Operator Generic Fundamentals
Nuclear Physics - Fission
© Copyright 2017 – Rev 3
Operator Generic Fundamentals
2
Terminal Learning 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.
© Copyright 2017 – Rev 3
TLOs
Operator Generic Fundamentals
3
Neutron Interactions
TLO 1 – Describe neutron interactions with matter.
1.1 Describe the following neutron scattering interactions, including
conservation principles:
a. Elastic scattering
b. Inelastic scattering
1.2 Describe the following reactions where a neutron is absorbed in a
nucleus:
a. Radiative capture
b. Particle ejection
c.
Fission
© Copyright 2017 – Rev 3
TLO 1
Operator Generic Fundamentals
4
Neutron Scattering Interactions
ELO 1.1 – Describe the following neutron scattering interactions, including
conservation principles: elastic scattering and inelastic scattering.
• A neutron scattering reaction occurs when a nucleus, after having
been struck by a neutron, emits a single neutron
– In some cases the initial and final neutrons are not the same
– Free neutron "bounced off," or scattered
• Two categories of scattering reactions
– Elastic
– Inelastic
• Important for thermal or low energy neutron availability
© Copyright 2017 – Rev 3
ELO 1.1
Operator Generic Fundamentals
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Elastic Scattering
Figure: Elastic Scattering
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Elastic Scattering
• Conservation of momentum (𝑚𝑣)
𝑚𝑛 𝑣𝑛,𝑖 + 𝑚 𝑇 𝑣𝑇,𝑖 = 𝑚𝑛 𝑣𝑛,𝑓 + 𝑚 𝑇 𝑣𝑇,𝑓
• Conservation of kinetic energy
1
2
𝑚𝑛 𝑣𝑛,𝑖
2
+
1
2
𝑚 𝑇 𝑣𝑇,𝑖
2
=
1
𝑚𝑣 2
2
1
2
𝑚𝑛 𝑣𝑛,𝑓
2
+
1
2
𝑚 𝑇 𝑣𝑇,𝑓
2
)
• Where:
𝑚𝑛 = mass of the neutron
𝑚 𝑇 = 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
© Copyright 2017 – Rev 3
ELO 1.1
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7
Inelastic Scattering
• Some KE transferred to target nucleus as excitation energy
• Total KE outgoing neutron and nucleus is less than KE of incoming
neutron
Figure: Inelastic Scattering
© Copyright 2017 – Rev 3
ELO 1.1
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Neutron Scattering Interactions
Knowledge Check
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
Correct answer is A.
© Copyright 2017 – Rev 3
ELO 1.1
Operator Generic Fundamentals
9
Neutron Absorption Reactions
ELO 1.2 - Describe the following reactions where a neutron is absorbed
in a nucleus: Radiative capture, Particle ejection, Fission.
• Most absorption reactions result in:
– The loss of a neutron
– The production of charged particles
– Gamma ray(s)
– Fission fragments
– The release of neutrons and energy
• Three types of absorption reactions: radiative capture, particle
ejection, and fission
© Copyright 2017 – Rev 3
ELO 1.3
Operator Generic Fundamentals
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Radiative Capture
• Incident neutron interacts with the target nucleus forming a
compound nucleus
• Compound nucleus, with the neutron added - decays to its ground
state via gamma emission
1
238
239 ∗ 239
0
𝑛+
𝑈→
𝑈 →
𝑈+ 𝛾
0
92
92
92
0
* Represents an excited nucleus
© Copyright 2017 – Rev 3
ELO 1.3
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Particle Ejection
• A compound nucleus is formed when the incident neutron interacts
with the target nucleus
• New compound nucleus, is excited to a high enough energy level for
it to:
– Eject a new particle with the incident neutron remaining in the
nucleus
– Ejected particles can be alpha particles, protons, etc
• After the new particle is ejected, the nucleus may or may not exist in
an excited state depending upon the mass-energy balance of the
reaction
1
10
11
𝑛+ 𝐵→
𝐵
0
5
5
© Copyright 2017 – Rev 3
∗
7
4
→ 𝐿𝑖 + 𝛼
3
2
ELO 1.3
Operator Generic Fundamentals
12
Fission
• The nucleus absorbs the incident neutron resulting in nucleus
splitting into two smaller nuclei, called fission fragments
• Fission reaction typically produces
– Two fission fragments
– 2 to 3 neutrons
– Considerable amount of energy (e.g. kinetic and gamma rays)
1
236 ∗ 140
93
1
235
𝑛+
𝑈→
𝑈 →
𝐶𝑠 + 𝑅𝑏 + 3 𝑛
0
92
55
37
0
92
© Copyright 2017 – Rev 3
ELO 1.3
Operator Generic Fundamentals
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Neutron Absorption Reactions
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
Correct answer is C.
© Copyright 2017 – Rev 3
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Neutron Fission
TLO 2 – Describe the process of nuclear fission and the types of material
which can undergo fission.
2.1 Define the following terms:
a. Excitation energy (Eexc)
b. Critical energy (Ecrit)
2.2 Explain the fission process using the liquid drop model of a
nucleus.
2.3 Define the following terms:
a. Fissile material
b. Fissionable material
c. Fertile material
d. Thermal neutrons
2.4 Explain binding energy per nucleon
2.5 Explain the shape of the binding energy per nucleon versus mass
number curve including its significance to fission energy.
© Copyright 2017 – Rev 3
TLO 2
Operator Generic Fundamentals
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Excitation Energy
ELO 2.1 – Define the following terms: Excitation energy (Eexc), Critical
energy (Ecrit).
Excitation Energy
• The measure of how far the energy level of a nucleus is above its
ground state is called the excitation energy (Eexc)
Critical Energy
• The minimum excitation energy for a specific nuclide to fission
© Copyright 2017 – Rev 3
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Critical Energy
• Many neutron reactions can cause an increase in the excitation
energy of a nucleus
• When an incident neutron strikes a target nucleus, the reaction
excites the target nucleus by an amount equal to the sum of
– Binding energy of the neutron
– The neutron’s kinetic energy (KE)
• If the binding energy is less than the required critical energy for the
nucleus
– Additional energy is required to cause the nucleus to fission
– Could be in the form of KE from the incident neutron
• Neutrons of low KE cannot cause fission with some types of isotopes
used in nuclear reactor fuels
© Copyright 2017 – Rev 3
ELO 2.1
Operator Generic Fundamentals
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Excitation Energy
Knowledge Check
Excitation must be at least equal to critical energy for fission to occur.
A. True
B. False
Correct answer is A.
© Copyright 2017 – Rev 3
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Fission Process
ELO 2.2 - Explain the fission process using the liquid drop model of a
nucleus.
• In a fission reaction the incident neutron is absorbed in the heavy
target nucleus
• Creates a compound nucleus that is excited at a high-energy level
(Eexc > Ecrit)
• Nucleus "splits" (fissions) into two large fragments plus some
neutrons
• In addition a large amount of energy is released in the form of
fragment kinetic energy and radiation
© Copyright 2017 – Rev 3
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Fission Process
• The nucleus is held together by the attractive nuclear force between
nucleons
– Resists the opposing electrostatic forces within the nucleus
• Characteristics of the nuclear force are:
– Very short range attractive force, with essentially no strength
beyond nuclear scale dimensions (≈10-13 cm)
– Stronger than repulsive electrostatic forces within the nucleus
– Independent of nucleon pairing
– Saturable – a nucleon can attract only a few of its nearest
neighbors
© Copyright 2017 – Rev 3
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Fission Process
• One theory considers fission similar to the splitting of a liquid drop
– Liquid drop is held together by molecular forces that tend to make
the drop spherical in shape and resist deformation in the same
manner as nuclear forces are thought to hold the nucleus together
– By considering the nucleus as a liquid drop the fission process is
better understood and visualized
© Copyright 2017 – Rev 3
ELO 2.2
Operator Generic Fundamentals
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Fission Process
• The undisturbed nucleus in the ground state is undistorted
Figure: Liquid Drop Model
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Fission Process
• When the incident neutron strikes the target nucleus and is absorbed
an excited a compound nucleus is formed
Figure: Liquid Drop Model
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Fission Process
• If the excitation energy is greater than the critical energy
– oscillations may cause the compound nucleus to become
dumbbell-shaped as it starts to come apart
Figure: Liquid Drop Model
© Copyright 2017 – Rev 3
ELO 2.2
Operator Generic Fundamentals
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Fission Process
• When the repulsive electrostatic forces exceed the attractive nuclear
forces, nuclear fission occurs
Figure: Liquid Drop Model
© Copyright 2017 – Rev 3
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Fission Process
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
Correct answer is C.
© Copyright 2017 – Rev 3
ELO 2.2
Operator Generic Fundamentals
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Fission Process
ELO 2.3 - Define the following terms: Fissile material, Fissionable
material, Fertile material, Thermal neutrons.
Fissile Material
• Fissile materials are nuclides for which fission is possible with
neutrons of any energy level
– Desired thermal reactors
– Can fission with thermal neutrons
– Change in binding energy (BE) from neutron alone is sufficient to
exceed the critical energy
• Thermal neutrons – very low KE levels, add essentially no KE to the
reaction
• Examples:
– Uranium-235
– Uranium-233
– Plutonium-239
© Copyright 2017 – Rev 3
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Fissionable Material
Fissionable Material
• Nuclides for which fission is possible
– All fissile nuclides fall into this category
– And nuclides that fission only from high-energy neutrons
• Change in binding energy causes excitation energy level insufficient
to reach critical energy for fission
• Examples (requiring high-energy neutrons):
– Thorium-232
– Uranium-238
– Plutonium-240
© Copyright 2017 – Rev 3
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Fissionable Material
Critical Energies Compared to Binding Energy (BE) of Last
Neutron
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
© Copyright 2017 – Rev 3
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Fertile Materials
• Transmutation – all of the neutron absorption reactions that do not
result in fission lead to the production of new nuclides
• Nuclides can
– Be transmuted again
– Undergo radioactive decay to produce nuclides known as
transmutation products
• Some fissile nuclides do not exist in nature, only produced by nuclear
reactions (transmutation)
• Fertile materials are materials (nuclides) that can undergo
transmutation to become fissile materials
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Operator Generic Fundamentals
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Fertile Materials
Figure: Transmutation Mechanisms
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Thermal Neutrons
• Thermal neutrons are neutrons with very low KE levels
• Basically in equilibrium with the thermal energy of surrounding
materials
• Undergo scattering collisions where they are gaining and losing equal
amounts of energy in successive collisions
• Add essentially no KE to a neutron absorption reaction
• In a commercial PWR, as a thermal reactor, the goal is to thermalize
as many neutrons as possible
© Copyright 2017 – Rev 3
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Thermal Neutrons
• Thermal neutrons are standardized in table format for cross-section
data at 68°F (20°C).
• At 68°F, thermal neutron
– Energy is 0.025 eV
– Velocity of 2.2 X 105 cm/sec
𝐸𝑝 = 0.025 𝑒𝑉
𝑇
𝑇𝑜
1
2
𝑉𝑝 = 2.2 × 105
𝑐𝑚
𝑠𝑒𝑐
𝑇
𝑇𝑜
1
2
• Where:
𝐸𝑝 = Most probable energy (eV)
𝑉𝑝 = Most probable velocity (cm/sec)
𝑇𝑜 = Table temp 68°F (528°R)
𝑇 = New temperature in °R (°F + 460°)
© Copyright 2017 – Rev 3
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Fission Process
Knowledge Check
What is the difference between a fissionable material and a fissile
material?
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 kinetic energy in
order to fission, whereas fissile materials will fission with a neutron
that has zero kinetic energy.
Correct answer is D.
© Copyright 2017 – Rev 3
ELO 2.3
Operator Generic Fundamentals
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Binding Energy per Nucleon
ELO 2.4 - Explain binding energy per nucleon.
• Binding energy per nucleon (BE/A) equals the average energy
required to remove a nucleon from a specific nucleus
• This is important from the standpoint of a thermal neutron’s ability to
cause a fission
© Copyright 2017 – Rev 3
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Binding Energy per Nucleon
• Determined by dividing the total binding energy of a nuclide by the
total number of nucleons in its nucleus.
Example
• Given that the total binding energy (BE) for a U-238 nucleus is
𝐵𝐸
1,804.3 MeV, calculate the binding energy per nucleon ( ) for U-238.
𝐴
Solution:
𝐵𝐸 1,804.3 𝑀𝑒𝑉
=
= 7.6 𝑀𝑒𝑉
𝐴
238
© Copyright 2017 – Rev 3
ELO 2.4
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Binding Energy per Nucleon
Knowledge Check
Binding energy per nucleon is independent of the specific nuclide.
A. True
B. False
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 2.4
Operator Generic Fundamentals
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Binding Energy per Nucleon Curve
ELO 2.5 - Explain the shape of the binding energy per nucleon versus
mass number curve including its significance to fission energy.
𝐵𝐸
• The relationship of
as the mass number changes is important for
𝐴
understanding how energy is released during fission
• It also explains the theory of energy release from fusion
• As the number of nucleons in a nucleus increases,
– Total binding energy increases
– Rate of BE increase is not linear as mass number increases
• Variation in the binding energy per nucleon is seen when plotted
against the atomic mass numbers (A)
© Copyright 2017 – Rev 3
ELO 2.5
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Binding Energy per Nucleon Curve
Figure: Binding Energy per Nucleon versus Mass Number
© Copyright 2017 – Rev 3
ELO 2.5
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Binding Energy per Nucleon Curve
Knowledge Check
Generally, less stable nuclides have a higher
ones.
𝐵𝐸
𝐴
than the more stable
A. True
B. False
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 2.5
Operator Generic Fundamentals
40
Neutron Fission
TLO 3 – Explain the production of heat from fission.
3.1 Describe the average total amount of energy released per fission
event including:
a. Energy released immediately from fission
b. Delayed fission energy
3.2 Describe which fission products nuclides are most likely to result
from fission.
3.3 Describe the energy released from fission by the following
methods:
a. Change in binding energy
b. Conservation of mass – energy
c. Decay energy
3.4 Describe how heat is produced as a result of fission.
© Copyright 2017 – Rev 3
TLO 3
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Energy Release Per Fission
ELO 3.1 – Describe the average total amount of energy released per
fission event including: energy released immediately from fission and
delayed fission energy.
• Total energy released per fission varies, depending on what fission
products are formed
– Average total energy released from U-235 fission with a thermal
neutron is approximately 200 MeV
• The majority of this energy (approximately 83 percent) is from the KE
of the fission fragments
• Specific isotope undergoing fission has a small impact on the energy
released
– Thermal fission of U-235 and fast fission of U-238 yield nearly
identical energy
© Copyright 2017 – Rev 3
ELO 3.1
Operator Generic Fundamentals
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Energy Release Per Fission
Instantaneous Energy
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
Kinetic Energy of Beta Particles
7 MeV
Decay Gamma Rays
6 MeV
Neutrinos
10 MeV*
Total Energy Released
200 MeV
© Copyright 2017 – Rev 3
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Delayed Fission Energy (Decay Heat)
• Of the 200 MeV released per fission, about seven percent (13 MeV)
is released sometime after the instant of fission
• Decay heat – energy released from the decay of fission products
after the reactor is shut down and fissions mostly cease
– Represents about seven percent of reactor heat production during
reactor operation
– Once the reactor is shutdown decay heat production “decays” off
but remains a significant source of heat for a very long time
© Copyright 2017 – Rev 3
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Energy Release Per Fission
Knowledge Check
Which of the following statements correctly describes the amount of energy
released from a single fission event?
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).
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 3.1
Operator Generic Fundamentals
45
Fission Fragment Yield
ELO 3.2 - Describe which fission product nuclides are most likely to result
from fission.
• Fissions do not produce identical results on each occurrence
– Both the number of neutrons and the resultant fission fragments
vary
• Scientific experiments have developed a yield curve of fission product
probabilities
© Copyright 2017 – Rev 3
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Most Probable Fission Fragments
• Resultant fission fragments have
masses that vary widely
• Most probable pair of fission
fragments for a thermal neutron
fission of uranium-235 have
masses of about 95 and 140
• Cesium-140 and rubidium-93 are
very likely to result from fission
• An example of one fission
fragment yield is:
236 ∗
235
𝑈+𝑛 →
𝑈
140
94
→
𝑋𝑒 + 𝑆𝑟 + 2𝑛
• 2 neutrons result from this fission
© Copyright 2017 – Rev 3
ELO 3.2
Figure: Uranium-235 Fission Yield for Fast and
Thermal Neutrons versus Mass Number
Operator Generic Fundamentals
47
Most Probable Fission Fragments
Isotope
Average Neutrons released per fission (v)
U-233
2.492
U-235
2.418
Pu-239
2.871
Pu-241
2.927
© Copyright 2017 – Rev 3
ELO 3.2
Operator Generic Fundamentals
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Fission Fragment Yield
Knowledge Check
List the two most likely fission fragments resulting from fission of
uranium-235.
Correct answer is cesium and rubidium.
© Copyright 2017 – Rev 3
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Energy Release Calculation
ELO 3.3 - Describe the energy released from fission by the following
methods: change in binding energy, conservation of mass – energy, and
decay energy.
• Nuclear fission releases enormous quantities of energy
• Amount of energy release can be calculated
• Consider a typical fission reaction:
1
236
235
𝑛+
𝑈→
𝑈
0
92
92
© Copyright 2017 – Rev 3
∗
→
140
93
1
𝐶𝑠 + 𝑅𝑏 + 3 𝑛
55
37
0
ELO 3.3
Operator Generic Fundamentals
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Change in Binding Energy Method
𝐵𝐸
• Use
curve to determine the
𝐴
amount of energy released by a
fission
• An increase in total BE
indicates greater stability by the
release of the equivalent
energy
• In a fission process, the energy
liberated is equal to the
increase in the total BE of the
system
∗
1
236
235
𝑛+
𝑈→
𝑈
0
92
92
140
93
1
→
𝐶𝑠 + 𝑅𝑏 + 3 𝑛
55
37
0
© Copyright 2017 – Rev 3
ELO 3.3
Figure: Change in Binding Energy for Typical Fission
Operator Generic Fundamentals
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Change in Binding Energy Demonstration
• Total BE for a nucleus is found by multiplying the BE/A by the number
of nucleons
∆𝐵𝐸 = 𝐵𝐸𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐵𝐸𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
= 𝐵𝐸𝑅𝑏−93 + 𝐵𝐸𝐶𝑠−140 − 𝐵𝐸𝑈−235
= 809 𝑀𝑒𝑉 + 1,176 𝑀𝑒𝑉 − 1,786 𝑀𝑒𝑉
= 199 𝑀𝑒𝑉
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
© Copyright 2017 – Rev 3
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Conservation of Mass-Energy Method
• During fission a decrease in the system mass occurs equal to the
energy liberated
• This calculation method is more accurate than the change in BE
method since it considers all mass changes
– EInst, the instantaneous energy, is the energy released
immediately after the fission process
• This calculation includes:
1. Adding the nucleon masses of the fuel isotope including the
incident neutron mass (reactants)
2. Adding the nucleon masses of the fission products and
released neutrons (products)
3. Subtracting the mass of the products from the reactants
4. Multiple the mass calculated by 931.5 MeV to convert to energy
© Copyright 2017 – Rev 3
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Conservation of Mass-Energy
Demonstration
Mass of the Reactants
235
𝑈 235.043924 amu
92
1
1.008665 amu
𝑛
0
Mass of the Products
93
92.91699 amu
𝑅𝑏
37
140
𝐶𝑠 139.90910 amu
55
1
3.02599 amu
3 𝑛
0
235.85208 amu
Totals 236.052589 amu
Mass Difference
Energy Equivalent
= 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 − 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
= 236.052589 𝑎𝑚𝑢 − 235.85208 𝑎𝑚𝑢
= 0.200509 𝑎𝑚𝑢
= 𝑀𝑎𝑠𝑠 × 931.5 𝑀𝑒𝑉/𝑎𝑚𝑢
= 0.200509 𝑎𝑚𝑢 × 931.5 𝑀𝑒𝑉/𝑎𝑚𝑢
= 186.8 𝑀𝑒𝑉
© Copyright 2017 – Rev 3
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Estimation of Decay Energy
Decay energy
• Additional energy released when the fission fragments decay by βemission
• Added to the instantaneous energy release,
• EDecay
𝛽−
𝛽−
𝛽−
𝛽−
93
93
93
93
93
𝑅𝑏 → 𝑆𝑟 → 𝑌 → 𝑍𝑟 → 𝑁𝑏
37
38
39
40
41
𝛽−
𝛽−
𝛽−
140
140
140
140
𝐶𝑠 →
𝐵𝑎 →
𝐿𝑎 →
𝐶𝑒
55
56
57
58
© Copyright 2017 – Rev 3
ELO 3.3
Operator Generic Fundamentals
55
Estimation of Decay Energy
• The energy released during the decay for each chain is equivalent to:
– The mass difference between the original fission product, and
– The sum of the final stable nuclide and emitted beta particles
• The total decay energy is the sum of both chains
© Copyright 2017 – Rev 3
ELO 3.3
Operator Generic Fundamentals
56
Estimation of Decay Energy
Demonstration
• The energy released in the decay chain of rubidium-93 is:
𝐸𝐷𝑒𝑐𝑎𝑦 = 𝑚𝑅𝑏−93 − 𝑚𝑁𝑏−93 + 4 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
= 92.91699 𝑎𝑚𝑢 − 92.90638 𝑎𝑚𝑢 + 4 0.0005486 𝑎𝑚𝑢
= 0.008416 𝑎𝑚𝑢
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
= 7.84 𝑀𝑒𝑉
© Copyright 2017 – Rev 3
ELO 3.3
Operator Generic Fundamentals
57
Estimation of Decay Energy
Demonstration
• The energy released in the decay chain of cesium-140 is:
𝐸𝐷𝑒𝑐𝑎𝑦 = 𝑚𝐶𝑠−140 − 𝑚𝐶𝑒−140 + 3 𝑚𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
= 139.90910 𝑎𝑚𝑢 − 139.90543 𝑎𝑚𝑢 + 3 0.0005486 𝑎𝑚𝑢
= 0.000202 𝑎𝑚𝑢
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
= 1.89 𝑀𝑒𝑉
• The total decay energy is the sum of the energies of the two chains,
or 9.73 MeV
© Copyright 2017 – Rev 3
ELO 3.3
Operator Generic Fundamentals
58
Energy Release Calculation
Knowledge Check
The change in binding energy per nucleon between the fissile nuclide
and the fission products, is used to calculate _________.
A. fission fragment speed
B. decay heat
C. energy released by fission
D. increase in mass
Correct answer is C.
© Copyright 2017 – Rev 3
ELO 3.3
Operator Generic Fundamentals
59
Fission Heat Production
ELO 3.4 – Describe how heat is produced as a result of fission.
• Majority of the energy liberated in the fission process is released
immediately after the fission occurs
• This energy appears as
– KE of the fission fragments and neutrons
– Instantaneous gamma rays
• Remaining energy is released over a period of time after the fission
occurs and appears as KE of the decay products
– Beta
– Decay gammas
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
60
Fission Heat Production
• All of the energy released in fission, with the exception of the neutrino
energy, is transformed into heat
• Fission fragments, with a positive charge and KE, cause ionization
directly as they rip orbital electrons from the surrounding atoms
– Provides biggest increase in fuel temperature
• Beta particles and gamma rays also give up energy also through
ionization
• Fission neutrons interact and lose their energy through scattering
– Primarily elastic scattering
© Copyright 2017 – Rev 3
ELO 3.4
Operator Generic Fundamentals
61
Fission Heat Production
Knowledge Check
Which of the following is NOT a method by which heat is produced from
fission?
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.
Correct answer is D.
© Copyright 2017 – Rev 3
ELO 3.4
Operator Generic Fundamentals
62
Neutron Fission
TLO 4 – Describe intrinsic and installed neutron sources and their
contribution to source neutron strength over core life.
4.1 Describe the purpose and importance of source neutrons.
4.2 Describe, including examples, of each of the following types of
intrinsic neutron sources:
a. Spontaneous fission
b. Photo-neutron reactions
c. Alpha-neutron reactions
d. Transuranic elements
4.3 Describe the purpose and type of installed neutron sources.
4.4 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
© Copyright 2017 – Rev 3
TLO 4
Operator Generic Fundamentals
63
Source Neutrons Introduction
ELO 4.1 – Describe the purpose and importance of source neutrons.
• Sources neutrons – Non-fission produced neutrons
• Help monitor the fission process during reactor startup
• Ensure neutron population during shutdown conditions remains high
enough for indication on source range nuclear instrumentation
• Important during reactor shutdown and startup conditions:
– To confirm instrument operability
– Monitoring of the reactor’s neutron population changes
– Subcritical multiplication
• Source neutron sources are classified as either
– Intrinsic
– Installed neutron sources
© Copyright 2017 – Rev 3
ELO 4.1
Operator Generic Fundamentals
64
Source Neutrons
Knowledge Check
Source neutrons are important because they:
A. Extend the neutron lifetime allowing for 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.
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 4.1
Operator Generic Fundamentals
65
Intrinsic Neutron Sources
ELO 4.2 - Describe, the following types of intrinsic neutron sources:
spontaneous fission, photo-neutron reactions, alpha-neutron reactions,
and transuranic elements.
• Intrinsic neutron sources are neutron-producing reactions that occur
in reactor core fuel related materials
• The following types of neutron reactions are capable of producing
source neutrons and transuranic contributions
– Spontaneous fission
– Photo-neutron reactions
– Alpha-neutron reactions
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
66
Intrinsic Neutron Sources
• A limited number of neutrons will be present, even in a reactor core
that has never been operated
• Due to spontaneous fission of some heavy nuclides present in the
fuel. Examples:
– Uranium-238
– Uranium-235
– Plutonium-239
• Later in core life the transuranic elements provide more spontaneous
fission neutrons
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
67
Intrinsic Neutron Sources
In a reactor that has been operated at power, source neutrons come
from:
• Photo-neutron reactions provide a significant source
– Interaction of a gamma ray and a deuterium nucleus
• Interactions between alpha particles and various isotopes in the
reactor
• Alpha particles from the decay of heavy nuclides, interacts with
oxygen-18 and boron-11 in the core
• Transuranic elements building in the core
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
68
Spontaneous Fission
• The four isotopes that contribute most to the source neutron
population via spontaneous fission are:
– Uranium-235
– Uranium-238
– Curium-242
– Curium-244
• Uranium-235 and 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
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
69
Spontaneous Fission
• Prior to a new core reactor startup, uranium spontaneous sources are
significant contributors to the source neutron population
• From startup to ≈ 20,000 MWD/MTU of core irradiation
– curium-242 becomes a major producer of source neutrons from
spontaneous fission
• Beyond 20,000 MWD/MTU
– curium-244 becomes a more predominant source neutron
producer from spontaneous fission
• As an example, one ton of spent nuclear fuel will contain on the order
of 20 grams of curium
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
70
Photo-Neutron Reactions
• In a reactor that has been operated at power, source neutrons from
photo-neutron reactions become significant
1
2
1
𝛾+ 𝐻→ 𝐻+ 𝑛
1
1
0
• Referred to as a photo-neutron reaction because it is initiated by
electromagnetic (gamma photon) radiation and results in the
production of a neutron
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
71
Photo-Neutron Reactions
• After the reactor has operated for a short time there is an abundant
supply of high-energy gammas (2.22 MeV or greater)
– Moderator/coolant has some deuterium present because the
naturally occurring atom percentage of deuterium is 0.015 percent
• This is sufficient deuterium for production of photo-neutrons following
power operation
• Quantity of gamma rays decreases with time after shutdown as
gamma ray emitters decay off
– Photo-neutron production rate decreases with reactor shutdown
time
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
72
Alpha-Neutron Reactions
• Alpha particles occur from the decay of heavy elements in the fuel
• They interact with naturally occurring oxygen-18 and boron-11
• Transuranic elements produced during power operation also provide
source neutrons via alpha-neutron reaction
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
73
Alpha-Neutron Reactions – Oxygen 18
This reaction does not contribute significantly to the source neutron
population because
• Uranium does not produce many alpha particles
• Abundance of oxygen-18 in the reactor core is low
18
21
1
4
𝐻𝑒 + 𝑂 → 𝑁𝑒 + 𝑛
2
8
10
0
Nuclide
© Copyright 2017 – Rev 3
t½ (Fission)
t½ (α-decay)
235
92U
1.8 x 1017 years
6.8 x 108 years
238
92U
8.0 x 1015 years
4.5 x 109 years
239
94Pu
5.5 x 105 years
2.4 x 104 years
240
94Pu
1.2 x 1011 years
6.6 x 103 years
ELO 4.2
Operator Generic Fundamentals
74
Alpha-Neutron Reactions – Boron 11
• Soluble boron is added to the reactor coolant system
• Soluble boron contains boron-11 (80.1 percent of natural boron)
• Boron 11 and alpha particles from heavy elements decay react to
produce source neutrons
11
1
4
14
𝐻𝑒 + 𝐵 → 𝑁 + 𝑛
2
5
7
0
• The boron-11 must be in very close proximity to the fuel for this
reaction because of the short travel path length of alpha particles
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
75
Alpha-Neutron Reactions – Transuranic
Elements
• Neutrons (transuranic) are produced from either spontaneous fission
or alpha-neutron reactions
• Curium and americium isotopes are the major transuranic elements
of interest for source neutrons
• Curium-242 production and neutron:
239
242
1
4
𝑃𝑢 + 𝐻𝑒 →
𝐶𝑚 + 𝑛
94
2
96
0
• Alpha production (163 day half-life):
242
238
4
𝐶𝑚 →
𝑃𝑢 + 𝐻𝑒
96
94
2
• In typical nuclear reactor core, transuranic neutron sources produce
– About 100 neutrons per second for every gram of fuel in the core
– Approximately 1 x 107 neutrons per second core wide
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
76
Alpha-Neutron Reactions – Transuranic
Elements
• One gram of curium-244 produces about
– 5 x 105 neutrons per second due to alpha-neutron reactions
– 1.2 x 107 neutrons per second due to spontaneous fission
• One gram of curium-242 produces about
– 2.5 x 107 neutrons per second from alpha-neutron reactions
– 2.3 x 107 neutrons per second from spontaneous fission
• One gram of americium-241 produces approximately
– 4 x 103 neutrons per second from alpha-neutron reactions
– No spontaneous fission
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
77
Intrinsic Neutron Sources
Knowledge Check
Select the three major contributors (types of reactions) to the intrinsic
neutron source in a nuclear reactor.
A. Photo-neutron
B. Alpha-Neutron
C. Spontaneous fission
D. Oxygen-18 neutron
Correct answers are A, B, and C.
© Copyright 2017 – Rev 3
ELO 4.2
Operator Generic Fundamentals
78
Installed Neutron Sources
ELO 4.3 - Describe, the purpose and types of installed neutron sources.
• Because intrinsic neutron sources can be relatively weak or
dependent upon the recent power history
– Many reactors have additional neutrons sources installed to
ensure a reliable and sufficient number of source neutrons
– May be especially important with all new fuel in the core
• Installed neutron sources ensure that shutdown neutron levels are
high enough to be detected by the nuclear instruments at all times
– Provides the operators with valid indication of the reactor’s status
– Installed neutron sources are assemblies placed within the
reactor for the sole purpose of producing source neutrons
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
79
Californium-252
• A strong source of neutrons
• Artificial nuclide
• Emits neutrons at the rate of about 2 x 1012 neutrons per second per
gram from spontaneous fission
• Not widely used as an installed neutron source in commercial PWRs
because of its high cost and short half-life (2.65 years)
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
80
Alpha-Neutron Beryllium Source
• Many installed neutron sources use an alpha-neutron reaction with
beryllium
• Composed of metallic beryllium (100 percent beryllium-9) with an
alpha particle emitter, such as a compound of radium, polonium, or
plutonium
13 ∗ 12
1
9
4
𝐵𝑒 + 𝛼 →
𝐶 → 𝐶+ 𝑛
4
2
6
6
0
• Is intimately (homogeneously) mixed with the alpha emitter and is
usually enclosed in a stainless steel capsule
© Copyright 2017 – Rev 3
ELO 4.3
Operator Generic Fundamentals
81
Photo- Neutron Beryllium Source
• Beryllium-9 - stable isotope with weakly attached neutron with a
binding energy of only 1.66 MeV
• A gamma ray with greater energy than 1.66 MeV can cause neutrons
to be ejected by the photo-neutron reaction
1
9
8
𝛾 + 𝐵𝑒 → 𝐵𝑒 + 𝑛
4
4
0
© Copyright 2017 – Rev 3
ELO 4.3
Operator Generic Fundamentals
82
Antimony-Beryllium Source
• Most common secondary installed source
• Antimony activated by absorbing neutron
123
1
124
𝑆𝑏 + 𝑛 →
𝑆𝑏 + 𝛾
51
0
51
• Radioactive antimony decays to Tellurium then emits high-energy
gamma (60-day half-life)
−
124 𝛽 124
0
𝑆𝑏
𝑇𝑒 +
𝑒+𝛾
51
52
−1
• Gamma ray has sufficient energy to interact with the beryllium to
produce a neutron
1
9
8
𝛾 + 𝐵𝑒 → 𝐵𝑒 + 𝑛
4
4
0
© Copyright 2017 – Rev 3
ELO 4.3
Operator Generic Fundamentals
83
Installed Neutron Sources
Knowledge Check
Which of the following would be used to provide a source neutron
population for a nuclear reactor core which has never been operated
(newly installed core)?
A. Photo-neutron source
B. Transuranic source
C. Startup source
D. Installed source
Correct answer is D.
© Copyright 2017 – Rev 3
ELO 4.3
Operator Generic Fundamentals
84
Intrinsic Source Neutrons over Core Life
ELO 4.4 - Describe the primary source of intrinsic source neutrons in the
reactor for the following conditions; at beginning and end of core life:
Immediately following a reactor shutdown, Several weeks after reactor
shutdown.
• Installed neutron sources provide a steady source of neutrons
throughout core life
– Over long periods their strength does decay
• Intrinsic neutrons, however, do change in importance over core life
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
85
Photo-neutron Strength over Core Life
• Changes significantly over core life
• At BOL with no fission products present in the core (no high-energy
gamma rays) little source neutron contribution
– Deuterium concentration is also low BOL
• Both factors change to increase source neutron production with
increasing core life
• After some power history - photo-neutron reactions are the largest
contributor
– Depending on amounts of new or reused fuel present
– Exists for several days following shutdown from power operations
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
86
Transuranic Source Strength versus Core
Life
• Greater past middle-of-life (MOL) after several weeks of reactor
shutdown from power operations
– Transuranic alpha-neutron intrinsic neutron source tends to
contribute the greatest number of source neutrons
– Due to the photo-neutron source strength decaying off after a
couple of weeks
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
87
Intrinsic Neutron Source Strength at Core
BOL
• Immediately following reactor shutdown from power:
1. Photo-neutron sources
2. Spontaneous fission sources
3. Alpha-neutron (transuranic) sources
• Several weeks following reactor shutdown from power:
1. Spontaneous fission sources
2. Photo-neutron sources
3. Alpha-neutron (transuranic) sources
© Copyright 2017 – Rev 3
ELO 4.4
Operator Generic Fundamentals
88
Intrinsic Neutron Source Strength at Core
EOL
• Immediately following reactor shutdown from power:
1. Photo-neutron sources
2. Alpha-neutron (transuranic) sources
3. Spontaneous fission sources
• Several weeks following reactor shutdown from power:
1. Alpha-neutron (transuranic) sources
2. Spontaneous fission sources
3. Photo-neutron sources
© Copyright 2017 – Rev 3
ELO 4.4
Operator Generic Fundamentals
89
Intrinsic Source Neutrons over Core Life
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 shutdown?
A. Photo-neutron source
B. Transuranic source
C. Spontaneous fission source
D. Californium source
Correct answer is A.
© Copyright 2017 – Rev 3
ELO 4.4
Operator Generic Fundamentals
90
Neutron Reaction Rates
TLO 5 – Explain the relationship between neutron flux, microscopic and
macroscopic cross-sections; and their affect on neutron reaction rates.
5.1 Explain the following terms, include 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
5.2 Define the following neutrons:
a. Fast
b. Intermediate
c. Slow
© Copyright 2017 – Rev 3
TLO 5
Operator Generic Fundamentals
91
Enabling Learning Objectives for TLO 5
5.3 Describe how the absorption and scattering cross-section of typical
nuclides varies with neutron energies in the 1/v region and the
resonance absorption region.
5.4 Given appropriate nuclei information, calculate the macroscopic
cross-section and mean free path at various temperatures.
5.5 Describe radial and axial neutron flux distribution.
5.6 Describe how changes in neutron flux and macroscopic crosssection, affect reaction rates.
5.7 Describe the relationship between neutron flux and reactor power.
© Copyright 2017 – Rev 3
TLO 5
Operator Generic Fundamentals
92
Neutron Reaction Terms
ELO 5.1 - Explain the following terms, include any mathematical
relationships: atomic density, neutron flux, fast neutron flux, thermal
neutron flux, microscopic cross-section, barn, macroscopic cross-section,
and mean free Path.
Atomic Density
• Atomic density is the number of atoms of a given type per unit
volume of the material.
𝜌𝑁𝐴
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
𝑁=
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
93
Atomic Density
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:
𝜌𝑁𝐴
𝑁=
𝑀
𝑔
23 𝑎𝑡𝑜𝑚𝑠
6.022
×
10
𝑚𝑜𝑙𝑒
𝑐𝑚3
=
𝑔
26.9815
𝑚𝑜𝑙𝑒
𝑎𝑡𝑜𝑚𝑠
22
= 6.024 × 10
𝑐𝑚3
2.699
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
94
Neutron Flux
• For calculations we need to know:
– Number of neutrons existing in one cubic centimeter at any one
instant
– Total distance they travel each second while in that cubic
centimeter
• Neutron density (n)
– Number of neutrons existing in a cm3 of material at any instant
– Units of neutrons/cm3
• The total distance these neutrons travel each second is determined
by the speed of the neutron
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Neutron Flux
• Number of neutrons passing through the unit area (cm2) per unit time
– Total path length covered by all neutrons in one cubic centimeter
during one second
• Units are neutrons per square centimeter per second (n/cm2-sec)
• Symbol Φ
• Mathematically,
Φ = 𝑛𝑣
• Where:
Φ = neutron flux (neutron/cm2-sec)
𝑛 = neutron density (neutrons/cm3)
𝑣 = neutron velocity (cm/sec)
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
96
Neutron Flux
• Since the atoms in a reactor do not have a preference for neutrons
coming from any particular direction
– All of the directional beams contribute to the total reaction rate
– At any given point within a reactor, neutrons are traveling in all
directions
• In nuclear reactor, neutron flux is classified as either
– Thermal flux
– Fast flux
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
97
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
– Expressed in terms of neutrons (thermal) per square centimeter
per second (n/cm2/sec)
• Neutron flux is omni-directional
– Neutrons can enter a particular square centimeter or reactor
material from any direction
• Therefore, Φth is viewed as the total distance of all thermal neutrons
diffused (moved) in a particular unit volume in one second
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Fast Neutron Flux
• Fast neutron flux (Φf) is number of fast neutrons crossing a unit area
in a given amount of time
– Expressed in terms of neutrons (fast) per square centimeter per
second (n/cm2/sec)
• Fast neutron flux, Φf is viewed as the total distance that fast neutrons
diffuse (move) in a particular unit volume in one second
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
99
Cross-Sections
• The probability of a neutron interacting with a nucleus for a particular
reaction is dependent on
– The nucleus involved
– The energy of the neutron
• Absorption of a thermal neutron in most materials is much more
probable than the absorption of a fast neutron
• Probability of the neutron interaction varies with the type of reaction
involved
• Cross-section is based on the characteristics of the nucleus, not on
the size of nucleus
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
100
Microscopic Cross-Section (σ)
• Probability of a particular reaction occurring between a neutron and a
nucleus varies with the energy of the neutron
• May be regarded as the effective area the nucleus presents to the
neutron for the particular reaction
– Measured in barns
– The larger the effective area, the greater the probability for
reaction
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
101
Total Microscopic Cross-Section (σT)
• A neutron interacts with an atom in two basic ways
– Scattering
– Absorption reaction
• Probability of a neutron being absorbed is the microscopic crosssection for absorption (σa)
• Probability of a neutron scattering is the microscopic cross-section for
scattering (σs)
• Sum of these two microscopic cross-sections is the total microscopic
cross-section (σT)
𝜎𝑇 = 𝜎𝑎 + 𝜎𝑠
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
102
Microscopic Cross-Section for Scattering
(σS)
• Both the absorption and the scattering microscopic cross-sections
can be further divided
• Scattering cross-section is the sum of
– Elastic scattering cross-section (σse)
– The inelastic scattering cross-section (σsi)
𝜎𝑠 = 𝜎𝑠𝑒 + 𝜎𝑠𝑖
© Copyright 2017 – Rev 3
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Microscopic Cross-Section for
Absorption (σa)
• Includes all reactions except scattering
• For most purposes it is separated into two categories,
– Fission (σf) and
– Capture (σc)
𝜎𝑎 = 𝜎𝑓 + 𝜎𝑐
• The Chart of Nuclides uses two different conventions to represent
microscopic cross-section for capture
– If a gamma ray results, use σγ
– If an alpha particle results, use σα
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Barns
• Microscopic cross-sections are expressed in units of area - square
centimeters
• 1 barn = 10-24 cm2
• Boron-10
– Has a relatively low mass with an absorption cross-section of
3838 barns that results in an ejected alpha particle
– Presents a very large effective area for neutron interaction
• Lead-208
– Relatively high mass and has an absorption cross-section of 8
microbarns (8  10-6 barns) for the same reaction
– Presents a very small effective area for neutron interaction
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
105
Macroscopic Cross-Section (Σ)
• Whether or not a neutron interacts with a certain volume of material
depends
– On its microscopic cross-section and
– The number of nuclei within that volume (atomic density)
• Macroscopic cross-section is the probability of a given reaction
occurring per unit travel of the neutron
– Related to microscopic cross-section (σ) by the following:
Σ = 𝑁𝜎
• Where:
Σ = macroscopic cross-section (cm-1)
𝑁 = atomic density (atoms/cm3)
σ = microscopic cross-section (cm2)
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
106
Macroscopic Cross-Section (Σ)
Microscopic Cross-Section Calculation
• Most materials are composed of several elements, and most
elements are composed of several isotopes;
– Therefore, most materials involve many cross-sections, one for
each isotope involved
• To include all the isotopes for a material, each macroscopic crosssection is determined, then added together:
Σ = 𝑁1𝜎1 + 𝑁2𝜎2 + 𝑁3𝜎3 + ⋯ … . . 𝑁𝑛𝜎𝑛
• Where:
𝑁𝑛 = the number of nuclei per cm3 of the nth element
𝜎𝑛 = the microscopic cross-section of the nth element
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
107
Macroscopic Cross-Section (Σ)
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.
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
108
Macroscopic Cross-Section (Σ)
• Step 1: Using Equation for atomic density, calculate the atomic
density of iron.
𝑔
23 𝑎𝑡𝑜𝑚𝑠
7.68
6.022
×
10
𝜌𝑁𝐴
𝑎𝑡𝑜𝑚𝑠
3
𝑚𝑜𝑙𝑒
𝑐𝑚
22
𝑁=
=
= 8.48 × 10
𝑔
𝑀
𝑐𝑚3
55.847
𝑚𝑜𝑙𝑒
• Step 2: Using the atomic density and given microscopic crosssection, calculate the macroscopic cross-section.
Σ𝑎 = 𝑁𝜎𝑎 = 8.48 ×
1022
𝑎𝑡𝑜𝑚𝑠
2.56 𝑏𝑎𝑟𝑛𝑠
3
𝑐𝑚
1 × 10−24 𝑐𝑚2
1 𝑏𝑎𝑟𝑛
Σ𝑎 = 0.217 𝑐𝑚−1
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
109
Microscopic versus Macroscopic CrossSection
Review
• Microscopic cross-section (σ) represents the effective target area that
a single nucleus presents to a bombarding particle (neutron)
– The units are in barns or cm2
• Macroscopic cross-section (Σ) represents the effective target area
that is presented by all of the nuclei contained in 1 cm3 of the material
– The units are 1/cm or cm-1
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
110
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
• Mean free path for interaction = average distance of travel by a
neutron before interaction
– Symbol λ
1
𝜆=
Σ
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
111
Mean Free Path
Mean Free Path Calculation 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. Properties of aluminum
and silicon are shown below.
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:
1. Calculate the atomic densities for aluminum and silicon.
2. Determine absorption and scattering macroscopic crosssections for thermal neutrons.
3. Calculate the mean free paths for absorption and scattering.
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Mean Free Path
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 =
𝜌𝐴𝑙 𝑁𝐴
𝑀𝐴𝑙
𝑁𝑆𝑖 =
𝑔
23 𝑎𝑡𝑜𝑚𝑠
6.022
×
10
𝑚𝑜𝑙𝑒
𝑐𝑚3
=
𝑔
26.9815
𝑚𝑜𝑙𝑒
𝑎𝑡𝑜𝑚𝑠
= 5.64 × 1022
𝑐𝑚3
𝑔
23 𝑎𝑡𝑜𝑚𝑠
6.022
×
10
𝑚𝑜𝑙𝑒
𝑐𝑚3
=
𝑔
28.0855
𝑚𝑜𝑙𝑒
𝑎𝑡𝑜𝑚𝑠
= 2.85 × 1021
𝑐𝑚3
0.95 2.66
© Copyright 2017 – Rev 3
𝜌𝑆𝑖 𝑁𝐴
𝑀𝑆𝑖
0.05 2.66
ELO 5.1
Operator Generic Fundamentals
113
Mean Free Path
Solution
• Step 2: The absorption and scattering macroscopic cross-sections
are calculated from the microscopic cross-sections and the combined
atomic densities (95 percent aluminum and 5 percent silicon).
Σ𝑎 = 𝑁𝐴𝑙 𝜎𝑎,𝐴𝑙 + 𝑁𝑆𝑖 𝜎𝑎,𝑆𝑖
= 5.64 × 1022
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
0.23 × 10−24 𝑐𝑚2 + 2.85 × 1021
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
0.16 × 10−24 𝑐𝑚2
1.49 × 10−24 𝑐𝑚2 + 2.85 × 1021
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
2.20 × 10−24 𝑐𝑚2
= 0.0134 𝑐𝑚−1
Σ𝑠 = 𝑁𝐴𝑙 𝜎𝑠,𝐴𝑙 + 𝑁𝑆𝑖 𝜎𝑠,𝑆𝑖
= 5.64 × 1022
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
= 0.0903 𝑐𝑚−1
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
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Mean Free Path
Solution
• Step 3: Mean free paths are determined from the calculated absorption
and scattering macroscopic cross-sections.
𝜆𝑎 =
1
Σ𝑎
1
0.01345 𝑐𝑚−1
= 74.3 𝑐𝑚
=
1
𝜆𝑠 =
Σ𝑠
1
=
0.0903 𝑐𝑚−1
= 11.1 𝑐𝑚
© Copyright 2017 – Rev 3
ELO 5.1
Operator Generic Fundamentals
115
Neutron Reaction Terms
Knowledge Check:
_______________ is the total path length traveled by all neutrons in
one cubic centimeter of material during one second.
A. Mean free path
B. Neutron flux
C. Gamma flux
D. Atomic density
Correct answer is B.
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
116
Neutron Energy Terms
ELO 5.2 - Define the following neutrons: fast, intermediate, and slow.
• Neutrons are classified by energy levels:
– Fast
– Intermediate
– Slow
• Neutrons in energy equilibrium with their surrounding are called
thermal neutrons.
– These are in the slow category
– Most important where thermal reactors are concerned
© Copyright 2017 – Rev 3
ELO 5.2
Operator Generic Fundamentals
117
Neutron Energy Terms
< 1 ev
> 1 ev and < .1 Mev
Born fast - > .1 Mev
Figure: Neutron Energy
© Copyright 2017 – Rev 3
ELO 5.2
Operator Generic Fundamentals
118
Neutron Energy Terms
Knowledge Check
Thermal neutrons are classified in the intermediate neutron energy
range.
A. True
B. False
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 5.2
Operator Generic Fundamentals
119
Neutron Energies versus Cross-Sections
ELO 5.3 - Describe how the absorption and scattering cross-section of
typical nuclides varies with neutron energies in the 1/v region and the
resonance absorption region.
• Neutron cross-sections, both absorption and scattering, are affected
by neutron energies
• In general, the higher the energy of the neutron the lower the
probability of interaction
• Resonance peaks at certain specific energies possess very high
cross-sections
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
120
Neutron Energies versus Cross-Sections
• Absorption cross-sections have three unique regions of probability
related to neutron energy
– 1/v region
– Resonance region
– Fast neutron region
• Each with a different relationship to changing neutron energy
• Neutron scattering cross-sections are different
– Resonance elastic scattering and inelastic scattering crosssections do have resonance peaks smaller than absorption peaks
– Potential elastic scattering cross-sections are not greatly affected
by neutron energy
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
121
Neutron Scattering Cross-Section versus
Incident Neutron Energy
• With the exception of hydrogen, the elastic scattering cross-sections
are generally small
– Typically 5 barns to 10 barns
– Close to the magnitude of the actual geometric cross-sectional
area expected for atomic nuclei
• In potential elastic scattering
– Cross-section is essentially constant
– Independent of neutron energy up to around 1 MeV
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
122
Neutron Scattering Cross-Section versus
Incident Neutron Energy
• Resonance elastic scattering and inelastic scattering exhibit
resonance peaks similar to those for absorption cross-sections
– Resonances occur at lower energies for heavy nuclei than for
light nuclei
– Variations in scattering cross-sections are very small when
compared to the variations that occur in absorption cross-sections
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
123
Neutron Scattering Cross-Section versus
Incident Neutron Energy
• In the thermal reactor scattering reactions are primarily of concern in
the moderator where neutrons are thermalized
• Water (the moderator) are light nuclei, and elastic scatterings are
most probable in light nuclei, therefore, σs = σselastic
Figure: Elastic Scattering Cross-Section versus Neutron Energy
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
124
Neutron Absorption Cross-Section
versus Incident Neutron Energy
• Variation of absorption cross-sections with neutron energy is complex
• Absorption cross-sections are small for many elements, ranging from
a fraction of a barn to a few barns for thermal neutrons
• For a considerable number of nuclides the variation of absorption
cross-sections with incident neutron energy reveals three distinct
regions:
– 1/v region
– Resonance peaks region
– Fast neutron region
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
125
Neutron Absorption Cross-Section
versus Neutron Energy - 1/v Region
Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
126
Neutron Absorption Cross-Section vs.
Neutron Energy - Resonance Region
Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
127
Neutron Absorption Cross-Section
versus Incident Neutron Energy
Example
• Assuming that uranium-235 has a nuclear quantum energy level at
6.8 MeV above its ground state, calculate the kinetic energy a
neutron must possess to undergo resonant absorption in uranium235 at this resonance energy level.
Solution
𝑀𝑒𝑉
𝐵𝐸 = 𝑀𝑎𝑠𝑠
𝑈 + 𝑀𝑎𝑠𝑠 𝑛𝑒𝑢𝑡𝑟𝑜𝑛 − 𝑀𝑎𝑠𝑠
𝑈 × 931
𝑎𝑚𝑢
𝑀𝑒𝑉
𝐵𝐸 = 235.043925 + 1.008665 − 236.045563 × 931
𝑎𝑚𝑢
𝑀𝑒𝑉
𝐵𝐸 = 0.007025 𝑎𝑚𝑢 × 931
= 6.54 𝑀𝑒𝑉
𝑎𝑚𝑢
6.8 𝑀𝑒𝑉 − 6.54 𝑀𝑒𝑉 = 0.26 𝑀𝑒𝑉
236
235
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
128
Neutron Absorption Cross-Section
versus Incident Neutron Energy
Figure: Typical Neutron Absorption Cross-Section versus Neutron Energy
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
129
Neutron Energies versus Cross-Sections
Knowledge Check
At low neutron energies (<1 eV) the absorption cross-section for a
material is ___________ proportional to the neutron velocity. (Fill in the
blank).
Correct answer is inversely.
© Copyright 2017 – Rev 3
ELO 5.3
Operator Generic Fundamentals
130
Temperature Effects to Macroscopic
Cross-Sections
ELO 5.4 - Describe the macroscopic cross-section and mean free path at
various temperatures.
• Microscopic absorption cross-sections vary significantly as thermal
neutron energy varies
– Values given on most charts and tables are at an ambient
temperature of 68°F
– Equals a neutron velocity of 2,200 meters/second or 0.025 eV
energy
– At higher temperatures, microscopic absorption cross-sections
must be corrected for the new temperature value
• A temperature corrected microscopic cross-section allows corrected
macroscopic cross-sections and mean free paths
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
131
Temperature Effects to Macroscopic
Cross-Sections
Note
• 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
scatterings occurring in the moderator
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
132
Temperature Affects to Macroscopic
Cross-Sections
• Determine the microscopic cross-section by obtaining the standard (at
68°F) table value
• Convert temperatures to either Rankine or Kelvin absolute temperature
scales
– °R = °F + 460 or °K = °C + 273
1
• Calculate new microscopic cross-section using: 𝜎 =
𝑇 2
𝜎𝑜 𝑇𝑜
• 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
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
133
Temperature Affects to Macroscopic
Cross-Sections
• Calculate new macroscopic cross-section
Σ = 𝑁𝜎
• Where:
Σ = macroscopic cross-section (cm-1)
𝑁 = atomic density (atoms/cm3)
𝜎 = microscopic cross-section (cm2)
• Calculate new mean free path (λ)
1
𝜆=
Σ
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
134
Temperature Affects to Macroscopic
Cross-Sections
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 is 7.03 x 1020 atoms/cm3
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
135
Temperature Affects to Macroscopic
Cross-Sections
Solution
• Steps 1, 2 and 3. Determine microscopic cross-section
𝜎𝑓 = 𝜎𝑓,𝑜
1
𝑇𝑜 2
𝑇
68℉ + 460
= 583 𝑏𝑎𝑟𝑛𝑠
500℉ + 460
1
2
= 432 𝑏𝑎𝑟𝑛𝑠
• Step 4. Determine macroscopic cross-section for fission
2
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚
Ʃ𝑓 = 𝑁𝜎 = 7.03 × 1020
× 432 𝐵𝑎𝑟𝑛𝑠 × 1 × 10−24
3
𝑐𝑚
𝐵𝑎𝑟𝑛
Ʃ𝑓 = 0.304 𝑐𝑚−1
• Step 5. Determine the mean free path for fission
1
1
𝜆= =
= 3.29 𝑐𝑚
Σ
0.304 𝑐𝑚−1
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
136
Temperature Affects to Macroscopic
Cross-Sections
Macroscopic Cross-Section versus Temperature
• Macroscopic cross-section is inversely proportional to temperature
– As temperature increases, macroscopic cross-section decreases
Mean Free Path versus Temperature
• Mean free path shows that it is inverse to the macroscopic crosssection
– Directly proportional to temperature
– As temperature increases, the mean free path increases
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
137
Temperature Affects to Macroscopic
Cross-Sections
Knowledge Check
If an isotope has an absorption microscopic cross-section of 5,000
barns at 68°F, what happens to its absorption microscopic crosssection if temperature is increased to 1,000°F?
A. Increases
B. Decreases
C. No change
D. Need more data to answer
Correct answer is B.
© Copyright 2017 – Rev 3
ELO 5.4
Operator Generic Fundamentals
138
Neutron Flux Distribution
ELO 5.5 - Describe radial and axial neutron flux distribution.
• Neutron flux distribution is very important and is monitored by not
only the operators but also by reactor engineering
• Limiting uneven flux distribution ensures fuel cladding integrity
© Copyright 2017 – Rev 3
ELO 5.5
Operator Generic Fundamentals
139
Neutron Flux Distribution
• Neutrons interact with all materials in a reactor, absorptions,
scattering
• The type of material found in particular locations of the core affects
the neutron flux in that area
• Examples:
– Fast neutron flux in materials with large scattering cross-sections
causes fast neutrons to slow down to lower energy levels more
quickly
– This reduces the fast neutron flux
– Intermediate and thermal neutron flux increases as a larger
percentage of fast neutrons slow down to these energy levels
© Copyright 2017 – Rev 3
ELO 5.5
Operator Generic Fundamentals
140
Axial and Radial Neutron Flux
Commercial PWR nuclear reactor
cores are cylindrical in shape.
• Axial and radial flux are used to
describe neutron flux profiles
from top to bottom and across
the core.
Figure: Axial and Radial Neutron Flux in a Reactor Core
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
141
Axial and Radial Neutron Flux
• The axial and radial neutron flux
distribution across a nuclear
reactor is a spatial representation
of neutron flux level throughout
the core
• Within the core boundaries,
neutrons are
– Being produced from fission
– Slowed down by the
moderator
– Captured by core materials
and fuel, and fissioned
Figure: Axial and Radial Neutron Flux in a Reactor Core
• These all affect the neutron flux
distribution in the core
© Copyright 2017 – Rev 3
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Operator Generic Fundamentals
142
Axial and Radial Neutron Flux
• 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 the
edges of the core are more likely
to leak out and not cause fission
– Reducing thermal neutron
flux in the outer boundaries
of the core
• Neutrons toward the center,
have less leakage and greater
probability of thermalizing to
cause more fissions
– Increasing neutron flux
levels toward the center of
the core
© Copyright 2017 – Rev 3
Figure: Axial and Radial Neutron Flux in a Reactor Core
ELO 5.5
Operator Generic Fundamentals
143
Self-Shielding
• Neutron flux levels may be lower in some localized areas than in
others
– Known as self-shielding
• For example, the interior of a fuel pin or pellet is exposed to a lower
average neutron flux level than the outer surfaces
– Caused by large fraction of neutrons having been absorbed in the
outer layers of the fuel, reducing neutron availability to the pellet
interior
• Concept of self-shielding is also important relating to neutron poisons
in the reactor core
© Copyright 2017 – Rev 3
ELO 5.5
Operator Generic Fundamentals
144
Neutron Flux Distribution
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 ________.
Correct answer is axial; radial.
© Copyright 2017 – Rev 3
ELO 5.5
Operator Generic Fundamentals
145
Reaction Rate
ELO 5.6 - Describe how changes in neutron flux and macroscopic crosssection, affect reaction rates.
• Neutron flux [Φ] times macroscopic cross-section [Σ] equals the
reaction rate
– Denoted by the symbol R
• The type of reaction rate calculated depends on the macroscopic
cross-section used in the calculation
• Normally, the reaction rate of greatest interest is the fission reaction
rate
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
146
Neutron Reaction Rates
• Fission neutrons are born at an average energy of about 2 MeV
– Interact with reactor core materials in various absorption and
scattering reactions
– Scattering reactions are useful for thermalizing neutrons
– Thermal neutrons absorbed for fission
– Absorption in fertile material for production of fissionable fuel
• Some neutrons are absorbed in structural components, reactor
coolant, and other non-fuel materials
– Resulting in the removal of neutrons from the fission process
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
147
Neutron Reaction Rates
• To determine these neutron interaction rates, it is necessary to
– Identify the number of neutrons available
– The probability of interaction
• To quantify neutron availability and reaction probabilities, terms
including
– Neutron flux
– Microscopic cross-section
– Macroscopic cross-section
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
148
Reaction Rate
Calculation of the reaction rate:
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
3. Using the following formula for reaction rate, calculate the
reaction rate
𝑅 = ΣΦ
Where:
𝑅 = reaction rate (reactions/cm3 sec)
Σ = macroscopic cross-section (cm-1)
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
149
Reaction Rate Demonstration
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
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
= 1 × 10
𝑐𝑚2 – 𝑠𝑒𝑐
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
= 1 × 1012
𝑐𝑚3 – 𝑠𝑒𝑐
13
© Copyright 2017 – Rev 3
0.1 𝑐𝑚−1
ELO 5.6
Operator Generic Fundamentals
150
Reaction Rate
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.
neutrons
cm2 –sec
A.
R = 2.0 × 1014
B.
R = 0.2 × 1013
C.
R = 2.0 × 1013
neutrons
cm3 –sec
D.
R = 20 × 1013
neutrons
cm3 –sec
neutrons
cm2 –sec
Correct answer is C.
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
151
Reaction Rate
Knowledge Check
If the microscopic cross section for fission increases in a nuclear
reactor, how will the reaction rate for fission be affected?
A. R will decrease
B. R will stay the same
C. R change can not be determined based on information provided.
D. R will increase
Correct answer is D.
© Copyright 2017 – Rev 3
ELO 5.6
Operator Generic Fundamentals
152
Neutron Flux and Reactor Power
ELO 5.7 - Describe the relationship between neutron flux and reactor
power.
• 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
• If the amount of energy involved in each reaction is known, the rate of
energy release (power) due to a certain reaction can be determined
© Copyright 2017 – Rev 3
ELO 5.7
Operator Generic Fundamentals
153
Neutron Flux and Reactor Power
• If average energy per fission is 200 MeV, the number of fissions to
produce one watt of power equals:
– 1 fission = 200 MeV
– 1 MeV = 1.602 x 10-6 ergs
– 1 erg = 1 x 10-7 watt-sec
1 𝑤𝑎𝑡𝑡
1 𝑒𝑟𝑔
1 × 10−7 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐
1 𝑀𝑒𝑉
1.602 × 10−6 𝑒𝑟𝑔
1 𝑓𝑖𝑠𝑠𝑖𝑜𝑛
𝑓𝑖𝑠𝑠𝑖𝑜𝑛
= 3.12 × 1010
200 𝑀𝑒𝑉
𝑠𝑒𝑐𝑜𝑛𝑑
• 3.12 x 1010 fissions release 1 watt-second of energy
© Copyright 2017 – Rev 3
ELO 5.7
Operator Generic Fundamentals
154
Neutron Flux and Reactor Power
Step
1.
2.
© Copyright 2017 – Rev 3
Action
Obtain the following reactor data:
• Volume of core (cm3)
𝑅 = ΣΦ
• Reaction rate
OR
• Fission macroscopic cross-section and average thermal neutron flux
Calculate reactor power using the following equation, substitute reaction
rate if known:
Φ𝑡ℎ Σ𝑓 𝑉
𝑃=
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010
𝑤𝑎𝑡𝑡– 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)
ELO 5.7
Operator Generic Fundamentals
155
Relationship between Neutron Flux and
Reactor Power
• In an operating reactor the volume of the reactor is constant
• Over a relatively short period of time (days and weeks), the number
(density) of fuel atoms is also relatively constant
• With a constant atomic density and microscopic cross-section, the
macroscopic cross-section is also constant
• If the reactor volume and macroscopic cross-section are constant,
then reactor power and the neutron flux are directly proportional
• If the fission macroscopic cross-section decreases from the reactor
fuel burnup, neutron flux must increase to maintain power constant
© Copyright 2017 – Rev 3
ELO 5.7
Operator Generic Fundamentals