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Revision 1.1
Sept 2016
Neutrons
Student Guide
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ii
Table of Contents
INTRODUCTION ..................................................................................................................... 2
TLO 1 ATOMS ...................................................................................................................... 3
Overview .......................................................................................................................... 3
ELO 1.1 Atomic Structure ............................................................................................... 4
ELO 1.2 Atomic Terms.................................................................................................... 7
ELO 1.3 Atomic Forces ................................................................................................. 10
TLO 1 Summary ............................................................................................................ 14
TLO 2 MASS DEFECT AND BINDING ENERGY ..................................................................... 15
Overview ........................................................................................................................ 15
ELO 2.1 Mass Defect and Binding Energy ................................................................... 16
ELO 2.2 Determining Mass Defect and Binding Energy .............................................. 17
ELO 2.3 Binding Energy per Nucleon ........................................................................... 22
ELO 2.4 Binding Energy Per Nucleon Curve ................................................................ 23
TLO 2 Summary ............................................................................................................ 24
TLO 3 NEUTRON INTERACTIONS WITH MATTER ................................................................. 25
Overview ........................................................................................................................ 25
ELO 3.1 Neutron Scattering Interactions....................................................................... 26
ELO 3.2 Neutron Absorption Reactions ........................................................................ 28
TLO 3 Summary ............................................................................................................ 29
TLO 4 FISSION PROCESS ..................................................................................................... 30
Overview ........................................................................................................................ 30
ELO 4.1 Excitation and Critical Energy ........................................................................ 30
ELO 4.2 Fissile and Fissionable Material ...................................................................... 31
ELO 4.3 Fission Process - Liquid Drop Model ............................................................. 34
TLO 4 Summary ............................................................................................................ 36
TLO 5 PRODUCTION OF HEAT FROM FISSION ...................................................................... 37
Overview ........................................................................................................................ 37
ELO 5.1 Energy Release Per Fission ............................................................................. 37
ELO 5.2 Fission Fragment Yield ................................................................................... 39
ELO 5.3 Fission Heat Production .................................................................................. 41
TLO 5 Summary ............................................................................................................ 42
TLO 6 PROMPT AND DELAYED NEUTRONS ......................................................................... 43
Overview ........................................................................................................................ 43
ELO 6.1 Production of Prompt and Delayed Neutrons ................................................. 43
ELO 6.2 Delayed Neutron Fraction ............................................................................... 46
ELO 6.3 Neutron Lifetimes and Generation Times ....................................................... 47
TLO 6 Summary ............................................................................................................ 49
TLO 7 NEUTRON MODERATION .......................................................................................... 50
Overview ........................................................................................................................ 50
ELO 7.1 Neutron Moderation ........................................................................................ 51
ELO 7.2 Moderator Characteristics ............................................................................... 54
TLO 7 Summary ............................................................................................................ 56
iii
TLO 8 NEUTRON REACTION RATES/PROBABILITIES .......................................................... 56
Overview ....................................................................................................................... 56
ELO 8.1 Neutron Reaction Terms ................................................................................. 58
ELO 8.2 Neutron Energy Terms ................................................................................... 64
ELO 8.3 Neutron Energies versus Cross-Sections ........................................................ 66
ELO 8.4 Reaction Rate .................................................................................................. 68
ELO 8.5 Neutron Flux and Reactor Power ................................................................... 71
TLO 8 Summary ............................................................................................................ 74
KNOWLEDGE CHECK ANSWER KEY...................................................................................... 1
ELO 1.1 Atomic Structure ............................................................................................... 1
ELO 1.2 Atomic Terms ................................................................................................... 1
ELO 1.3 Atomic Forces ................................................................................................... 2
ELO 2.1 Mass Defect and Binding Energy ..................................................................... 3
ELO 2.2 Determining Mass Defect and Binding Energy ................................................ 4
ELO 2.3 Binding Energy per Nucleon ............................................................................ 4
ELO 2.4 Binding Energy Per Nucleon Curve ................................................................. 5
ELO 3.1 Neutron Scattering Interactions ........................................................................ 5
ELO 3.2 Neutron Absorption Reactions ......................................................................... 5
ELO 4.1 Excitation and Critical Energy.......................................................................... 6
ELO 4.2 Fissile and Fissionable Material ....................................................................... 6
ELO 4.3 Fission Process - Liquid Drop Model ............................................................... 7
ELO 5.1 Energy Release Per Fission .............................................................................. 7
ELO 5.2 Fission Fragment Yield..................................................................................... 8
ELO 5.3 Fission Heat Production .................................................................................... 8
ELO 6.1 Production of Prompt and Delayed Neutrons ................................................... 9
ELO 6.2 Delayed Neutron Fraction................................................................................. 9
ELO 6.3 Neutron Lifetimes and Generation Times ...................................................... 10
ELO 7.1 Neutron Moderation........................................................................................ 11
ELO 7.2 Moderator Characteristics ............................................................................... 12
ELO 8.1 Nuclear Reaction Terms ................................................................................. 13
ELO 8.2 Neutron Energy Terms ................................................................................... 13
ELO 8.3 Neutron Energies versus Cross-Sections ........................................................ 14
ELO 8.4 Reaction Rate .................................................................................................. 14
ELO 8.5 Neutron Flux and Reactor Power ................................................................... 14
iv
Neutrons
Revision History
Revision
Date
9-29-2016
Rev 0.1
Version
Number
Purpose for Revision
Performed
By
0
New Module
OGF Team
0.1
Incorporated additional
reviewer comments
OGF Team
1
Introduction
Atoms are the building blocks of matter; they are comprised of a nucleus,
an orbiting field and a large volume of empty space. Atoms are the smallest
components of matter that retain the identifying properties of an element.
Each element is made up of atoms identified by a unique combination of
subatomic particles making up their nuclei and orbiting fields.
When an atom’s subatomic particle configuration is changed, the atom’s
elemental identification is changed. Understanding these subatomic
interactions is important to understanding the fission process that occurs in
a nuclear reactor.
In thermal reactors, the neutrons that cause fission are born during the
fission process at a much higher energy level than required. To make
fission more probable, these neutrons must be slowed down to what is
known as thermal energy. To reduce high-energy neutrons to thermal
energy levels a process known as moderation must take place. Pressurized
water reactors (PWRs) utilize water as a moderator for thermalizing
neutrons.
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 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 atoms, including components, structure, and nomenclature.
2. Describe mass defect and binding energy and their relationship to one
another.
3. Describe neutron interactions with matter.
4. Describe the process of nuclear fission and the types of material that
can undergo fission.
5. Explain the production of heat from fission.
6. Describe the production of prompt and delayed neutrons from fission
as well as the differences between them.
7. Describe the process of neutron moderation in a nuclear reactor and
the characteristics of desirable moderators.
8. Explain the relationship between neutron flux, microscopic and
macroscopic cross-sections, and their effect on neutron reaction rates.
Rev 0.1
2
TLO 1 Atoms
Overview
Chemist John Dalton first proposed the modern proof for the atomic nature
of matter in 1803. He theorized that unique atoms characterize each
element, that the unique atoms distinguish each element from all others, and
that the physical difference between different types of atoms is their weight.
Subatomic Particles
Because of technology limitations, it took almost 100 years to prove
Dalton’s theories. Initially, chemical experiments indicated that the atom
was indivisible. Later, electrical and radioactivity experimentation
indicated that particles of matter smaller than the atom do exist. In 1906,
Joseph John Thompson won the Nobel Prize in physics for establishing the
existence of electrons. In 1920, Earnest Rutherford named the hydrogen
nucleus a proton and in 1932, Sir James Chadwick confirmed the existence
of the neutron.
In the 1970s, the application of the standard model of particle physics
proved the existence of quarks demonstrating the complexity of the atom.
Many questions about the atom still exist. Experiments now in progress and
in the future will provide a greater understanding of the atom.
Atomic Properties Determine Nuclear Fuel
Good
Points
Knowledge of atoms is important because their
properties determine whether they would be a good
nuclear fuel. Understanding the characteristics of atoms
helps the student understand the nature of atomic power
and the forces that control it.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Using Bohr's model of an atom, describe the characteristics of the
following atomic particles, including mass, charge, and location
within the atom:
a. Proton
b. Neutron
c. Electron
2. Describe the nomenclature for identifying nuclides and isotopes of a
given nuclide.
3. Describe the three forces that act on particles within the nucleus, and
how they affect the stability of the nucleus.
Rev 0.1
3
ELO 1.1 Atomic Structure
Introduction
Physicist Ernest Rutherford postulated that the positive charge in an atom is
concentrated at the center of the atom with electrons orbiting around it.
Later, Niels Bohr, combining Rutherford's theory and the quantum theory of
Max Planck, proposed that orbiting electrons in discrete fixed distances
from the nucleus surround the atom’s center positive charge of protons. An
electron in one of these orbits, called shells, has a specific or discrete
quantity of energy (quantum). Electron movement between shells results in
an energy difference either emitted or absorbed in the form of a single
quantum of radiant energy called a photon.
Neutrons and Protons
Protons and neutrons are located in a tight cluster in the center of the atom
called the nucleus. Atoms comprise each element, having a unique number
of protons in their nuclei. Neutrons are electrically neutral and have no
electrical charge. Protons are electrically positive and exhibit an electrical
charge of +1. Protons give the nucleus its positive charge. A nucleus with
one proton has a +1, with two protons a charge of +2, and so on with
increasing numbers of protons.
Neutrons and protons are each essentially equal in mass; together they make
up the mass of the nucleus.
Figure: Simple Carbon-12 Atom
Rev 0.1
4
Electrons
Electrons are the particles that orbit the nucleus. They orbit the nucleus in
concentric orbits referred to as orbitals or shells. Electrons are small and
light, with a mass of only 1/1835 the mass of a proton or neutron.
Each electron exhibits an electrical charge of minus one (-1) and equals in
magnitude the charge of one (+1) proton. For the atom to be electrically
neutral, its normal state, the number of electrons orbiting the nucleus is
exactly equal to the number of protons in the nucleus. Electrons are bound
to the nucleus by electrostatic attraction because opposite electrical charges
attract. The atom remains neutral unless some external force causes a
change in the number of electrons.
Bohr's Model
The figure below shows Bohr’s model using a hydrogen atom. The figure
shows an electron that has dropped from the third shell to the first shell
releasing energy in the process. The energy is released as a photon
emission equal to h (h = Planck's constant - 6.63 x 10-34 J-s (Jouleseconds) and = frequency of the photon). Bohr's theory accounts for the
quantum energy levels as measured in the laboratory. Although Bohr's
atomic model specifically explains the hydrogen atom, it applies as first
generation model to all atoms.
Figure: Bohr's Model of the Hydrogen Atom
Rev 0.1
5
Atomic Measuring Units
Atoms are so small that normal measuring units are difficult to apply. Mass
and energy use universally accepted units of measure on the atomic scale to
standardize measurement units and calculations. It is possible to convert
values expressed in these atomic scale units to non-atomic scale units if
desired.
Atomic Mass Unit (amu): The unit of measure for mass is the amu. One
amu equals 1.66 x 10-24 grams. 6.02 x 1023 amu = 1 gram. Neutrons and
protons each have a mass close to one (1) amu.
Electron Volt (eV): The unit for energy is the electron volt (eV) or Megaelectron Volt (MeV). The electron volt is the amount of energy gained (or
lost) by a single electron moved across a potential difference of one volt.
One electron volt is equals 1.602 x 10-19 Joules (J) or 1.18 x 10-19 footpounds (lbf). A proton’s eV value is positive one (+ 1) and an electron’s eV
is negative one (-1).
The table below shows properties of the three subatomic particles:
Properties of Three Subatomic Particles
Particle
Location
Charge
Mass (Rest)
Neutron
Nucleus
None
1.008665 amu
Proton
Nucleus
+1 eV
1.007277 amu
Electron
Shells Around
Nucleus
-1 eV
0.0005486 amu
Knowledge Check (Answer Key)
Identify the particles included in the make-up of an atom.
(More than one answer may apply.)
Rev 0.1
A.
neutron
B.
electron
C.
gamma
D.
amu
6
ELO 1.2 Atomic Terms
Introduction
Atoms have characteristics describing their behavior. This section defines
and outlines those terms.
Nuclear Nomenclature
The following terms describe characteristics of atoms:





Atomic number
Neutron number
Mass number
Nuclide
Isotope
Atomic Number
The atomic number describes the number of protons in the atom’s nucleus
and identifies the element. It is the Z number in the Chart of the Nuclides
(atomic notation). The number of protons identifies the particular element;
therefore, the atomic number identifies a particular element.
For example, any atom having two protons in its nucleus has an atomic
number of two (2) and is identified as the element helium. Because the
number of electrons in an atom matches the number of protons (electrically
neutral), the atomic number equals the number of electrons in the atom.
Neutron Number
The symbol N denotes the number of neutrons in a nucleus, which is the
neutron number. Atomic notation does not include N; however, N is
determined by subtracting the atomic number (Z) from the atomic mass
number (A).
Mass Number
The mass number of the atom is equal to the total number of protons and
neutrons in the nucleus. In atomic notation, the mass number is the A
number. We calculate A as follows:
𝐴=𝑍+𝑁
Where:
•
Z = Atomic number
•
N = Neutron number
Rev 0.1
7
Nuclides
Nuclides are atoms that contain a unique combination of protons and
neutrons. Not all proton and neutron combinations can exist in nature or as
man-made or fabricated combinations. However, scientists have identified
about 2,500 specific nuclides. The figure below shows the atomic notation
of a nuclide with the chemical symbol (X letter) of the element, the atomic
number written as a subscript, and the mass number written as a superscript.
Isotope
Atoms of the same element always contain the same number of protons (Z),
but not always the same number of neutrons (N). This results in some
atoms of an element with different atomic mass numbers (A). These atoms
are isotopes. Isotopes of a particular element have different atomic mass
numbers (A); however, they have the same chemical characteristics with
different numbers of neutrons. This affects their radioactivity stability.
Most elements have a few stable isotopes and several unstable radioactive
isotopes. For example, oxygen has three stable isotopes found in nature
(oxygen-16, oxygen-17, and oxygen-18) and eight unstable radioactive
isotopes. Another example is hydrogen, which has two stable isotopes
(hydrogen-1 and hydrogen-2) and a single radioactive isotope (hydrogen-3).
Atomic Notation
A convention, known as atomic or standard notation, identifies elements
using the nuclear nomenclature previously described. The figure below
shows standard notation:
Figure: Nomenclature for Identifying Nuclides
Rev 0.1
8
Identifying a Nuclide
Each element has a unique chemical name, symbol, and atomic number.
Any one of the three identifies the element. The chemical name or symbol
followed by the mass number (for example, U-235 or uranium-235)
identifies an element. Another frequently used identification method is the
chemical symbol with a left superscript (for example, 235U).
The nuclide concept (referring to individual nuclear species) emphasizes
nuclear properties over chemical properties, while the isotope concept
(grouping all atoms of each element) emphasizes chemical over nuclear.
The neutron number has large effects on nuclear properties, but its effect on
chemical properties is negligible for most elements.
Knowledge Check (Answer Key)
What is the element and number of neutrons for the
following:
235
92𝑈
A.
Uranium; 143
B.
Uranium; 92
C.
Plutonium; 143
D.
Plutonium; 92
Knowledge Check (Answer Key)
In the table below, complete the columns for element, protons, electrons,
and neutrons.
Nuclide
Element
Protons
Electrons
Neutrons
1
1𝐻
10
5𝐵
16
8𝑂
235
92𝑈
239
94𝑃𝑢
Rev 0.1
9
ELO 1.3 Atomic Forces
Introduction
Electrical forces in the nucleus determine the way the atomic forces behave
and their respective electrical charge.
Atomic Forces Acting in the Nucleus
The nucleus consists of positively charged protons and electrically neutral
neutrons in the Bohr model of the atom. The protons and neutrons are
termed nucleons.
Electrostatic and Gravitational Forces
Two forces present in the nucleus are: (1) electrostatic
forces between charged particles, and (2) gravitational
forces between any two objects that have mass. It is
For More possible to calculate the magnitude of the gravitational
Information force and electrostatic force based on principles from
classical physics.
Gravitational Force
Newton’s law of universal gravitation states that gravitational force between
two bodies is directly proportional to the masses of the two bodies and
inversely proportional to the square of the distance between the bodies. The
equation below shows this relationship:
𝐹𝑔 =
𝐺𝑚2 𝑚2
𝑟2
Where:
•
Fg = gravitational force (Newtons)
•
m1 = mass of first body (kilograms)
•
m2 = mass of second body (kilograms)
•
G = gravitational constant (6.67 x 10-11 N-m2/kg2)
•
r = distance between particles (meters)
Within the nucleus the nucleon mass is small, but the distance is extremely
short. Calculating the gravitational force for two protons separated by a
distance of 10-13 meters is about 10-17 Newtons.
Rev 0.1
10
Electrostatic Force
We use Coulomb's Law to calculate the electrostatic force between two
protons. The electrostatic force is directly proportional to the electrical
charges of the two particles and inversely proportional to the square of the
distance between the particles. The equation below shows the Coulomb's
Law relationship:
𝐹𝑒 =
𝐾𝑄1 𝑄2
𝑟2
Where:
•
Fe = electrostatic force (Newtons)
•
K = electrostatic constant (9.0 x 109 N-m2/C2 [Coulombs squared])
•
Q1 = charge of first particle (Coulombs [C])
•
Q2 = charge of second particle (Coulombs)
•
r = distance between particles (meters [m])
Using this equation, the electrostatic force between two protons separated
by a distance of 10-13 meters is about 105 Newtons. Because the
electrostatic force (105 Newtons) is much greater than the gravitational
force (10-17 Newtons), the gravitational force can be neglected.
Nuclear Force
Without another explanation, it is impossible to have a stable nuclei
composed of protons and neutrons if only the electrostatic and gravitational
forces existed in the nucleus. The gravitational forces are much too weak to
hold the nucleons together.
Another force, called the nuclear force, is a strong attractive force
independent of charge. It acts equally between pairs of neutrons, pairs of
protons, or a neutron and a proton. Nuclear force acts over a short range
limited to distances approximately equal to the diameter of the nucleus
(10-13 meters). The attractive nuclear force between nucleons decreases
with distance much quicker than the repulsive electrostatic force between
protons.
Rev 0.1
11
Atomic Forces
Force
Interaction
Range
1.
Gravitational
Weak attractive force
between all nucleons
Relatively long
2.
Electrostatic
Strong repulsive
force between like
charged particles
(protons)
Relatively long
3.
Nuclear Force
Strong attractive
force between all
nucleons
Extremely short
Attractive and repulsive forces in the nucleus balance in stable atoms. If
unbalanced, the nucleus emits radiation in an attempt to achieve a stable
configuration. This phenomenon is discussed later in this module.
Knowledge Check (Answer Key)
Very weak attractive force between all nucleons
describes which of the forces listed below?
A.
Electrostatic
B.
Nuclear
C.
Gravitational
D.
Atomic
Neutron Proton Ratio
The neutron-proton ratio (N/Z ratio or nuclear ratio) is the ratio of the
number of neutrons to protons that make up the nucleus. Light elements up
to calcium (Z = 20) have stable isotopes with a neutron-proton ratio of one,
except for beryllium and every element with odd proton numbers from
fluorine (Z = 9) to potassium (Z = 19). Helium-3 is the only stable isotope
with a neutron-proton ratio under one. Uranium-238 has a high N/Z ratio of
any natural isotope at 1.59; lead-208 has the highest N/Z ratio of any known
stable isotope at 1.54.
Rev 0.1
12
The figure below shows the distribution of the stable nuclides plotted on the
same axes as the Chart of the Nuclides. The ratio of neutrons to protons in
the nucleus becomes larger as the mass numbers become higher. For
helium-4 (2 protons and 2 neutrons) and oxygen-16 (8 protons and 8
neutrons), this ratio is unity. For indium-115 (49 protons and 66 neutrons),
the ratio of neutrons to protons has increased to 1.35, and for uranium-238
(92 protons and 146 neutrons) the neutron to proton ratio is 1.59.
Figure: Neutron-Proton Plot of the Stable Nuclides
A nuclide existing outside of the band of stability can undergo alpha decay,
positron emission, electron capture, or beta emission to gain stability. For
example, following fission, the two resulting fragments have nuclei with
approximately the same high neutron-to-proton ratio as the original heavy
nucleus. This high neutron-to-proton ratio with lower proton numbers
places the fragments below and to the right of the stability line. Successive
beta emissions, each converting a neutron to a proton, create a more stable
neutron-to-proton ratio.
Knowledge Check (Answer Key)
Which of the following nuclides has the higher neutronproton ratio?
Rev 0.1
A.
Cobalt-60
B.
Selenium-79
C.
Silver-108
D.
Cesium-137
13
TLO 1 Summary
 Atoms consist of three basic subatomic particles:
— Protons: particles that have a positive charge and exist in the
nucleus. A proton has a mass of 1 amu.
— Neutrons: particles that have no electrical charge also exist in
the nucleus. A neutron has approximately the same mass as a
proton, about 1 amu.
— Electrons: particles that have a negative charge, orbit in shells
around the nucleus and have a mass about 1/1,800 the mass of a
proton.
 Bohr model of an atom: a dense nucleus of protons and neutrons
surrounded by orbiting electrons traveling in discrete orbits at fixed
distances.
 Nuclides are atoms containing certain numbers of protons and
neutrons.
 Isotopes are nuclides having same atomic number but with differing
numbers of neutrons. Isotopes have the same chemical properties.
 Atomic number of an atom is the number of protons in the nucleus.
 Mass number of an atom is the total number of nucleons (protons
and neutrons) in the nucleus.
 Atomic (standard) notation identifies a specific nuclide, shown below
in the graphic:
— Z represents the atomic number, which equals the number of
protons
— A represents the mass number, which equals the number of
nucleons
— X represents the chemical symbol of the element
— Number of protons = Z
— Number of neutrons = A - Z
 The different forces interacting within the nucleus determine the
nucleus’ stability.
— Gravitational force: a long-range, relatively weak attraction
between masses, and negligible compared to other forces.
— Electrostatic force: a relatively long-range, strong, and
repulsive force that acts between positively charged protons.
— Nuclear force: a short-range, attractive force between all
nucleons that is able to balance the repulsive electrostatic
force in a stable nucleus.
Rev 0.1
14

A high neutron-to-proton ratio places nuclides below and to the right
of the stability curve.
 Instability caused by excess neutrons is often fixed by successive beta
emissions, with each converting a neutron to a proton.
 Atom percent (a/o) is the percentage of the atoms of an element that
are of a particular isotope.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Using Bohr's model of an atom, describe the characteristics of the
following atomic particles, including mass, charge, and location
within the atom:
a. Proton
b. Neutron
c. Electron
2. Describe the nomenclature for identifying nuclides and isotopes of a
given nuclide.
3. Describe the three forces that act on particles within the nucleus and
how they affect the stability of the nucleus.
TLO 2 Mass Defect and Binding Energy
Overview
Binding energy and mass defect describe the energy associated with nuclear
reactions. Understanding mass defect and binding energy and their
relationship is important for understanding energies associated with atomic
reactions, including fission.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Define the relationship between mass defect and binding energy.
2. Given the atomic mass for a nuclide and the atomic masses of a
neutron, proton, and electron, calculate the mass defect and binding
energy of the nuclide.
3. Explain binding energy per nucleon.
4. Explain the shape of the binding energy per nucleon versus mass
number curve including its significance to fission energy.
Rev 0.1
15
ELO 2.1 Mass Defect and Binding Energy
Introduction
Although the laws of conservation of mass and conservation of energy hold
true, conversion between mass and energy occurs on a nuclear level.
Instead of two separate conservation laws, a single conservation law states
that the sum of mass and energy is conserved.
In Einstein’s theory of relativity, energy and mass are equivalent. In fact,
the rest energy and the mass are related by E = mc2, where c is the speed of
light in a vacuum. Therefore, a change in the rest energy of the system is
equivalent to a change in the mass of the system. The binding energy used
to disassemble the nucleus appears as extra mass of the separated and
stationary nucleons. In other words, the sum of the individual masses of the
separated protons and neutrons is greater by some amount than the mass of
the stable nucleus.
Mass Defect and Binding Energy
 Mass defect: experimental measurements show that the mass of a
particular atom is always slightly less than the sum of the atom’s
individual neutrons, protons, and electron masses. This difference is
the mass defect (∆m).
 Binding energy: a change of mass occurs from the conversion of
mass to binding energy (BE) during formation of a nucleus. Binding
energy is the amount of energy supplied to a nucleus to separate its
nuclear particles completely. Conversely, it is the amount of energy
released if separate particles formed the nucleus.
Knowledge Check (Answer Key)
_______________ is the amount of energy that must be
supplied to a nucleus to completely separate its nuclear
particles.
Rev 0.1
A.
Nuclear energy
B.
Binding energy
C.
Mass defect
D.
Separation energy
16
ELO 2.2 Determining Mass Defect and Binding Energy
Introduction
Binding energy is the expected reduction of potential energy in an attractive
force field. When the nucleus is divided into separated nucleons, energy is
required. The separated nucleons have a greater mass than the original
nucleus.
In atomic physics, mass defect is the difference in mass between the atom
and the sum of the masses of that atom’s respective protons, neutrons, and
electrons.
Calculating Mass Defect
The mass defect can be calculated using the below equation. It is important
to use the full accuracy of mass measurements in calculating the mass
defect because the difference in mass is small compared to the mass of the
atom. Rounding off the masses of atoms and particles to three or four
significant digits prior to the calculation results in a calculated mass defect
of zero (0). NOTE: The mass of the Hydrogen atom is the same as the mass
of the proton and electron used below in the equation.
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚
Where:
•
Δm = mass defect (amu)
•
mp = mass of a proton (1.007277 amu)
•
mn = mass of a neutron (1.008665 amu)
•
me = mass of an electron (0.000548597 amu)
•
matom = mass of nuclide 𝐴𝑍𝑋 (amu)
•
Z = atomic number (number of protons)
•
A = mass number (number of nucleons)
Rev 0.1
17
Steps for using the formula for mass defect:
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚
Step
Description
Action
1.
Determine the Z (atomic
number) and A (atomic
mass) of the nuclide.
Look up information in the Chart of
the Nuclides.
2.
Determine the mass of the
protons and electrons of
the nuclide.
Multiply Z times the mass of a proton
and the mass of an electron: 𝑍(mp +
me ).
Determine the mass of the
neutrons.
Subtract the atomic number (Z) from
the atomic mass (A) then multiply by
mass of a neutron: (𝐴 − 𝑍)mn .
4.
Add the mass of the
protons, electrons, and
neutrons.
Add the products determined in the
previous two steps: 𝑍(mp + me ) +
(𝐴 − 𝑍)mn .
5.
Determine the difference
between the atomic mass
of the nuclide and the mass
determined above.
Subtract the mass of the atom of the
nuclide: [𝑍( mp + me ) +
(𝐴 − 𝑍)mn ] − matom .
3.
Calculating Binding Energy
Binding energy is the energy equivalent to the mass defect. Calculate
binding energy using a conversion factor derived from Einstein's Theory of
Relativity. When the nucleus forms from its separate particles, mass defect
converts to binding energy.
Einstein's famous equation relating mass and energy is 𝐸 = 𝑚𝑐 2 , where c is
the velocity of light (c = 2.998 x 108 meters per second [m/sec]). The
energy equivalent of 1 amu is calculated by inserting this quantity of mass
into Einstein's equation and applying conversion factors.
Rev 0.1
18
𝐸 = 𝑚𝑐 2
1.6606 × 10−27 𝑘𝑔
𝑚
1𝑁
1𝐽
= 1 𝑎𝑚𝑢 (
) (2.998 × 108
)(
)(
)
𝑘𝑔– 𝑚 1𝑁 − 𝑚
1 𝑎𝑚𝑢
𝑠𝑒𝑐
1
𝑠𝑒𝑐 2
1 𝑀𝑒𝑉
= 1.4924 × 10−10 𝐽 (
)
1.6022 × 10−13 𝐽
= 931.5 𝑀𝑒𝑉
Conversion Factors:
1 𝑎𝑚𝑢 = 1.6606 × 10−27 𝑘𝑔
1 N𝑒𝑤𝑡𝑜𝑛 = 1
𝑘𝑔– 𝑚
𝑠𝑒𝑐 2
1 𝐽𝑜𝑢𝑙𝑒 = 1 𝑁𝑒𝑤𝑡𝑜𝑛– 𝑚𝑒𝑡𝑒𝑟
1 𝑀𝑒𝑉 = 1.6022 × 10−13 𝐽𝑜𝑢𝑙𝑒𝑠
Since 1 amu is equivalent to 931.5 MeV of energy, the binding energy can
be calculated from the following:
𝐵. 𝐸. = ∆𝑚 (
931.5 𝑀𝑒𝑉
)
1 𝑎𝑚𝑢
Steps to using the formula for binding energy:
𝐵. 𝐸. = ∆𝑚 (
931.5 𝑀𝑒𝑉
)
1 𝑎𝑚𝑢
Description
Action
1.
Determine the mass defect of the
nuclide.
Use the equation:
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) +
(𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚
2.
Use the binding energy equation to
calculate the binding energy.
Use the equation:
931.5 𝑀𝑒𝑉
𝐵. 𝐸. = ∆𝑚 (
)
1 𝑎𝑚𝑢
Calculate the binding energy.
Multiply the change in mass
from Step 1 by the energy
conversion for an amu.
3.
Rev 0.1
19
Calculating Mass Defect Example
Calculate the mass defect for lithium-7 given the mass of lithium-7 =
7.016003 amu.
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚
Step
Description
Action
1.
Determine the Z (atomic
number) and A (atomic
mass number) of the
nuclide.
Z = 3, A = 7
2.
Determine the mass of
the protons and electrons
of the nuclide.
3 (1.007826 amu + 0.000548597 amu)
= 3.02347979 amu
3.
Determine the mass of
the neutrons.
(7-3) (1.008665) = 4.03466 amu
4.
Add the mass of the
protons, electrons and
neutrons.
3.02347979 amu + 4.03466 amu =
7.058140 amu
5.
Determine the difference
between the atomic mass
of the nuclide and the
mass determined above.
7.058140 amu - 7.016003 amu =
0.042137 amu
Rev 0.1
20
Example: Calculating Binding Energy of Lithium
Calculate the binding energy for lithium-7:
Step
Description
Action
1.
Determine the mass
defect of the nuclide.
From above calculation: 0.042137
amu.
2.
Use the binding energy
equation to calculate the
binding energy.
3.
Calculate the binding
energy
931.5 𝑀𝑒𝑉
𝐵. 𝐸. = ∆𝑚 (
)
1 𝑎𝑚𝑢
BE = 0.042137 amu (
931.5 MeV
)
1 amu
= 39.2506 MeV
Knowledge Check (Answer Key)
Calculate the mass defect for uranium-235. One
uranium-235 atom has a mass of 235.043924 amu.
mp = mass of a proton (1.007277 amu)
mn = mass of a neutron (1.008665 amu)
me = mass of an electron (0.000548597 amu)
Rev 0.1
A.
1.86471 amu
B.
1.91517 amu
C.
0.191517 amu
D.
0.186471 amu
21
Knowledge Check (Answer Key)
Calculate the binding energy for uranium-235. One
uranium-235 atom has a mass defect of 1.9157 amu.
A.
1784 MeV
B.
178.4 MeV
C.
1783 MeV
D.
178.3MeV
ELO 2.3 Binding Energy per Nucleon
Introduction
Binding energy per nucleon (BE/A) is a measure of the stability of a
nucleus. The higher the binding energy per nucleon, the more stable the
nucleus. The lower the binding energy per nucleon the less stable the
nucleus. 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:
𝐵𝐸 1,804.3 𝑀𝑒𝑉
=
= 7.6 𝑀𝑒𝑉
𝐴
238
Knowledge Check (Answer Key)
True or False: Binding energy per nucleon is
independent of the specific nuclide.
Rev 0.1
A.
True
B.
False
22
ELO 2.4 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.
Rev 0.1
23
As the atomic number (number of protons) and the atomic mass number of
a nucleus increase, the repulsive electrostatic forces within the nucleus
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 (Answer Key)
True or False. Generally, less stable nuclides have a
higher BE/A than the more stable ones.
A.
True
B.
False
TLO 2 Summary
Binding energy and mass defect describe the energy associated with nuclear
reactions. The separated nucleons have a greater mass than the original
nucleus. Mass defect is the difference in mass between the atom and the
sum of the masses of that atom’s respective protons, neutrons, and
electrons.
Rev 0.1
24
The binding energy used to disassemble the nucleus appears as extra mass
of the separated and stationary nucleons. In other words, the sum of the
individual masses of the separated protons and neutrons is greater by some
amount than the mass of the stable nucleus.
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.
BE/A can be graphed resulting in a curve of binding energy per nucleon that
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.
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.
TLO 3 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. Describe the following reactions where a neutron is absorbed in a
nucleus:
a. Radiative capture
b. Particle ejection
c. Fission
Rev 0.1
25
ELO 3.1 Neutron Scattering Interactions
Introduction
A neutron scattering reaction occurs when a neutron strikes 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 is a mathematical illustration of the conservation of
momentum and kinetic energy during an elastic scattering reaction:
Conservation of momentum (𝑚𝑣)
(𝑚𝑛 𝑣𝑛,𝑖 ) + (𝑚 𝑇 𝑣𝑇,𝑖 ) = (𝑚𝑛 𝑣𝑛,𝑓 ) + (𝑚 𝑇 𝑣𝑇,𝑓 )
1
Conservation of kinetic energy (2 𝑚𝑣 2 )
1
1
1
1
2
2
2
2
( 𝑚𝑛 𝑣𝑛,𝑖
) + ( 𝑚 𝑇 𝑣𝑇,𝑖
) = ( 𝑚𝑛 𝑣𝑛,𝑓
) + ( 𝑚 𝑇 𝑣𝑇,𝑓
)
2
2
2
2
Rev 0.1
26
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.
Rev 0.1
27
Knowledge Check (Answer Key)
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 3.2 Neutron Absorption Reactions
Introduction
Most absorption reactions result in the loss of a neutron, the production of
charged particles, gamma ray(s), or fission fragments and the release of
neutrons and energy. Three types of absorption reaction are:



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+
𝑈→(
𝑈) →
𝑈+ 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.
1
10
11 ∗ 7
4
𝑛 + 𝐵 → ( 𝐵) → 𝐿𝑖 + 𝛼
0
5
5
3
2
Rev 0.1
28
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+
𝑈→(
𝑈) →
𝐶𝑠 + 𝑅𝑏 + 3 ( 𝑛)
0
92
55
37
0
92
Knowledge Check (Answer Key)
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 3 Summary
Neutron Interactions
 Interactions where a neutron scatters off a target nucleus are either
elastic or inelastic.
 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.
 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, there is conservation of momentum.
 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.
 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).
Rev 0.1
29

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
3. Describe the following reactions where a neutron is absorbed in a
nucleus:
a. Radiative capture
b. Particle ejection
c. Fission
TLO 4 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)
b. Critical energy (Ecrit)
2. Define the following terms:
a. Fissile material
b. Fissionable material
3. Explain the fission process using the liquid drop model of a nucleus.
ELO 4.1 Excitation and Critical Energy
Introduction
Excitation energy is the energy above ground state of a nucleus. Critical
energy is the required excitation energy for fission to occur.
Rev 0.1
30
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 was covered in TLO 2.
Knowledge Check (Answer Key)
True or False: Excitation must be at least equal to
critical energy for fission to occur.
A.
True
B.
False
ELO 4.2 Fissile and Fissionable 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:


Fissile material
Fissionable material
Rev 0.1
31
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.
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
Required
Kinetic Energy
233
92U
6.0 MeV
7.0 MeV
0 MeV
235
92U
6.5 MeV
6.8 MeV
0 MeV
239
94Pu
6.0 MeV
6.4 MeV
0 MeV
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.
Rev 0.1
32
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
Required
Kinetic Energy
238
92U
7.0 MeV
5.2 MeV
1.8 MeV
240
94Pu
6.2 Mev
5.2 MeV
1.0 MeV
242
94Pu
6.2 MeV
6.5 MeV
0.3 MeV
As seen in the tables 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 5.2 MeV BEn added by the neutron. Therefore, a fission
neutron must have at least 1.8 MeV of kinetic energy to cause fission of U238. This makes uranium-238 a fissionable material.
Uranium-238 is sometimes referred to as a “fertile” isotope. Fertile
material is a material that, although not itself fissionable by thermal
neutrons, can be converted into a fissile material (Pu-239) by neutron
absorption and subsequent nuclei conversions.
238 U  1n
92
0
Rev 0.1
 239
U 239 Np 239 Pu
92
93
94
33
Knowledge Check (Answer Key)
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.
ELO 4.3 Fission Process - Liquid Drop Model
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:

Very short range attractive force, with essentially no strength beyond
nuclear scale dimensions (≈10-13 cm)
 Stronger than the repulsive electrostatic forces within the nucleus
 The force is independent of nucleon pairing. Attractive forces
between pairs of neutrons, pairs of protons or neutron proton pairs are
identical.
 Saturable, that is, a nucleon can attract only a few of its nearest
neighbors.
Rev 0.1
34
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.
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 0.1
35
Knowledge Check (Answer Key)
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
TLO 4 Summary
Excitation energy is the energy above ground state of a nucleus. Critical
energy is the required excitation energy for fission to occur. 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.
There are two categories of nuclear fuel materials: those that can fission
with thermal neutrons (fissile materials) and those that require neutrons of
higher energies for fission to be possible (fissionable materials).
Fissionable materials include fissile nuclides as well as those nuclides that
fission only from high-energy neutrons.
Fissile materials are the desired nuclides for use in thermal nuclear reactors
because they can fission with neutrons possessing zero kinetic energy
(thermal neutrons).
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.
Rev 0.1
36
One theory, the Liquid Drop Model, 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. If enough
excitation energy is added, the nucleus becomes distorted and takes on a
dumbbell shape. 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. When the repulsive
electrostatic forces exceed the attractive nuclear forces, nuclear fission
occurs.
TLO 5 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 product nuclides are most likely to result from
fission.
3. Describe how heat is produced as a result of fission.
ELO 5.1 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.
Rev 0.1
37
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 rarely interact with materials inside the
reactor.
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. Fissions mostly cease with
the reactor shut down, but heat energy, referred to as decay heat, continues
to be released 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.
Rev 0.1
38
Isotope specifics, 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 (Answer Key)
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).
ELO 5.2 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
Note
Rev 0.1
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.
39
Figure: Uranium-235 Fission Yield for Fast and Thermal Neutrons versus
Mass Number
As shown in the above figure, the fission fragment yield varies
considerably. An example of one fission fragment yield is:
235
𝑈+𝑛 →
236
𝑈∗ →
140
𝑋𝑒 +
94
𝑆𝑟 + 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
Rev 0.1
40
Knowledge Check (Answer Key)
During full power operation, which of the following
combinations of isotopes are fission fragments likely to
be present in a fuel assembly?
A.
Oxygen-18, iron-59, and zirconium-95
B.
Hydrogen-2, iodine-131, and xenon-135
C.
Krypton-85, strontium-90, and iodine-135
D.
Hydrogen-2, hydrogen-3, and nitrogen-16
ELO 5.3 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.
Rev 0.1
41
Knowledge Check (Answer Key)
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.
TLO 5 Summary
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.
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.
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.
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 and capture gamma
rays. The remaining energy releases over time after the fission occurs and
appears as kinetic energy of the decay products.
Rev 0.1
42
TLO 6 Prompt and Delayed Neutrons
Overview
Not all neutrons are born immediately following fission. Fission releases
most neutrons virtually instantaneously; these are referred to as prompt
neutrons. The remaining neutrons (a very small fraction), are born after the
decay of certain fission products and are referred to as delayed neutrons.
Although delayed neutrons are a very small fraction of the total number of
neutrons, they play an extremely important role in controlling the reactor.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the origin and production of prompt and delayed neutrons.
2. State the approximate fraction of neutrons that are born as delayed
neutrons from the fission of the following fuels:
a. Uranium-235
b. Plutonium-239
3. Define prompt and delayed neutron lifetimes and their generation
times.
ELO 6.1 Production of Prompt and Delayed Neutrons
Introduction
There are two ways to classify neutrons. One is to categorize according to
energy, such as fast or thermal. Another is to classify according to birth
time relative to a fission event. Those neutrons born immediately after a
fission event are prompt neutrons, those born later from the decay of certain
fission products are delayed neutrons.
Rev 0.1
43
Glossary
Prompt Neutrons
Within about 10-14 seconds of a fission event, a majority
(≈ 99.36 percent) of the neutrons are released (born).
These are prompt neutrons. The number of prompt
neutrons emitted during a fission event depends on the
type of fuel used (U-235 averages 2.4 per fission event).
The most probable energy for a prompt neutron is
approximately 1 MeV, and the average energy is
approximately 2 MeV.
Delayed Neutrons
A small portion of the neutrons born of fission, are born
delayed from delayed neutron precursors. Delayed
neutrons are neutrons that are born significantly after the
fission process has taken place. On average, delayed
neutrons are born approximately 12.7 seconds after the
fission event. Delayed neutrons are born fast but at a
lower energy than prompt neutrons (≈ 0.5 MeV).
Delayed Neutron Precursors
A delayed neutron precursor refers to delayed neutrons emitted immediately
following the first beta decay of a fission fragment. An example of a
delayed neutron precursor is bromine-87 (Br), shown below. Br-87 is the
fission product; Kr-87 (Krypton) is the delayed neutron precursor, with Kr86 (Krypton) resulting from the delayed neutron birth.
86
𝐵𝑟
35
𝑛
𝛽−
87
86
→
𝐾𝑟
𝐾𝑟
→
36 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 36 𝑠𝑡𝑎𝑏𝑙𝑒
55.9 𝑠𝑒𝑐
Rev 0.1
44
Example
It is convenient to combine the known delayed neutron precursors into
groups with appropriately averaged half-life properties. These groups will
vary somewhat depending on the fuel or mixture of fuel in the reactor. The
table below lists the characteristics for the six delayed neutron precursor
groups resulting from the thermal fission of uranium-235.
Group
Half Life
(Seconds)
Delayed Neutron
Fraction
Average Energy
(MeV)
1
55.7
0.00021
0.25
2
22.7
0.00142
0.46
3
6.2
0.00127
0.41
4
2.3
0.00257
0.45
5
0.61
0.00075
0.41
6
0.23
0.00027
N/A
Total
N/A
0.0065
N/A
Knowledge Check (Answer Key)
Which one of the following types of neutrons has an
average neutron generation lifetime of 12.7 seconds?
Rev 0.1
A.
Prompt
B.
Delayed
C.
Fast
D.
Thermal
45
Knowledge Check (Answer Key)
Delayed neutrons are the neutrons that...
A.
have reached thermal equilibrium with the surrounding
medium.
B.
are expelled within 10-14 seconds of the fission event.
C.
are produced from the radioactive decay of certain fission
fragments.
D.
are responsible for the majority of U-235 fissions.
ELO 6.2 Delayed Neutron Fraction
Introduction
The fraction of all neutrons produced by each delayed neutron precursor is
the delayed neutron fraction for that precursor. The total fraction of all
neutrons born as delayed neutrons is the delayed neutron fraction (β). Each
nuclear fuel has different delayed neutron fraction (β).
Delayed Neutron Fraction (β)
The fraction of delayed neutrons (β) varies depending on the predominant
fissile nuclide in use. The delayed neutron fractions (β) for the nuclides of
most interest are as follows:




Uranium-233 β = 0.0026
Uranium-235 β = 0.0064
Uranium-238 β = 0.0148
Plutonium-239 β = 0.0021
It is significant to note that uranium-235 and plutonium-239 are the two
major fuels in use in PWRs. Although the β values are small, β for
uranium-235 is considerably larger than plutonium-239.
Over core life, uranium-235 concentration will decrease, while plutonium239 increases. This will result in lower delayed neutron fraction over core
life.
Rev 0.1
46
Knowledge Check (Answer Key)
What is the delayed neutron fraction of uranium-235?
A.
0.0064
B.
0.0148
C.
0.0021
D.
0.0026
ELO 6.3 Neutron Lifetimes and Generation Times
Introduction
Neutron lifetime is the time that a free neutron exists, from its birth until its
loss either from leakage or by absorption. Neutron generation time is the
time for a neutron from one generation to cause fission that produces the
next generation of neutrons.
Prompt Neutron Lifetime
Prompt neutron lifetime is the average time span from prompt neutron birth
until its loss from either leakage or absorption in another nucleus. This time
is the sum of thermalization time and diffusion time. Diffusion time relates
to absorption time of the thermal neutron and is a function of the absorption
mean free path, λa, divided by the average velocity of the thermal neutron.
Thermalization time is small (microseconds) compared to diffusion time
(milliseconds) and is usually ignored in calculations. Simply stated, prompt
neutron lifetime is the time from birth to loss by either leakage or
absorption.
Delayed Neutron Lifetime
Delayed neutron lifetime begins at its birth from a delayed neutron
precursor and ends at loss from leakage or absorption in another nucleus.
We calculate delayed neutron lifetime similarly to prompt neutron lifetime.
The key difference is the time of birth, which is not at the time of fission but
at the time of birth from decay of one of the delayed neutron precursors.
Prompt Neutron Generation Time
The generation time for prompt neutrons (ℓ* — pronounced ell-star) is the
total time from birth of a fast neutron in one generation to birth in the next
generation. Prompt neutron generation time is equal to the prompt neutron
lifetime added to the time required for a fissionable nucleus to emit a fast
neutron after absorption.
Rev 0.1
47
In water-moderated reactors, thermal neutrons exist for about 10-4 seconds
before absorption. Taking into account losses due to leakage, the prompt
neutron lifetime is equal to about 10-4 to 10-5 seconds (leakage occurs
quicker). Following absorption, fission, and the birth of fast neutron(s)
occurs in about 10-13 seconds — quickly. Therefore, in water-moderated
thermal reactors, ℓ* is about 10-4 seconds to 10-5 seconds.
Delayed Neutron Generation Time
Similar to prompt neutron generation time, delayed neutron generation time
equals the time of birth of a delayed neutron from a delayed neutron
precursor to the time of birth of a neutron(s) in the next generation.
The significant difference between prompt and delayed neutron generation
time is the delay in birth from their delayed neutron precursors. As
previously mentioned, the average time for decay of the delayed neutron
precursors from fission of uranium-235 is 12.7 seconds. With this relatively
large time interval, the delayed neutron lifetime is insignificant. Therefore,
the average delayed neutron generation time is equal to approximately 12.7
seconds. The significance of delayed neutrons in relation to average
generation time and reactor control is discussed in the next module –
Neutron Life Cycle.
Knowledge Check (Answer Key)
__________ begins when it is released from a precursor
and ends when it is absorbed in another nucleus.
Rev 0.1
A.
Delayed neutron lifetime
B.
Prompt neutron lifetime
C.
Fast neutron fraction
D.
Thermal neutron fraction
48
Knowledge Check (Answer Key)
Neutron generation time describes...
A.
time from one generation to the next generation of
neutrons.
B.
time that it takes for a neutron to become thermalized.
C.
time it takes for neutron precursors to emit neutrons.
D.
time that neutrons are born after a fission event.
Knowledge Check (Answer Key)
What effect on the average neutron generation time
would a smaller β value produce?
A.
The result would be a longer average generation time.
B.
The result would be a shorter average generation time.
C.
More information is needed to determine the effect on
average generation time.
D.
β does not have any effect on average generation time.
TLO 6 Summary
There are two ways to classify neutrons. One is to categorize according to
energy, such as fast or thermal. Another is to classify according to birth
time relative to a fission event. Those neutrons born immediately after a
fission event are prompt neutrons, those born later from the decay of certain
fission products are delayed neutrons.
Prompt Neutrons - Within about 10-14 seconds of a fission event, a
majority (≈ 99.36 percent) of the neutrons are released (born). These are
prompt neutrons. The number of prompt neutrons emitted during a fission
event depends on the type of fuel used (U-235 averages 2.4 per fission
event). The most probable energy for a prompt neutron is approximately 1
MeV, and the average energy is approximately 2 MeV.
Rev 0.1
49
Delayed Neutrons - A small portion of the neutrons born of fission, are
born delayed from delayed neutron precursors. Delayed neutrons are
neutrons that are born significantly after the fission process has taken place.
On average, delayed neutrons are born approximately 12.7 seconds after the
fission event. Delayed neutrons are born fast but at a lower energy than
prompt neutrons (≈ 0.5 MeV).
The significant difference between prompt and delayed neutron generation
time is the delay in birth from their delayed neutron precursors. Due to the
relatively large time interval for the decay of delayed neutron precursors
(12.7 seconds for U-235), the delayed neutron lifetime is insignificant.
Therefore, the average delayed neutron generation time is equal to
approximately 12.7 seconds.
TLO 7 Neutron Moderation
Overview
Fission neutrons are born at an average energy of 2 mega electron volt
(MeV) or fast neutrons and at high temperature. Fission neutrons
immediately begin to reduce their energy levels as they undergo numerous
scattering reactions with nuclei in the nuclear reactor core. A number of
collisions with nuclei reduce the neutron’s energy to approximately the
same average kinetic energy as its surrounding atoms or molecules. This
energy reduction occurs in a medium known as the moderator.
A neutron in energy equilibrium with its surrounding atoms is a thermal
neutron. Since the kinetic energy depends on temperature (molecular
movement), the energy of a thermal neutron also depends on temperature.
At 68° Fahrenheit (F), the energy of a thermal neutron is 0.025 electron volt
(eV). Energies less than 1eV yield neutrons designated or known as slow
neutrons.
Objectives
At the completion of this training session, you will be able to do the
following:
1. Describe the following:
a. Thermalization
b. Moderator
c. Moderating ratio
2. Describe the four desirable characteristics of a good moderator and
explain how moderator density affects neutron moderation.
Rev 0.1
50
ELO 7.1 Neutron Moderation
Introduction
Lowering the energy level of a neutron is essential because thermal fission
requires a neutron to be at thermal equilibrium. In a nuclear reactor, fast
energy neutrons born from a fission event must be slowed to the thermal
energy region to maintain a chain reaction. Thermalization or moderation is
the process of reducing neutron energy to the thermal range by elastic
scattering.
Thermalization
Thermalization or moderation is the process of reducing neutron energy to
the thermal range by elastic scattering.
Moderator
A moderator is the material used to thermalize neutrons. A desirable
moderator is one that reduces the velocity of the fission neutrons, using a
minimum number of scattering collisions with a low probability of neutron
absorption.
Slowing the neutrons in as few collisions as possible reduces their travel
distance, thereby reducing the number of neutrons that leak out of the core.
This shorter travel distance also reduces the number of resonance
absorptions in non-fuel materials. Neutron leakage and resonance
absorption are discussed in detail later in this module.
Average Logarithmic Energy Decrement
The average logarithmic energy decrement is the measure of neutron energy
loss per collision. This term is the average decrease per collision of the
logarithm of the change in neutron energy, denoted by the symbol ξ (Xi).
𝜉 = 𝑙𝑛
𝐸𝑖
𝐸𝑓
Where:
ξ = logarithmic energy decrement
Ei = initial energy level of neutron
Ef = final energy level of neutron
Rev 0.1
51
Macroscopic Slowing Down Power
The logarithmic energy decrement is a convenient measure of the ability of
a material to slow neutrons, but does not consider the probability of
collisions taking place. Another measure of a moderator is the macroscopic
slowing down power (MSDP), defined as the product of the logarithmic
energy decrement and the macroscopic cross-section for scattering in the
material. The equation below calculates the macroscopic slowing down
power.
𝑀𝑆𝐷𝑃 = 𝜉𝛴𝑠
Where:
ξ = logarithmic energy decrement
𝛴𝑠 = Macroscopic scattering cross-section
Moderating Ratio
Macroscopic slowing down power calculates how rapidly a neutron will
thermalize in the chosen moderator. However, it does not fully explain the
effectiveness of the material as a moderator. An element such as boron has
a high logarithmic energy decrement and good MSDP, but it is a poor
moderator because of its high probability of absorbing neutrons.
The moderating ratio considers absorbing probability as well as slowing
down power; and therefore, is a more complete measure of moderator
effectiveness. Moderating ratio is the ratio of the microscopic slowing
down power to the microscopic cross-section for absorption. The higher the
moderating ratio, the more effectively the material performs as a moderator.
This equation shows the calculation for the moderating ratio (MR):
𝑀𝑅 =
𝜉𝜎𝑠
𝜎𝑎
The moderating ratio characterizes the effectiveness of a material as a
moderator. It considers the ratio between the absorption and scattering
cross-sections factoring in neutron energy levels. It does not consider the
density of the moderator. The table below compares moderating properties
of different materials.
Rev 0.1
52
Moderating properties of different materials are compared in the table.
Material
ξ
Number of
Collisions to
Thermalize
Microscopic
Cross-Sections
σa–σs
Moderating
Ratio
Water (H2O)
also known as
light water
0.948
19.0
0.66 – 103.0
148.0
Deuterium
Oxide (D2O)
also known as
heavy water
0.570
35.0
0.001 – 13.6
7,752.0
Beryllium (Be)
0.209
86.0
0.0092 – 7.0
159.0
Carbon (C)
0.158
114.0
0.003 – 4.8
253.0
A good moderator is one with a high moderating ratio. Any of the materials
shown in the table above make good moderators; however, commercial
nuclear power applications often use light water (H2O) because it is
plentiful, easily obtained, and inexpensive. Another benefit of using water
is that water can serve as both a moderator and a coolant. Heavy water
(D2O) is by far the best performing moderator; however, its higher cost
precludes its use in U.S. commercial PWRs.
Knowledge Check (Answer Key)
A _______________ is a material within a reactor which
is responsible for thermalizing neutrons.
Rev 0.1
A.
fuel rod
B.
moderator
C.
poison
D.
reflector
53
Knowledge Check (Answer Key)
The process of reducing the energy level of a neutron
from the energy level at which it is produced to an
energy level in the thermal range is known as
_______________.
A.
moderating ratio
B.
resonance absorption
C.
thermalization
D.
inelastic scattering
Knowledge Check (Answer Key)
The average logarithmic energy decrement is important
because...
A.
it can be used to determine if a material is a good
moderator.
B.
it can be used to determine the amount of energy released
from fission.
C.
it accounts for the change in binding energy change
during fission.
D.
it accounts for the change in mass during fission.
ELO 7.2 Moderator Characteristics
Introduction
The ideal moderator requires the following nuclear properties:




Large scattering cross-section
Small absorption cross-section
Large energy loss per collision
High atomic density
Rev 0.1
54
Desirable Moderator Properties
A material with a mass equal to a neutron, but with a large scattering crosssection is desirable as a moderator. A substance having a high absorption
cross-section, acts as a poison, removing available neutrons for fission. A
moderator with a low atomic density has few atoms available to thermalize
neutrons.
Carbon, hydrogen, beryllium, and water (both heavy and light) are
moderators used in nuclear reactors. In the U.S., most reactors use light
water as the moderator. Water is a good moderator because the hydrogen
atoms in water are close in mass (good for elastic scattering) to the
thermalized neutrons.
A neutron loses more energy in a collision with an atom of nearly its own
mass than in a collision with an atom whose mass is much greater than the
incident neutron (inelastic scattering — billiard ball effect). Water
moderators, rather than other materials, thermalize neutrons in fewer
collisions, reducing both the time and distance required for thermalization.
Moderator Density Effects
Temperature changes affect moderator density in an operating reactor.
These density changes affect the moderator’s ability to thermalize neutrons.
This in turn affects the number of neutrons available for fission.
Consider a PWR (light water moderator and coolant) as an example. If the
moderator temperature increases, the density of the water decreases.
Decreasing water density means there are fewer water atoms per unit
volume to thermalize neutrons. The neutrons will have to travel further to
thermalize, and this larger travel distance will take longer. Because of this
and other factors discussed later, there is now an increased chance that these
neutrons will not be available for fission. Fewer neutrons available for
fission, means fewer fissions, resulting in less reactor power.
Knowledge Check (Answer Key)
Which of the items below is not a desirable property for
a neutron moderator?
Rev 0.1
A.
Large absorption cross-section
B.
Large scattering cross-section
C.
Large energy loss per collision
D.
High atomic density
55
Knowledge Check (Answer Key)
The density of a moderator is important because...
A.
it can affect the number of target nuclei available for
collisions.
B.
it can affect the number of target nuclei available for
absorption.
C.
it can affect the number of collisions.
D.
it can affect the energy loss per collision.
TLO 7 Summary
A moderator is the material used to thermalize neutrons. A desirable
moderator is one that reduces the velocity of the fission neutrons, using a
minimum number of scattering collisions with a low probability of neutron
absorption.
Slowing the neutrons in as few collisions as possible reduces their travel
distance, thereby reducing the number of neutrons that leak out of the core.
This shorter travel distance also reduces the number of resonance
absorptions in non-fuel materials. Neutron leakage and resonance
absorption are discussed in detail later in this module.
The ideal moderator requires the following nuclear properties:




Large scattering cross-section
Small absorption cross-section
Large energy loss per collision
High atomic density
TLO 8 Neutron Reaction Rates/Probabilities
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.
Rev 0.1
56
To determine these neutron interaction rates, it is necessary to identify the
number of neutrons available and the probability of 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 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.
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. Microscopic cross-section
c. Barn
d. Macroscopic cross-section
e. Mean free path
f. Neutron flux
g. Fast neutron flux
h. Thermal neutron flux
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/V region, and the
resonance absorption region.
4. Describe how changes in neutron flux and macroscopic cross-section
affect reaction rates.
5. Describe the relationship between neutron flux and reactor power.
Rev 0.1
57
ELO 8.1 Neutron Reaction Terms
Introduction
The following terms are important for understanding the theoretical
concepts of nuclear reactors and are foundational for future lessons:








Atomic density
Microscopic cross-section
Barn
Macroscopic cross-section
Mean free path
Neutron flux
Fast neutron flux
Thermal neutron flux
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.
𝑁=
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
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=
𝜌𝑁𝐴
𝑀
𝑔
𝑎𝑡𝑜𝑚𝑠
(6.022 × 1023
)
3
𝑚𝑜𝑙𝑒
𝑐𝑚
=
𝑔
26.9815
𝑚𝑜𝑙𝑒
𝑎𝑡𝑜𝑚𝑠
= 6.024 × 1022
𝑐𝑚3
2.699
Rev 0.1
58
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).
𝜎𝑇 = 𝜎𝑎 + 𝜎𝑠
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).
𝜎𝑠 = 𝜎𝑠𝑒 + 𝜎𝑠𝑖
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).
𝜎𝑎 = 𝜎𝑓 + 𝜎𝑐
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.


If a gamma ray results from the capture, σγ 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.
Rev 0.1
59
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 𝑏𝑎𝑟𝑛 = 10−24 𝑐𝑚2
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 = 𝑁𝜎
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.
Rev 0.1
60
Macroscopic 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 𝜎𝑛
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=
𝜌𝑁𝐴
𝑀
7.68
=
𝑔
𝑎𝑡𝑜𝑚𝑠
(6.022 × 1023
)
𝑚𝑜𝑙𝑒
𝑐𝑚3
𝑔
55.847
𝑚𝑜𝑙𝑒
= 8.48 × 1022
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3
Step 2:
Using the atomic density and given microscopic cross-section, calculate the
macroscopic cross-section.
Σ𝑎 = 𝑁𝜎𝑎
= 8.48 × 1022
𝑎𝑡𝑜𝑚𝑠
1 × 10−24 𝑐𝑚2
(2.56
𝑏𝑎𝑟𝑛𝑠)
(
)
𝑐𝑚3
1 𝑏𝑎𝑟𝑛
= 0.217 𝑐𝑚−1
Rev 0.1
61
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
The average distance of travel by a neutron before interaction is the mean
free path. (λ).
1
𝜆=
𝛴
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.
An average neutron travels three mean free paths (considering all materials
available to scatter it).
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 centimeter squared per second (n/cm2/sec). The equation below shows
the relationship between neutron flux, neutron density, and neutron
velocity.
Rev 0.1
62
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.
Rev 0.1
63
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.
Knowledge Check (Answer Key)
_______________ 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
ELO 8.2 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.
Rev 0.1
64
Neutron Energy 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:

Fast neutrons - Fission neutrons are born as fast neutrons. They are
categorized with energy levels greater than 0.1 MeV (105 eV)
 Intermediate neutrons - neutrons with energy levels between 1 eV and
0.1 MeV
 Slow neutrons - neutrons with energy levels less than 1 eV
 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.
Knowledge Check (Answer Key)
Thermal neutrons are classified in the intermediate
neutron energy range.
Rev 0.1
A.
True
B.
False
65
Knowledge Check (Answer Key)
A SLOW neutron is defined as one where its energy is:
A.
Less than 1 ev
B.
Between 1 eV and 100,000 ev
C.
Greater than 0.1 MeV
D.
At rest with relationship to other particles
ELO 8.3 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.
Rev 0.1
66
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
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).
Rev 0.1
67
Resonance Region
Beyond the 1/v region, there occurs the "resonance region" in which the
cross-sections rise sharply to high value peaks. These "resonance peaks"
occur with neutrons in the intermediate energy (epithermal) range.
Resonance peaks result with affinity of the nucleus for neutrons with
energies that match its discrete, quantum (preferred) energy levels. Peaks
occur when neutron’s BE plus its KE are exactly equal to the amount of
energy needed to raise a compound nucleus from ground state to a quantum
level energy level. Resonance absorption occurs as these match.
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 (Answer Key)
At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to
the neutron velocity. (Fill in the blank).
ELO 8.4 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.
Rev 0.1
68
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 = 𝑁𝜎
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
Where:
R = reaction rate (reactions/cm3-sec)
Σ = macroscopic cross-section (cm-1)
Φ = neutron flux (neutrons/cm2-sec)
Note
Note
Rev 0.1
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.
69
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
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
) (0.1 𝑐𝑚−1 )
𝑐𝑚2 – 𝑠𝑒𝑐
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
𝑐𝑚3 – 𝑠𝑒𝑐
Reaction Rate Over Core Life
Reaction rate required to obtain 100% power does NOT change over core
life, however, the amount of fuel (N) decreases over core life. Microscopic
cross-section is a constant value. Therefore, flux must increase over core
life to have the same reaction rate.
At the beginning of core life (BOL) reaction rate is calculated as follows:
R = 𝜎𝑁∅
To maintain 100% power over core life:
R ↔ (required to obtain 100% power)
𝜎 ↔ (constant value)
𝑁 ↓, due to fuel depletion, therefore ∅ must ↑
Rev 0.1
70
Knowledge Check (Answer Key)
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
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
cm3 –sec
𝑐𝑚3 –𝑠𝑒𝑐
ELO 8.5 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:



1 fission = 200 MeV
1 MeV = 1.602 x 10-6 ergs
1 erg = 1 x 10-7 watt-sec
1 𝑒𝑟𝑔
1 𝑀𝑒𝑉
1 𝑓𝑖𝑠𝑠𝑖𝑜𝑛
1 𝑤𝑎𝑡𝑡 (
)(
)(
)
−7
−6
1 × 10 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐 1.602 × 10 𝑒𝑟𝑔 200 𝑀𝑒𝑉
𝑓𝑖𝑠𝑠𝑖𝑜𝑛
= 3.12 × 1010
𝑠𝑒𝑐𝑜𝑛𝑑
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.
Rev 0.1
71
Step
1.
Action
Obtain the following reactor data:



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=
Φ𝑡ℎ Σ𝑓 𝑉
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)
Rev 0.1
72
Neutron Flux and Reactor Power Demonstration
Example:
Calculate reactor power given the following:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑟𝑒 (𝑐𝑚3 ) = 20,000 𝑐𝑚3
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = 1 × 1015
𝑐𝑚3 –𝑠𝑒𝑐

Solution:
Step 1:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑟𝑒 (𝑐𝑚3 ) = 20,000 𝑐𝑚3
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 = 1 × 1015
𝑐𝑚3 –𝑠𝑒𝑐

Step 2:
𝑃=
Φ𝑡ℎ Σ𝑓 𝑉
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010 𝑤𝑎𝑡𝑡– 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 𝑤𝑎𝑡𝑡𝑠 = 641 𝑀𝑒𝑔𝑎𝑤𝑎𝑡𝑡𝑠𝑡ℎ𝑒𝑟𝑚𝑎𝑙
Knowledge Check (Answer Key)
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.
Rev 0.1
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
(10,000 𝑐𝑚3 )
𝑐𝑚3 – 𝑠𝑒𝑐
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐
2 × 1013
𝑃=
C.
Φ𝑡ℎ Σ𝑓 𝑉
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐
𝑃 = 6.41 × 106 𝑤𝑎𝑡𝑡𝑠
73
TLO 8 Summary
This section reviewed important terms for understanding the theoretical
concepts of nuclear reactors:








Atomic density: number of atoms of a given type per unit volume of
material
Microscopic cross-section: probability of an interaction between a
neutron and a target nucleus
Barn: unit of microscopic cross-session (1 barn = 10-24 cm2)
Macroscopic cross-section: the probability of a given interaction
occurring per unit travel of the neutron
Mean free path: average distance of travel by a neutron before
interaction
Neutron flux: number of neutrons passing through the unit area
(cm2) per unit time
Fast neutron flux: the number of fast neutrons crossing a unit area in
a given amount of time
Thermal neutron flux: the number of thermal neutrons crossing a
unit area in the reactor in a given amount of time.
Neutrons are also classified by energy levels:
 Fast (Prompt and Delayed neutrons are born fast) - >0.1 Mev
 Intermediate (Epithermal) - >1ev but < 0.1 Mev
 Slow - <1 ev
 Thermal - 0.025ev @ 68oF
Absorption cross-sections have three unique regions of probability related
to neutron energy
 1/v region
 Resonance region
 Fast neutron region
Rev 0.1
74
Scattering cross-sections are not really affected by energy levels of neutrons
 U-238 scattering cross section is ≈ 10 Barns
Reactor power can be determined using the following equation:
𝑃=
Φ𝑡ℎ Σ𝑓 𝑉
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)
Review Knowledge Check (Answer Key)
In a comparison between a delayed neutron and a prompt
neutron produced from the same fission event, the
prompt neutron is more likely to...
Rev 0.1
A.
require a greater number of collisions to become a
thermal neutron.
B.
be captured by U-238 at a resonance energy peak
between 1 eV and 1,000 eV.
C.
be expelled with a lower kinetic energy.
D.
cause thermal fission of a U-235 nucleus.
75
Neutrons Knowledge Check Answer Key
Knowledge Check Answer Key
ELO 1.1 Atomic Structure
Knowledge Check - Answer
Identify the particles included in the make-up of an atom.
(More than one answer may apply.)
A.
neutron
B.
electron
C.
gamma
D.
amu
ELO 1.2 Atomic Terms
Knowledge Check - Answer
Match the term to its definition.
1 Protons + neutrons
A. Deuterium
2 One neutron
B. Neutron number
3 Number of neutrons
C. Atomic number
4 Number of protons
D. Mass number
1.
2.
3.
4.
D – Mass number
A – Deuterium
B – Neutron number
C – Atomic number
Rev 0.1
1
Neutrons Knowledge Check Answer Key
Knowledge Check - Answer
What is the element and number of neutrons for the
following:
235
92𝑈
A.
Uranium; 143
B.
Uranium; 92
C.
Plutonium; 143
D.
Plutonium; 92
Knowledge Check – Answer
Nuclide
Element
Protons
Electrons
Neutrons
1
1𝐻
Hydrogen
1
1
0
10
5𝐵
Boron
5
5
5
16
8𝑂
Oxygen
8
8
8
235
92𝑈
Uranium
92
92
143
Plutonium
94
94
145
239
94𝑃𝑢
ELO 1.3 Atomic Forces
Knowledge Check - Answer
Very weak attractive force between all nucleons
describes which of the forces listed below?
Rev 0.1
A.
Electrostatic
B.
Nuclear
C.
Gravitational
D.
Atomic
2
Neutrons Knowledge Check Answer Key
Knowledge Check – Answer
Which of the following nuclides has the higher neutronproton ratio?
A.
Cobalt-60
B.
Selenium-79
C.
Silver-108
D.
Cesium-137
ELO 2.1 Mass Defect and Binding Energy
Knowledge Check - Answer
_______________ is the amount of energy that must be
supplied to a nucleus to completely separate its nuclear
particles.
Rev 0.1
A.
Nuclear energy
B.
Binding energy
C.
Mass defect
D.
Separation energy
3
Neutrons Knowledge Check Answer Key
ELO 2.2 Determining Mass Defect and Binding Energy
Knowledge Check - Answer
Calculate the mass defect for uranium-235. One
uranium-235 atom has a mass of 235.043924 amu.
mp = mass of a proton (1.007277 amu)
mn = mass of a neutron (1.008665 amu)
me = mass of an electron (0.000548597 amu)
A.
1.86471 amu
B.
1.91517 amu
C.
0.191517 amu
D.
0.186471 amu
Knowledge Check - Answer
Calculate the binding energy for uranium-235. One
uranium-235 atom has a mass defect of 1.9157 amu.
A.
1784 MeV
B.
178.4 MeV
C.
1783 MeV
D.
178.3MeV
ELO 2.3 Binding Energy per Nucleon
Knowledge Check - Answer
True or False. Generally, less stable nuclides have a
higher BE/A than the more stable ones.
Rev 0.1
A.
True
B.
False
4
Neutrons Knowledge Check Answer Key
ELO 2.4 Binding Energy Per Nucleon Curve
Knowledge Check - Answer
True or False. Generally, less stable nuclides have a
higher BE/A than the more stable ones.
A.
True
B.
False
ELO 3.1 Neutron Scattering Interactions
Knowledge Check - Answer
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 3.2 Neutron Absorption Reactions
Knowledge Check - Answer
What type of neutron interaction has occurred when a
nucleus absorbs a neutron and ejects proton?
Rev 0.1
A.
Fission
B.
Fusion
C.
Particle ejection
D.
Radiative capture
5
Neutrons Knowledge Check Answer Key
ELO 4.1 Excitation and Critical Energy
Knowledge Check - Answer
True or False: Excitation must be at least equal to
critical energy for fission to occur.
A.
True
B.
False
ELO 4.2 Fissile and Fissionable Material
Knowledge Check - Answer
What is the difference between a fissionable material
and a fissile material?
Rev 0.1
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
kinetic energy in order to fission, whereas fissile
materials will fission with a neutron that has zero
kinetic energy.
6
Neutrons Knowledge Check Answer Key
ELO 4.3 Fission Process - Liquid Drop Model
Knowledge Check - Answer
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 5.1 Energy Release Per Fission
Knowledge Check - Answer
Which of the following statements correctly describes
the amount of energy released from a single fission
event?
Rev 0.1
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).
7
Neutrons Knowledge Check Answer Key
ELO 5.2 Fission Fragment Yield
Knowledge Check - Answer
During full power operation, which of the following
combinations of isotopes are fission fragments likely to
be present in a fuel assembly?
A.
Oxygen-18, iron-59, and zirconium-95
B.
Hydrogen-2, iodine-131, and xenon-135
C.
Krypton-85, strontium-90, and iodine-135
D.
Hydrogen-2, hydrogen-3, and nitrogen-16
(C is correct since they appear near the tops of the two curves of the Fission
Product Yield curve)
ELO 5.3 Fission Heat Production
Knowledge Check - Answer
Which of the following is NOT a method by which
heat is produced from fission?
Rev 0.1
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.
8
Neutrons Knowledge Check Answer Key
ELO 6.1 Production of Prompt and Delayed Neutrons
Knowledge Check - Answer
Which one of the following types of neutrons has an
average neutron generation lifetime of 12.5 seconds?
A.
Prompt
B.
Delayed
C.
Fast
D.
Thermal
Knowledge Check - Answer
Delayed neutrons are the neutrons that...
A.
have reached thermal equilibrium with the surrounding
medium.
B.
are expelled within 10-14 seconds of the fission event.
C.
are produced from the radioactive decay of certain
fission fragments.
D.
are responsible for the majority of U-235 fissions.
ELO 6.2 Delayed Neutron Fraction
Knowledge Check - Answer
What is the delayed neutron fraction of uranium-235?
Rev 0.1
A.
0.0064
B.
0.0148
C.
0.0021
D.
0.0026
9
Neutrons Knowledge Check Answer Key
ELO 6.3 Neutron Lifetimes and Generation Times
Knowledge Check – Answer
__________ begins when it is released from a precursor
and ends when it is absorbed in another nucleus.
A.
Delayed neutron lifetime
B.
Prompt neutron lifetime
C.
Fast neutron fraction
D.
Thermal neutron fraction
Knowledge Check - Answer
Neutron generation time describes...
A.
time from one generation to the next generation of
neutrons.
B.
time that it takes for a neutron to become thermalized.
C.
time it takes for neutron precursors to emit neutrons.
D.
time that neutrons are born after a fission event.
Knowledge Check - Answer
What effect on the average neutron generation time
would a smaller β value produce?
Rev 0.1
A.
The result would be a longer average generation time.
B.
The result would be a shorter average generation time.
C.
More information is needed to determine the effect on
average generation time.
D.
β does not have any effect on average generation time.
10
Neutrons Knowledge Check Answer Key
ELO 7.1 Neutron Moderation
Knowledge Check - Answer
A _______________ is a material within a reactor which
is responsible for thermalizing neutrons.
A.
fuel rod
B.
moderator
C.
poison
D.
reflector
Knowledge Check - Answer
The process of reducing the energy level of a neutron
from the energy level at which it is produced to an
energy level in the thermal range is known as
_______________.
Rev 0.1
A.
moderating ratio
B.
resonance absorption
C.
thermalization
D.
inelastic scattering
11
Neutrons Knowledge Check Answer Key
Knowledge Check – Answer
The average logarithmic energy decrement is important
because...
A.
it can be used to determine if a material is a good
moderator.
B.
it can be used to determine the amount of energy released
from fission.
C.
it accounts for the change in binding energy change
during fission.
D.
it accounts for the change in mass during fission.
ELO 7.2 Moderator Characteristics
Knowledge Check - Answer
Which of the items below is not a desirable property for
a neutron moderator?
A.
Large absorption cross-section
B.
Large scattering cross-section
C.
Large energy loss per collision
D.
High atomic density
Knowledge Check - Answer
The density of a moderator is important because...
Rev 0.1
A.
it can affect the number of target nuclei available for
collisions.
B.
it can affect the number of target nuclei available for
absorption.
C.
it can affect the number of collisions.
D.
it can affect the energy loss per collision.
12
Neutrons Knowledge Check Answer Key
ELO 8.1 Nuclear Reaction Terms
Knowledge Check - Answer
_______________ 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
ELO 8.2 Neutron Energy Terms
Knowledge Check - Answer
Thermal neutrons are classified in the intermediate
neutron energy range.
A.
True
B.
False
Knowledge Check - Answer
A SLOW neutron is defined as one where its energy is:
Rev 0.1
A.
Less than 1 ev
B.
Between 1 eV and 100,000 ev
C.
Greater than 0.1 MeV
D.
At rest with relationship to other particles
13
Neutrons Knowledge Check Answer Key
ELO 8.3 Neutron Energies versus Cross-Sections
Knowledge Check - Answer
At low neutron energies (<1 eV) the absorption crosssection for a material is ___________ proportional to
the neutron velocity. (Fill in the blank).
A.
inversely
ELO 8.4 Reaction Rate
Knowledge Check - Answer
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.
E. R = 2.0 × 1014
F. R = 0.2 × 1013
neutrons
cm2 –sec
neutrons
cm2 –sec
G. R = 2.0 × 1013
neutrons
H. 𝑅 = 20 × 1013
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
cm3 –sec
𝑐𝑚3 –𝑠𝑒𝑐
ELO 8.5 Neutron Flux and Reactor Power
Knowledge Check - Answer
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.
B.
C.
Rev 0.1
Φ𝑡ℎ Σ𝑓 𝑉
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐
𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠
3
2 × 1013
3 – 𝑠𝑒𝑐 (10,000 𝑐𝑚 )
𝑐𝑚
𝑃=
𝑓𝑖𝑠𝑠𝑖𝑜𝑛𝑠
3.12 × 1010 𝑤𝑎𝑡𝑡– 𝑠𝑒𝑐
𝑃 = 6.41 × 106 𝑤𝑎𝑡𝑡𝑠
𝑃=
14
Neutrons Knowledge Check Answer Key
Neutron Topic Review Question
Knowledge Check - Answer
In a comparison between a delayed neutron and a prompt
neutron produced from the same fission event, the
prompt neutron is more likely to...
Rev 0.1
A.
require a greater number of collisions to become a
thermal neutron.
B.
be captured by U-238 at a resonance energy peak
between 1 eV and 1,000 eV.
C.
be expelled with a lower kinetic energy.
D.
cause thermal fission of a U-235 nucleus.
15