Revision 1
December 2014
Fission
Student Guide
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Table of Contents
INTRODUCTION................................................................................................................. 1
TLO 1 NEUTRON INTERACTIONS WITH MATTER .............................................................. 2
Overview ..................................................................................................................... 2
ELO 1.1 Neutron Scattering Interactions .................................................................... 3
ELO 1.2 Neutron Elastic Scattering Interactions ........................................................ 5
ELO 1.3 Neutron Absorption Reactions ..................................................................... 6
TLO 1 Summary ......................................................................................................... 7
TLO 2 FISSION PROCESS .................................................................................................. 8
Overview ..................................................................................................................... 8
ELO 2.1 Excitation Energy ......................................................................................... 9
ELO 2.2 Fission Process Definition .......................................................................... 10
ELO 2.3 Fissile Material ........................................................................................... 12
ELO 2.4 Fission Materials ........................................................................................ 16
ELO 2.5 Binding Energy Per Nucleon Curve ........................................................... 17
TLO 2 Summary ....................................................................................................... 19
TLO 3 PRODUCTION OF HEAT FROM FISSION ................................................................. 20
Overview ................................................................................................................... 20
ELO 3.1 Energy Release Per Fission ........................................................................ 21
ELO 3.2 Fission Fragment Yield .............................................................................. 23
ELO 3.3 Calculate Fission Energy: Binding Energy and Conservation of Mass ..... 25
ELO 3.4 Fission Heat Production ............................................................................. 29
TLO 3 Summary ....................................................................................................... 30
TLO 4 INTRINSIC AND INSTALLED NEUTRON SOURCES ................................................. 31
Overview ................................................................................................................... 31
ELO 4.1 Source Neutrons ......................................................................................... 31
ELO 4.2 Intrinsic Source Neutrons ........................................................................... 32
ELO 4.4 Intrinsic Source Neutrons Over Core Life ................................................. 39
TLO 4 Summary ....................................................................................................... 41
TLO 5 RELATIONSHIP BETWEEN NEUTRON FLUX, MICROSCOPIC, AND MACROSCOPIC
CROSS-SECTIONS ........................................................................................................... 43
Overview ................................................................................................................... 43
ELO 5.1 Neutron Reaction Terms ............................................................................ 44
ELO 5.2 Neutron Energy Terms ............................................................................... 53
ELO 5.3 Neutron Energies versus Cross-Sections .................................................... 54
ELO 5.4 Temperature Affects to Macroscopic Cross-Sections ................................ 57
ELO 5.5 Neutron Flux Distribution .......................................................................... 60
ELO 5.6 Reaction Rate ............................................................................................. 62
ELO 5.7 Neutron Flux and Reactor Power ............................................................... 64
TLO 5 Summary ....................................................................................................... 67
FISSION SUMMARY ......................................................................................................... 68
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Fission
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/14/2014
0
New Module
OGF Team
12/10/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Introduction
This module focuses on neutron interactions, including fission, building on
the atomic structure module. To provide complete knowledge of fission, it
is important to understand all of the various types and probabilities of
neutron interactions that exist. This module will cover neutron scattering
and absorption reactions, materials used for nuclear fuel, the production of
heat from fission, neutron sources for shutdown and startup conditions, and
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1
the various terms used for describing and measuring neutron intensity,
reaction probabilities, and reaction rates. You will then apply this
background knowledge to understand the calculations for determining
reactor thermal power output.
Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80 percent
or higher on the following Terminal Learning Objectives (TLOs):
1. Describe neutron interactions with matter.
2. Describe the process of nuclear fission and the types of material that can
undergo fission.
3. Explain the production of heat from fission.
4. Describe intrinsic and installed neutron sources and their contribution to
source neutron strength over core life.
5. Explain the relationship between neutron flux, microscopic and
macroscopic cross-sections; and their effect on neutron reaction rates.
TLO 1 Neutron Interactions with Matter
Overview
Neutrons can cause many different types of interactions. The neutron may
simply scatter off the nucleus, which can result in a reduction of the
neutron's energy, or be absorbed within the nucleus and removed from the
system. When neutrons are absorbed into the nucleus, the nucleus becomes
excited. This results in emission of gamma rays and/or particles, or it can
result in fission.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the following neutron scattering interactions, including
conservation principles:
a. Elastic scattering
b. Inelastic scattering
2. Define the following:
a. Resonance elastic scattering
b. Potential elastic scattering
3. Describe the following reactions where a neutron is absorbed in a
nucleus:
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Rev 1
a. Radiative capture
b. Particle ejection
c. Fission
ELO 1.1 Neutron Scattering Interactions
Introduction
A neutron scattering reaction occurs after by a neutron striking a nucleus
and emits a single neutron. The net effect of the reaction is that the free
neutron has merely bounced off, or scattered as it interacts with the nucleus.
In some cases, the initial and final neutrons are not the same. There are two
categories of scattering reactions, elastic, and inelastic scattering. These
collisions are of great importance for the process of making thermal or low
energy neutrons available for a thermal reactor.
Neutron Scattering Interactions
Elastic Scattering
Elastic scattering, also referred to as the billiard ball effect, is most probable
with light nuclides. In an elastic scattering reaction, the neutron does not
contribute to nuclear excitation of the target nucleus because of a
conservation of momentum. Energy completely transfers from the neutron
to the target nucleus, conserving the kinetic energy of the system. The
target nucleus gains an amount of kinetic energy equal to the kinetic energy
the neutron loses. This kinetic energy transfer is dependent on target size
and angle. The figure below illustrates an elastic scattering reaction.
Figure: Elastic Scattering
The following mathematically illustrates the conservation of momentum
and kinetic energy during an elastic scattering reaction:
Conservation of momentum (π‘šπ‘£)
(π‘šπ‘› 𝑣𝑛,𝑖 ) + (π‘š 𝑇 𝑣𝑇,𝑖 ) = (π‘šπ‘› 𝑣𝑛,𝑓 ) + (π‘š 𝑇 𝑣𝑇,𝑓 )
1
Conservation of kinetic energy (2 π‘šπ‘£ 2 )
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3
1
1
1
1
2
2
2
2
( π‘šπ‘› 𝑣𝑛,𝑖
) + ( π‘š 𝑇 𝑣𝑇,𝑖
) = ( π‘šπ‘› 𝑣𝑛,𝑓
) + ( π‘š 𝑇 𝑣𝑇,𝑓
)
2
2
2
2
Where:
mn = mass of the neutron
mT = mass of the target nucleus
vn,i = initial neutron velocity
vn,f = final neutron velocity
vT,i = initial target velocity
vT,f = final target velocity
Inelastic Scattering
In inelastic scattering, the target nucleus absorbs the incident neutron by
forming a compound nucleus and transfers the kinetic energy into nuclear
excitation. The compound nucleus emits a neutron of lower kinetic energy
than the incident neutron, leaving the nucleus in an excited state. A gamma
emission occurs if the excitation state is not at a preferred level, dropping
excess energy to reach ground state. Inelastic scattering is most probable
with heavy nuclides and neutrons above 1 MeV in energy. The figure
below illustrates inelastic scattering.
Figure: Inelastic Scattering
Some amount of kinetic energy transfers into excitation energy of the target
nucleus. The total kinetic energy of the outgoing neutron and nucleus is
less than the kinetic energy of the incoming neutron. Therefore, there is no
conservation of kinetic energy; however, momentum is conserved.
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Rev 1
Knowledge Check
True/False: The difference between elastic and
inelastic scattering where neutrons are concerned is
elastic scattering involves no energy being transferred
into excitation energy of the target nucleus. Inelastic
scattering involves a transfer of kinetic energy into
excitation energy of the target nucleus.
A. True
B.
False
ELO 1.2 Neutron Elastic Scattering Interactions
Introduction
Elastic scattering of neutrons by nuclei can occur in two ways:
ο‚·
ο‚·
Resonance elastic scattering
Potential elastic scattering
Resonance Elastic Scattering
Resonance elastic scattering, also referred to as compound elastic scattering
is the least likely type of scattering event to occur. The nucleus absorbs the
incident neutron, temporarily forming a compound nucleus. The nucleus
also emits a neutron, returning the nucleus to its ground state and
conserving total kinetic energy.
This type of elastic scattering is dependent upon the initial kinetic energy
possessed by the neutron.
Potential Elastic Scattering
Potential elastic scattering is the most likely type of scattering event to
occur. The interaction between neutrons in potential elastic scattering
referred to as the billiard ball effect, resembles two hard billiard balls
bouncing off each other.
Potential elastic scattering takes place with incident neutrons of up to 1
MeV. The incident neutron does not touch the nucleus nor does a
compound nucleus form in this type of scattering. Instead, the incident
neutron, as it comes close to the target nucleus, scatters because of the
short-range nuclear forces of the nucleus. The result is no energy transfer
into nuclear excitation, conservation of momentum, and conservation of
kinetic energy of the system. The target nucleus gains the amount of kinetic
energy that the incident neutron loses.
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5
Knowledge Check
Which of the following statements accurately describes
resonance elastic scattering of a neutron?
A.
A nucleus absorbs a neutron, and a gamma ray is
ejected.
B.
A nucleus absorbs a neutron and breaks into two
smaller particles.
C.
An approaching neutron is repelled by the target
nucleus.
D.
A nucleus absorbs a neutron and a neutron is reemitted.
ELO 1.3 Neutron Absorption Reactions
Introduction
Most absorption reactions result in the loss of a neutron, the production of
charged particles, gamma ray, or fission fragments and the release of
neutrons and energy. Three types of absorption reaction are:
ο‚·
ο‚·
ο‚·
Radiative capture
Particle ejection
Fission
Radiative Capture
In radiative capture, the incident neutron interacts with the target nucleus
forming a compound nucleus. The compound nucleus, with the additional
neutron, then decays to its ground state via gamma emission. The equation
below shows an example of radiative capture with uranium-238.
1
238
239 ∗ 239
0
𝑛+
π‘ˆ→(
π‘ˆ) →
π‘ˆ+ 𝛾
0
92
92
92
0
Particle Ejection
In particle ejection, an incident neutron interacts with the target nucleus,
forming a compound nucleus. The new compound nucleus excites to a high
enough energy level for it to eject a new particle with the incident neutron
remaining in the nucleus. The nucleus may or may not exist in an excited
state depending upon the mass-energy balance of the reaction after ejection
of the new particle. The equation below shows a common example of this,
specifically the boron-10 neutron reaction resulting in a lithium nucleus and
alpha particle.
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Rev 1
1
10
11 ∗ 7
4
𝑛 + 𝐡 → ( 𝐡) → 𝐿𝑖 + 𝛼
0
5
5
3
2
Fission
Fission is a very important absorption reaction. An incident neutron
interacts with the target nucleus, the nucleus absorbs the incident neutron,
and the nucleus splits into two smaller nuclei, called fission fragments.
This reaction releases multiple neutrons and a considerable amount of
energy in addition to the fission fragments. A fission reaction typically
produces two fission fragments, 2 to 3 neutrons, and energy (in the form of
kinetic energy and gamma rays). The following shows the fission of a
uranium-235 atom.
1
236 ∗ 140
93
1
235
𝑛+
π‘ˆ→(
π‘ˆ) →
𝐢𝑠 + 𝑅𝑏 + 3 ( 𝑛)
0
92
55
37
0
92
Knowledge Check
What type of neutron interaction has occurred when a
nucleus absorbs a neutron and ejects proton?
A.
Fission
B.
Fusion
C.
Particle ejection
D.
Radiative capture
TLO 1 Summary
Neutron Interactions
ο‚· Interactions where a neutron scatters off a target nucleus are either
elastic or inelastic.
ο‚· Elastic scattering - there is conservation of both kinetic energy and
momentum, and no energy is transferred into excitation energy of the
target nucleus. This is most probable with light nuclides.
- Resonance elastic scattering involves absorption of the neutron
by the target nucleus, forming a compound nucleus, followed by
the re-emission of a neutron.
- In potential elastic scattering, the neutron does not touch the
nucleus and does not form a compound nucleus. The neutron is
scattered by the short-range nuclear forces when it approaches
close to the nucleus.
ο‚· Inelastic scattering - some amount of kinetic energy transfers into
excitation energy of the target nucleus. The total kinetic energy of the
outgoing neutron and nucleus is less than the kinetic energy of the
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7
incoming neutron. Therefore, there is no conservation of kinetic
energy; however, there is conservation of momentum.
ο‚· Radiative capture is the absorption of a neutron by the target nucleus,
resulting in an excited nucleus that subsequently (typically within a
small fraction of a second) releases its excitation energy in the form
of a gamma ray.
ο‚· Particle ejection occurs when a target nucleus, absorbs a neutron,
resulting in an excited compound nucleus. The compound nucleus
immediately ejects a particle (for example, alpha, or proton).
ο‚· Fission - an incident neutron adds sufficient energy to the target
nucleus that the target nucleus splits apart, releasing two fission
fragments, several neutrons, and energy.
Now that you have completed this lesson, you should be able to:
1. Describe the following neutron scattering interactions, including
conservation principles:
a. Elastic scattering
b. Inelastic scattering
2. Define the following:
a. Resonance elastic scattering
b. Potential elastic scattering
3. Describe the following reactions where a neutron is absorbed in a
nucleus:
a. Radiative capture
b. Particle ejection
c. Fission
TLO 2 Fission Process
Overview
Nuclear fission is the splitting apart of a heavy nuclide into two fission
products with the release of energy and additional neutrons. The release of
neutrons causes additional fissions to occur, causing a self-sustaining
fission rate capable of producing sufficient heat for power production.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the following terms:
a. Excitation energy (Eexc)
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Rev 1
b. Critical energy (Ecrit)
2. Explain the fission process using the liquid drop model of a nucleus.
3. Define the following terms:
a. Fissile material
b. Fissionable material
c. Fertile material
d. Thermal neutrons
4. Explain binding energy per nucleon.
5. Explain the shape of the Binding Energy per nucleon versus mass
number curve including its significance to fission energy release.
ELO 2.1 Excitation Energy
Introduction
Excitation energy is the energy above ground state of a nucleus. Critical
energy is the required excitation energy for fission to occur.
Excitation Energy
Excitation energy (Eexc) is the measure of how far the energy level of a
nucleus is above its ground state. The last topic discussed the many neutron
reactions that can cause an increase in the excitation energy of a nucleus.
Critical Energy
The excitation energy must be above a specific minimum value for that
nuclide for fission to occur. The critical energy (Ecrit) is the minimum
excitation energy required for fission to occur for a particular nuclide.
The reaction excites the target nucleus by an amount equal to binding
energy of the neutron plus the neutron's kinetic energy when an incident
neutron strikes a target nucleus. If the binding energy is less than the
required critical energy for the nucleus, additional energy is required to
cause the nucleus to fission. This energy could be in the form of kinetic
energy from the incident neutron.
For this reason, neutrons of low kinetic energy cannot cause fission with
some types of isotopes used in nuclear reactor fuels. The discussion of
binding energy in this course covers this concept in more detail.
Rev 1
9
Knowledge Check
True or False: Excitation must be at least equal to
critical energy for fission to occur.
A.
True
B.
False
ELO 2.2 Fission Process Definition
Introduction
In a fission reaction, the incident neutron interacts with the target nucleus,
and the target nucleus absorbs the incident neutron. This creates a
compound nucleus that is excited at such a high energy level (Eexc > Ecrit)
that the compound nucleus splits (fissions) into two large fragments plus
some neutrons. In addition, the fission process releases a large amount of
energy in the form of radiation and fragment kinetic energy.
Fission Definition
As previously discussed, the attractive nuclear force between nucleons
holds the nucleus together. In doing so, it is resisting the opposing
electrostatic forces within the nucleus. Characteristics of the attractive
nuclear force are:
ο‚·
Very short range attractive force, with essentially no strength beyond
nuclear scale dimensions (≈10-13 cm)
ο‚· Stronger than the repulsive electrostatic forces within the nucleus
ο‚· The force is independent of nucleon pairing. Attractive forces
between pairs of neutrons, pairs of protons or neutron proton pairs are
identical.
ο‚· Saturable, that is, a nucleon can attract only a few of its nearest
neighbors.
Fission Process Example
One theory of fission considers the fission of a nucleus similar to the
splitting of a liquid drop. Molecular forces hold a liquid drop together,
making the drop spherical in shape and resist deformation. Nuclear forces
hold the nucleus together in the same manner. The next figure illustrates
the liquid drop model.
10
Rev 1
Figure: Liquid Drop Model
Liquid Drop Model Steps
The steps illustrated above for the liquid drop model are as follows:
The undisturbed nucleus in the ground state is undistorted, and its attractive
nuclear forces are greater than the repulsive electrostatic forces between the
protons within the nucleus.
The nucleus becomes an excited compound nucleus when the incident
neutron strikes the target nucleus and is absorbed. This compound nucleus
temporarily contains all the charge and mass involved in the reaction, and
exists in an excited state. The excitation energy added to the compound
nucleus is equal to the binding energy contributed by the incident neutron
plus its kinetic energy. The excitation energy may cause the nucleus to
oscillate and distort in shape.
If the excitation energy is greater than an isotope's specific critical energy,
the oscillations may cause the compound nucleus to become dumbbellshaped as it starts to come apart. As this occurs, the attractive nuclear
forces (short-range) in the neck area reduce due to saturation, while the
repulsive electrostatic forces (long-range) remain almost as strong. Less
force holds the nucleus together.
When the repulsive electrostatic forces exceed the attractive nuclear forces,
nuclear fission occurs - the nucleus breaks apart into two fission fragments
and releases neutrons and energy.
Rev 1
11
Knowledge Check
In the Liquid Drop Model of fission, the nucleus
absorbs a neutron, becomes distorted into a dumbbell
shape, and splits into two nuclei. Which of the
following forces is responsible for the nucleus
splitting?
A.
Liquid drop force
B.
Gravitational force
C.
Electrostatic force
D.
Fission force
ELO 2.3 Fissile Material
Introduction
There are two categories of nuclear fuel materials: those that can fission
with thermal neutrons and those that require neutrons of higher energies for
fission to be possible. It is possible to convert some non-fissionable
materials into materials capable of fission. This section will explain the
following terms:
ο‚·
ο‚·
ο‚·
ο‚·
Fissile material
Fissionable material
Fertile material
Thermal neutrons
Fissile Material
A fissile material consists of nuclides that will fission with incident
neutrons of any energy level. They are the desired nuclides for use in
thermal nuclear reactors. The advantage of fissile materials is that they can
fission with neutrons possessing zero kinetic energy (thermal neutrons).
Thermal neutrons have very low kinetic energy levels because they are in
approximate equilibrium with the thermal energy of surrounding materials.
They add essentially no kinetic energy to the reaction, however add enough
BE to reach the critical energy, Ecrit. Fissile materials will fission after
absorbing a thermal neutron.
It is possible for fissile materials to fission with thermal neutrons because
the change in binding energy supplied by the neutron addition alone is
sufficient to exceed the critical energy. Examples of fissile materials are
uranium-235, uranium-233, and plutonium-239.
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Rev 1
Fissionable Material
A fissionable material is composed of nuclides for which fission is possible.
All fissile nuclides fall into this category as well as those nuclides that
fission only from high-energy neutrons.
The change in binding energy that occurs from neutron absorption causes an
excitation energy level insufficient to reach critical energy for fission in
fissionable materials. Therefore, the incident neutron must supply
additional excitation energy by adding kinetic energy to the reaction.
Experimental observation shows the reason for this difference between
fissile and fissionable materials. A nucleus with an even number of
neutrons or protons is more stable than a nucleus with an odd number.
Adding a neutron to an odd numbered nucleus (changing it to an even
numbered nucleus) produces higher binding energy than adding a neutron to
an even numbered nucleus.
The nuclides thorium-232, uranium-238, and plutonium-240 are fissionable
materials that require high-energy neutrons to cause fission.
The table below lists the critical energy (Ecrit) and the binding energy
change for an added neutron (BEn) to target nuclei of interest. The change
in binding energy plus the kinetic energy must equal or exceed the critical
energy (ΔBE + KE ≥ Ecrit) for fission to be possible.
Target
Nucleus
Critical Energy
Ecrit
Binding Energy of
Last Neutron BEn
BEn - Ecrit
232
90Th
7.5 MeV
5.4 MeV
-2.1 MeV
238
92U
7.0 MeV
5.5MeV
-1.5 MeV
235
92U
6.5 MeV
6.8 MeV
+0.3 MeV
233
92U
6.0 MeV
7.0 MeV
+1.0 MeV
234
94Pu
5.0 MeV
6.6 MeV
+1.6 MeV
As seen in the table above, uranium-235 can fission with thermal neutrons
because its BEn is greater than its critical energy. This makes U-235 a
fissile material.
Uranium-238, on the other hand, has a critical energy of 7.0 MeV, which is
greater than the 5.5 MeV BEn added by the neutron. Therefore, a fission
Rev 1
13
neutron must have at least 1.5 MeV to cause fission of U-238. This makes
uranium-238 a fissionable material.
Fertile Materials
All of the neutron absorption reactions that do not result in fission lead to
the production of new nuclides through the process known as transmutation.
These nuclides can be transmuted again or undergo radioactive decay to
produce nuclides known as transmutation products. Nuclear reactions can
produce transmutation products because several fissile nuclides do not exist
in nature.
Fertile materials are materials (nuclides) that can undergo transmutation to
become fissile materials. The figure below shows the chain of
transmutation mechanisms from which two fertile nuclides, thorium-232
and uranium-238, produce uranium-233 and plutonium-239, respectively.
Figure: Transmutation Mechanisms
Breeding and Conversion
If a reactor contains fertile material in addition to its fissile fuel, depletion
of original fuel produces new fuel in a process called conversion.
Breeder Reactors
Glossary
Breeder reactors are reactors specifically designed to
produce fissionable fuel.
In such reactors, the amount of fissionable fuel produced is greater than the
amount of fuel depleted. The process is called conversion if less fuel is
produced than used, and the reactor is termed a converter. Commercial
Pressurized Water Reactors (PWRs) would be converters.
Thermal Neutrons
Thermal neutrons are neutrons with very low kinetic energy levels
(essentially zero) because they are in equilibrium with the thermal energy of
surrounding materials. In reality, thermal neutrons undergo scattering
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Rev 1
collisions, gaining and losing equal amounts of energy in successive
collisions. They add essentially no kinetic energy to a neutron absorption
reaction. The goal is to thermalize as many neutrons as possible in a
commercial PWR, as a thermal reactor.
Tables that summarize data for thermal neutrons' cross-sections at 68°F
(20°C) allow comparisons under similar temperatures. At 68°F thermal
neutron energy is 0.025 eV with a velocity of 2.2 X 105 cm/sec. As
temperature increases, thermal neutron energy and velocity increase by the
relationships shown below:
1
𝑇 2
𝐸𝑝 = (0.025 𝑒𝑉) ( )
π‘‡π‘œ
1
π‘π‘š 𝑇 2
𝑉𝑝 = (2.2 × 105
)( )
𝑠𝑒𝑐 π‘‡π‘œ
Where:
Ep = Most probable energy (eV)
Vp = Most probable velocity (cm/sec)
To = Table temperature of 68°F (528°R)
T = New temperature in °R (°F + 460°)
Knowledge Check
What is the difference between a fissionable material
and a fissile material?
Rev 1
A.
There is no difference between a fissionable and a
fissile material, except for the number of protons and
neutrons located in the nuclei of the particular
materials.
B.
A fissionable material can become fissile by capturing
a neutron with zero kinetic energy, whereas a fissile
material can become fissionable by absorbing a neutron
that has some kinetic energy.
C.
Fissile materials require a neutron with some kinetic
energy in order to fission, whereas fissionable
materials will fission with a neutron that has zero
kinetic energy.
D.
Fissionable materials require a neutron with some
15
kinetic energy in order to fission, whereas fissile
materials will fission with a neutron that has zero
kinetic energy.
ELO 2.4 Fission Materials
Introduction
Binding energy per nucleon (BE/A) equals the average energy required to
remove a nucleon from a specific nucleus. The total binding energy related
to the mass defect and the binding energy related to the addition of a single
neutron to a nucleus determines a thermal neutron's ability to cause fission.
The next two sections go into more detail about BE/A.
Binding Energy per Nucleon
Binding energy per nucleon (BE/A) is equal to the average energy required
to remove a single nucleon from a specific nucleus. It is determined by
dividing the total binding energy of a nuclide by the total number of
nucleons in its nucleus. The example below illustrates this calculation.
Example:
Given that the total binding energy (BE) for a U-238 nucleus is 1,804.3
MeV, calculate the binding energy per nucleon (BE/A) for U-238.
Solution:
𝐡𝐸 1,804.3 𝑀𝑒𝑉
=
= 7.6 𝑀𝑒𝑉
𝐴
238
Knowledge Check
True or False: Binding energy per nucleon is
independent of the specific nuclide.
16
A.
True
B.
False
Rev 1
ELO 2.5 Binding Energy Per Nucleon Curve
Introduction
This lesson describes the relationship of binding energy per nucleon as the
mass number changes.
As the number of nucleons in a nucleus increases, the total binding energy
also increases. Many scientific experiments have determined that the rate of
increase is not linear. This non-linear relationship results in a variation of
the binding energy per nucleon for nuclides of different mass numbers.
Binding Energy per Nucleon Curve Example
Plotting the average BE/A against the atomic mass numbers (A) shows the
variation in the binding energy per nucleon, as shown in the figure below.
Figure: Binding Energy per Nucleon versus Mass Number
The figure above shows that as the atomic mass number (A) increases,
BE/A increases, until A reaches about 60, and BE/A decreases as A
increases above 60. The BE/A curve reaches a maximum value of 8.79
MeV at A = 56 and decreases to about 7.6 MeV for A = 238. No stable
nuclei exist with A greater than 209.
The general properties of nuclear forces determine the general shape of the
BE/A curve. Very short-range attractive forces existing between nucleons
hold the nucleus together and long range repulsive electrostatic (coulomb)
forces existing between positively charged protons in the nucleus are
forcing nucleons apart.
As the atomic number (number of protons) and the atomic mass number of
a nucleus increase, the repulsive electrostatic forces within the nucleus
Rev 1
17
increase due to the greater number of protons. Increasing the proportion of
neutrons in the nucleus overcomes this increased repulsion. The slope of
the nuclide stability curve illustrates the effect of increasing repulsive
forces. The increase in the neutron-to-proton ratio only partially
compensates for the increasing proton-proton repulsive force in heavier
naturally occurring elements. Less energy is required, on average, to
remove a nucleon from the nucleus because repulsive forces are increasing.
Therefore, the BE/A decreases.
In the case of fissile materials (heaviest nuclei), only a small distortion from
a spherical shape (a small energy addition) is required for the relatively
large electrostatic forces attempting to force the two halves of the nucleus
apart to overcome the attractive nuclear forces holding the two halves
together. Consequently, the heaviest nuclei are more easily fissionable than
lighter nuclei.
The BE/A of a nucleus is also an indication of its stability. Generally, the
higher the BE/A, the more stable the nuclide is. The increase in the BE/A
as the atomic mass number decreases from 260 to 60 is the main reason for
energy liberation in the fission process as the fuel isotope is split into two
fragments of lower A values. The fission arrow in the figure above denotes
this area. A later section will discuss delta binding energy.
However, the increase in the BE/A as the atomic mass number increases
from 1 to 60 is the reason that energy release is possible in a fusion event
that combines rather than splits atoms. The fusion arrow in the figure above
denotes this area.
Knowledge Check
True or False. Generally, less stable nuclides have a
higher BE/A than the more stable ones.
18
A.
True
B.
False
Rev 1
TLO 2 Summary
Nuclear Fission
ο‚· Nuclear fission liquid drop model of a nucleus
- In the ground state, the nucleus is nearly spherical in shape.
- After absorbing a neutron, the nucleus will be in an excited state,
start to oscillate, and become distorted.
- If distortion causes the nucleus to become dumbbell-shaped, the
repulsive electrostatic forces may overcome the short-range
attractive nuclear forces causing the nucleus to split in two.
ο‚· Excitation energy is the amount of energy a nucleus has above its
ground state.
ο‚· Critical energy is the minimum excitation energy that a nucleus must
have before it can fission.
ο‚· Fissile material is material for which fission is possible with neutrons
that have zero kinetic energy.
ο‚· Fissionable material is material for which fission caused by neutron
absorption is possible, but the binding energy from the additional
neutron is not sufficient for excitation above critical energy.
Therefore, the neutron must have additional kinetic energy to make
fission possible.
ο‚· Fertile material is fissionable material that can undergo transmutation
to become fissile material.
ο‚· Thermal neutrons have very low kinetic energy levels (essentially
zero) because they are in energy equilibrium with the thermal motion
of surrounding materials.
ο‚· Transmutation is the process of neutron absorption and subsequent
decay, which changes one nuclide to another nuclide.
ο‚· Conversion is the process of transmuting fertile material into fissile
material where the amount of fissile material produced is less than the
fissile material consumed.
ο‚· Breeding is the process of transmuting fertile material into fissile
material where the amount of fissile material produced is more than
the amount of fissile material consumed.
ο‚· The curve of binding energy per nucleon increases quickly through
the light nuclides and reaches a maximum at a mass number of about
56. The curve decreases slowly for mass numbers greater than 60.
ο‚· The heaviest nuclei are easily fissionable because they require only a
small distortion from the spherical shape allowing the electrostatic
forces to overcome the attractive nuclear forces, causing the nucleus
to split.
ο‚· Fission of uranium-235 versus fission of uranium-238:
- Uranium-235 fissions with thermal neutrons because the binding
energy released by the absorption of a neutron is greater than the
critical energy.
- The binding energy released by uranium-238 absorbing a
neutron is less than the critical energy, so the neutron must have
additional kinetic energy for fission to be possible.
Now that you have completed this lesson, you should be able to:
Rev 1
19
1.
Define the following terms:
a. Excitation energy (Eexc)
b. Critical energy (Ecrit)
2.
Explain the fission process using the liquid drop model of a nucleus.
3.
Define the following terms:
a. Fissile material
b. Fissionable material
c. Fertile material
d. Thermal neutrons
4.
Explain binding energy per nucleon.
5.
Explain the shape of the Binding Energy per nucleon versus mass
number curve including its significance to fission energy release.
TLO 3 Production of Heat from Fission
Overview
Fission of heavy nuclides converts a small amount of mass into a large
amount of energy. There are two ways to calculate the energy released by
fission: you can base computations on the change in mass that occurs during
the fission or by the difference in binding energy per nucleon between the
fissile nuclide and the fission products. This TLO discusses the process of
fission and heat production.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the average total amount of energy released per fission event
including:
a. Energy released immediately from fission
b. Delayed fission energy
2. Describe which fission products nuclides are most likely to result from
fission.
3. Describe the energy released from fission by the following methods:
a. Change in binding energy
b. Conservation of mass-energy
c. Decay energy
4. Describe how heat is produced as a result of fission.
20
Rev 1
ELO 3.1 Energy Release Per Fission
Introduction
Total energy released per fission varies from one fission event to the next,
depending on what fission products the reaction forms, but the average total
energy released from uranium-235 fission with a thermal neutron is
approximately 200 MeV. The majority of this energy (approximately 83
percent) is from the kinetic energy of the fission fragments.
Energy Release per Fission Details
The table below shows the average energy distribution for the energy
released in a U-235 thermal fission, approximately 200 MeV.
Instantaneous Energy
Value
Kinetic Energy of Fission Fragments
165 MeV
Kinetic Energy of Fission Neutrons
5 MeV
Instantaneous Gamma Rays
7 MeV
Capture Gamma Ray Energy
10 MeV
Delayed Energy
Value
Kinetic Energy of Beta Particles
7 MeV
Decay Gamma Rays
6 MeV
Neutrinos**
10 MeV**
Total Energy Released
200 MeV
**Not included in total
Some fission neutrons undergo radiative capture reactions, producing
gamma ray emissions as their compound nuclei drop to ground state. This
provides the additional 10 MeV of instantaneous energy from Capture
Gamma Ray energy shown in the table above.
The total energy released per fission does not include the delayed energy
from neutrinos, because neutrinos do not interact with materials inside the
reactor.
Rev 1
21
Delayed Fission Energy (Decay Heat)
About seven percent (13 MeV) releases sometime after the instant of fission
of the 200 MeV released per fission. Fissions mostly cease with the reactor
shut down, but heat energy, referred to as decay heat, continues to release
from the reactor because of the decay of fission products.
After reactor shutdown, decay heat production tapers off but remains a
significant source of heat for a very long time. Systems exist to remove this
decay heat after reactor shutdown in order to prevent damage to the reactor
core. The Three Mile Island event is an example of what can happen if this
cooling source is unavailable following a shutdown. The reactor operations
module discusses decay heat in detail.
Isotope specifies, including the particular isotopes that are undergoing
fission, and their classification as fissionable or fissile material have a small
effect on the amount of energy released. For instance, the fission of U-235
by a slow neutron yields nearly identical energy to that of a U-238 fission
by a fast neutron.
Knowledge Check
Which of the following statements correctly describes
the amount of energy released from a single fission
event?
22
A.
Approximately 200 MeV are released during fission.
13 MeV are released instantaneously, and 187 MeV are
released later (delayed).
B.
Approximately 200 MeV are released during fission.
187 MeV are released instantaneously, and 13 MeV are
released later (delayed).
C.
Approximately 200 eV are released during fission. 13
eV are released instantaneously, and 187 eV are
released later (delayed).
D.
Approximately 200 eV are released during fission. 187
eV are released instantaneously, and 13 eV are released
later (delayed).
Rev 1
ELO 3.2 Fission Fragment Yield
Introduction
Fissions do not produce identical results on each occurrence. In fact, both
the number of neutrons and the resultant fission fragments vary. Scientific
experiments have developed a yield curve of fission product probabilities.
Most Probable Fission Fragments
Resultant fission fragments have masses that vary widely. The figure below
shows the percent yield of various atomic mass numbers. The most
probable pair of fission fragments for a thermal neutron fission of uranium235 have masses of about 95 and 140.
NOTE: The vertical axis of the fission yield curve is a logarithmic scale
further amplifying the higher probability for mass numbers of 95 and 140.
Rubidium-93 and cesium-140 are very likely to result from fission.
Note
Note
The vertical axis of the fission yield curve is a logarithmic
scale further amplifying the higher probability for mass
numbers of 95 and 140. Rubidium-93 and cesium-140 are
very likely to result from fission.
Figure: Uranium-235 Fission Yield for Fast and Thermal Neutrons versus
Mass Number
Rev 1
23
As shown in the above figure, the fission fragment yield varies
considerably. An example of one fission fragment yield is:
235
π‘ˆ+𝑛 →
236
π‘ˆ∗ →
140
𝑋𝑒 +
94
π‘†π‘Ÿ + 2𝑛
Notice in the example above that two neutrons result from this fission.
Normally, one fission event releases two or three fast neutrons. The figure
below shows average number of neutrons released per fission event for
various fuels.
Isotope
Average Neutrons Released per Fission (v)
U-233
2.492
U-235
2.418
Pu-239
2.871
Pu-241
2.927
Knowledge Check
List the two most likely fission fragments resulting
from fission of uranium-235.
24
Rev 1
ELO 3.3 Calculate Fission Energy: Binding Energy and
Conservation of Mass
Introduction
Nuclear fission releases enormous quantities of energy; to compute these
energies, start with the fission reaction equation. Consider first a typical
fission reaction as shown below to begin these calculations.
1
236 ∗ 140
93
1
235
𝑛+
π‘ˆ→(
π‘ˆ) →
𝐢𝑠 + 𝑅𝑏 + 3 ( 𝑛)
0
92
55
37
0
92
When the compound nucleus splits in the equation above, it breaks into two
fission fragments, cesium-140, rubidium-93, and three neutrons. Both
fission products will decay by multiple β- emissions (not shown) because of
the high neutron-to-proton ratios they possess.
Change in Binding Energy
From the binding energy per nucleon curve (BE/A) we can determine the
amount of energy released by a "typical" fission by plotting the uranium235 fission reaction on the curve below and calculating the change in
binding energy (ΔBE) between the reactants on the left-hand side of the
fission equation and the products on the right-hand side.
Figure: Change in Binding Energy for Typical Fission
The figure above shows that the binding energy per nucleon for the
products (rubidium-93 and cesium-140) is greater than that for the reactant
(uranium-235).
The system has become more stable by releasing energy equal to the
increase in total binding energy of the system when there is an increase in
Rev 1
25
the total binding energy of a system. The energy liberated is equal to the
increase in the total binding energy of the system in a fission process.
Example:
We find the total binding energy for a nucleus by multiplying the binding
energy per nucleon by the number of nucleons, as shown in the table below.
Nuclide
BE per Nucleon
(BE/A)
Mass Number
(A)
Binding Energy
(BE/A) x (A)
93
37Rb
8.7 MeV
93
809 MeV
140
55Cs
8.4 MeV
140
1,176 MeV
235
92U
7.6 MeV
235
1,786 MeV
Note that the “A” symbols in the above table are not the same and do not
cancel out in the binding energy calculation of the last column.
The energy released will be equivalent to the difference in binding energy
(BE) between the reactants and the products.
βˆ†π΅πΈ = π΅πΈπ‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘  − π΅πΈπ‘Ÿπ‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘π‘ 
= (𝐡𝐸𝑅𝑏−93 + 𝐡𝐸𝐢𝑠−140 ) − (π΅πΈπ‘ˆ−235 )
= (809 𝑀𝑒𝑉 + 1,176 𝑀𝑒𝑉) − (1,786 𝑀𝑒𝑉)
= 199 𝑀𝑒𝑉
Conservation of Mass-Energy
During the fission process a decrease in the system mass occurs. The loss
of mass equals the energy liberated. The conservation of mass-energy
method considers this relationship between mass and energy. This method
is more accurate than the ΔBE method since it considers all mass changes
immediately occurring from fission. Therefore, measurements that require
more precision should use this method.
The instantaneous energy (Einst) is the energy released immediately after the
fission process and is equal to the energy equivalent of the mass lost. This
calculation method includes:
ο‚·
Add the nucleon masses of the fuel isotope including the incident
neutron mass (reactants).
ο‚· Add the nucleon masses of the fission products and released neutrons
(products).
ο‚· Subtract the mass of the products from the reactants.
ο‚· Multiply the mass calculated by 931.5 MeV to convert to energy.
26
Rev 1
The table below facilitates this calculation.
Mass of the Reactants
235
π‘ˆ
92
1
𝑛
0
Totals
235.043924 amu
1.008665 amu
Mass of the Products
93
𝑅𝑏
37
140
𝐢𝑠
55
1
3 𝑛
0
236.052589 amu
92.91699 amu
139.90910 amu
3.02599 amu
235.85208 amu
Mass Difference
= π‘€π‘Žπ‘ π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘π‘  − π‘€π‘Žπ‘ π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘ƒπ‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘ 
= 236.052589 π‘Žπ‘šπ‘’ − 235.85208 π‘Žπ‘šπ‘’
= 0.200509 π‘Žπ‘šπ‘’
Energy Equivalent = π‘€π‘Žπ‘ π‘  × 931.5 𝑀𝑒𝑉/π‘Žπ‘šπ‘’
= 0.200509 π‘Žπ‘šπ‘’ × 931.5 𝑀𝑒𝑉/π‘Žπ‘šπ‘’
= 186.8 𝑀𝑒𝑉
Estimation of Decay Energy
Decay energy (EDecay) is the additional energy added to the instant fission
reaction when the fission fragments decay by β- emission. The equations
below show the decay chains for rubidium-93 and cesium-140.
𝛽−
𝛽−
𝛽−
𝛽−
93
93
93
93
93
𝑅𝑏 → π‘†π‘Ÿ → π‘Œ → π‘π‘Ÿ → 𝑁𝑏
37
38
39
40
41
𝛽−
𝛽−
𝛽−
140
140
140
140
𝐢𝑠 →
π΅π‘Ž →
πΏπ‘Ž →
𝐢𝑒
55
56
57
58
The energy released during the decay for each chain will be equivalent to
the mass difference between the original fission product and the sum of the
final stable nuclide and the beta particles emitted. The total decay energy is
the sum of the energies of the two chains. The example below shows the
steps to calculate the total decay energy.
Example
The energy released the decay chain of rubidium-93 is:
Rev 1
27
931.5 𝑀𝑒𝑉
πΈπ·π‘’π‘π‘Žπ‘¦ = [π‘šπ‘…π‘−93 − (π‘šπ‘π‘−93 + 4 π‘šπ‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘› )] (
)
π‘Žπ‘šπ‘’
= [92.91699 π‘Žπ‘šπ‘’
− (92.90638 π‘Žπ‘šπ‘’ + 4 (0.0005486 π‘Žπ‘šπ‘’))] (
931.5 𝑀𝑒𝑉
)
π‘Žπ‘šπ‘’
931.5 𝑀𝑒𝑉
= 0.008416 π‘Žπ‘šπ‘’ (
)
π‘Žπ‘šπ‘’
= 7.84 𝑀𝑒𝑉
The following steps show the calculation of energy released in the decay
chain of cesium-140.
931.5 𝑀𝑒𝑉
πΈπ·π‘’π‘π‘Žπ‘¦ = [π‘šπΆπ‘ −140 − (π‘šπΆπ‘’−140 + 3 π‘šπ‘’π‘™π‘’π‘π‘‘π‘Ÿπ‘œπ‘› )] (
)
π‘Žπ‘šπ‘’
= [139.90910 π‘Žπ‘šπ‘’
931.5 𝑀𝑒𝑉
− (139.90543 π‘Žπ‘šπ‘’ + 3 (0.0005486 π‘Žπ‘šπ‘’))] (
)
π‘Žπ‘šπ‘’
931.5 𝑀𝑒𝑉
= 0.000202 π‘Žπ‘šπ‘’ (
)
π‘Žπ‘šπ‘’
= 1.89 𝑀𝑒𝑉
The total decay energy is the sum of the energies of the two chains, or 9.73
MeV.
Knowledge Check
The change in binding energy per nucleon between the
fissile nuclide and the fission products, is used to
calculate _________.
28
A.
fission fragment speed
B.
decay heat
C.
energy released by fission
D.
increase in mass
Rev 1
ELO 3.4 Fission Heat Production
Introduction
The majority of the energy liberated in the fission process releases
immediately after the fission occurs. This energy appears as kinetic energy
of the fission fragments and neutrons, and instantaneous gamma rays. The
remaining energy releases over time after the fission occurs and appears as
kinetic energy of the decay products.
Fission Heat Production
All of the energy released in fission, with the exception of the neutrino
energy, transforms into heat via ionization and scattering. Fission
fragments, with a positive charge and kinetic energy, cause ionization
directly as they remove orbital electrons from the surrounding atoms
through this ionization process. Kinetic energy transfers to the surrounding
atoms of the fuel material, increasing temperature. Beta particles and
gamma rays also give up energy through ionization, and fission neutrons
interact and lose their energy through scattering.
Knowledge Check
Which of the following is NOT a method by which
heat is produced from fission?
Rev 1
A.
Fission fragments causing direct ionizations resulting
in an increase in temperature.
B.
Beta particles and gamma rays causing ionizations
resulting in increased temperature.
C.
Neutrons interacting and losing their energy through
scattering, resulting in increased temperature.
D.
Neutrinos interacting and losing their energy through
scattering and ionizations resulting in increased
temperatures.
29
TLO 3 Summary
Fission Energy Release
ο‚· A typical fission event releases approximately 200 MeV of energy.
- Approximately 187 MeV of energy releases instantaneously.
- Approximately 13 MeV appears as delayed energy.
ο‚· Fission products generally decay by β- emission.
ο‚· Fission product yields are near mass numbers of 95 and 140.
ο‚· The energy released by fission equals the difference in mass between
the reactants before fission and the fission fragments and fission
neutrons after fission, multiplied by 931.5 MeV per amu.
ο‚· Another method to determine the energy released by fission considers
the change in binding energy per nucleon between the fissile nuclide
and the fission products.
ο‚· Heat is produced from fission by:
- Fission fragments cause direct ionizations resulting in an
increase in temperature.
- Beta particles and gamma rays cause ionizations resulting in
increased temperature.
- Neutrons interact and lose their energy through scattering,
resulting in increased temperature.
ο‚· Decay heat is a result of the decay of fission products. During reactor
operation, decay heat accounts for approximately 7 percent of the heat
produced from fission. In a shutdown reactor, decay heat becomes a
significant source of heat and requires removal to prevent core
damage (Three Mile Island, as an example).
Objectives
Now that you have completed this lesson, you should be able to:
1. Describe the average total amount of energy released per fission event
including:
a. Energy released immediately from fission
b. Delayed fission energy
2. Describe which fission products nuclides are most likely to result from
fission.
3. Describe the energy released from fission by the following methods:
a. Change in binding energy
b. Conservation of mass-energy
c. Decay energy
4. Describe how heat is produced as a result of fission.
30
Rev 1
TLO 4 Intrinsic and Installed Neutron Sources
Overview
Neutrons from a variety of sources are always present in a reactor core,
even when the reactor is shut down. Some neutrons result from naturally
occurring (intrinsic) neutron sources, while others are from fabricated
(installed) neutron sources designed for the reactor. Neutrons produced by
sources other than normal neutron-induced fission are source neutrons.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the purpose and importance of source neutrons.
2. Describe, including examples, each of the following types of intrinsic
neutron sources:
a.
b.
c.
d.
Spontaneous fission
Photo-neutron reactions
Alpha-neutron reactions
Transuranic elements
3. Describe the purpose and type of installed neutron sources.
4. Describe the primary source of intrinsic source neutrons in the reactor
for the following conditions:
a. At beginning and end of core life
b. Immediately following a reactor shutdown
c. Several weeks after reactor shutdown
ELO 4.1 Source Neutrons
Introduction
In addition to fission, other reactions also produce neutrons. The term
source neutrons refers to these non-fission produced neutrons. Source
neutrons are important because they help to monitor the fission process
during a nuclear reactor startup.
Source Neutrons
Source neutrons ensure that the neutron population during shutdown
conditions remains high enough to allow the operators visible indication of
the source range nuclear instrumentation neutron level. This is important
during reactor shutdown and startup conditions to confirm instrument
operability and to monitor the reactor's neutron population changes. There
are two classes of source neutrons: intrinsic and installed neutron sources.
The following sections discuss these two classes.
Rev 1
31
Knowledge Check
Source neutrons are important because they:
A.
Extend the neutron lifetime allowing a nuclear chain
reaction to occur.
B.
Allow for visible indication of neutron level in a
shutdown nuclear reactor.
C.
Shorten the neutron generation time allowing for
operational control of a nuclear reactor.
D.
Contribute a larger percentage of the thermal neutron
population than the fast neutron population in an
operating nuclear reactor.
ELO 4.2 Intrinsic Source Neutrons
Introduction
Intrinsic neutron sources are nuclei that yield neutron-producing reactions,
and that occur in reactor core fuel related materials. This section discusses
the following types of neutron reactions capable of producing source
neutrons and transuranic contributions:
ο‚·
ο‚·
ο‚·
Spontaneous fission
Photo-neutron reactions
Alpha-neutron reactions
Intrinsic Neutron Sources
A limited number of neutrons will be present, even in a reactor core with no
operation history. This is due to spontaneous fission of some heavy
nuclides present in the fuel. Uranium-238, uranium-235, and plutonium239 are examples of these heavy nuclides. Later in core life, the transuranic
elements provide more spontaneous fission neutrons.
Photo-neutron reactions provide a significant source of neutrons in a reactor
that operated at power. The interaction of a gamma ray and a deuterium
nucleus produce these neutrons.
Interactions between alpha particles and various isotopes in the reactor core
also result in source neutron production. Alpha particles from the decay of
heavy nuclides interact with oxygen-18 and boron-11 in the core to produce
neutrons. Transuranic elements building in the core also produce source
neutrons from these reactions.
32
Rev 1
Spontaneous Fission
A limited number of neutrons will always be present, even in a reactor core
that has never been operated, due to spontaneous fission of some heavy
nuclides that are present in the fuel. Uranium-238, uranium-235, and
plutonium-239 undergo spontaneous fission to a limited extent. Uranium238, for example, yields almost 60 neutrons per hour per gram through
spontaneous fission.
The table below shows a comparison of the rate at which different heavy
nuclides produce neutrons by spontaneous fission.
t½ (α-decay)
Nuclide
t½ (Fission)
Neutrons/sec/gram
235
92U
1.8 x 1017 years
6.8 x 108 years
8.0 x 10-4
238
92U
8.0 x 1015 years
4.5 x 109 years
1.6 x 10-2
239
94Pu
5.5 x 105 years
2.4 x 104 years
3.0 x 10-2
240
94Pu
1.2 x 1011 years
6.6 x 103 years
1.0 x 103
252
98Cf
66.0 years
2.65 years
2.3 x 1012
The four isotopes in a reactor that contribute most to the source neutron
population via spontaneous fission are:
ο‚·
ο‚·
ο‚·
ο‚·
Uranium-235
Uranium-238
Curium-242
Curium-244
Uranium-235 and uranium-238 from the core fuel load contribute
approximately 1 x 106 neutrons per second (n/sec) to the source neutron
population. Curium (a transuranic element) isotopes produced during
reactor power operation are also a source of neutrons via spontaneous
fission.
Prior to a new core reactor startup, uranium spontaneous sources are
significant contributors to the source neutron population. Curium-242
becomes a major producer of source neutrons via spontaneous fission after
power operation and up to approximately 20,000 MWd/T of core
irradiation. Beyond 20,000 MWd/T, curium-244 becomes a more
predominant source neutron producer from spontaneous fission. One ton of
spent nuclear fuel will contain on the order of 20 grams of curium.
Rev 1
33
Photo-Neutron Reactions
Source neutrons from photo-neutron reactions become significant in a
reactor with an operational history at power. The equation below describes
the interaction between a gamma ray and a deuterium nucleus. This
reaction is referred to as a photo-neutron reaction because it is initiated by
electromagnetic (gamma photon) radiation and results in the production of a
neutron - photo-neutron. The high-energy gammas for this reaction come
from the decay of fission products.
1
2
1
𝛾+ 𝐻→ 𝐻+ 𝑛
0
1
1
Once the reactor has been operated for a short time there is an abundant
supply of high-energy gammas (2.22 MeV or greater). Commercial PWRs
have some deuterium present in the moderator/coolant because the atom
percentage of naturally occurring deuterium is 0.015 percent. This is
sufficient deuterium for production of photo-neutrons following power
operation. However, after shutdown, gamma ray emitters decay off and the
quantity of gamma rays decreases with time; therefore, the photo-neutron
production rate also decreases with reactor shutdown time.
Alpha Neutron Reactions
Interactions between alpha particles and various isotopes in the reactor core
also result in the production of source neutrons. Alpha particles occur from
the decay of heavy elements in the fuel. They interact with naturally
occurring oxygen-18 and boron-11, from the soluble boron added to the
reactor coolant system, to produce neutrons. Transuranic elements
produced during power operation also provide source neutrons via alphaneutron reaction.
Oxygen-18
Oxygen-18 is naturally occurring in small quantities (0.2 percent) in the
reactor coolant system and in uranium oxide fuel (U3O8). Uranium-235,
uranium-238, curium-242, and curium-244 all undergo alpha decay. The
following table lists the half-lives (alpha decay) for uranium and plutonium.
The heaviest of these elements have long half-lives; therefore, they will
produce only limited alpha particles.
t½ (α-decay)
Nuclide
t½ (Fission)
235
92U
1.8 x 1017 years
6.8 x 108 years
8.0 x 10-4
238
92U
8.0 x 1015 years
4.5 x 109 years
1.6 x 10-2
239
94Pu
5.5 x 105 years
2.4 x 104 years
3.0 x 10-2
34
Neutrons/sec/gram
Rev 1
t½ (α-decay)
Nuclide
t½ (Fission)
Neutrons/sec/gram
240
94Pu
1.2 x 1011 years
6.6 x 103 years
1.0 x 103
252
98Cf
66.0 years
2.65 years
2.3 x 1012
This reaction does not contribute significantly to the source neutron
population because uranium does not produce many alpha particles and the
abundance of oxygen-18 in the reactor core is low. The equation below
describes this reaction.
4
18
21
1
𝐻𝑒 + 𝑂 → 𝑁𝑒 + 𝑛
8
10
0
2
Boron-11
Another intrinsic neutron source is a reaction involving natural boron and
fuel. Consider reactor coolant with added soluble boron. The soluble boron
contains boron-11 (80.1 percent of natural boron). Boron-11 and available
alpha particles from the heavy elements react to produce source neutrons.
The equation below describes this reaction.
4
11
14
1
𝐻𝑒 + 𝐡 → 𝑁 + 𝑛
5
0
2
7
The boron-11 must be in very close proximity to the fuel for this reaction to
be possible because of the short travel length of alpha particles.
Transuranic Elements
A transuranic element is one that is beyond uranium (atomic number greater
than 92). These elements build up in the core during reactor power
operations due to various fuel isotopes undergoing neutron capture
reactions. The equation below describes the example of uranium-238
undergoing neutron capture to produce plutonium-239.
238
1
239
π‘ˆ+ 𝑛→
π‘ˆ
92
0
92
Transuranic elements contribute a significant number of source neutrons in
highly exposed reactor fuel. Either spontaneous fission or alpha-neutron
reactions produce these types of neutrons. Isotopes providing the major
contribution to these source neutron-producing reactions are the transuranic
elements curium and americium. The following reactions illustrate the
production of curium and alpha particles.
Curium-242 production:
Rev 1
35
239
4
242
1
𝑃𝑒 + 𝐻𝑒 →
πΆπ‘š + 𝑛
94
96
0
2
Alpha production (163-day half-life):
242
238
4
πΆπ‘š →
𝑃𝑒 + 𝐻𝑒
96
94
2
One gram of curium-244 produces about 5 x 105 neutrons per second
due to alpha-neutron reactions and about 1.2 x 107 neutrons per
second due to spontaneous fission.
ο‚· One gram of curium-242 produces about 2.5 x 107 neutrons per
second from alpha-neutron reactions and about 2.3 x 107 neutrons per
second from spontaneous fission.
ο‚· One gram of americum-241 produces approximately 4 x 103 neutrons
per second from alpha-neutron reactions. This isotope does not
undergo spontaneous fission.
ο‚·
Transuranic neutron sources produce about 100 neutrons per second for
every gram of fuel in the core or approximately 1 x 107 neutrons per second
core wide in a typical nuclear reactor core.
Knowledge Check
Select the three major contributors (types of reactions)
to the intrinsic neutron source in a nuclear reactor.
36
A.
Photo neutron
B.
Alpha neutron
C.
Spontaneous fission
D.
Oxygen-18 neutron
Rev 1
Knowledge Check
Some heavy nuclides will produce neutrons directly
from a reaction in the reactor referred to as
_______________ _____________.
A.
photo neutron
B.
alpha neutron
C.
spontaneous fission
D.
oxygen-18 alpha
ELO 4.3 Installed Neutron Sources
Introduction
Many reactors have additional neutron sources installed to ensure a reliable
and sufficient number of source neutrons because intrinsic neutron sources
can be relatively weak or dependent upon the recent power history of the
reactor. This may be especially important with all new fuel in the core.
Installed Neutron Sources
These installed neutron sources ensure that shutdown neutron levels are
high enough for the nuclear instruments to detect at all times. This provides
the operators with valid indication of the reactor's status. Installed neutron
sources are assemblies in the reactor for the sole purpose of producing
source neutrons.
Californium-252
The artificial nuclide californium-252 is a strong source of neutrons that
emits neutrons at the rate of about 2 x 1012 neutrons per second per gram
from spontaneous fission. It is not widely used as an installed neutron
source in commercial PWRs because of its high cost and short half-life
(2.65 years).
Beryllium Sources
Alpha-Neutron Beryllium Source
Many installed neutron sources use an alpha-neutron reaction with
beryllium. These sources are composed of metallic beryllium (100 percent
beryllium-9) with an alpha particle emitter, such as a compound of radium,
polonium, or plutonium. The reaction that occurs is below.
4
13 ∗ 12
1
9
𝐡𝑒 + 𝛼 → ( 𝐢) → 𝐢 + 𝑛
6
6
0
4
2
Rev 1
37
The beryllium is intimately (homogeneously) mixed with the alpha emitter
and is usually enclosed in a stainless steel capsule.
Photo-neutron Beryllium Source
Another installed neutron source that is commonly used is the photoneutron reaction with beryllium. Beryllium is a good source because its
stable isotope beryllium-9 has a weakly attached neutron with a binding
energy of only 1.66 MeV. Thus, a gamma ray with greater energy than 1.66
MeV can cause neutrons to eject because of the photo-neutron reaction
shown below.
1
9
8
𝛾 + 𝐡𝑒 → 𝐡𝑒 + 𝑛
0
4
4
Antimony-Beryllium Source
The most common installed source is antimony-beryllium (Sb-Be). Many
startup neutron sources use antimony and beryllium because after activation
with neutrons, the radioactive antimony becomes an emitter of high-energy
gammas. The reactions below show the reactions resulting in gamma
emissions.
123
1
124
𝑆𝑏 + 𝑛 →
𝑆𝑏 + 𝛾
51
0
51
𝛽 − 124
124
0
𝑆𝑏 →
𝑇𝑒 +
𝑒+𝛾
51
52
−1
The activated antinomy also decays with a 60-day half-life to produce a
gamma ray of sufficient energy to interact with the beryllium to produce a
neutron.
1
9
8
𝛾 + 𝐡𝑒 → 𝐡𝑒 + 𝑛
0
4
4
Antimony-beryllium (Sb-Be) photo-neutron sources of this type are
constructed differently from the alpha-neutron types. One design
incorporates a capsule of irradiated antimony enclosed in a beryllium
sleeve. Stainless steel cladding then encases the entire assembly. A PWR
may have several neutron sources of this type installed.
38
Rev 1
Knowledge Check
Which of the following would be used to provide a
source neutron population for a nuclear reactor core
that has never been operated (newly installed core)?
A.
Photo-neutron source
B.
Transuranic source
C.
Startup source
D.
Installed source
ELO 4.4 Intrinsic Source Neutrons Over Core Life
Introduction
Installed neutron sources provide a steady source of neutrons throughout
core life, although over long periods their strength does decay. Intrinsic
neutrons, however, do change in importance over core life.
Intrinsic Source Neutrons over Core Life
Photo-neutron source strength changes significantly over core life.
Initially, the core contains no fission products, and therefore there are no
high-energy gamma rays available to initiate photo-neutron reactions
resulting in source neutron production. In addition, deuterium
concentration, which is low in the beginning of life, limits source neutron
production. Both of these factors change and increase source neutron
production as core life increases.
At times past approximately core middle-of-life (MOL), photo-neutron
reactions are the largest contributor to the source neutron population
immediately following reactor shutdown. There are variations depending
on amounts of new or reused fuel present. Photo-neutron reactions
contribute large amounts of source neutrons for several days following
shutdown from power operations.
Transuranic Source Strength versus Core Life
The transuranic alpha-neutron intrinsic neutron source tends to contribute
the greatest number of source neutrons (than other intrinsic sources) in a
reactor core past middle-of-life (MOL) conditions, after several weeks of
reactor shutdown from power operations. The photo-neutron source
strength decays off after a couple of weeks.
Whether the photo-neutron or transuranic source provides the greatest
source neutron strength will depend on time in core life and the fuel mix new or partially reused. Once-burned and twice-burned fuel will possess
Rev 1
39
transuranic alpha-neutron sources as the reactor core's dominant intrinsic
neutron source.
Intrinsic Neutron Source Strength at Core Beginning of Life
At the beginning of core life (BOL) for a reactor, the relative strengths of
the intrinsic sources should be as follows (strongest to weakest):
Immediately following reactor shutdown from power:
ο‚·
ο‚·
ο‚·
Photo-neutron sources
Spontaneous fission sources
Alpha-neutron (transuranic) sources
Several weeks following reactor shutdown from power:
ο‚·
ο‚·
ο‚·
Spontaneous fission sources
Photo-neutron sources
Alpha-neutron (transuranic) sources
Note
Note
The alpha neutron (transuranic source) does not produce a
significant number of source neutrons until later in core life
or if once or twice burned fuel is used.
Intrinsic Neutron Source Strength at Core End of Life
As a reactor core approaches end-of-life (EOL) conditions, the relative
strengths of the intrinsic sources would be as follows (strongest to weakest):
Immediately following reactor shutdown from power:
ο‚·
ο‚·
ο‚·
Photo-neutron sources
Alpha-neutron (transuranic) sources
Spontaneous fission sources
Several weeks following reactor shutdown from power:
ο‚·
ο‚·
ο‚·
40
Alpha-neutron (transuranic) sources
Spontaneous fission sources
Photo-neutron sources
Rev 1
Knowledge Check
Which of the following reactions contributes the most
to the source neutron population in a nuclear reactor
core that is early in core life, shortly after the reactor is
shut down?
A.
Photo-neutron source
B.
Transuranic source
C.
Spontaneous fission source
D.
Californium source
TLO 4 Summary
Source Neutrons
ο‚· Source neutrons help initiate the fission process during a nuclear
reactor startup.
- They also ensure that the neutron population remains high
enough to allow a visible indication of neutron level during
shutdown and startup conditions.
ο‚· Intrinsic neutron sources are sources of neutrons from materials that
are in the reactor for other purposes such as fuel, burnable poison, or
moderator.
ο‚· Installed neutron sources are materials or components placed in the
reactor specifically to produce source neutrons.
ο‚· Intrinsic neutron sources are:
- Spontaneous fission
- Photo-neutron reactions
- Alpha-neutron reactions
ο‚· Spontaneous fission - results in production of fission fragments and
neutrons
- Uranium-235
- Uranium-238
- Curium-242
- Curium-244
ο‚· Photo-neutron reaction:
1
2
1
𝛾+ 𝐻→ 𝐻+ 𝑛
0
1
1
ο‚· After approximately MOL, the photo-neutron reaction is largest
contributor of intrinsic neutron sources.
ο‚· Alpha-neutron reactions:
4
18
21
1
𝐻𝑒 + 𝑂 → 𝑁𝑒 + 𝑛
8
10
0
2
4
11
14
1
𝐻𝑒 + 𝐡 → 𝑁 + 𝑛
5
0
2
7
Rev 1
41
ο‚·
Transuranic elements - contribute a significant number of source
neutrons in highly exposed reactor fuel. Either spontaneous fission or
alpha-neutron reactions produce these types of neutrons.
ο‚· Intrinsic neutron source strength at core beginning of life - at BOL the
relative strengths of the intrinsic sources are (strongest to weakest):
- Immediately following reactor shutdown from power:
o Photo-neutron sources
o Spontaneous fission sources
o Alpha-neutron (transuranic) sources
- Several weeks following reactor shutdown from power:
o Spontaneous fission sources
o Photo-neutron sources
o Alpha-neutron (transuranic) sources
- Note that the alpha neutron (transuranic source) does not
produce a significant number of source neutrons until later in
core life or if once or twice burned fuel is used.
ο‚· Intrinsic neutron source strength at core end of life - At EOL
conditions, the relative strengths of the intrinsic sources are (strongest
to weakest):
- Immediately following reactor shutdown from power:
o Photo-neutron sources
o Alpha-neutron (transuranic) sources
o Spontaneous fission sources
- Several weeks following reactor shutdown from power:
o Alpha-neutron (transuranic) sources
o Spontaneous fission sources
o Photo-neutron sources
ο‚· Examples of installed neutron sources are:
- Spontaneous fission of californium-252 results in fission
fragments and free neutrons
- Beryllium alpha-neutron reaction (alpha from the decay of
plutonium, polonium, or radium):
4
13 ∗ 12
1
9
𝐡𝑒 + 𝛼 → ( 𝐢) → 𝐢 + 𝑛
6
6
0
4
2
- Beryllium photo-neutron reaction (high-energy gamma from
decay of antimony-124):
1
9
8
𝛾 + 𝐡𝑒 → 𝐡𝑒 + 𝑛
0
4
4
- Antimony-beryllium source – this is the most common installed
source. The activated antinomy also decays with a 60-day halflife to produce a gamma ray of sufficient energy to interact with
the beryllium to produce a neutron.
123
1
124
𝑆𝑏 + 𝑛 →
𝑆𝑏 + 𝛾
51
0
51
𝛽 − 124
124
0
𝑆𝑏 →
𝑇𝑒 +
𝑒+𝛾
51
52
−1
42
Rev 1
Objectives
Now that you have completed this lesson, you should be able to:
1. Describe the purpose and importance of source neutrons.
2. Describe, including examples, each of the following types of intrinsic
neutron sources:
a.
b.
c.
d.
Spontaneous fission
Photo-neutron reactions
Alpha-neutron reactions
Transuranic elements
3. Describe the purpose and type of installed neutron sources.
4. Describe the primary source of intrinsic source neutrons in the reactor
for the following conditions:
a. At beginning and end of core life
b. Immediately following a reactor shutdown
c. Several weeks after reactor shutdown
TLO 5 Relationship between Neutron Flux,
Microscopic, and Macroscopic Cross-Sections
Overview
Fission neutrons are born at an average energy of about 2 MeV and interact
with reactor core materials in various absorption and scattering reactions.
Scattering reactions are useful for thermalizing neutrons. Thermal neutrons
may be absorbed by fissile nuclei to produce fission or absorbed in fertile
material resulting in the production of fissionable fuel. Additionally, some
neutrons are absorbed in structural components, reactor coolant, and other
non-fuel materials resulting in the removal of neutrons from the fission
process.
To determine these neutron interaction rates, it is necessary to identify the
number of neutrons available and the probability interaction. To assist in
quantifying neutron availability and reaction probabilities, we use terms
including neutron flux, microscopic and macroscopic cross-section.
The complexity of designing a reactor requires predicting of the various
reaction rates, both in specific portions of the core and averages throughout
the core. One example is the calculation of the reactor's thermal output,
knowing the fission rate and core volume.
Rev 1
43
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Explain the following terms, including any mathematical relationships:
a. Atomic density
b. Neutron flux
c. Fast neutron flux
d. Thermal neutron flux
e. Microscopic cross-section
f. Barn
g. Macroscopic cross-section
h. Mean free path
2. Define the following neutron classes:
a. Fast
b. Intermediate
c. Slow
3. Describe how the absorption and scattering cross-section of typical
nuclides varies with neutron energies in the 1 eV region, and the
resonance absorption region.
4. Describe the macroscopic cross-section and mean free path at various
temperatures.
5. Describe radial and axial neutron flux distribution.
6. Describe how changes in neutron flux and macroscopic cross-section
affect reaction rates.
7. Describe the relationship between neutron flux and reactor power.
ELO 5.1 Neutron Reaction Terms
Introduction
The following terms are important for understanding the theoretical
concepts of nuclear reactors and are foundational for future lessons:
ο‚·
ο‚·
ο‚·
ο‚·
ο‚·
ο‚·
Atomic density
Neutron flux
Fast neutron flux
Thermal neutron flux
Microscopic cross-section
Mean free path
Atomic Density
An important property of a material is its atomic density. The atomic
density is the number of atoms of a given type per unit volume of the
material. Use the following equation to calculate the atomic density of a
substance.
44
Rev 1
𝑁=
πœŒπ‘π΄
𝑀
Where:
N = Atomic density (atoms/cm3)
ρ = density (g/cm3)
NA = Avogadro’s number (6.022 x 1023 atoms/mole)
M = gram atomic weight or gram molecular weight
Example:
A block of aluminum has a density of 2.699 g/cm3. If the gram atomic
weight of aluminum is 26.9815 g, calculate the atomic density of the
aluminum.
Solution:
𝑁=
πœŒπ‘π΄
𝑀
𝑔
23 π‘Žπ‘‘π‘œπ‘šπ‘ 
)
3 (6.022 × 10
π‘šπ‘œπ‘™π‘’
π‘π‘š
=
𝑔
26.9815
π‘šπ‘œπ‘™π‘’
π‘Žπ‘‘π‘œπ‘šπ‘ 
= 6.024 × 1022
π‘π‘š3
2.699
Neutron Flux
Macroscopic cross-sections for neutron reactions with a specific material
determine the probability of one neutron undergoing a specific reaction per
centimeter of its travel through that material. It is necessary to determine
how many neutrons travel through the material, and the distance (in
centimeters) the neutrons travel each second, to determine how many
reactions actually occur in a field of neutrons. These calculations are based
on the number of neutrons existing in one cubic centimeter at any one
instant and the total distance they travel each second while in that cubic
centimeter.
The number of neutrons existing in a cm3 of material at any instant is the
neutron density, represented by the symbol n with units of neutrons/cm3.
The neutrons' velocities determine the total distance these neutrons travel
each second.
Neutron flux is the number of neutrons passing through the unit area (cm2)
per unit time. Neutron flux (Φ) is the total path length covered by all
neutrons in one cubic centimeter during one second. Its units are neutrons
per square centimeter per second (n/cm2/sec). The equation below shows
Rev 1
45
the relationship between neutron flux, neutron density, and neutron
velocity.
Φ = 𝑛𝑣
Where:
Φ = neutron flux (neutron/cm2-sec)
n = neutron density (neutrons/cm3)
v = neutron velocity (cm/sec)
The term neutron flux refers to parallel beams of neutrons traveling in a
single direction. The intensity (I) of a neutron beam is the product of the
neutron density and the average neutron velocity. The directional beam
intensity is equal to the number of neutrons per unit area and time
(neutrons/cm2-sec) falling on a surface perpendicular to the direction of the
beam.
Neutron flux is modeled as many neutron beams traveling in various
directions in a nuclear reactor. The neutron flux becomes the scalar sum of
these directional flux intensities (added as numbers and not vectors),
expressed as follows: 𝛷 = 𝐼1 + 𝐼2 + 𝐼3 + 𝐼𝑛 . All of the directional beams
contribute to the total reaction rate since the atoms in a reactor do not prefer
neutrons coming from any particular direction. In reality, at any given point
within a reactor, neutrons are traveling in all directions.
There are two classes of neutron flux: thermal neutron flux and fast neutron
flux in a nuclear reactor. The next section defines these neutron energies.
Thermal Neutron Flux
Thermal neutron flux (Φth) is the number of thermal neutrons crossing a unit
area in the reactor in a given amount of time. The units are neutrons
(thermal) per square centimeter per second (n/cm2/sec) as with total neutron
flux.
Remember that neutron flux is omni-directional, meaning neutrons can
enter a particular square centimeter of reactor material from any direction.
Therefore, Φth also equals the total distance of all thermal neutrons diffused
(moved) in a particular unit volume in one second.
Fast Neutron Flux
Fast neutron flux (Φf) is number of fast neutrons crossing a unit area in a
given amount of time. The units are neutrons (fast) per square centimeter
per second (n/cm2/sec) as with thermal neutron flux. Fast neutron flux also
equals the distance that fast neutrons diffuse (move) in a particular unit
volume in one second. The next section defines fast neutrons.
46
Rev 1
Cross-Sections
The probability of a neutron interacting with a nucleus for a particular
reaction depends on the nucleus involved and, importantly, the energy of
the neutron. Most materials are more likely to absorb a thermal neutron
than a fast neutron. Additionally, the probability of the neutron interaction
varies with the type of reaction involved.
Note
Note
Cross-section depends on the characteristics of the nucleus,
not on the size of nucleus.
Microscopic Cross-Section (σ)
The probability of a particular reaction occurring between a neutron and a
nucleus is the microscopic cross-section (σ) of the nucleus. This crosssection varies with the energy of the neutron, and is the effective area the
nucleus presents to the neutron for the particular reaction. Barn is the unit
of measure for this area. A larger effective area yields a greater probability
for reaction.
Total Microscopic Cross-Section (σT)
A neutron interacts with an atom in two basic ways: scattering or absorption
reaction. The probability of a neutron being absorbed is the microscopic
cross-section for absorption (σa). The probability of a neutron scattering is
the microscopic cross-section for scattering (σs). The sum of these two
microscopic cross-sections is the total microscopic cross-section (σT).
πœŽπ‘‡ = πœŽπ‘Ž + πœŽπ‘ 
Microscopic Cross-Section for Scattering (σs)
Both the absorption and the scattering microscopic cross-sections have two
components. For instance, the scattering cross-section is the sum of the
elastic scattering cross-section (σse) and the inelastic scattering cross-section
(σsi).
πœŽπ‘  = πœŽπ‘ π‘’ + πœŽπ‘ π‘–
Microscopic Cross-Section for Absorption (σa)
The microscopic absorption cross-section (σa) includes all reactions except
scattering. However, for most purposes it has two categories, fission (σf)
and capture (σc).
πœŽπ‘Ž = πœŽπ‘“ + πœŽπ‘
Rev 1
47
The Chart of Nuclides uses two different conventions to represent
microscopic cross-section for capture, depending on the type of particle
ejected from the nucleus after the capture reaction.
ο‚·
ο‚·
If a gamma ray results from the capture, σs is used
If an alpha particle results from the capture, σα is used
A later section discusses fission and radiative capture of neutrons in detail.
Barns
Microscopic cross-sections are expressed in units of area - usually square
centimeters. A square centimeter is very large compared to the effective
area of a nucleus. The old story goes that a physicist once referred to the
measure of a square centimeter as being "as big as a barn" when viewed on
a nuclear level. The name has persisted, so now microscopic cross-sections
have units of barns. The conversion to cm2 is below.
1 π‘π‘Žπ‘Ÿπ‘› = 10−24 π‘π‘š2
Consider boron-10, for example. This nuclide has a relatively low mass
with an absorption cross-section of 3,838 barns resulting in an ejected alpha
particle. A boron-10 nucleon presents a very large effective area for
neutron interaction. Lead-208 is a nuclide of relatively high mass and only
has an absorption cross-section of 8 microbarns (8 x 10-6 barns) for the same
reaction. Lead-208 presents a very small effective area for neutron
interaction.
Macroscopic Cross-Section (Σ)
Whether or not a neutron interacts with a certain volume of material
depends not only on its microscopic cross-section but also on the number of
nuclei within that volume (atomic density). Macroscopic cross-section (Σ)
is the probability of a given reaction occurring per unit travel of the neutron.
Macroscopic cross-section is relates to microscopic cross-section (σ) by the
following relationship.
Σ = π‘πœŽ
Where:
Σ = macroscopic cross-section (cm-1)
N = atomic density (atoms/cm3)
σ = microscopic cross-section (cm2)
The number of target nuclei per unit volume increases and the probability of
interaction increases if atomic density increases.
48
Rev 1
Microscopic Calculation
Most materials are composed of several elements. Most elements are
composed of several isotopes; therefore, most materials involve many
cross-sections, one for each isotope involved. Each macroscopic crosssection is determined to include all the isotopes for a material, which are
added together as follows:
𝛴 = 𝑁1 𝜎1 + 𝑁2 𝜎2 + 𝑁3 𝜎3 +. . . . . . .. 𝑁𝑛 πœŽπ‘›
Where:
Nn = the number of nuclei per cm3 of the nth element
σn = the microscopic cross-section of the nth element
Example:
Find the macroscopic thermal neutron absorption cross-section for iron,
which has a density of 7.86 g/cm3. The iron microscopic cross-section for
absorption is 2.56 barns and the gram atomic weight is 55.847 g.
Solution:
Step 1:
Using the equation for atomic density, calculate the atomic density of iron.
𝑁=
πœŒπ‘π΄
𝑀
7.68
=
𝑔
π‘Žπ‘‘π‘œπ‘šπ‘ 
(6.022 × 1023
)
π‘šπ‘œπ‘™π‘’
π‘π‘š3
𝑔
55.847
π‘šπ‘œπ‘™π‘’
= 8.48 × 1022
π‘Žπ‘‘π‘œπ‘šπ‘ 
π‘π‘š3
Step 2:
Using the atomic density and given microscopic cross-section, calculate the
macroscopic cross-section.
Σπ‘Ž = π‘πœŽπ‘Ž
= 8.48 × 1022
π‘Žπ‘‘π‘œπ‘šπ‘ 
1 × 10−24 π‘π‘š2
(2.56
π‘π‘Žπ‘Ÿπ‘›π‘ )
(
)
π‘π‘š3
1 π‘π‘Žπ‘Ÿπ‘›
= 0.217 π‘π‘š−1
Rev 1
49
Microscopic versus Macroscopic Cross-Section
The difference between the microscopic and macroscopic cross-sections is
extremely important and deserves restatement. The microscopic crosssection (σ) represents the effective target area that a single nucleus presents
to a bombarding particle (neutron). The units are in barns or cm2.
The macroscopic cross-section (Σ) represents the total effective target area
of the nuclei contained in 1 cm3 of the material. The units are 1/cm or cm-1.
Mean Free Path
A neutron has a certain probability of undergoing a particular interaction in
one centimeter of travel (Σ) in a material. The inverse of this probability
describes how far the neutron will travel (average) before undergoing an
interaction. This average distance of travel by a neutron before interaction
is the mean free path for interaction and represented by the symbol λ. The
equation below expresses the mathematical relationship between the mean
free path (λ) and the macroscopic cross-section (Σ):
πœ†=
1
Σ
Mean Free Path Calculation
The macroscopic cross-section must be determined to calculate the mean
fee path. The following example will demonstrate a mean free calculation
for a material composed of both aluminum and silicon.
Example
An alloy is composed of 95 percent aluminum and 5 percent silicon (by
weight). The density of the alloy is 2.66 g/cm3. The table below shows the
properties of aluminum and silicon.
Element
Gram Atomic
Weight
σa (barns)
σs (barns)
Aluminum
26.9815
0.23
1.49
Silicon
28.0855
0.16
2.20
Perform the following:
Calculate the atomic densities for aluminum and silicon.
Determine absorption and scattering macroscopic cross-sections for thermal
neutrons.
50
Rev 1
Calculate the mean free paths for absorption and scattering.
Solution:
Step 1:
The density of aluminum is 95 percent of the total density. Using the
formula for atomic density, calculate the atomic densities for aluminum and
silicon.
𝑁𝐴𝑙 =
πœŒπ΄π‘™ 𝑁𝐴
𝑀𝐴𝑙
𝑔
π‘Žπ‘‘π‘œπ‘šπ‘ 
) (6.022 × 1023
)
3
π‘šπ‘œπ‘™π‘’
π‘π‘š
=
𝑔
26.9815
π‘šπ‘œπ‘™π‘’
π‘Žπ‘‘π‘œπ‘šπ‘ 
= 5.64 × 1022
π‘π‘š3
0.95 (2.66
𝑁𝑆𝑖 =
πœŒπ‘†π‘– 𝑁𝐴
𝑀𝑆𝑖
𝑔
π‘Žπ‘‘π‘œπ‘šπ‘ 
) (6.022 × 1023
)
3
π‘šπ‘œπ‘™π‘’
π‘π‘š
=
𝑔
28.0855
π‘šπ‘œπ‘™π‘’
π‘Žπ‘‘π‘œπ‘šπ‘ 
= 2.85 × 1021
π‘π‘š3
0.05 (2.66
Step 2:
Calculate the absorption and scattering macroscopic cross-sections from the
microscopic cross-sections and the combined atomic densities (95 percent
aluminum and 5 percent silicon).
Σπ‘Ž = 𝑁𝐴𝑙 πœŽπ‘Ž,𝐴𝑙 + 𝑁𝑆𝑖 πœŽπ‘Ž,𝑆𝑖
π‘Žπ‘‘π‘œπ‘šπ‘ 
) (0.23 × 10−24 π‘π‘š2 )
π‘π‘š3
π‘Žπ‘‘π‘œπ‘šπ‘ 
+ (2.85 × 1021
) (0.16 × 10−24 π‘π‘š2 )
π‘π‘š3
= (5.64 × 1022
= 0.0134 π‘π‘š−1
Σ𝑠 = 𝑁𝐴𝑙 πœŽπ‘ ,𝐴𝑙 + 𝑁𝑆𝑖 πœŽπ‘ ,𝑆𝑖
Rev 1
51
π‘Žπ‘‘π‘œπ‘šπ‘ 
) (1.49 × 10−24 π‘π‘š2 )
π‘π‘š3
π‘Žπ‘‘π‘œπ‘šπ‘ 
+ (2.85 × 1021
) (2.20 × 10−24 π‘π‘š2 )
π‘π‘š3
= (5.64 × 1022
= 0.0903 π‘π‘š−1
Step 3:
Determine mean free paths from the calculated absorption and scattering
macroscopic cross-sections.
πœ†π‘Ž =
=
1
Σπ‘Ž
1
0.01345 π‘π‘š−1
= 74.3 π‘π‘š
πœ†π‘  =
=
1
Σ𝑠
1
0.0903 π‘π‘š−1
= 11.1 π‘π‘š
Conclusions:
ο‚· A neutron must travel an average of 74.3 cm to interact by absorption
in this alloy.
ο‚· A neutron must travel an average of 11.1 cm to interact by scattering
in this alloy.
Knowledge Check
_______________ is the total path length traveled by
all neutrons in one cubic centimeter of material during
one second.
52
A.
Mean free path
B.
Neutron flux
C.
Gamma flux
D.
Atomic density
Rev 1
ELO 5.2 Neutron Energy Terms
Introduction
There are three classes of neutron energy levels: fast, intermediate and slow.
Neutrons that are in energy equilibrium with their surrounding are thermal
neutrons. They are the most important for thermal reactors.
Neutron Energies Terms
The figure below illustrates the relative energy levels of neutrons and their
flux distribution in a typical thermal reactor.
Figure: Neutron Energy
Definitions of the three classes of neutron energy levels are as follows:
ο‚·
Fast neutrons - Fission neutrons are born as fast neutrons. They are
categorized with energy levels greater than 0.1 MeV (105 eV)
ο‚· Intermediate neutrons - neutrons with energy levels between 1 eV and
0.1 MeV
ο‚· Slow neutrons - neutrons with energy levels less than 1 eV
ο‚· Thermal Neutrons – neutrons with an energy level of 0.025 eV (2.2 x
105 cm/sec. velocity) at 68° F. Velocity and energy increase with
temperature.
Rev 1
53
Knowledge Check
Thermal neutrons are classified in the intermediate
neutron energy range.
A.
True
B.
False
ELO 5.3 Neutron Energies versus Cross-Sections
Introduction
Neutron energies affect both neutron absorption and neutron scattering
cross-sections. A higher energy neutron has a lower probability of
interaction in general. Resonance peaks at certain specific energies possess
very high cross-sections.
Neutron Energies versus Cross-Sections
Neutron absorption cross-sections have three unique regions of resonance
probability related to neutron energy. These are the 1/v region, the
resonance region, and the fast neutron region. Each region has a different
relationship to changing neutron energy.
Neutron scattering cross-sections are somewhat different. Resonance
elastic scattering and inelastic scattering cross-sections do have resonance
peaks, but they are smaller than absorption peaks. Neutron energy has little
effect on potential elastic scattering cross-sections. Scattering reactions are
good since they do not lose neutrons but cause them to thermalize for use as
fission neutrons.
Neutron Scattering Cross-Section versus Incident Neutron
Energy
The elastic scattering cross-sections are generally small, typically 5 barns to
10 barns, with the exception of hydrogen. This is close to the magnitude of
the actual geometric cross-sectional area expected for atomic nuclei. The
cross-section is essentially constant and independent of neutron energy up
to around 1 MeV in potential elastic scattering.
Resonance elastic scattering and inelastic scattering exhibit resonance peaks
similar to those associated with absorption cross-sections. The resonances
occur at lower energies for heavy nuclei than for light nuclei. In general,
the variations in scattering cross-sections are very small when compared to
the variations that occur in absorption cross-sections.
Neutron scattering cross-sections are most important in the moderator
where neutrons are thermalized in a thermal reactor. Elastic scatterings are
most probable in light nuclei, σs = σselastic since water molecules (the
moderator) are light nuclei. The following figure illustrates the relative
54
Rev 1
consistency of σs throughout the neutron energy ranging from 0.025 eV to
0.1 MeV. The standard elastic scattering cross-section table is based on a
value of 0.025 eV at 68 degrees F.
Figure: Elastic Scattering Cross-Section versus Neutron Energy
Neutron Absorption Cross-Section versus Incident Neutron
Energy
The variation of absorption cross-sections with neutron energy is complex.
The absorption cross-sections are small, ranging from a fraction of a barn to
a few barns for slow and thermal neutrons for many elements.
For a considerable number of nuclides of moderately high to high mass
numbers the variation of absorption cross-sections with incident neutron
energy reveals three distinct regions on an absorption cross-section versus
neutron energy curve. These regions are the 1/v region, resonance peaks
region and fast neutron region. The figure below illustrates these regions.
Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy
Rev 1
55
1/v Region
In the 1/v region, the cross-section decreases steadily with increasing
neutron energy; this low energy region includes thermal neutrons (< 1 eV).
In this region the absorption cross-section, which is often high, is inversely
proportional to the velocity (v). This region is frequently referred to as the
"1/v region" because of the inverse relationship to velocity or energy
(velocity and energy are directly proportional, but both are inversely
proportional to cross-section).
Resonance Region
Beyond the 1/v region, there is a resonance region where the cross-sections
rise sharply to high value peaks. These are resonance peaks with neutrons
in the intermediate energy (epithermal neutrons) range. These energies are
resonance peaks and are a result of the affinity of the nucleus for neutrons
whose energies closely match its discrete, quantum (preferred) energy
levels. These peaks occur where the neutron's binding energy plus its
kinetic energy are exactly equal to the amount of energy required to raise a
compound nucleus from its ground state to a quantum level energy state.
Resonance absorption occurs when these peaks match energy required to
reach a quantum energy state.
The following example problem further illustrates the absorption in the
resonance region.
Example:
Calculate the kinetic energy a neutron must possess to undergo resonant
absorption in uranium-235 at this resonance energy level assuming that
uranium-235 has a nuclear quantum energy level at 6.8 MeV above its
ground state.
Solution:
𝐡𝐸 = [π‘€π‘Žπ‘ π‘  (
235
π‘ˆ) + π‘€π‘Žπ‘ π‘  (π‘›π‘’π‘’π‘‘π‘Ÿπ‘œπ‘›) − π‘€π‘Žπ‘ π‘  (
235
π‘ˆ)] × 931
𝐡𝐸 = (235.043925 + 1.008665 − 236.045563) × 931
𝐡𝐸 = (0.007025 π‘Žπ‘šπ‘’) × 931
𝑀𝑒𝑉
π‘Žπ‘šπ‘’
𝑀𝑒𝑉
π‘Žπ‘šπ‘’
𝑀𝑒𝑉
= 6.54 𝑀𝑒𝑉
π‘Žπ‘šπ‘’
6.8 𝑀𝑒𝑉 − 6.54 𝑀𝑒𝑉 = 0.26 𝑀𝑒𝑉
The difference between the binding energy and the quantum energy level
equals the amount of kinetic energy the neutron must possess. The typical
heavy nucleus will have many closely spaced resonances starting in the low
energy (eV) range. This is because heavy nuclei are complex and have
more possible configurations and corresponding energy states. Light nuclei,
56
Rev 1
being less complex, have fewer possible energy states and fewer resonances
distributed at higher energy levels.
Fast Neutron Region
The absorption cross-section steadily decreases as the energy of the neutron
increases for higher neutron energies. This is the fast neutron region. The
absorption cross-sections are usually less than 10 barns in this region. This
explains the low probability for fast fissioning of neutrons in this region.
Knowledge Check
At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to
the neutron velocity. (Fill in the blank).
ELO 5.4 Temperature Affects to Macroscopic Cross-Sections
Introduction
As neutron energy varies, microscopic absorption cross-sections vary
significantly. The microscopic cross-section values given on most charts
and tables are at an ambient temperature of temperature of 68°F. This
equates to a neutron velocity of 2,200 eV of energy. You must correct the
absorption cross-section for the new temperature value if a material is at
higher temperatures.
Calculate the macroscopic cross-section and mean free path following the
determination of the temperature corrected microscopic cross-section.
Note
Note
Rev 1
Temperature corrections apply to any cross-section
involving absorption (for example, σa, σc, σf); however,
they do not apply to scattering cross-sections, as neutron
energies up to 1 MeV have little effect on the predominant
elastic scatters occurring in the moderator. (ELO 5.3)
57
Temperature Effects on Microscopic Cross-Sections
The following calculations illustrate how macroscopic cross-section and
mean free path vary with temperatures.
Step
1.
Action
Obtain the standard (at 68°F) table value of the desired
microscopic cross-section.
2.
Convert temperatures to °R or °K.
°π‘… = °πΉ + 460
°πΎ = °πΆ + 273
3.
Calculate new microscopic cross-section using:
1
π‘‡π‘œ 2
𝜎 = πœŽπ‘œ ( )
𝑇
Where:
σ = microscopic cross-section corrected for temperature
σo = microscopic cross-section at reference temperature (68°F or
20°C)
To = reference temperature (68°F) in degrees Rankine (°R) or
Kelvin (°K)
T = temperature for which corrected value is being calculated
Temperature Effects to Macroscopic Cross-Sections
Demonstration
Example:
What is the fission macroscopic cross-section and mean free path for
uranium-235 for thermal neutrons at 500°F? Given:
Uranium-235 has a σf of 583 barns at 68°F.
Atomic density of uranium-235 equals 7.03 x 1020 atoms/cm3 at 500°F
58
Rev 1
Solution:
Steps 1, 2, and 3:
1
π‘‡π‘œ 2
πœŽπ‘“ = πœŽπ‘“,π‘œ ( )
𝑇
1
68℉ + 460 2
= 583 π‘π‘Žπ‘Ÿπ‘›π‘  (
)
500℉ + 460
= 432 π‘π‘Žπ‘Ÿπ‘›π‘ 
Step 4:
Σ = π‘πœŽ
πœŽπ‘“ = 7.03 × 1020
π‘Žπ‘‘π‘œπ‘šπ‘ 
π‘π‘š2
−24
×
432
π΅π‘Žπ‘Ÿπ‘›π‘ 
×
(1
×
10
)
π‘π‘š3
π΅π‘Žπ‘Ÿπ‘›
πœŽπ‘“ = 0.304 π‘π‘š−1
Step 5:
πœ†=
πœ† =
1
Σ
1
0.304 π‘π‘š−1
πœ† = 3.29 π‘π‘š
Note
Note
Rev 1
As temperature increases, the microscopic cross-section for
fission decreases (less probability for fission). The fission
macroscopic cross-section dropped from 0.41 to 0.30 cm-1
and the fission mean free path increased from 2.44 cm to
3.29 cm.
59
Knowledge Check
If an isotope has an absorption microscopic crosssection of 5,000 barns at 68°F, what happens to its
absorption microscopic cross-section if temperature is
increased to 1,000°F?
A.
Increases
B.
Decreases
C.
No change
D.
Need more data to answer
ELO 5.5 Neutron Flux Distribution
Introduction
Neutron flux distribution is very important; both operator and reactor
engineering sections monitor it. This topic is important for safe operation
of the reactor. This section provides initial information on neutron flux
distribution for the operator.
Neutron Flux Distribution
Neutrons interact with all materials in a reactor, absorptions, scattering, etc.
The type of material found in particular locations of the core affects the
neutron flux in that area. For example, fast neutron flux in materials with
large scattering cross-sections will cause fast neutrons to slow down to
lower energy levels more quickly. This reduces the fast neutron flux. At
same time, intermediate and thermal neutron flux increases as a larger
percentage of fast neutrons slow down to these energy levels.
Similarly, thermal neutron flux in a material with large absorption crosssections decreases as a larger number of thermal neutrons are absorbed in
material thereby reducing the thermal neutron flux levels.
Axial and Radial Neutron Flux
Commercial PWR nuclear reactor cores are cylindrical in shape. Axial and
radial flux, describe neutron flux profiles from top to bottom (axial) and
across (radial) the core. Axial flux is the side view of the core (top to
bottom) and radial flux is the top view of the core (side to side). The next
figure graphically illustrates axial and radial flux (assuming a uniformly
distributed fuel with no zoning at the BOL).
60
Rev 1
Figure: Axial and Radial Neutron Flux in a Reactor Core
The axial and radial neutron flux distribution across a nuclear reactor is a
spatial representation of neutron flux level throughout the core. Fission
produces neutrons, the moderator slows them down, the core materials and
fuel within the core boundaries capture neutrons, and the neutrons undergo
fission, leading to more of these same processes (the cycle repeats). All of
these neutron processes affect the neutron flux distribution in the core.
Neutron flux distribution is generally highest in the center of the core and
drops off toward the core boundaries (top and bottom, sides). Neutrons
produced near any edge (top, bottom, sides) of the core are more likely to
leak out of the core and not cause fission. This reduces thermal neutron
flux in the outer boundaries of the core. Neutrons produced toward the
center of the core have less leakage and a greater probability of thermalizing
and causing more fission events; therefore, neutron flux levels are higher
toward the center of the core.
Self-Shielding
The neutron flux level may be lower in some localized areas than in others
because of self-shielding. For example, the interior of a fuel pin receives a
lower average neutron flux level than the outer surfaces since an
appreciable fraction of the neutrons will have been absorbed in the outer
layers of the fuel, thereby reducing neutron availability to the fuel pellet
interior.
The concept of self-shielding is also important relating to neutron poisons in
the reactor core; a later section of the course will cover self-shielding in
detail.
Rev 1
61
Knowledge Check
The neutron flux that varies from the top to bottom of
the core is called the ________ flux and the neutron
flux that varies from across the top is ________.
ELO 5.6 Reaction Rate
Introduction
Two factors are required in order to calculate the number of interactions
taking place in a cubic centimeter in one second. These two factors are the
total path length of all the neutrons in a cubic centimeter per second
(neutron flux [Φ]), and the probability of an interaction per centimeter path
length (macroscopic cross-section [Σ]). Multiply these two factors together
to get the number of interactions taking place in that cubic centimeter in one
second. The resulting value is the reaction rate, denoted by the symbol R.
The type of reaction rate calculated will depend on the macroscopic crosssection used in the calculation. Normally, the reaction rate is of greatest
interest is the fission reaction rate.
Calculate the Reaction Rate
Step
Action
1.
Obtain the average thermal neutron flux in the reactor.
2.
Obtain the fuel macroscopic cross-section for the particular
material and reaction of interest. Determine the microscopic
cross-section and atomic density for material and calculate the
macroscopic cross-section if the macroscopic cross is unknown
using:
Σ = π‘πœŽ
Where:
Σ = macroscopic cross-section (cm-1)
N = atomic density (atoms/cm3)
σ = microscopic cross-section (cm2)
3.
Calculate the reaction rate using the following formula:
𝑅 = ΣΦ
62
Rev 1
Step
Action
Where:
R = reaction rate (reactions/cm3-sec)
Σ = macroscopic cross-section (cm-1)
Φ = neutron flux (neutrons/cm2-sec)
Note
This reaction rate is only calculating the reactions occurring
in one cubic centimeter. This unit value requires
multiplication by the core volume to determine the
reactions in the entire core. The next session will discuss
the conversion.
Note
Reaction Rate
Example:
A reactor has a macroscopic fission cross-section of 0.1 cm-1, and thermal
neutron flux of 1013 neutrons/cm2-sec, what is the fission rate in that cubic
centimeter?
Solution:
𝑅𝑓 = ΦΣf
= (1 × 1013
= 1 × 1012
π‘›π‘’π‘’π‘‘π‘Ÿπ‘œπ‘›π‘ 
) (0.1 π‘π‘š−1 )
π‘π‘š2 – 𝑠𝑒𝑐
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
π‘π‘š3 – 𝑠𝑒𝑐
Neutron Flux and Macroscopic Cross-Section Versus Reaction
Rate
Increasing either the neutron flux or the macroscopic cross-section increases
the reaction rate. Macroscopic cross-section decreases with fuel burnup and
decreasing the reaction rate. An operator can increase neutron flux to
compensate and maintain the reaction rate.
Rev 1
63
Knowledge Check
Calculate the reaction rate (fission rate) in a one cubic
centimeter section of a reactor that has a macroscopic
fission cross-section of 0.2 cm-1, and a thermal neutron
flux of 1014 neutrons/cm2-sec.
A. R = 2.0 × 1014
B. R = 0.2 × 1013
neutrons
cm2 –sec
neutrons
cm2 –sec
C. R = 2.0 × 1013
neutrons
D. 𝑅 = 20 × 1013
π‘›π‘’π‘’π‘‘π‘Ÿπ‘œπ‘›π‘ 
cm3 –sec
π‘π‘š3 –𝑠𝑒𝑐
ELO 5.7 Neutron Flux and Reactor Power
Introduction
Multiplying the reaction rate per unit volume by the total volume of the core
equals the total number of reactions occurring in the core per unit time. It is
possible to calculate the rate of energy release (power) due to a certain
reaction given the amount of energy involved in each reaction.
Neutron Flux and Reactor Power Step-by-Step Tables
The number of fissions to produce one watt of power requires the following
conversion factors in a reactor where average energy per fission is 200
MeV:
ο‚·
ο‚·
ο‚·
1 fission = 200 MeV
1 MeV = 1.602 x 10-6 ergs
1 erg = 1 x 10-7 watt-sec
1 π‘’π‘Ÿπ‘”
1 𝑀𝑒𝑉
1 π‘“π‘–π‘ π‘ π‘–π‘œπ‘›
1 π‘€π‘Žπ‘‘π‘‘ (
)(
)(
)
−7
−6
1 × 10 π‘€π‘Žπ‘‘π‘‘– 𝑠𝑒𝑐 1.602 × 10 π‘’π‘Ÿπ‘” 200 𝑀𝑒𝑉
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›
= 3.12 × 1010
π‘ π‘’π‘π‘œπ‘›π‘‘
This is equivalent to stating that 3.12 x 1010 fissions release 1 watt-second
of energy. You can use this equation to calculate the power released in a
reactor. Multiply the reaction rate by the volume of the reactor to obtain the
total fission rate for the entire reactor. Divide the total fission rate by the
number of fissions per watt-sec to obtain the power released by fission in
the reactor in watts. Solve for reactor thermal power (watts or Megawatts)
using the following steps based on this equation.
64
Rev 1
Step
1.
Action
Obtain the following reactor data:
ο‚·
ο‚·
ο‚·
Volume of core (cm3)
Reaction rate
Fission macroscopic cross-section and average thermal
neutron flux
or,
𝑅 = ΣΦ
2.
Calculate reactor power using the following equation, substitute
reaction rate if known:
𝑃=
Φπ‘‘β„Ž Σ𝑓 𝑉
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
3.12 × 1010 π‘€π‘Žπ‘‘π‘‘– 𝑠𝑒𝑐
Where:
P = power (watts)
Φth = thermal neutron flux (neutrons/cm2 - sec)
Σf = macroscopic cross-section for fission (cm-1)
V = volume of core (cm3)
Neutron Flux and Reactor Power Demonstration
Example:
Calculate reactor power given the following:
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘œπ‘Ÿπ‘’ (π‘π‘š3 ) = 20,000 π‘π‘š3
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
ο‚· π‘…π‘’π‘Žπ‘π‘‘π‘–π‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘’ = 1 × 1015
π‘π‘š3 –𝑠𝑒𝑐
ο‚·
Solution:
Step 1:
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘œπ‘Ÿπ‘’ (π‘π‘š3 ) = 20,000 π‘π‘š3
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
ο‚· π‘…π‘’π‘Žπ‘π‘‘π‘–π‘œπ‘› π‘Ÿπ‘Žπ‘‘π‘’ = 1 × 1015
π‘π‘š3 –𝑠𝑒𝑐
ο‚·
Step 2:
Rev 1
65
𝑃=
Φπ‘‘β„Ž Σ𝑓 𝑉
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
3.12 × 1010 π‘€π‘Žπ‘‘π‘‘– 𝑠𝑒𝑐
Substitute: Reaction rate for thermal neutron flux multiplied by the
macroscopic cross-section for fission
𝑃 =
1 × 1015 × 20,000
3.12 × 1010
𝑃 = 6.41 × 108 π‘€π‘Žπ‘‘π‘‘π‘  = 641 π‘€π‘’π‘”π‘Žπ‘€π‘Žπ‘‘π‘‘π‘ π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™
Relationship between Neutron Flux and Reactor Power
The volume of the reactor is constant in an operating reactor. The number
(density) of fuel atoms is also relatively constant over a relatively short
period (days and weeks). With a constant atomic density and microscopic
cross-section, the macroscopic cross-section is also constant.
By examining the equation for power, we see that reactor power and the
neutron flux are directly proportional if the reactor volume and macroscopic
cross-section are constant. However, if the fission macroscopic crosssection decreases from the reactor fuel, the neutron flux must increase to
maintain power constant. This will deplete (macroscopic cross-section =
atomic density time microscopic cross-section) and atomic density will
decrease as a result.
Knowledge Check
With a reaction rate of 2 x 1013 neutrons/cm3-sec what is the
reactor power level? Assume the entire volume of the core is
10,000 cm3.
A.
𝑃=
B.
66
π‘›π‘’π‘’π‘‘π‘Ÿπ‘œπ‘›π‘ 
(10,000 π‘π‘š3 )
π‘π‘š3 – 𝑠𝑒𝑐
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
3.12 × 1010 π‘€π‘Žπ‘‘π‘‘– 𝑠𝑒𝑐
2 × 1013
𝑃=
C.
Φπ‘‘β„Ž Σ𝑓 𝑉
π‘“π‘–π‘ π‘ π‘–π‘œπ‘›π‘ 
3.12 × 1010 π‘€π‘Žπ‘‘π‘‘– 𝑠𝑒𝑐
𝑃 = 6.41 × 106 π‘€π‘Žπ‘‘π‘‘π‘ 
Rev 1
TLO 5 Summary
Neutron Reaction Rates Summary
ο‚· Atomic density (N) is the number of atoms of a given type per unit
volume of material.
ο‚· Microscopic cross-section (σ) is the probability of a given reaction
occurring between a single neutron and a single nucleus.
- Microscopic cross-sections are measured in units of barns, where
1 barn = 10-24 cm2.
ο‚· Macroscopic cross-section (Σ) is the probability of a given reaction
occurring per unit length of travel of the neutron.
- The units for macroscopic cross-section are cm-1.
- Calculate the macroscopic cross-section for a material using the
equation below:
𝛴 = π‘πœŽ
- Calculate the macroscopic cross-section for a mixture of
materials using the equation below:
𝛴 = 𝑁1 𝜎1 + 𝑁2 𝜎2 + 𝑁3 𝜎3 +. . .. 𝑁𝑛 πœŽπ‘›
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Rev 1
The absorption cross-section for a material usually has three distinct
regions.
- At low neutron energies (<1 eV) the cross-section is inversely
proportional to the neutron velocity.
- Resonance peaks exist at intermediate (epithermal) energy
levels. Resonance absorption occurs when the sum of the kinetic
energy of the neutron and its binding energy is equal to amount
of energy required to raise a compound nucleus from its ground
state to a quantum level energy state.
- For higher neutron energies, the absorption cross-section
steadily decreases as the neutron energy increases.
The mean free path (λ) is the average distance that a neutron travels in
a material between interactions.
- The mean free path equals 1/Σ.
Neutron flux (Φ) is the total path length traveled by all neutrons in
one cubic centimeter of material during one second.
- Thermal neutron flux (Φth) is number of thermal neutrons
crossing a unit area in the reactor in a given amount of time:
n/cm2/sec.
- Fast neutron flux (Φf) is number of fast neutrons crossing a unit
area in a given amount of time.
The neutron flux distribution is generally highest in the center of the
core and drops off toward the core boundaries (top and bottom, sides).
The magnitude of axial flux is a side view of the core flux (top to
bottom).
The magnitude of radial flux is a top view of the reactor core flux
(side to side).
Self-shielding is where the local neutron flux is depressed within a
material due to neutron absorption near the surface of the material.
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ο‚·
The reaction rate is the number of interactions of a particular type
occurring in a cubic centimeter of material in a second.
ο‚· Calculate the reaction rate using the equation below.
𝑅 = 𝛷𝛴
ο‚· Fission reaction rate multiplied by the core volume is directly
proportional to reactor power.
Objectives
Now that you have completed this lesson, you should be able to:
1. Explain the following terms, including any mathematical
relationships:
a. Atomic density
b. Neutron flux
c. Fast neutron flux
d. Thermal neutron flux
e. Microscopic cross-section
f. Barn
g. Macroscopic cross-section
h. Mean free path
2. Define the following neutron classes:
a. Fast
b. Intermediate
c. Slow
3. Describe how the absorption and scattering cross-section of typical
nuclides varies with neutron energies in the 1 eV region, and the
resonance absorption region.
4. Describe the macroscopic cross-section and mean free path at
various temperatures.
5. Describe radial and axial neutron flux distribution.
6. Describe how changes in neutron flux and macroscopic crosssection affect reaction rates.
7. Describe the relationship between neutron flux and reactor power.
Fission Summary
This module focused on the fission process and energy released through
fission.
TLO 1 introduced neutron interactions, covering resonance and potential
elastic scattering where kinetic energy is conserved, and inelastic scattering
where there is momentum conservation. TLO 1 also covered radiative
capture, where a target nucleus absorbs a neutron to become an excited
nucleus, releasing excitation energy in the form of a gamma ray.
Alternatively, the excited nucleus may eject an alpha or proton particle.
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Rev 1
Finally, if the nucleus has gained enough energy, it may split apart,
releasing two fission fragments, neutrons, and energy.
TLO 2 focused on the processes included in nuclear fission, definitions of
key terms involved in the fission process and classification of different
materials with respect to their fission capability.
TLO 3 covered the energy released during a fission event, including
methods of calculating energy released in fission, and the heat generated in
the fission process, as well as the heat generated during decay of fission
products.
TLO 4 focused on neutron sources, both those that are intrinsic, and those
installed specifically to produce source neutrons. This TLO described the
variation of source neutrons over the life of the core, including which
sources provide the highest source of neutrons at varying points in the core
life.
TLO 5 presented a summary of neutron reaction rates, discussing
microscopic and macroscopic cross-sections and calculation methods, as
well as neutron flux and its relationship to the above cross-sections, and the
effect these parameters have on neutron reaction rates.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Describe neutron interactions with matter.
2. Describe the process of nuclear fission and the types of material that
can undergo fission.
3. Explain the production of heat from fission.
4. Describe intrinsic and installed neutron sources and their
contribution to source neutron strength over core life.
5. Explain the relationship between neutron flux, microscopic and
macroscopic cross-sections, and their effect on neutron reaction
rates.
Rev 1
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