energy ~ 6.2 MeV so EN = 0 induces fission
 The same calculation for 238U gives E(excitation energy) = 4.8 MeV <<6.6MeV
hence it requires a neutron with ~ 1 MeV to induce fission.
The primary reason for this difference lies in the pairing effect in nuclei.
Why does Fission occur?
Recall the Semi-Empirical Mass Formula
B = aVA – aSA2/3 – aCZ(Z-1) – aA(A – 2Z) +/- 
A1/3
A
Volume Energy
Surface energy
Coulomb energy
Asymmetry energy
Pairing energy
It turns out that pairing plays a significant role in determining which nuclei are
suitable as fuel in a nuclear reactor. It arises because like nucleons prefer to form
spin zero pairs in the same spatial state. The extra binding then comes from the
strong spatial overlap of their spatial wave functions.
 is +ve if N and Z are even and all nucleons are coupled to zero spin.
 is –ve if N and Z are both odd.
 is zero if N or Z is odd i.e. A is odd.
Note that all the terms vary smoothly except the pairing term. A common value for
the pairing term is  = 12/A1/2 MeV [ 0.78 MeV for A = 235]
Differences between Even and Odd Nuclei
The difference in the cross-section lies in difference between their excitation
energies. Why are they different?

236U*
=
235U
EEX
+n
239U*
=
238U
δ
δ
EEX
239U
+n
The binding energy of 236U is increased
i.e. the g.s energy is lowered by an
amount δ ( ~ 0.56 MeV ). The excitation
energy is correspondingly increased by
Δ over what it would be in the absence
of pairing.
δ
235U
238U
Overall the difference in excitation energy
between the two cases is  2δ or 1.1 MeV
which accounts for the different behaviour.
In the case of 238U the energy of the g.s.
before capture is lowered by δ, which,
in turn, lowers the energy of the capture
State. The excitation energy is therefore
reduced by δ relative to its value without
the pairing effect.
The Mass Distribution of Fission frgaments
A typical result of thermal neutron induced fission in 235U leads to the result
93Rb + 141Cs + 2n
n + 235U
This is not a unique split and there is a wide distribution of masses for the two
resulting fission fragments. Since every heavy fragment must have a corresp.
light fragment the distribution is symmetric about an equal split.
 Fission into almost equal mass
fragments is less probable than
the maximum yield by  600.
 Fission induced by fast particles
shows a different mass
distribution, with more yield
around A/2.
Typical Emitted Neutron Energies in Fission
See B.E. Watt Phys. Rev. 87 (1952) 1037
i.e., typically 2.5 neutrons per fission event (thermal or fast) on 235U;
higher (~2.9 neutrons per fission) for 239Pu at similar average neutron energies.
39
140
Plot of % of total events as a function of kinetic energy of the fragment
The Energy Released in Fission
This distribution in kinetic energy is consistent with the mass distribution observed in
fission.
Overall the two fragments carry about 80% of the total energy released.
Where does the rest go?
Decay
Secondary Fission Fragments
These secondary fragments are radioactive and represent the decay of the primary
Fragments back to the line of stability. They produce  7% of the power when the
Reactor is in operation and the decays continue after shutdown.
Delayed Neutron Emission
I n addition to the prompt neutrons emitted in fission there are also delayed neutrons
associated with beta decay.
AX
N
SN
AX
N-1
Z
Z+1
AX
N-2
Z+1
For neutron-rich nuclei the most
general decay mode is β- - decay
but in some cases if the neutron
separation energy is low enough it
may populate levels above SN.
Such levels will then emit a neutron
leading to a level in the nucleus
which is an isotope with 1 neutron
fewer.
We see below an example where the
primary fragment is 87Br, which beta decays
to levels in 87Kr, some of which are above
the neutron separation energy. These levels
decay by emitting a neutron.
Neutron Emission in Fission
 Consider again the common split in thermal fission – 95Rb + 140Cs
Here the two fragments share 92 protons. These two fragments are rich in neutrons
and they shed a few neutrons at the instant of fission ( within 10-16 s ) to get rid of this
excess. These are known as prompt neutrons.
The average no. of prompt neutrons = n is characteristic of a given fission process.
[Greek nu]
We find that for thermal neutron induced fission:Nucleus
n
233U
2.48
235U
2.42
239Pu
2.86
The distribution about the mean shows the
statistical behaviour expected for an
evaporation process.
Fission Cross-sections
Here we see the cross-sections for neutron
-induced fission and neutron capture on
235U and 238U
Thermal
In the thermal (low energy) part of the
spectrum the cross-section falls with energy
as 1/ v (v = velocity of the neutron).
For 235U the cross-section for fission =
584 b, much larger than for scattering 9 b
or for capture 97 b. Overall the cross-section
falls by 3 orders-of-magnitude from thermal
to fast energies.
For 238U no fission occurs until the neutron
reaches an energy of  1 MeV.
Part 4: Criticality & Chain Reactions
The Idea of a Chain Reaction
Each neutron inducing fission results in the production of several neutrons, typically
2.5 as we saw earlier. Each of these neutrons is capable of initiating fission in
another nucleus with the emission of another 2.5 neutrons on average.
Clearly if this continues the number of fissions and neutrons will increase very rapidly.
This process is described as a chain reaction.
The chain reaction is characterised by the neutron multiplication factor k, which
is defined as the ratio of the number of neutrons in one generation to the number
in the preceding generation.
If k < 1 then the number of neutrons decreases with time and the process stops.
In the context of a reactor it would be said to be sub-critical.
If k > 1 then the number of neutrons increases with time and the chain reaction
diverges. A reactor would be said to be super-critical. Effectively we have a bomb.
If k = 1everything proceeds at a steady rate. A reactor in this state would be said to
be critical.
The Idea of a Chain Reaction
If we try to use natural Uranium ( now 0.7% 235U ) we cannot get k above 1
because the dominant 238U absorbs too many neutrons.
As we can see from the cross-section
curve as a function of energy most of
the neutrons end up by being
captured.
However if we can enrich the sample highly in 235U then we can get k > 1 and the
chain reaction can be sustained. This is the basis of the fast reactor, namely a
reactor that operates with fission being initiated by neutrons that have not been
slowed down much following their production. If the enrichment is high enough it is
also the basis of an explosive weapon.
The Idea of a Chain Reaction
However looking at the 235U cross-section we see that it is three orders-of-magnitude
higher at thermal neutron energies than
at 1 MeV. If we can efficiently moderate
the neutron energies down to thermal
values then we can arrange that k > 1
for natural Uranium.
Thus it is possible to make a reactor
using natural Uranium but it demands
efficient moderation to ensure that the
neutrons have thermal energies.
Such a reactor is referred to as a thermal reactor.
The Oklo reactor is interesting in itself but it is also highly relevant to the discussion
of dealing with present day waste. Neither the fission fragments nor the Pu migrated
from the site in 2 x 109 y.
A natural Reactor in Oklo, Gabon
A. The power generation industry has meant that very careful measurements have
been made of the fraction of 235U in natural Uranium from all the deposits we have
found. It is found to be very precisely
0.00720 +/- 0.00001
with the error reflecting not only the measurement error but also the variation
found in samples from all the places where uranium is mined.
B. This is, of course, a ratio which will vary on a geological timescale. The mass 235
and 238 isotopes have half-lives of 7.0 x 108 y and 4.5 x 109 y resp.
Note:- 2 x 109 y ago the ratio would have been ~ 3%
C. 1972 – a sample was found in Gabon with an abundance of 0.00717. Further
samples from the same place were as low as 0.00440
D. Hypothesis was that 2 x 109 y ago a natural reactor operated in Oklo.
Moderation would have been by groundwater. Before this could happen it needed
uranyl ions to be transported in the ground water. Prior to 2 x 109 y ago there was
not enough oxygen in the water to form these ions.
E. The reactor would have worked intermittently with the heat boiling off the water and
it would then start again when enough water accumulated.
The Four Factor Formula
 A nuclear reactor is simply a controlled chain reaction – a system
arranged so that it is stable and produces constant power.
Initially we consider an infinitely large system.
- In effect we ignore the loss of neutrons from the surface.
We define a Neutron Reproduction Factor k∞
k∞ =
No.of neutrons in a generation
No.of neutrons in preceding generation
In other words it gives the net change in the number of thermal neutrons from
one generation to the next.
On average each thermal neutron produces k∞ new thermal neutrons
If a chain reaction is to continue we must have k∞ ≥ 1
However although 2.5 neutrons are produced on average in each fission they
are fast neutrons.
The Four Factor Formula
Why are fast neutrons ”bad” in this context?
The answer lies in the cross-section curve
we saw earlier. The cross-section for fission
is three orders-of-magnitude larger for
thermal neutrons compared with the fast
neutrons created in fission. It increases from
a few barns to 580 b
It is essential to moderate the initial neutron energy so that the average energy drops
from a few MeV to 0.025eV. Of course there is a price to pay. In the moderation
process we will lose neutrons either by absorption or by some other process.
We already learned how to moderate neutron energies. The energy is most
easily lost in elastic collisions with nuclei. The best choice for a moderating material
is a light element because then the neutron transfers the largest possible energy in
an elastic collision with a light nucleus.
The Four Factor Formula
In terms of energy loss alone hydrogen is clearly the best moderator but there are
other considerations.
- It must be light
- best if it is a solid both in terms of constructing a reactor and the high density
of atoms.
- It should be inexpensive
- It should be easy to handle
- It should have a small capture cross-section
Carbon in the form of graphite fulfils all of these criteria.
The simplest way to make a reactor or
chain-reacting pile as it was called is to make
a lattice of alternating blocks of Uranium and
graphite
The neutrons are produced in the U and then
when they enter the graphite their energies
are moderated.
The picture shows Enrico Fermi and the first
pile built in a squash court at the University
of Chicago in 1942.
The Four Factor Formula
If we are to have a self-sustaining reaction in the pile the reproduction factor
k must equal 1. In other words to get a steady stable flow of energy the pile must
be critical.
 Still assuming the pile is infinitely large we can calculate k by following what
happens to neutrons as they go from one generation to the next.
We begin with N thermal neutrons
Each of them should produce n  2.5 fast neutrons per fission.
However not all of them induce fission. In particular some are captured.
We define  = Mean Number of fission neutrons produced per initial thermal neutron.
Note  < n since some of the original neutrons do not cause fission.
If sf = fission cross-section for thermal neutrons
sa = cross-section for other absorptive processes
then  = n
sf
sf + sa
The Four Factor Formula
 For 235U sf = 584 b and sa = 97 b so  = 2.08 fast neutrons per thermal neutron
 For 238U sf = 0 for thermal neutrons and sa = 2.75 b
 Natural U contains 0.72% 235U
For natural Uranium we have effective cross – sections for fission and capture
sf =
0.72
99.28
sf (235) +
sf (238) = 4.20 b
100
100
sa =
0.72
99.28
sa (235) +
s (238) = 3.43 b
100
100 a
So the effective value of  for natural U is 1.33
This number is rather close to 1 so we have to do something to ensure we have a
critical system.
One “simple” thing we can do is enrich the U to 3% in 235U
This increases  to 1.84
The Four Factor Formula
Moderation:- Of our original N neutrons some have been absorbed and some
cause fission so we now have N fast neutrons which we must arrange to lose
energy and become thermal neutrons-in order to take advantage of the high thermal
cross-section.
This is achieved in the collisions with nuclei in the moderator. However in the
course of their random walk through the pile they may encounter 238U nuclei
which have a small cross-section ( ~1 b ) for fission by fast neutrons.This means
they are “lost” to the overall process and we have to introduce another factor ε the fast fission factor which takes account of this effect.
The number of fast neutrons is now Nε
Typically ε has a value of about 1.03 for natural Uranium.
The Four Factor Formula
Nucleus
H
2H
4He
12C
238U
No.of collisions to reach thermalisation
18
25
43
110
2200
If the moderator is graphite then as we can see from the table it takes about
100 collisions on average before an MeV neutron is thermalised.
In going from MeV to 0.025 eV energy the neutron
will pass through a range where its
energy is between 10 and 100 eV. In this energy
range 238U has many resonances with large
cross-sections ( ~ 1000 b) for neutron capture.
The Four Factor Formula
Unless we can ensure that the resonances
are “avoided” we will lose all our potential
thermal neutrons.
For example if we have a fine powder of U
and graphite then any given neutron will only
scatter a few times before encountering 238U.
Then a significant fraction will be captured in
one of the resonances.
How do we get round this?
The answer is to make the lumps of graphite larger so that the neutrons can be
completely thermalised whilst in the graphite. This would then avoid the resonance
region.
The average distance in graphite to achieve thermalisation is 19 cm. So we construct
the pile of U fuel elements separated by ~ 19 cm of graphite.
This does not eliminate the problem completely. Some neutrons stray near the
graphite surface and escape into U and are lost. We account for this by introducing the
the resonance escape probability p.
The Four Factor Formula
Now the number of neutrons we have left after thermalisation is
Nεp
A typical value of p is about 0.9
It is worth noting that the size of the graphite must not be too large. Once thermalised
we want the neutrons to be in the U to induce fission. One of the reasons we chose
graphite was the low capture cross-section but the neutron encounters a lot of it and
there will be some losses to capture in this and other reactor components. To take
account of these losses we introduce the thermal utilisation factor f which gives
The fraction of thermal neutrons that are actually available to the U.
Typically f is about 0.9.
The Four Factor Formula
Now the number of neutrons we have left after thermalisation is
Nεp
A typical value of p is about 0.9
It is worth noting that the size of the graphite must not be too large. Once thermalised
we want the neutrons to be in the U to induce fission. One of the reasons we chose
graphite was the low capture cross-section but the neutron encounters a lot of it and
there will be some losses to capture in this and other reactor components. To take
account of these losses we introduce the thermal utilisation factor f which gives
The fraction of thermal neutrons that are actually available to the U.
Typically f is about 0.9.
The Four Factor Formula
Overall the number of neutrons which survive the moderation process is
Nεpf
It is whether this number is greater or smaller then the original number N that
Decides whether a reactor is critical or not.
The neutron reproduction factor is
k∞ = Nεpf/N
= εpf
This is called the
The Four Factor Formula
A real reactor.
What lessons are there for us in terms of designing a real reactor?
There are a number of factors we can control.
A. In particular the enrichment of the fuel.
B. Secondly the choice of moderator.
C. Thirdly the geometry of the pile.
In this third case there are three factors involved, namely ε, p and f
One has to consider the relative geometry of U and graphite. Neutrons of energy
10-100 eV do not travel far in U before absorption so resonant capture basically
occurs on the surface of the U. As far as the centre of a piece of U is concerned
this effect is of no real relevance. So the U has to be sufficiently large to
minimise this surface effect but not so large that all the thermal fission occurs
at the surface too and so the centre of the U fuel sees fewer neutrons.
A real reactor.
If we take reasonable values we can calculate k∞
For Natural Uranium  = 1.33
ε = 1.03
p = 0.9
f = 0.9
So
k∞ = 1.33 x 1.03 x 0.9 x 0.9
= 1.11
A real reactor.
 If we take reasonable values of the four factors εpf
Then we find k∞ = 1.11
This would be sufficient to sustain a chain reaction in a reactor.
However we have been discussing an infinitely large reactor. Real reactors are
finite in size.
All of the four factors we have discussed are still appropriate but now we have
to take into account the fact that we will lose neutrons from the surface. In a real
reactor there is leakage of fast and slow neutrons.
If we now introduce LF and LS as the fractions of fast and slow neutrons that are
lost from the reactor then the complete formula for the reproduction factor is
k = εpf( 1-LF )(1-LS )
This is called the
SIX FACTOR FORMULA
Example Question:
Comment?
A real reactor.
Clearly one way to minimise the leakage factors is to make the reactor as large as
possible. In this way we minimise the surface to volume ratio. If the two leakage
factors are kept small then
k∞ - k  k(LF + LS )
So increasing the volume will mean this difference decreases.
The longer a neutron travels in the moderator before it is absorbed the more
chance we have of losing it from the surface.
There are two contributions to this distance which is called the migration length - M.
For thermal neutrons, where they diffuse through the medium, we have L(D). This
diffusion length is the distance a thermal neutron can travel on average before
absorption. For fast neutrons we have a slowing down distance, the distance over
which it travels in slowing down to thermal energies, L(S).
M = (L(D)2 + L(S)2)1/2
For graphite L(S) = 18.7 cm and L(D) = 50.8 cm.
A real reactor.
From a dimensional analysis we can get some idea of how big our reactor made
of natural U and graphite has to be.
Let us take R as the characteristic dimension – this could be the radius if it is a
sphere or it could be the length of one side if it is a cube.
It is reasonable to assume that the difference in k for an infinite and a finite reactor
is proportional to the surface area. Thus
k∞ -k ~ M2/R2
The critical size when k = 1 is given by RC, where
RC ~
M
(k∞ - 1)1/2
The constant of proportionality is of order 1 but for a sphere
RC =
M
(k ∞ - 1)1/2
The result for a U-graphite reactor is R ~ 5m. In other words a sphere of U and
graphite of 5m radius would go critical. There are some measures one can take to
Reduce the size. For example by putting a neutron ‘reflector’ round the pile.
Moderating the Neutrons
Moderation is the process of the reduction of the initial (high) energy of the emitted
(fast) neutrons from fission to lower (ideally thermal) energies, where they have the
highest cross-section for causing fission (after capture on 235U).
In the case of elastic collision between a neutron (mass 1u) and a nucleus with mass
Au. For a head-on, elastic collision it can be shown from conservation of KE and linear
momentum that:
which leads to
More generally we need to take into account both head on and glancing elastic
collisions. In these cases, using 2-body kinematics, the mean logarthmic neutron
energy reduction (or decrement) per collision, x, can be defined. This can be
calculated using the expression:
x=ln (E0/E) = 1 + [ (A-1)2 / 2A ] . ln [ (A-1)/(A+1)] ~ 2 / A+1
From this, the number of collisions, n, between the neutron and a nucleus of mass A
are expected on average to reduce the neutron kinetic energy from E0 to E is given by
n = 1/x .(ln E0 – ln E)
Choice of Moderator Materials
• The ideal moderator materials have
– Low –A values (to maximise energy loss per collision for
elastic scattering)
– Small neutron absorption cross-sections (sa) to minimize
the neutron lost from the flux by absorption in the
moderator itself.
• We can define the MODERATING EFFICIENCY, which
incorporates both aspects and is given by x.(ss /sa) ,i.e.,
the product of the mean logarthmic neutron energy reduction
per collision and the ratio of the scattering and absorption
cross-sections for neutrons.
Controlling the Reactivity.
A. There is a delicate balance between neutron production and absorption. We can
sum it up by saying the power level is proportional to the number of neutrons
available in the reactor.
B. It is not possible to design a reactor in which the number of neutrons in successive
generations is exactly constant.
C. Instead we add extra fissile material to the fuel and use what are called “control
rods” to absorb some of the neutrons. These rods can be pushed in or taken out
to control the reactivity. Control rods are made of neutron absorbing materials such
as cadmium (with silver and indium), boron and hafnium.
D. We need the extra fissile material anyway since over a long period of operation
enough fissile material is destroyed that the reactor would turn off.
E. In addition operation produces many fission products that absorb neutrons and
this would also reduce k.
Can we actually do it?
Control Rods
We need control rods but can they be made to work?
Remember the timescales involved. If we had only prompt neutrons it would all
happen so quickly that it would not be possible (see later).
However the small fraction of delayed neutrons lengthens the timescale and this
alters the rate at which the neutron population changes giving us time to insert
or withdraw control rods.
Typical control uses rods loaded with elements such as cadmium or boron that
absorb neutrons. The rods can be moved in or out of the core to stabilise the
number of neutrons.
There may be “burnable poisons” as part of the core. Once they capture they
are neutralised.
Sometimes as a safety measure poisons are added to the cooling fluid. For
example in emergency shutdowns (SCRAM) the operators can inject solutions
containing poisons into the coolant e.g.sodium polyborate or gadolinium nitrate
Timescales
As in any physical process it takes time for the multiplication of neutrons. What is
the timescale?
A. We have not only to design so that k = 1 but also that we take account of any
changes with time. We associate a time τ with the process.
B. There are essentially two steps or stages in the process.
- slowing down to thermal energies
- diffusion of thermal neutrons before absorption.
- they have characteristic times 10-6 and 10-3 s
C. If we have N neutrons at time t then on average we have kN after t+ τ
and k2N after t + 2 τ etc
D. If we now consider a small time interval dt we can write
dN = (kN – N) dt
τ
So N(t) = N0 exp ((k-1)t/ τ)
Timescales
A. N(t) = N0 exp ((k-1)t/ τ)
B. If k = 1 then there is no variation with time and we would happily operate
our reactor.
C. Even if k deviates only slightly from 1 then the number of neutrons grows or decays
exponentially with a time constant τ/(k-1)
D.If we take the case of k = 1.01 then the reactor is super-critical and the time
constant is ~ τ/(k-1) ~0.001s/(1-1.01) = 0.1 s
Thus in 1 second the number of neutrons would grow by exp (10)  22,000 and the
reactor can not be controlled on this time scale.
This is a very rapid growth even for a small increase in k. In these circumstances it
would not be possible to control the reactor.
Fortunately Beta-Delayed neutrons come to the rescue.
Timescales
A. About a fraction d = 0.0065 of fission neutrons are from beta decay and are
delayed- in other words they follow the beta decay half lives. Because of this our
previous formula for the number of neutrons as a function of time must be modified.
B. We take q = (k – 1) and d is our fraction of delayed neutrons, τ is the time constant for
the prompt neutrons and τD is the time constant associated with the delayed
neutrons.
C. Now .
N(t) = N0
[
d
d-q
exp
qt
((d – q) τ )
D
q
-
d-q
exp (
-(d – q)t
ττ
)]
This equation reverts to our simple case with no delayed neutrons if d = 0
Timescales
A.
N(t) = N0
[
d
d-q
exp
qt
((d – q) τ )
D
q
-
exp
d-q
(
------------(1)
-(d – q)t
τ
) ]
Provided q = (k – 1) is always less than d then the rate of change of N(t) will be
Governed by the mean delay time τD which is of the order of 12.5 s.
B.
If q << d then we can approximate
N(t) ~ N0 ( qt/(d – q) τD ) --------------(2)
C. We must keep q = (k – 1) less than d or second term in (1) takes over and we get
rapid rise in neutrons.
D. If we take our graphite-U reactor with q = 0.001 then eqn.(2) gives  exp(t/70) so
increase is by a factor e = 2.7 in 70 secs (and the reactor is controllable).
Reactor Poisons
When a reactor is in operation it produces many fission fragments. They will all
absorb neutrons and hence decrease k. Those with large cross-sections are
referred to as reactor poisons.
135Cs
T1/2 = 9.14 h
135Xe
136Xe
135Xe
+n
136Xe
With σC = 2.75 x 106 b
T1/2 = 6.57 h
135I
After shutdown the amount of Xe continues to increase because the main part of
Ist production comes from the decay of I. The amount of compensation one can
supply with the control rods is limited (typically ~ 5%). As a result if the reactor
was stopped and not restarted quickly it would be impossible to restart for 1-2 days
until the Xe decays.
Reactor Poisons
Xe is the most significant poison but there are others.
149Sm
also has a large cross-section but it is stable. It reaches an equilibrium value
after about 500 hours and this essentially stays constant during the reactor
operation.
Numerous other fission products – individually they have a small effect but lumped
together they are significant.
In practice the buildup of poisons in the fuel leads to a loss of efficiency and
sometimes instability. It is this build up that limits the lifetime of the fuel. Long
before all possible fissions have taken place buildup of long-lived, neutron-absorbing
Fission products kills off the chain reaction process.
Reprocessing is essential to separate these activities chemically and since spent
fuel has 99% of the original fissile material which is in fresh fuel it allows it to be
used again.
The main elements of a reactor.
1. Fuel – pellets of UO2 (1cm diam. by 1.5 cm long) arranged in tubes to form fuel rods.
They are usually formed into fuel assemblies in the core.
2. Moderator – usually water but may be graphite or heavy water.
3.Control rods – Made with neutron absorbing material included so that inserting or
withdrawing the rod controls or halts the rate of reaction.
Note:-Secondary shutdown systems involve adding other absorbers of neutrons,
usually in the primary cooling system.
4.Coolant- Liquid or gas circulating in the core to carry away heat. In light water
reactors the coolant acts as moderator and coolant.
5.Pressure vessel – Usually a robust steel vessel containing the core and
moderator/coolant but it may be a series of tubes holding the fuel and
conveying the coolant through the moderator.
6.Steam generator – Part of cooling system where the reactor heat is used to make
steam to drive the turbines.
7.Containment –Structure round core to protect it from intrusion and protect the outside
from radiation in case of a major malfunction.
Part 5 : Breeder Reactors
Breeder Reactors






A more advanced kind of reactor is the breeder reactor, which
produces more fissionable fuel than it consumes.
The chain reaction is:
The plutonium is easily separated from uranium by chemical means.
Fast breeder reactors have been built that convert 238U to 239Pu. The
reactors are designed to use fast neutrons.
Breeder reactors hold the promise of providing an almost unlimited
supply of fissionable material.
One of the downsides of such reactors is that plutonium is highly
toxic, and there is concern about its use in unauthorised weapons
production.
The re-cycling of spent fuel
Normal reaction
235U
+ slow n  fission fragments + 2(fast)n
“Parasite” reaction
238U
+ fast n 
239U*

Pu/U fuel cycle
239Np
Process “used” fuel to extract
(or to make bomb!)
(neptunium) 
239Pu
239Pu
to be mixed with U
Other parasite reaction
232Th
+ fast n 
Th/U/Pu cycle
233Th*
(fissile)
 233Pa (protactinium) 
Lace natural Uranium with Thorium to extract
injected into natural Uranium.
233U
239Pu
(fissile)
and
233U
to be re-
Additional Slides for Reference
Making Fuel rods
Most reactors use enriched fuel- enriched in mass 235.
The yellowcake is converted to UF6 – a gas- which is enriched either by gas diffusion
or in a centrifuge.
The former relies on the different diffusion rates of masses 235 and 238.
In the latter the gas passes through spinning cylinders and the centrifugal force
causes the mass 238 move to the outside leaving a higher mass 235 concentration
on the inside.
Uranium dioxide pellets are then made form the enriched material.
The pellets are then encased in long metal tubes, usually made of zirconium alloy
(zircalloy) or stainless steel to form fuel rods. The rods are sealed and assembled
in clusters to form fuel assemblies for use in reactors.
Fuel Economy in Thermal and Fast Reactors
Note:- the much higher fraction of neutrons captured in mass 238 in the FBR
Energy Transfer
The
most common method is to pass hot water
heated by the reactor through some form of heat
exchanger.
In boiling water reactors (BWRs) the
moderating water turns into steam, which drives
a turbine producing electricity.
In pressurised water reactors (PWRs) the
moderating water is under high pressure and
circulates from the reactor to an external heat
exchanger where it produces steam, which
drives a turbine.
Boiling water reactors are inherently simpler
than pressurized water reactors. However, the
possibility that the steam driving the turbine may
become radioactive is greater with the BWR.
The two-step process of the PWR helps to
isolate the power generation system from
possible radioactive contamination.
Boiling Water Reactor
Pressurised Water Reactor
Years
Used Fuel
A. Used fuel emits radiation and heat.
B. It is unloaded into a storage pond adjacent to the reactor to allow it to decay.
C. It can be stored there for long periods. It can also be stored in dry stores cooled
by air.
D. Both kinds of store are intended to be temporary. It will be reprocessed or sent to
final disposal. The longer it is stored the easier it is to handle.
E. Main options for long term – reprocessing to recover useful fuel
- storage and final disposal
F. Reprocessing – separates U and Pu from waste products by chopping up rods
and dissolving them in acid to separate the various materials.
G. Typically used fuel is 95% 238U, 1% 238U, 1% Pu and 3% fission products including
other transuranics.
H. Reprocessing enables recycling of fuel and produces a significantly reduced waste
volume.
Where does the Uranium come from?
Uranium is relatively common – found in seawater and rocks.
Half the world’s production is in Canada and Australia in open pit or relatively
shallow mines
It is then milled – the ore is crushed to form a fine slurry and it is leached with
sulphuric acid to produce concentrated U3O8 – which is called yellowcake and
generally has more than 80% U compared with the original 0.1%
Underground mines cause less disturbance but one needs very good ventilation to
protect against airborne radiation exposure.
Tailings are radioactive with long-lived activities in low concentrations and also
contain heavy metals. They have to be isolated.
Increasingly the mining industry uses in-situ leaching. Here oxygenated groundwater
is circulated through the U deposit underground to dissolve the U and bring it to
the surface.
Coolant: to extract heat from core
Need to remove heat from core and make use of this heat
Desired properties of “heat extractor”
- Can be circulated (gas, liquid)
- Low neutron capture (to economize neutron and to prevent
too much [radio]activation)
- Chemically stable
- High thermal conductivity
H2O, D2O, He, liquid Sodium
Then heat is exchanged to a steam boiler to power turbines