Revision 0
November 2014
Atomic Structure
Student Guide
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ii
Table of Contents
INTRODUCTION ..................................................................................................................... 2
TLO 1 ATOMS ...................................................................................................................... 3
Overview .......................................................................................................................... 3
ELO1.1 Atomic Structure ................................................................................................ 4
ELO 1.2 Atomic Terms.................................................................................................... 7
ELO 1.3 Atomic Forces ................................................................................................. 11
TLO 1 Summary ............................................................................................................ 14
TLO 2 CHART OF THE NUCLIDES ........................................................................................ 15
Overview ........................................................................................................................ 15
ELO 2.1 Chart of the Nuclides ...................................................................................... 16
ELO 2.2 Neutron Proton Ratio ...................................................................................... 19
ELO 2.3 Atomic Quantities ........................................................................................... 20
ELO 2.4 Enrichment and Depletion............................................................................... 23
TLO 2 Summary ............................................................................................................ 24
TLO 3 MASS DEFECT AND BINDING ENERGY ..................................................................... 25
Overview ........................................................................................................................ 25
ELO 3.1 Mass Defect and Binding Energy ................................................................... 25
ELO 3.2 Determining Mass Defect and Binding Energy .............................................. 26
ELO 3.3 Gamma Rays and X-Rays ............................................................................... 31
TLO 3 Summary ............................................................................................................ 34
TLO 4 NUCLEAR STABILITY ............................................................................................... 35
Overview ........................................................................................................................ 35
ELO 4.1 Conservation Principles .................................................................................. 36
ELO 4.2 Decay Processes .............................................................................................. 38
ELO 4.3 Stability Curve ................................................................................................ 45
ELO 4.4 Decay Chains .................................................................................................. 48
TLO 4 Summary ............................................................................................................ 50
TLO 5 RADIATION EMITTED ............................................................................................... 51
Overview ........................................................................................................................ 51
ELO 5.1 Charged Versus Uncharged Particles .............................................................. 52
ELO 5.2 Radioactive Interactions .................................................................................. 54
ELO 5.3 Shielding ......................................................................................................... 59
TLO 5 Summary ............................................................................................................ 61
TLO 6 RADIOACTIVE DECAY .............................................................................................. 62
Overview ........................................................................................................................ 62
ELO 6.1 Define Terms ................................................................................................... 63
ELO 6.2 Convert Between Half-Life and Decay Constant............................................ 64
ELO 6.3 Calculating Activity Over Time ...................................................................... 68
ELO 6.4 Equilibrium ..................................................................................................... 77
TLO 6 Summary ............................................................................................................ 81
ATOMIC STRUCTURE SUMMARY ......................................................................................... 82
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Atomic Structure
Revision History
Revision
Date
Version
Number
Purpose for Revision
Performed
By
11/18/2014
0
New Module
OGF Team
12/10/2014
1
Added signature of OGF
Working Group Chair
OGF Team
Rev 1
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.
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. Use the Chart of the Nuclides to obtain information on specific
nuclides.
3. Describe Mass Defect and Binding Energy and their relationship to
one another.
4. Describe the processes by which unstable nuclides achieve stability.
5. Describe how radiation emitted by an unstable nuclide interacts with
matter and materials typically used to shield against this radiation.
6. Describe radioactive decay terms and perform calculations to
determine activity levels, half-lives and decay constants and
radioactive equilibrium.
2
Rev 1
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. Define the following terms and given the standard notation for a
given nuclide identify its nucleus and electron makeup:
a. Nuclide
b. Isotope
c. Atomic number
d. Mass number
Rev 1
3
3. Describe the three forces that act on particles within the nucleus, and
how they affect the stability of the nucleus.
ELO1.1 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.
Atomic Structure Details
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 Atom
4
Rev 1
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 hv (h = Planck's constant - 6.63 x 10-34 J-s (Joule-seconds)
and nu = 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
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
Rev 1
5
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. 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
Neutron
Nucleus
None
1.008665 amu
Proton
Nucleus
+1 eV
1.007277 amu
Electron
Shells Around
Nucleus
-1 eV
0.0005486 amu
Knowledge Check
Identify the particles included in the make-up of an atom.
(More than one answer may apply.)
6
A.
neutron
B.
electron
C.
gamma
D.
amu
Rev 1
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
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 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).
Isotopes of Hydrogen
Some hydrogen isotopes are unique in that they have a unique name instead
of the common element name. Hydrogen-1 (with no neutrons) isotopes are
normally called hydrogen. Hydrogen-2 (with one neutron) isotopes are
commonly called deuterium, and are symbolized by 21𝐷. Hydrogen-3 (with
two neutrons) isotopes are commonly called tritium, and are symbolized by
3
3
2
1𝑇 . This text will use the symbols 1𝐻 and 1𝐻 for deuterium and tritium,
respectively.
Atomic Notation
A convention, known as atomic or standard notation, identifies elements
using the nuclear nomenclature previously described. The figure below
shows standard notation:
8
Rev 1
Figure: Nomenclature for Identifying Nuclides
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).
Knowledge Check
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
Rev 1
9
Knowledge Check
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
In the table below, complete the columns for element, protons, electrons,
and neutrons.
Nuclide
Element
Protons
Electrons
Neutrons
1
1𝐻
10
5𝐵
14
7𝑁
114
48𝐶𝑑
239
94𝑃𝑢
10
Rev 1
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)
Rev 1
11
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 10-20 meters is about 10-24 Newtons.
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 10-20 meters is about 1012 Newtons. Because the
electrostatic force (1012 Newtons) is much greater than the gravitational
force (10-24 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 (1013
cm). The attractive nuclear force between nucleons decreases with
distance much quicker than the repulsive electrostatic force between
protons.
12
Rev 1
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
Very weak attractive force between all nucleons
describes which of the forces listed below?
Rev 1
A.
Electrostatic
B.
Nuclear
C.
Gravitational
D.
Atomic
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.
14
Rev 1
— Nuclear force: a short-range, attractive force between all
nucleons that is able to balance the repulsive electrostatic
force in a stable nucleus.
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. Define the following terms and given the standard notation for a
given nuclide identify its nucleus and electron makeup:
a. Nuclide
b. Isotope
c. Atomic number
d. Mass number
3. Describe the three forces that act on particles within the nucleus and
how they affect the stability of the nucleus.
TLO 2 Chart of the Nuclides
Overview
The Chart of the Nuclides is a convenient format for presenting a large
amount of scientific information in an organized manner. This chart is
important because it gives information about the characteristics of each
elemental isotope. Using the Chart of the Nuclides, we can determine how
an unstable atom becomes stable, the type and strength of radiation emitted,
and decay chains including probabilities of interactions; for example,
absorption or neutron capture.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the information for stable and radioactive isotopes found on
the Chart of the Nuclides.
2. Describe how an element’s neutron to proton ratio affects its stability.
3. Explain the difference between atom percent, atomic weight and
weight percent, and given the atom percent and the atomic masses for
isotopes of a particular element, calculate the atomic weight of the
element.
4. Describe the following terms:
a. Enriched uranium
b. Depleted uranium
Rev 1
15
ELO 2.1 Chart of the Nuclides
Introduction
The Chart of the Nuclides is a two-dimensional graph plotting the number
of neutrons on one axis and the number of protons on the other. Each point
plotted on the graph represents a nuclide of a real or hypothetical element.
This chart provides a map of the radioactive behavior of isotopes of the
chemical elements. The Chart of the Nuclides contrasts with a periodic
table, which maps only chemical behavior.
Chart of the Nuclides
The Chart of the Nuclides provides large amounts of pertinent information
regarding stable and unstable nuclides. The figure below shows a small
portion of the chart. This chart has a box for each individual nuclide with
the number of protons (Z) on the vertical axis and the number of neutrons
(N = A - Z) on the horizontal axis.
The following information is available on a full-scale copy of the Chart of
the Nuclides:











Symbol and mass number
Percent abundance
Thermal neutron and resonance cross sections
Isotopic mass
Half-life values
Mode and energy of decay (in Mega electron Volts [MeV])
Beta disintegration energy in MeV
Isomeric states
Indication if it is a fission product
Nuclear transmutations
Other information beyond the scope of this module
Figure: Nuclide Chart for Atomic Numbers
16
Rev 1
Stable and Unstable Nuclides
Only 254 isotopes are stable or naturally occurring radioactive forms. Gray
colored boxes on the Chart of the Nuclides denote naturally occurring and
stable nuclides. When viewing a complete chart, all the gray boxes
comprise a group of stable nuclides. A line of stability can be plotted using
only the data from the stable nuclide boxes, as shown below in the
simplified figure. Stable nuclides exist along this line of stability. The
stable nuclide boxes contain:






Chemical symbol
Number of nucleons
Percentage abundance in nature
Isotopic mass
Capture cross section in barns
Indication if it is a fission product
The figure below shows a typical block for a stable nuclide from the Chart
of the Nuclides.
Figure: Stable Nuclide from the Chart of the Nuclides
Rev 1
17
Figure: Line of Stability
Unstable nuclides are white or color boxes outside of the line of stability.
These boxes contain:








Chemical symbol
Number of nucleons
Half-life of the nuclide
Mode and energy of decay (in MeV)
Beta disintegration energy in MeV
Isomeric states
Indication if it is a fission product
Nuclear transmutations
The figure below shows a typical block for an unstable nuclide from the
Chart of the Nuclides. Charts show decay modes and half-lives in colors.
Figure: Unstable Nuclide from the Chart of the Nuclides
18
Rev 1
Knowledge Check
On the Chart of the Nuclides, a stable isotope is indicated
by a …
A.
white box
B.
gray box
C.
red box
D.
black box
ELO 2.2 Neutron Proton Ratio
Introduction
The neutron-proton ratio (N/Z ratio or nuclear ratio) is the ratio of the
number of protons to neutrons 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 the highest 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.
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.
Rev 1
19
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
Which of the following nuclides has the higher neutronproton ratio?
A.
Cobalt-60
B.
Selenium-79
C.
Silver-108
D.
Cesium-137
ELO 2.3 Atomic Quantities
Introduction
Isotopic calculations determine the relative amount of isotopes in a given
quantity of an element. These calculations use the terms atom percent,
atomic weight, and weight percent.
20
Rev 1
Naturally Abundant Isotopes
The relative natural abundance of a specific isotope within an element is
relatively constant and is shown on the Chart of the Nuclides.
The following terms provide quantitative measures of nuclides:

Atom percent (a/o): the percentage of atoms of an element that are
from a particular isotope. Abbreviated as a/o. For example, a cup of
water containing 8.23 x 1024 atoms of oxygen, if the a/o of oxygen-18
is 0.20 percent, then there are 1.65 x 1024 atoms of oxygen-18 in the
cup.
 Atomic weight: the average atomic weight of all isotopes of the
element.
 Weight percent (w/o): the percentage of weight of a particular
isotope of an element. Abbreviated as w/o. For example, a sample of
material contains 100 kilograms (kg) of uranium, if the w/o of
uranium-235 is 28, then 28 kg of uranium-235 is present in the
sample.
Atomic Weight Calculation
To calculate atomic weight, multiply each isotope present in the sample by
the atomic mass of that isotope. Then, add the calculated products.
Step
Action
1.
Determine the abundance of each isotope present (Chart Of The
Nuclides)
2.
For each isotope determine the atomic mass (Chart Of The
Nuclides)
3.
For each isotope multiply its abundance times its atomic mass
4.
Sum the products of each isotope calculation
Example: Calculating Atomic Weight
The contribution from each isotope present in an element must be included
in calculating the atomic weight. For example, we calculate the atomic
weight for the element lithium as follows:
Rev 1
21
Step
Action
Result
1.
Determine the
abundance of each
isotope present (Chart
Of The Nuclides)
Lithium-6 (Li-6): 7.5 percent; Lithium-7
(Li-7): 92.5 percent
For each isotope
determine the atomic
mass (Chart Of The
Nuclides)
Lithium-6: 6.015122 amu; Lithium-7:
7.016003 amu
2.
3.
4.
For each isotope,
multiply abundance
times atomic mass
Sum the products of
each isotope
calculation
Li-6: (0.075)(6.015) = 0.4511 amu;
Li-7: (0.925)(7.016) = 6.4898 amu
0.4511 amu + 6.4898 amu
= 6.9409 amu
Knowledge Check
What is the abundance of beryllium-9?
22
A.
100 a/o
B.
10 a/o
C.
9.012182 a/o
D.
5 a/o
Rev 1
Knowledge Check
Calculate the atomic weight for the element silver with
the following stable isotopes …
Ag-107, abundance 51.84 percent, Mass 106.905097
amu
Ag-109, abundance 48.16 percent, Mass 108.904752
amu
A.
107.8886 amu
B.
1078.8860 amu
C.
2907.8109 amu
D.
2.9507 amu
ELO 2.4 Enrichment and Depletion
Introduction
Natural uranium from the earth contains isotopes of uranium-238, uranium235, and uranium-234, with the majority (99.2745 percent) of uranium
existing as uranium-238. With the remaining isotopes, 0.72 percent are
uranium-235 with a slight trace (0.0055 percent) of uranium-234. These
isotopes of uranium have significantly different nuclear properties. For
reasons discussed later uranium-235 is the desired material for use in
reactors.
Enriched and Depleted Uranium
Enrichment is the complex and expensive process of separating isotopes
from natural elements. The details of this process are beyond the scope of
this training module. In pressurized water reactors (PWRs), uranium
enriched with uranium-235 is used for fuel. For uranium, the enrichment
process results in enriched uranium (used as fuel) and depleted uranium.
Enriched Uranium
Enriched uranium is uranium that has a higher concentration of uranium235 than found in natural uranium.
Depleted Uranium
Depleted uranium is a by-product of the enrichment process. Depleted
uranium is uranium where the isotope uranium-235 has a lower
concentration than its natural value (0.72 percent). Although depleted
Rev 1
23
uranium is a by-product of the enrichment process, it does have important
uses in both commercial and defense industries.
Knowledge Check
Depleted uranium will have ___________ atomic weight
than natural uranium.
A.
less
B.
the same
C.
greater
D.
much less
TLO 2 Summary
 In the Chart of the Nuclides, gray squares indicate stable naturally
occurring isotopes. Those in white or color squares are radioactive.
— Stable isotopes: symbol, atomic mass number, isotopic
percentage in the naturally occurring element, thermal neutron
activation cross section and the mass (amu) are given.
— Unstable isotopes: symbol, mode of decay, for example, β- or α,
disintegration energy in MeV, mass (amu) when available, and
half-life are provided.
 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.
 Atomic weight for an element is the average atomic weight of the
isotopes of the element.
 Weight percent (w/o) is the percent weight of an element that is a
particular isotope.
 Enriched uranium is uranium in which the isotope uranium-235 has
a concentration greater than its natural value of 0.7 percent.
 Depleted uranium is uranium in which the isotope uranium-235 has
a concentration less than its natural value of 0.7 percent.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the information for stable and radioactive isotopes found on
the Chart of the Nuclides.
2. Describe how an element’s neutron to proton ratio affects its stability.
24
Rev 1
3. Explain the difference between atom percent, atomic weight and
weight percent, and given the atom percent and the atomic masses for
isotopes of a particular element, calculate the atomic weight of the
element.
4. Describe the following terms:
a. Enriched uranium
b. Depleted uranium
TLO 3 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 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 the difference between an x-ray and a gamma ray and their
effects on the atom. Include an explanation for ionization, ionization
energy, nucleus energy, and application of the nuclear energy level
diagram.
ELO 3.1 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 (𝐸 =
𝑚𝑐 2 ). Instead of two separate conservation laws, a single conservation law
states that the sum of mass and energy is conserved. A mass decrease
results in a corresponding energy increase and vice-versa.
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.
Rev 1
25
Knowledge Check
_______________ is the amount of energy that must be
supplied to a nucleus to completely separate its nuclear
particles.
A.
Nuclear energy
B.
Binding energy
C.
Mass defect
D.
Separation energy
ELO 3.2 Determining Mass Defect and Binding Energy
Introduction
Binding energy supplied by atomic forces holds a stable nucleus together.
When the nucleus is divided into separated nucleons, energy is required.
The separated nucleons have a greater mass than the original nucleus. The
more stable the nucleus, the greater energy required to break it apart.
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 resulst in a calculated mass defect
of zero (0).
∆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)
26
Rev 1
•
Z = atomic number (number of protons)
•
A = mass number (number of nucleons)
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 1
27
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
1.
Use the equation:
Determine the mass defect of the
nuclide.
2.
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) +
(𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚
Use the equation:
Use the binding energy equation to
calculate the binding energy.
28
Action
931.5 𝑀𝑒𝑉
𝐵. 𝐸. = ∆𝑚 (
)
1 𝑎𝑚𝑢
Rev 1
Description
Action
Calculate the binding energy.
Multiply the change in mass
from Step 1 by the energy
conversion for an amu.
3.
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 1
29
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
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)
30
A.
1.86471 amu
B.
1.91517 amu
C.
0.191517 amu
D.
0.186471 amu
Rev 1
Knowledge Check
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 3.3 Gamma Rays and X-Rays
Introduction
Radiation sources distinguish these two types of radiation. Emitted
electrons are the source of X-rays, while emissions from the nucleus are the
source of gamma rays. They are similar in that they are both photons and
undergo similar interactions. An overview of ionization, ionization energy,
and the nuclear energy level diagram explaining gamma ray and X-ray
characteristics are presented here.
Energy Levels of Atoms
Electrons move in defined orbits around the nucleus. The binding forces
keeping an electron in its orbit depend on the orbit location. For example,
only 7.38 electron volts (eV) is required to remove an outermost electron
from a lead atom, while 88,000 eV are required to remove an innermost
electron. The attractive force between a positive proton in the nucleus and a
negative electron depends on the distance between the two. As the
electrons orbit further from the nucleus this attraction weakens. Therefore,
less energy is required to remove electrons farther from the nucleus.
Ionization is the process of removing an electron from its orbit. Ionization
energy is the energy required to remove an electron from its orbit.
Rev 1
31
Ground State
In a neutral atom (number of electrons = Z) the electrons
are in defined orbital shells each with a different energy
level. The ground state is the normal lowest energy state
for that electron.
Exited State
For More
Information An excited state means that an atom possesses more
energy than its ground state energy. An atom cannot stay
in the excited state for an indefinite period of time.
X-Rays and Gamma Rays
X-rays emitted from an electron cloud and gamma rays emitted from the
nucleus are identified by wavelengths; gamma wavelengths are shorter than
X-ray wavelengths. Electromagnetic radiation emitted by X-ray tubes has a
longer wavelength and lower photon energy than the radiation emitted by
radioactive nuclei (for example, gamma rays).
X-Ray Production
An electron in an excited state eventually transitions to
either a lower-energy excited state, or directly to its
ground state, by emitting a discrete bundle of
electromagnetic energy called an X-ray. The energy of
the X-ray(s) is equal to the difference in the energy level
between the excited state and the ground state and
typically ranges from several electron volts (eV) to
100,000 eVs.
For More Gamma Ray Production
Information
Similar to excited electrons, an excited nucleus
transitions to its lowest energy configuration by emitting
a gamma ray. The only differences between X-rays and
gamma rays are their energy levels and whether they are
emitted from the electron shell or from the nucleus.
Nuclear Energy Level Diagram
Nucleons in the nucleus of an atom, similar to electrons, exist in shells that
correspond to energy states. The nucleus energy shells are less defined and
less understood than electron shells. Discrete energy states for electrons
range from eV to kilo-electron volt (keV); the energy levels within the
nucleus are higher and typically in MeV.
32
Rev 1
A nuclear energy level diagram shows the ground state and the excited
states of a nucleus. This diagram consists of a stack of horizontal bars with
one bar for each excited state of the nucleus. The ground state of a nuclide
has zero (0) excitation energy. The excitation energy of the excited state is
the difference in energy between the ground state and the excited state. The
figure below is the energy level diagram for nickel-60.
Figure: Energy Level Diagram- Nickel-60
Knowledge Check
In order for uranium-238 to be stable, there must be
_______ electrons orbiting the nucleus.
Rev 1
A.
238
B.
235
C.
146
D.
92
33
Knowledge Check
What will the state of excitation be if a 2.506 MeV
gamma is emitted using the energy level diagram below
and the nucleus is at a 2.506 MeV level of excitation?
A.
0 MeV
B.
2.158 MeV
C.
1.332 MeV
D.
1.174 MeV
TLO 3 Summary
 Mass defect: difference between the mass of the atom and the sum of
the masses of its parts.
 Binding energy: amount of energy that must be supplied to a nucleus
to completely separate its nuclear particles. Binding energy is the
energy equivalent of the mass defect.
 Mass defect can be calculated by using the equation below:
∆𝑚 = [𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 ] − 𝑚𝑎𝑡𝑜𝑚

Binding energy can be calculated by multiplying the mass defect by
the factor of 931.5 MeV per amu from Einstein’s equation.
 The differences between x-rays and gamma rays include the
following:
— Energy levels
34
Rev 1
— X-rays are emitted from the electron shell
— Gamma rays are emitted from the nucleus
 Ionization is the process of removing an electron from an atom.
 Ionization energy is the energy required to remove electron from an
atom.
 A nuclear energy-level diagram is used to depict the ground state
and the excited states of a nucleus.
Now that you have completed this lesson, you should be able to do the
following:
1. Define 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 the difference between an x-ray and a gamma ray and their
effects on the atom. Include an explanation for ionization, ionization
energy, nucleus energy, and application of the nuclear energy level
diagram.
TLO 4 Nuclear Stability
Overview
Most naturally occurring atoms are stable and do not emit particles or
energy or change state. However, some atoms are unstable and emit
radiation to achieve a more stable configuration.
Unstable Nuclides Can Achieve Stability
Good
Points
It is important to understand nuclear stability because
unstable nuclides achieve stability by emitting the
following:


High-energy photons
High-energy particles
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the conservation principles that must be observed during
radioactive decay. Include an explanation of neutrinos.
2. Describe the following radioactive decay processes:
a. Alpha decay
b. Beta-minus decay
c. Beta-plus decay
d. Electron capture
e. Gamma ray emission
f. Internal conversions
Rev 1
35
g. Isomeric transitions
h. Neutron emission
3. Given the stability curve on the Chart of the Nuclides, determine the
type of radioactive decay that the nuclides in each region of the chart
will typically undergo.
4. Given a Chart of the Nuclides, describe the radioactive decay chain
for a nuclide.
ELO 4.1 Conservation Principles
Introduction
For stable nuclides, as the mass number increases the ratio of neutrons to
protons increases. Unstable nuclei with an excess or shortage of neutrons
undergo a transformation process known as beta (β) decay. Unstable nuclei
also undergo other processes such as alpha (α) or neutron (n) decay. The
final nucleus is in a more stable configuration resulting from these decay
processes.
Some naturally occurring heavy elements, such as uranium or thorium and
their unstable decay chain elements, emit radiation. Uranium and thorium,
both present since creation, have an extremely slow decay rate. All
naturally occurring nuclides with atomic numbers greater than 82 are
radioactive.
Conservation Principles
Studies of radioactive decay processes have identified the following
conservation principles applying to the decay of radionuclides:
Principle
Description
Conservation of
Electric Charge
Conservation of electric charge implies that
charges are neither created nor destroyed.
Single positive and negative charges may
neutralize each other. It is possible for a
neutral particle to produce one charge of each
sign.
Conservation of Mass
Number
Conservation of mass number does not allow a
net change in the number of nucleons.
However, the conversion of one type of
nucleon to another type (proton to a neutron
and vice versa) is allowed.
36
Rev 1
Principle
Description
Conservation of Mass
and Energy
Conservation of mass and energy implies that
the total of the kinetic energy and the energy
equivalent of the mass in a system must be
conserved in all decays and reactions. Mass
can be converted to energy and energy can be
converted to mass, but the sum of mass and
energy must be constant.
Conservation of
Momentum
Conservation of momentum is responsible for
the distribution of the available kinetic energy
among product nuclei, particles, and/or
radiation. The total amount is the same before
and after the reaction, even though it may be
distributed differently among entirely different
nuclides and/or particles.
Example
Xenon-135 is a radioactive isotope. To achieve stability, xsnon-135 decays
by emitting a beta particle, resulting from a neutron converting to a proton,
which illustrates the conservation of mass and energy. The beta is ejected
from the nucleus and no longer contributes to the atomic mass of the
resultant isotope. Although no longer in the nucleus, the beta particle
accounts for any mass difference between the proton and the neutron.
Xenon-135 Decay
Resultant Isotope and Energy
134.90720 amu
Cesium-135
54 protons
134.905977 amu
81 neutrons
55 protons
81 neutrons
∆Mass = 0.001235 amu
Mass is accounted for in the beta particle and
energy of the gammas emitted.
Rev 1
37
Knowledge Check
Charges are neither created nor destroyed describes
which of the following conservation principle?
A.
of mass
B.
of electrical charge
C.
of momentum
D.
of thermal energy
ELO 4.2 Decay Processes
Introduction
To attain stability, nuclei emit radiation by a spontaneous disintegration
process known as radioactive decay or nuclear decay. This radiation may
be electromagnetic radiation, particles, or both.
38
Rev 1
Radioactive Decay Processes
Alpha Decay
Decay
What Happens
Alpha Decay (α)
Alpha decay is the emission of alpha particles
(helium nuclei). When an unstable nucleus ejects
an alpha particle, the atomic number is reduced
by two (2) and the mass number decreased by
four (4).
234
92𝑈
→
230
90𝑇ℎ
+ 42𝛼 + 𝛾 + 𝐾𝐸
An example is uranium-234, which decays to
Thorium-230 by the ejection of an alpha particle
and emission of a 0.068 MeV gamma.
The combined kinetic energy of the daughter
nucleus (thorium-230) and α particle are
designated as KE. The sum of the KE and the
gamma energy equals the difference in mass
between the original nucleus (uranium-234) and
the final particles (binding energy released). The
alpha particle carries off as much as 98 percent of
the kinetic energy, and in most cases can be
considered to carry off all the kinetic energy.
Figure: Alpha Decay
Rev 1
39
Beta Decay
Decay
What Happens
Beta Decay (β)
Beta decay is the emission of electrons of nuclear
rather than orbital origin. These particles are
electrons expelled by radioactive nuclei and may
have a charge of either sign (β- or β+).
Beta Minus Decay
Negative electron emission, from the conversion
of a neutron to a proton, increases the atomic
number by one (1) and leaves the mass number
unchanged.
This is a common decay mode for nuclei with
excess neutrons, such as fission fragments and to
the right of the stability curve. An example is
shown below in both the equation and the
graphic:
239
93𝑁𝑝
→
239
94𝑃𝑢
+ −10𝛽 + 00𝑣̅
The symbol on the end represents an antineutrino. More information about neutrinos is
provided later in this module.
Figure: Beta Decay
40
Rev 1
Decay
What Happens
Beta Plus Decay
Positively charged electrons (beta-plus particles)
are positrons. With the exception of their charge,
positrons they are identical to electrons. They are
represented by the following example:
0
+1𝑒
𝑜𝑟 +10𝛽
𝑒 + 𝑜𝑟 𝛽 +
Positron emission decreases the atomic number
by one and leaves the mass number unchanged by
converting a proton into a neutron, shown below
in the example:
13
7𝑁
→
13
6𝐶
+ +10𝛽 + 00𝑣
An example of typical positron decay is shown
below in the graphic.
Figure: Beta Plus Decay
Rev 1
41
Electron Capture
Decay
What Happens
Electron Capture
(EC, K-capture)
Nuclei with excess protons may capture an inner
orbit electron that immediately combines with a
proton to form a neutron and a neutrino. The
electron is often captured from the innermost
orbit (K-shell); therefore, this process is also
called K-capture. The following example depicts
electron capture:
7
4𝐵𝑒
+ −10𝑒 → 73𝐿𝑖 + 00𝑣
Resulting from beta decays, a neutrino is formed
and its energy conserving momentum. Any
energy available due to the atomic mass of the
product being less than that of the parent appears
as gamma radiation. Characteristic x-rays are
emitted when an electron from another shell fills
the vacancy in the K-shell.
Figure: Electron Capture or K-Capture
Electron capture and positron emission exist as competing processes. They
both result in production of the same daughter product. For positron
emission, the mass of the daughter product must be at least the mass of two
electrons less than the mass of the parent. This mass difference accounts
for the ejected positron and for the daughter having one less electron than
the parent does. If these requirements are not met, then electron capture
occurs and positron emission does not.
42
Rev 1
Gamma Emission (γ)
Gamma radiation is high-energy electromagnetic radiation originating in the
nucleus. It is emitted in the form of photons that are discrete bundles of
energy with both wave and particle properties.
A daughter nuclide from decay often remains in an excited state. The
nucleus drops to the ground state by the emitting gamma radiation,
resolving the excited state.
Gamma rays are penetrating, often requiring several inches of metal or a
couple of feet of concrete to stop or shield.
Internal Conversion
Normally an excited nucleus goes from the excited state to the ground state
by emission of a gamma ray.
In some cases, the released gamma ray interacts with one of the innermost
orbital electrons. This transfers the gamma’s energy to the electron,
referred to as undergoing internal conversion. This energized electron is
ejected from the atom with KE equal to the gamma energy minus the BE of
the electron. An orbital electron then drops to a lower energy state to fill
the vacancy with the emission of X-rays.
Isometric Transition
A nucleus may remain in an excited state for a measurable time before
dropping to ground state. A nucleus in an excited state is a nuclear isomer
because it differs in energy and behavior from other nuclei with identical
atomic and mass numbers.
Isomeric transition happens when the excited nuclear isomer drops to a
lower energy level. Isomeric transition commonly occurs immediately after
particle emission; however, the isomer may remain in an excited state for a
measurable time before dropping to ground state.
It is also possible for the excited isomer to decay by some alternate means.
An example of delayed gamma emission accompanying beta emission is
illustrated below by the decay of nitrogen-16.
16
7𝑁
→ ( 168𝑂) + 01𝛽 + 00𝛾
( 168𝑂) →
16
8𝑂
+ 00𝛾
Neutron Emission
Non-stable nuclei may also emit neutrons (n) to become more stable. An
example of neutron decay is shown below:
Rev 1
43
87
35𝐵𝑟
𝛽−
→
55.9 𝑠𝑒𝑐
𝑛
87
86
36𝐾𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠→ 36𝐾𝑟𝑠𝑡𝑎𝑏𝑙𝑒
Neutrons emitted from the nucleus of a radioactive atom possess a great
deal of KE. Because of their small size, these neutrons can penetrate many
materials.
Neutron production and interaction with matter is of great importance in
nuclear physics and will be discussed in greater detail later in this module.
Neutrinos
Neutrinos are uncharged particles that have an extremely weak interaction
with matter, no mass, and travel at the speed of light. For all practical
purposes, neutrinos pass through all materials with so few interactions that
the energy the neutrino possesses is not recovered.
Neutrinos and antineutrinos are included in this text because they carry a
portion of the KE that otherwise belongs to the beta particle. Therefore,
both neutrinos and antineutrinos are considered for the conservation of
energy and momentum. Neutrinos are usually ignored since they are not
significant in the context of nuclear reactor applications.
Knowledge Check
Which of the following statements accurately describes
alpha decay?
44
A.
A neutron is converted to a proton and an electron. The
electron is ejected from the nucleus.
B.
A neutron is converted to a proton and a positron. The
positron is ejected from the nucleus.
C.
A particle is emitted from a nucleus containing two (2)
neutrons and two (2) protons.
D.
A particle is emitted from a nucleus containing two (2)
electrons and two (2) protons.
Rev 1
Knowledge Check
_______________ occurs when a gamma ray, emitted by
the nucleus as it goes from the excited state to the ground
state, interacts with one of the innermost electrons of the
same atom. The electron is ejected from the atom.
A.
Isomeric transition
B.
Internal conversion
C.
Gamma decay
D.
Electron ejection
Knowledge Check
Which of the following statements accurately describes
beta-minus decay?
A.
A neutron is converted to a proton and an electron. The
electron is ejected from the nucleus.
B.
A neutron is converted to a proton and a positron. The
positron is ejected from the nucleus.
C.
A particle is emitted from a nucleus containing 2
neutrons and 2 protons.
D.
A particle is emitted from a nucleus containing 2
electrons and 2 protons.
ELO 4.3 Stability Curve
Introduction
Radioactive nuclides decay, resulting in a daughter nuclide with a neutronproton ratio closer to the line of stability on the Chart of the Nuclides.
Knowing the decay process helps predict the type of decay a nuclide
undergoes based on its location relative to the line of stability.
Predicting Type of Decay
The figure below illustrates possible decay methods for nuclides in different
regions of the Chart of the Nuclides.
Rev 1
45

Nuclides below and to the right of the line of stability usually undergo
beta-minus (β-) decay.
 Nuclides above and to the left of the line of stability usually undergo
either beta-plus (β+) decay or electron capture.
 Nuclides in the upper right hand region are likely to undergo alpha (α)
decay.
These are general rules with exceptions, especially in the region of the
heavy nuclides.
Figure: Types of Radioactive Decay Relative to Line of Stability
Stable isotopes are isotopes that are not radioactive; for example, they do
not decay spontaneously. Stable isotopes are on the line of stability.
Example
Of the known elements, 80 have at least one stable nuclide. These are
found in the first 82 elements from hydrogen to lead. There are two
exceptions: technetium-43 and promethium-61, neither of which has any
stable nuclides.
There are approximately a total of 254 known stable nuclides. Please note
that different texts may show different numbers of stable isotopes,
depending on the publication date. In this instance, stable means a nuclide
that has not been observed to decay against the natural background. These
elements are not radioactive or have half-lives too long to measure.
Stable isotopes include the following:






46
1 element (tin) has 10 stable isotopes
1 element (xenon) has eight (8) stable isotopes
Four (4) elements have seven (7) stable isotopes apiece
Eight (8) elements have six (6) stable isotopes apiece
10 elements have five (5) stable isotopes apiece
Nine (9) elements have four (4) stable isotopes apiece
Rev 1



Five (5) elements have three (3) stable isotopes apiece
16 elements have two (2) stable isotopes apiece
26 elements have one (1) single stable isotope
Knowledge Check
Match the four (4) areas on the curve with the correct
description.
Figure: Areas of Radioactive Decay Relative to the Line
of Stability
A Alpha
B Line of stability
C Beta +
D Beta-
Rev 1
47
ELO 4.4 Decay Chains
Introduction
When an unstable nucleus decays, the resulting daughter nucleus is not
necessarily stable. If unstable, the daughter nucleus undergoes an
additional decay, which is particularly common with the larger nuclides.
Decay Chains
Using the Chart of the Nuclides, the decay chain can be traced from the
unstable parent through multiple decays to achieve stability. The sequence
of decay from the original unstable nuclide, the intermediary nuclides, and
the final stable nuclide is the decay chain. A common method for stating
the decay chain is illustrated on the next page with the decay chains for
rubidium-91 and actinium-215.
The daughter nuclide of a decay event may be unstable, resulting in another
daughter that may also be unstable, which leads to a sequence of several
decay events. Eventually, a stable nuclide results or is produced.
For example, the following steps are the natural decay chain of U-238:













48
U-238 decays through alpha-emission, with a half-life of 4.5 billion
years to thorium-234
thorium-234 decays through beta-emission, with a half-life of 24 days
to protactinium-234
protactinium-234 decays through beta-emission, with a half-life of 1.2
minutes to uranium-234
uranium-234 decays through alpha-emission, with a half-life of 240
thousand years to thorium-230
thorium-230 decays through alpha-emission, with a half-life of 77
thousand years to radium-226
radium-226 decays through alpha-emission, with a half-life of 1.6
thousand years to radon-222
radon-222 decays through alpha-emission, with a half-life of 3.8 days
to polonium-218
polonium-218 decays, through alpha-emission, with a half-life of 3.1
minutes to lead-214
lead-214 decays through beta-emission, with a half-life of 27 minutes
to bismuth-214
bismuth-214 decays, through beta-emission, with a half-life of 20
minutes to polonium-214
polonium-214 decays through alpha-emission, with a half-life of 160
microseconds to lead-210
lead-210 decays through beta-emission, with a half-life of 22 years to
bismuth-210
bismuth-210 decays through beta-emission, with a half-life of 5 days
to polonium-210
Rev 1

polonium-210 decays through alpha-emission, with a half-life of 140
days to lead-206, a stable nuclide
The standard 𝐴𝑍𝑋 format is often employed to describe the decay. Arrows
used between nuclides indicate where decays occur, with the type of decay
indicated above the arrow and the half-life below the arrow. Decay chains
continue until a stable nuclide or a nuclide with a half-life greater than 1 x
106 years is reached.
The following decay chains for rubidium-91 and actinium-215 are provided
as examples:
91
𝑅𝑏
37
𝛽
𝛽
𝛽
91
91
91
𝑆𝑟
𝑌
𝑍𝑟
→
→
→
38
39
40
58.0 𝑠
9.5 ℎ𝑟𝑠
58.5 𝑑
𝛼
𝛼
𝛽
211
207
207
215
→
→
𝐴𝑡
𝐵𝑖
𝑇𝑙
𝑃𝑡
→
82
85 0.10 𝑚𝑠 83 2.14 𝑚𝑖𝑛 81
4.77 𝑚𝑖𝑛
Knowledge Check
When an unstable nucleus decays, the resulting daughter
nucleus _____________________.
Rev 1
A.
is always stable
B.
is never stable
C.
has more nucleons
D.
is not necessarily stable
49
TLO 4 Summary
 Conservation principles observed during radioactive decay:
— Conservation of electric charge: charges are neither created
nor destroyed.
— Conservation of mass number: shows no net change in the
number of nucleons.
— Conservation of mass and energy: total of the KE and the BE
equivalent to the mass in a system is conserved in all decays and
reactions.
— Conservation of momentum: distribution of the available KE
among product nuclei, particles, and/or radiation.
 Alpha decay: emission of an alpha particle (2 protons and 2 neutrons)
from an unstable nucleus daughter nuclide includes the following
— Atomic number two (2) less than parent nuclide
— Mass number four (4) less than parent nuclide
— Daughter releases its excitation energy by gamma emission
 Beta-minus decay effectively converts a neutron to a proton and an
electron, which is immediately ejected from the nucleus.
— Daughter nuclide has its atomic number increased by one (1) and
the same mass number compared to the parent.
 Beta-plus decay converts a proton to a neutron with a positron
ejected.
— Daughter nuclide has its atomic number decreased by one (1)
and the same mass number.
 In electron capture, the nucleus absorbs an electron from innermost
orbit that combines with a proton to form a neutron.
 Gamma radiation is a high-energy electromagnetic radiation
originating in the nucleus.
 Internal conversion: when a gamma ray, emitted by the nucleus as it
goes from the excited state to the ground state interacts with an
innermost electron of the same atom to eject it from the atom.
 An isomeric transition: decay of an excited nucleus to a lowerenergy level by the emission of a gamma ray.
 Neutron emission: non-stable nuclei may also emit neutrons (n) to
become more stable.
 Neutrinos: uncharged particles that have weak interaction with
matter, no mass, and travel at the speed of light.
 Many modes of radioactive decay result in a daughter that has an
energy level above ground state.
— This excitation energy is often released as a gamma ray.
 The type of decay that a nuclide typically undergoes is determined by
its relationship to the line of stability.
— Beta-minus decay: below and to the right of the line
— Beta-plus decay or electron capture: above and to the left of
the line
— Most alpha emitters: upper, right-hand corner of the chart
 Decay chains are found by tracing the steps an unstable atom goes
through while trying to achieve stability.
50
Rev 1
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the conservation principles that must be observed during
radioactive decay. Include an explanation of neutrinos.
2. Describe the following radioactive decay processes:
a. Alpha decay
b. Beta-minus decay
c. Beta-plus decay
d. Electron capture
e. Gamma ray emission
f. Internal conversions
g. Isomeric transitions
h. Neutron emission
3. Given the stability curve on the Chart of the Nuclides, determine the
type of radioactive decay that the nuclides in each region of the chart
will typically undergo.
4. Given a Chart of the Nuclides, describe the radioactive decay chain
for a nuclide.
TLO 5 Radiation Emitted
Overview
Radiation is comprised of photons or energy waves and particles that
originate in either the nucleus or the electron shells of atoms. Photons and
radiation particles have energy and interact with matter, transferring their
energy in the process. The way radiation reacts with matter depends on the
mass and energy of these photons and particles.
It is important to understand the physical properties of the radiation emitted
from atoms in order to understand the potential hazards and methods for
protection.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the difference between charged and uncharged particle
interaction with matter. Include an explanation of specific ionization.
2. Describe interactions of the following types of particles with matter:
a. Alpha particle
b. Beta particle
c. Positron
d. Neutron
3. Describe the type of material that can be used to stop (shield) the
following types of radiation:
a. Alpha particle
b. Beta particle
Rev 1
51
c. Neutron
d. Gamma ray
ELO 5.1 Charged Versus Uncharged Particles
Introduction
Interactions with matter vary considerably according to the different types
of radiation. Large, massive, charged alpha particles have very limited
penetration capabilities. Neutrinos, the other extreme, have a low
probability of interacting matter, and a large penetrating capability.
Charged Versus Uncharged Particles
Radiation is classified into two general groups, charged and uncharged.
These two groups also exhibit different interactions with matter.


Charged particles directly ionize the media they pass through
Uncharged particles and photons only cause ionization indirectly or
by secondary radiation
Example
Charged Particle Interaction
Charged particles have surrounding electrical fields that interact with the
atomic structure of the medium through which they are traveling. This
interaction slows the particle and accelerates electrons in the atoms of the
medium. The accelerated electrons acquire enough energy to escape from
their parent atoms causing ionization of the affected atom.
Uncharged Particle Interaction
Uncharged moving particles do not have an electrical field. They can only
lose energy and cause ionization by direct collisions or scattering. A photon
loss energy by photoelectric effect, Compton scattering, or pair production.
Specific Ionization
Ionizing radiation creates ion-pairs (+ and – charged). The intensity of
ionization, called specific ionization, is the number of ion-pairs formed per
centimeter of travel in a given material.
The amount of ionization produced by a charged particle per unit path
length, a measure of its ionizing power, is roughly proportional to the
particle's mass and the square of its charge as shown below in the equation:
𝐼=
𝑚𝑧 2
𝐾. 𝐸.
Where:
52
Rev 1
•
I = ionizing power
•
m = mass of particle
•
z = number of unit charges particle carries
•
K.E. = kinetic energy of particle
Since m (mass) for an alpha particle is about 7,300 times as large as m
(mass) for a beta particle, and z (charge) is twice as great, an alpha particle
produces considerably more ionizations per unit path length than a beta
particle with the same energy.
Knowledge Check
Charged particles ionize the media they pass through.
A.
Directly
B.
Indirectly
C.
Never
D.
Sometimes
Knowledge Check
_______________ is term used to describe the number of
ion pairs formed by a charged particle per centimeter of
travel in a given material.
Rev 1
A.
Ionization
B.
Radioactive decay
C.
Specific ionization
D.
Activity
53
ELO 5.2 Radioactive Interactions
Introduction
The way radiation reacts with matter depends on the type of radiation. The
following types of radiation interact with matter in a specifically predictable
manner:
a. Alpha particle
b. Beta particle
c. Gamma
d. Positron
e. Neutron
Alpha Interaction
Alpha radiation is originates from the radioactive decay of heavy nuclides
and certain nuclear reactions. The alpha particle consists of two (2)
neutrons and two (2) protons, the same as a helium atom. With no
electrons, the alpha particle has a charge of positive two (+2). This positive
charge causes the alpha particle to strip electrons from the orbits of atoms in
its vicinity. Alpha particles have a high specific ionization.
As an alpha particle passes through material it interacts and removes
electrons from the atoms it passes near. Electron removal requires energy.
The alpha particle’s energy decreases with each reaction. Ultimately, the
alpha particle expends its KE, gains two (2) electrons, and becomes a
helium atom.
Beta Interaction
A beta-minus particle originates from an electron that was ejected at a high
velocity from an unstable nucleus. Electrons have a small mass and an
electrical charge of minus one (-1). Beta particles cause ionization by
displacing electrons from atomic orbits.
Beta-minus ionization occurs from interaction with collisions of orbiting
electrons. Each collision removes KE from the beta particle, slowing it
down. After a few collisions, the beta particle slows enough to allow it to
be captured as an orbiting electron in an atom.
Positron Interaction
Positrons originate from positively charged electrons. Except for the
positive charge, positrons are identical to beta-minus particles and interact
similarly with matter.
54
Rev 1
Positron interaction is short-lived and quickly annihilates via interactions
with negatively charged electrons. This produces two gamma rays with
energy equal to the rest mass of the electrons.
0.000549 𝑎𝑚𝑢 931.5 𝑀𝑒𝑉
2 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 (
)(
) = 1.02 𝑀𝑒𝑉
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
𝑎𝑚𝑢
These gamma rays interact with matter by the photoelectric effect, Compton
scattering or pair production, described later in this module.
Neutron Interactions
Neutrons originate primarily from nuclear reactions, such as fission, but
also result from the decay of radioactive nuclides. With no charge, the
neutron is difficult to stop and has a high penetrating power.
Neutrons are attenuated (reduced in energy and numbers) by three major
interactions:



Elastic scatter
Inelastic scatter
Absorption
Elastic Scatter
A neutron collides with a nucleus and bounces away from the atom’s
nucleus. This action transmits some of the KE of the neutron to the
nucleus, resulting in the neutron slowing and the atom gaining KE. Elastic
scatter is often referred to as the billiard ball effect.
Inelastic Scatter
The same neutron and/or nucleus collision occurs as in elastic scatter.
However, with reaction to inelastic scatter, the nucleus receives some
internal energy as well as KE. This slows the neutron and leaves the
nucleus in an excited state. When the nucleus decays to its original energy
level, it generally emits a gamma ray. The gamma ray emitted goes on to
interact with matter via the photoelectric effect, Compton scattering, or pair
production.
Absorption
In this instance, the neutron is absorbed into the nucleus of an atom. The
captured neutron leaves the atom in an excited state. If the nucleus emits
one or more gamma rays to reach a stable level, the process is radiative
capture. This reaction is more probable at lower energy levels. The
gammas rays emitted interact with matter via the photoelectric effect,
Compton scattering, or pair production.
Neutron absorption may also result in nuclear fission, which splits the atom
into two smaller atoms. A number of neutrons are released and one or more
Rev 1
55
gamma rays may also be emitted as a fission result. Fission fragments may
create additional neutrons or gamma radiation as they decay to stability.
Gamma Interactions
Gamma radiation is electromagnetic radiation, referred to as a gamma ray, it
is similar to an X-ray. Gamma rays result from the decay of excited nuclei
and also from nuclear reactions. Because the gamma ray has no mass and
no charge, it is difficult to stop and has a high penetrating power. Gamma
rays can pass through several feet of concrete or several meters of water. A
few inches of lead provide effective shielding.
There are three methods of attenuating or reducing the energy level of
gamma rays:



Photoelectric effect
Compton scattering
Pair production
Photoelectron Effect
The photoelectric effect occurs when a low energy gamma ray strikes an
orbital electron. The total energy of the gamma ray is expended ejecting the
electron from its orbit.
The result is ionization of the atom and expulsion of a high-energy electron
result. The photoelectric effect is most predominant with low-energy
gamma rays, and rarely occurs with gamma rays that have more than one
(1) MeV of energy.
Figure: Photoelectron Effect
Compton Scattering
Compton scattering is an elastic collision between an electron and a photon.
In this instance, the photon has more energy than is required to eject the
electron from orbit, or is unable to give up all of its energy in a collision
56
Rev 1
with a free electron. Not all of the energy from the gamma is transferred,
and the photon is scattered. The scattered photon has less energy or a
longer wavelength. The results are ionization of the atom, a high energy
beta, and a reduced energy gamma ray.
Compton scattering is most predominant with gammas at an energy level of
1.0 to 2.0 MeV.
Figure: Compton Scattering
Pair Production
At higher energy levels, pair production is the most likely gamma ray
interaction. When a high-energy gamma ray passes close enough to a heavy
nucleus, the gamma ray disappears, and its energy reappears in the form of
an electron and a positron. This transformation of energy into mass must
take place near a particle, such as a nucleus, to conserve momentum. The
KE of the recoiling nucleus is small; therefore, all of the photon’s energy in
excess of that needed to supply the mass of the pair appears as KE of the
pair. For this reaction to occur, the original gamma must have at least 1.02
MeV of energy.
Figure: Pair Production
Rev 1
57
Knowledge Check
Which of the following is NOT a method by which
neutrons interact with matter?
A.
Inelastic scattering
B.
Elastic scattering
C.
Ionization
D.
Absorption
Knowledge Check
Eventually the _______ particle will be slowed enough
to allow it to be captured as an orbiting __________ in
an atom.
A.
beta; electron
B.
beta; neutron
C.
alpha; electron
D.
alpha; neutron
Knowledge Check
Which one of the following interactions is not one that
gammas undergo?
58
A.
Compton scattering
B.
Photo-electric
C.
Pair production
D.
Inelastic scattering
Rev 1
ELO 5.3 Shielding
Introduction
Shielding describes material placed around a radiation source used to
attenuate or reduce the radiation level. Shielding effectiveness depends on
the material used and the type of radiation. Where one material may be
effective at attenuating neutrons, it may be ineffective at attenuating gamma
rays.
When used in this context, attenuation is the gradual loss in intensity of any
kind of radiation flux through a medium. For instance, sunlight is
attenuated by dark glasses; X-rays are attenuated by lead shielding; and
neutrons are attenuated by water.
Shielding properties for attenuating or reducing the energy level of gamma
rays, based on their type of radiation:




Alpha
Beta particle
Neutron
Gamma ray
Shielding Properties
Alpha Radiation
Because of its strong positive charge and large mass, the alpha particle
deposits a large amount of energy in a short distance, which means it loses
energy quickly and has limited penetrating power. A few centimeters of air
or a sheet of paper will stop the most energetic alpha particles.
Beta Particle
Beta particles are more penetrating than alpha particles, but are still
relatively easy to stop. A thin layer of metal stops the most energetic beta
radiation.
Neutron
With no electrical charge, the neutron is difficult to stop and has high
penetrating power.
Neutrons are attenuated by three major interactions:



Elastic scattering
Inelastic scattering
Absorption
The most effective means of reducing neutron flux is to have elements with
a similar mass available for elastic collisions.
Rev 1
59
A hydrogenous material such as water effectively attenuates neutrons.
Twelve (12) inches of water is an effective shield for neutrons.
Gamma Ray
Gamma rays have no mass and no charge, which gives them high
penetrating power and makes them difficult to stop.
Heavy nuclei such as lead provide large targets for gamma rays to interact
within one of the three gammas interactions. Although dependent on
gamma energies, several meters of concrete or water or a few inches of lead
provide effective shielding for gamma rays.
The figure below illustrates the effect various materials have on types of
radiation.
Figure: Effects Various Materials Have on Types of Radiation
Shielding Thickness
Shielding thickness is referred to as half thickness or
tenth thickness. The thickness is the amount of material
required to reduce the original radiation field strength to
half or a tenth respectively.
For More
Information For example:
The half thickness of lead for gammas is 0.4 inches.
60
Rev 1
Knowledge Check
Which of the following materials would provide the best
shielding against neutrons?
A.
Water
B.
Lead
C.
Paper
D.
Thin sheet of steel
TLO 5 Summary
 Charged particles interact with matter by ionization.
 Uncharged particles only lose energy and cause ionization indirectly
by collisions or scattering.
 Specific ionization: number of ion-pairs formed per unit of travel.
 Alpha particles deposit a large amount of energy in a short distance
of travel due to their large mass and charge.
 Beta-minus particles interact with the electrons orbiting the nucleus,
displacing the electrons to ionize the atom. When a beta particle loses
enough energy, it is captured in the orbital shells of an atom.
 Positrons interact with matter similarly to beta-minus particles. After
a positron has lost most of its energy, it is annihilated by interaction
with an electron. The electron-positron pair disappears and is
replaced by two gamma rays, each with the energy equivalent of the
mass of an electron (0.51 MeV).
 Neutrons interact with matter by elastic scattering, inelastic
scattering, or absorption.
 Gamma rays interact with matter in the following ways:
— Photoelectric effect: interaction with an electron. The entire
energy of the gamma ray transfers to the electron, ejecting the
electron.
— Compton scattering: only part of the gamma energy transfers to
the electron. The electron is ejected from its orbit, and the
gamma is scattered at a lower energy.
— Pair production: gamma rays interact with the electrical field of
a nucleus and are converted into an electron-positron pair. The
gamma must have energy greater than 1.02 MeV for this to
occur.
 Shielding materials for the following types of radiation include:
— Alpha particle: a sheet of paper
— Beta (+ or -): a thin sheet of metal
— Neutrons: a hydrogenous material, such as water
— Gamma rays: several meters of concrete, or water, or a few
inches of lead.
Rev 1
61
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the difference between charged and uncharged particle
interaction with matter. Include an explanation of specific ionization.
2. Describe interactions of the following types of particles with matter:
a. Alpha particle
b. Beta particle
c. Positron
d. Neutron
3. Describe the type of material that can be used to stop (shield) the
following types of radiation:
a. Alpha particle
b. Beta particle
c. Neutron
d. Gamma ray
TLO 6 Radioactive Decay
Overview
The decay rate of a sample of radioactive material is not constant. As
individual atoms of the material decay, fewer atoms remain. Since the
decay rate is directly proportional to the number of atoms, the decay rate
decreases as the number of atoms decreases. Knowledge of radioactive
decay is important for calculating reactivity poisons in the reactor as well as
for understanding personnel hazards.
Objectives
Upon completion of this lesson, you will be able to do the following:
1. Describe the following radioactive terms:
a. Radioactivity
b. Radioactive decay constant
c. Activity
d. Curie
e. Becquerel
f. Radioactive half-life
2. Convert between the half-life and decay constant for a nuclide.
3. Given the nuclide, number of atoms, half-life or decay constant,
determine current and future activity levels.
4. Describe the following:
a. Radioactive equilibrium
b. Transient radioactive equilibrium
c. Secular radioactive equilibrium
62
Rev 1
ELO 6.1 Define Terms
Introduction
Knowledge of the terms used to describe the decay rate of relationships is
required to gain an understanding of radioactive decay. The following
terms are described in this section:






Radioactivity
Radioactive decay constant
Activity
Curie
Becquerel
Radioactive half-life
Radioactive Decay Terms
 Radioactivity: the process whereby certain nuclides spontaneously
emit particles or gamma radiation, a process called radioactivity
decay. Radioactive decay occurs randomly because individual
radioactive emissions cannot be predicted. However, the average
behavior of a large sample can be accurately determined using
statistical methods.
 Radioactive Decay Constant (λ): In a specific time interval, a
specific fraction of a given sample will decay. This probability per
unit time that an atom of a specific nuclide will decay is known as the
radioactive decay constant, λ (lambda). Units for radioactive decay
constants are inverse time such as 1/second, 1/minute, 1/hour, or
1/year. They are expressed as second-1, minute-1, hour-1, and year-1.
 Activity (A): The decay rate of that sample. This decay rate is
measured by the number of disintegrations taking place per second.
In a sample containing millions of atoms, the activity is the product of
the decay constant (λ), and the number of atoms present in the sample
(N). This is shown by the following equation:
𝐴 = 𝜆𝑁
Where:
• A = Activity of the nuclide (disintegrations/second)
• λ = Decay constant of the nuclide (second-1)
• N = Number of atoms of the nuclide in the sample
Since λ is a constant, the activity and the number of atoms are always
proportional.

Rev 1
Radioactive half-life: measures how quickly a nuclide decays.
Radioactive half-life the amount of time required for the activity to
decrease to half of its original value.
63
Measurement Units for Radioactivity
Two common units to measure the activity of a substance are the Curie (Ci)
and the Becquerel (Bq). The Curie is more widely used in the United
States.

Curie: the Curie measures the rate of radioactive decay. One Curie
equals 3.7 x 1010 disintegrations per second. This is approximately
equivalent to the number of disintegrations that one gram of radium226 undergoes in one (1) second.
 Becquerel: a Becquerel also measures of the rate of radioactive
decay, and equals one (1) disintegration per second. The conversion
between a curie and a Becquerel is as follows:
1 𝐶𝑢𝑟𝑖𝑒 = 3.7 × 1010 𝐵𝑒𝑐𝑞𝑢𝑒𝑟𝑒𝑙𝑠
Knowledge Check
Match the following:
1 The decay of unstable atoms by the
emission of particles and
electromagnetic radiation.
A. Curie
2 Unit of radioactivity equal to 3.7 x
1010 disintegrations per second.
B. Radioactivity
3 Unit of radioactivity equal to 1
disintegration per second.
C. Becquerel
4 Probability per unit time that an atom
will decay.
D. Radioactive decay
constant
ELO 6.2 Convert Between Half-Life and Decay Constant
Introduction
Once the decay constant or half-life is known, calculations can be
performed to determine such things as number of atoms and activity level.
The equation below shows the relationship between half-life and the decay
constant:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
64
Rev 1
Half-life is calculated by solving the equation for time (t), when the current
activity (A) equals half the initial activity Ao, as follows:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
𝐴
= 𝑒 −𝜆𝑡
𝐴𝑜
𝐴
ln ( ) = −𝜆𝑡
𝐴𝑜
𝑡=
𝐴
− ln (𝐴 )
𝑜
𝜆
𝑡1 =
1
− ln (2)
𝜆
2
𝑡1 =
2
𝑡1 =
2
ln(2)
𝜆
0.693
𝜆
Converting Between Half-Life and Decay Constant
From the previous derivations, half-life or decay constants may be
determined. From half-lives or decay constants, activity levels and numbers
of atoms may be calculated for any time in an isotope’s decay process to
stability.
𝑡1 =
2
𝜆=
0.693
𝜆
0.693
𝑡1
2
Step
Action
1.
Determine the half-life if
decay constant is known.
Solution
Use the equation:
𝑡1 =
2
Rev 1
0.693
𝜆
65
Step
Action
Solution
2.
Determine the decay
constant if half-life is
known.
Use the equation:
𝜆=
0.693
𝑡1
2
Example
Determine the decay constant of cesium-136, half-life of 13.16 days:
Step
Action
Method
Calculation
1.
Determine the halflife if decay
constant is known.
Use the equation:
N/A*
𝑡1 =
2
2.
Determine the
decay constant if
half-life is known.
0.693
𝜆
Use the equation:
𝜆=
0.693
𝑡1
𝜆=
0.693
13.16 𝑑𝑎𝑦𝑠
𝜆 = 0.0527−1 𝑑𝑎𝑦𝑠
2
*N/A means not applicable
Example
Determine the half-life of potassium-44, decay constant of 0.0313 minutes-1
(min):
Step
Action
Method
1.
Determine the
half-life if decay
constant is
known.
Use the equation:
66
𝑡1 =
2
Calculation
0.693
𝜆
𝑡1 =
2
0.693
0.03131−𝑚𝑖𝑛
𝑡1 = 22.13 𝑚𝑖𝑛
2
Rev 1
Step
Action
Method
Calculation
2.
Determine the
decay constant if
half-life is
known.
Use the equation:
N/A*
𝜆=
0.693
𝑡1
2
*N/A means not applicable
Demonstration
The following graph shows the basic features of radionuclide sample decay:
Figure: Radioactive Decay as a Function of Time in Units of Half-Life
Assuming an initial number of atoms No, the population and activity are
seen to decrease by one-half of No in the time of one half-life. Additional
decreases of half occur in each half-life time.
After five half-lives have elapsed, only 1/32, or 3.1 percent, of the original
number of atoms remain. After seven half-lives, only 1/128, or 0.78
percent, of the atoms remains. The number of atoms existing after five (5)
to seven (7) half-lives is negligible.
Rev 1
67
Knowledge Check
What is the decay constant for plutonium-239, which has
a half-life of 24110 years?
A.
2.874 x 10-5 years
B.
2.874 x 105 years
C.
1.67 x 10-4 years
D.
1.67 x 104 years
ELO 6.3 Calculating Activity Over Time
Introduction
The relationship between activity (A), the number of atoms present (N), and
the decay constant (Greek letter lambda [λ]) are fundamental to
understanding radioactive decay. It is important to estimate the strength or
the amount of radiation that a sample of material can emit following a
specified time.
Calculating Activity Over Time
Decay rate for a given decay constant in a radionuclide sample is stated in
the following equation:
𝐴 = 𝜆𝑁
The following expressions (derived) are used to calculate the change in the
number of atoms present or activity over a period of time:
For the number of atoms present:
𝑁 = 𝑁𝑜 𝑒 −𝜆𝑡
Where:
•
N = number of atoms present at time t
•
No = number of atoms initially present
•
λ = decay constant (time-1)
•
t = time
Since activity and the number of atoms are proportional, the following
equation works for activity by substitution:
68
Rev 1
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
Where:
•
A = activity present at time t
•
Ao = activity initially present
•
λ = decay constant (time-1)
•
t = time
Calculating Activity
Step
Action
1.
Determine the
number of atoms
present in the
mass of the
isotope.
2.
Solution
Use the following equation:
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁 = 𝑚𝑎𝑠𝑠 (
)(
)
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
Use the following equation:
Determine the
decay constant.
𝜆=
0.693
𝑡1
2
3.
Determine the
activity
Use the following equation:
𝐴 = 𝜆𝑁
Example
A sample of material contains 20 micrograms of californium-252 with a
half-life of 2.638 years. Calculate the following:
(a) The number of californium-252 atoms initially present.
(b) The activity of the californium-252 in Curies.
First, determine the number of atoms present in the mass of the isotope.
Use the following equation:
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁 = 𝑚𝑎𝑠𝑠 (
)(
)
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
Rev 1
69
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁𝛼−252 = 𝑚𝑎𝑠𝑠 (
)(
)
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
1 𝑚𝑜𝑙𝑒
6.022 × 1023 𝑎𝑡𝑜𝑚𝑠
= (20 × 10−6 𝑔) (
)(
)
252.08 𝑔
1 𝑚𝑜𝑙𝑒
= 4.78 × 1016 𝑎𝑡𝑜𝑚𝑠
Second, determine the decay constant. Use the following equation:
𝑡1 =
2
𝑡1 =
2
=
0.693
𝜆
0.693
𝜆
0.693
2.638 𝑦𝑒𝑎𝑟𝑠
= 0.263 𝑦𝑒𝑎𝑟 −1
Finally, determine the activity. Use the following equation:
𝐴 = 𝜆𝑁
= (0.263 𝑦𝑒𝑎𝑟 −1 )(4.78
1 𝑦𝑒𝑎𝑟
1 𝑑𝑎𝑦
1 ℎ𝑜𝑢𝑟
× 1016 𝑎𝑡𝑜𝑚𝑠) (
)(
)(
)
365.25 𝑑𝑎𝑦𝑠 24 ℎ𝑜𝑢𝑟𝑠 3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
= (3.98 × 108
𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠
)(
𝑠𝑒𝑐𝑜𝑛𝑑
3.7 × 1010
1 𝑐𝑢𝑟𝑖𝑒
)
𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠
(
)
𝑠𝑒𝑐𝑜𝑛𝑑
= 0.0108 𝐶𝑢𝑟𝑖𝑒𝑠
Variation of Radioactivity Over Time
Once a few key pieces of information are known, we can predict the activity
level of a quantity of an isotope using the following expression:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
70
Rev 1
Where:
•
A = Activity at time t
•
Ao = Activity initially present
•
λ = decay constant
•
t = time
Step
Action
Equation
1.
If the initial activity is
unknown, determine
the number of atoms
present in the mass of
the isotope.
Use the following equation:
Determine the decay
constant if necessary.
Use the following equation:
2.
𝑁 = 𝑚𝑎𝑠𝑠 (
𝜆=
1 𝑚𝑜𝑙𝑒
𝑁𝐴
)(
)
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
0.693
𝑡1
2
3.
Determine the initial
activity.
Use the following equation:
𝐴 = 𝜆𝑁
4.
Determine the new
activity.
Use the following equation:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
Calculate Activity Level
A sample of material contains 20 micrograms of californium-252 with an
activity of 0.0108 Curies. Californium-252 half-life is 2.638 years.
Determine the activity level after 12 years.
First, if the initial activity is not known, determine the number of atoms
present in the mass of the isotope. Use the following equation (Not
necessary):
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁 = 𝑚𝑎𝑠𝑠 (
)(
)
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
Rev 1
71
Second, determine the decay constant if necessary. Use the following
equation:
𝜆=
0.693
𝑡1
2
𝜆=
0.693
𝑡1
2
=
0.693
2.638 𝑦𝑒𝑎𝑟𝑠
= 0.263 𝑦𝑒𝑎𝑟 −1
Third, determine the initial activity. Use the following equation:
𝐴 = 𝜆𝑁
Given: 0.0108 Curies
Finally, determine the new activity. Use the following equation:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
𝐴 = 0.0108 𝑒
−(
0.263
)(12𝑦𝑟)
𝑦𝑟
= 0.00046 𝐶𝑢𝑟𝑖𝑒𝑠
Plotting Radioactive Decay
For visual indication or planning purposes, plotting activity decay may be
useful. Either a linear or a logarithmic scale may be used for plotting
activity.
Step
Action
Method
1.
Calculate the decay
constant of the
isotope.
Use the equation:
𝜆=
0.693
𝑡1
2
2.
72
Use the decay constant Use the equation: 𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
to calculate the
activity at various
times.
Rev 1
Step
Action
Method
3.
Develop a table of
Use the above equations.
values from the
calculations performed
above.
4.
Plot the points from
the table on linear and
semi log scales.
Using the correct graph paper, plot the
points.
Demonstration
Plot the radioactive decay curve for nitrogen-16 over a period of 100
seconds. Initial activity is 142 curies and nitrogen-16 half-life is 7.13
seconds. Plot the curve on both linear rectangular coordinates and a semilog scale.
Step 1: Calculate the decay constant for a 7.13-second half-life using the
below equation:
𝑡1 =
2
𝜆=
0.693
𝜆
0.693
𝑡1
2
𝜆=
0.693
7.13 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝜆 = 0.0972 𝑠𝑒𝑐𝑜𝑛𝑑−1
Step 2: Using the calculated decay constant, calculate the activity at various
times:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
Time
Activity
0 seconds
142.0 Ci
20 seconds
20.3 Ci
40 seconds
2.91 Ci
Rev 1
73
Time
Activity
60 seconds
0.416 Ci
80 seconds
0.0596 Ci
100 seconds
0.00853 Ci
Step 3: Plot the calculated data points on both linear and semi log scales:
Figure: Semi-Log and Linear Plots of Nitrogen-16 Decay
Plotting Multiple Nuclides
For a substance with more than one radioactive nuclide, the total activity is
the sum of the individual activities. For example, consider a sample of
material that contains the following:
1 x 106 atoms of iron-59 that has a half-life of 44.51 days (λ = 1.80 x
10-7 sec-1)
 1 x 106 atoms of manganese-54 that has a half-life of 312.2 days (λ =
2.57 x 10-8 sec-1)
 1 x 106 atoms of cobalt-60 that has a half-life of 1,925 days (λ = 4.17
x 10-9 sec-1)

The initial activity for each nuclide is the product of the number of atoms
and the decay constant:
𝐴𝐹𝑒–59 = 𝑁𝐹𝑒–59 𝜆𝐹𝑒–59
𝐴𝐹𝑒–59 = (1 × 106 𝑎𝑡𝑜𝑚𝑠)(1.80 × 10−7 𝑠𝑒𝑐 −1 )
𝐴𝐹𝑒–59 = 0.180 𝐶𝑖
𝐴𝑀𝑛–54 = 𝑁𝑀𝑛–54 𝜆𝑀𝑛–54
𝐴𝑀𝑛–54 = (1 × 106 𝑎𝑡𝑜𝑚𝑠)(2.57 × 10−8 𝑠𝑒𝑐 −1 )
𝐴𝑀𝑛–54 = 0.0257 𝐶𝑖
74
Rev 1
𝐴𝐶𝑜–60 = 𝑁𝐶𝑜–60 𝜆𝐶𝑜–60
𝐴𝐶𝑜–60 = (1 × 106 𝑎𝑡𝑜𝑚𝑠)(4.17 × 10−9 𝑠𝑒𝑐 −1 )
𝐴𝐶𝑜–60 = 0.00417 𝐶𝑖
Plotting Multiple Nuclides
Plotting the decay activities for each of the three nuclides illustrates the
relative activities for each of the nuclides in the sample and the combined
total over time. In this example, shown below in the figure, initially the
activity of the shortest-lived nuclide (iron-59) dominates the total activity,
then manganese-54. After most of the iron and manganese have decayed
away, the only contributor to activity is cobalt-60.
Figure: Combined Decay of Iron-59, Manganese-54, and Cobalt-60
Knowledge Check
A sample contains 100 grams of xenon-135. Half-life is
9.14 hours and an atomic mass 134.907 amu. Calculate
the decay constant of xenon-135 and sample activity.
Rev 1
A.
0.0758 hour-1 (hr); 2.54 x 10-8 Curies
B.
0.0758 hr-1; 9.15 x 10-11 Curies
C.
6.334 hr-1; 2.54 x 10-8 Curies
D.
6.334 hr-1; 9.15 x 10-11 Curies
75
Knowledge Check
A sample of cobalt-60 contains 10 curies of activity. It
has a half-life of 5.274 years. What will the activity be
in 7.5 years?
A.
3.73 Curies
B.
5 Curies
C.
1.999 Curies
D.
2.68 Curies
Knowledge Check
The two plots below are different in shape because ...
76
A.
One is on a linear scale and the other is on a logarithmic
one.
B.
They are of two different nuclides.
C.
They have different decay constants.
D.
The time intervals for the activity levels are different.
Rev 1
ELO 6.4 Equilibrium
Introduction
Radioactive equilibrium describes the combined characteristics of parent
and daughter nuclides as they reach stability. Understanding equilibrium
allows operators to predict the effects of important nuclides such as iodine
and xenon on reactor operation. Three terms, listed below, describe
equilibrium:
Radioactive equilibrium exists when radioactive nuclide decay and
production rates are equal. With production and decay rates equal, the
number of atoms present remains constant over time.
Transient radioactive equilibrium happens with parent and daughter
nuclides decay at essentially the same rate. The half-life of the daughter is
shorter than that of the parent.
Secular equilibrium occurs with a parent having an extremely long halflife. In this instance, the equilibrium activities are established by the halflife of the original parent. The only exception is the final stable element at
the end of the chain. Its number of atoms constantly increases.
Radioactive Equilibrium Example
An example of radioactive equilibrium is the concentration of sodium-24 in
a sodium-cooled nuclear reactor. Assume that the sodium-24 production
rate is 1 x 106 atoms per second. If the sodium-24 were stable and not
decaying, the amount of sodium-24 present after some time could be
calculated by multiplying the production rate by the amount of time. The
figure below shows a plot of sodium-24 increasing over time.
Figure: Cumulative Production of Sodium-24 Over Time
Sodium-24 is not stable, and it decays at a half-life of 14.96 hours. Assume
that no sodium-24 is present initially and production starts at a rate of 1 x
Rev 1
77
106 atoms per second, the decay rate initially starts at zero (0) because there
is no sodium-24 present to decay. The rate of decay increases as the
amount of sodium-24 increases.
The amount of sodium-24 present initially increases rapidly, and then the
rate slows down as decay increases, until the rate of decay is equal to the
rate of production. The amount of sodium-24 present at equilibrium is
calculated by setting the production rate (R) equal to the decay rate (λ N),
shown below in the equation:
𝑅 = 𝜆𝑁
𝑁=
𝑅
𝜆
Where:
•
R = production rate (atoms/second)
•
λ = decay constant (sec-1)
•
N = number of atoms
𝜆=
0.693
𝑡1
2
=
0.693
1 ℎ𝑜𝑢𝑟
(
)
14.96 ℎ𝑜𝑢𝑟𝑠 3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
= 1.287 × 10−5 𝑠𝑒𝑐𝑜𝑛𝑑 −1
𝑁=
𝑅
𝜆
𝑎𝑡𝑜𝑚𝑠
𝑠𝑒𝑐
=
1.287 × 105 𝑠𝑒𝑐𝑜𝑛𝑑 −1
1 × 106
= 7.77 × 1010 𝑎𝑡𝑜𝑚𝑠
The equation develops to calculate how the amount of sodium-24 changes
over time as it approaches the equilibrium value; however, that is beyond
the scope of this text. Nevertheless, the equation is presented below:
𝑅
(1 − 𝑒 −𝜆𝑡 )
𝜆
This equation is used to calculate the amount of sodium-24 present at
different times. As the time increases, the exponential term approaches zero
(0), and the number of atoms present approaches R/λ. The figure below
shows a plot of the approach of sodium-24 to equilibrium.
𝑁=
78
Rev 1
Figure: Approach of Sodium-24 to Equilibrium
Transient Radioactive Equilibrium
For transient equilibrium to occur, the parent must have a long half-life
compared to the daughter. An example of this type of decay process is
barium-140, which decays by beta emission to lanthanum-140, which in
turn decays by beta emission to stable cerium-140.
𝛽−
𝛽−
140
140
140
→
→
𝐵𝑎
𝐿𝑎
𝐶𝑒
56
57
12.75 𝑑𝑎𝑦𝑠
1.678 𝑑𝑎𝑦𝑠 58
The decay constant for barium-140 is considerably smaller than the decay
constant for lanthanum-140. However, the decay rate of both the parent and
daughter is represented as λN. Although the decay constant for barium-140
is smaller, the actual rate of decay (λN) is initially larger than lanthanum140 because of the initial larger concentration. As the concentration of the
daughter increases, the decay rate of the daughter catches up and eventually
matches the decay rate of the parent. When this occurs, parent and daughter
are both said to be in transient equilibrium.
A plot of the barium-lanthanum-cerium decay chain reaching transient
equilibrium is shown below in the graphic.
Rev 1
79
Figure: Transient Equilibrium in the Decay of Barium-14
Secular Radioactive Equilibrium
Secular equilibrium occurs when the parent has an extremely long half-life.
In a long decay chain for a naturally radioactive element, such as thorium232, where all of the elements in the chain are in secular equilibrium, each
descendant has built up to an equilibrium amount and all decay is at the rate
set by the original parent. The only exception is the final stable element on
the end of the chain. Its number of atoms is constantly increasing.
Knowledge Check
Parent nuclide has an extremely long half-life is a
description of ___________
80
A.
transient equilibrium
B.
secular equilibrium
C.
stable equilibrium
D.
unstable equilibrium
Rev 1
TLO 6 Summary
 Radioactivity is the decay of unstable atoms by the emission of
particles and electromagnetic radiation.
 Curie (Ci): unit of radioactivity equal to 3.7 x 1010 disintegrations per
second.
 Becquerel (Bq): unit of radioactivity equal to one (1) disintegration
per second.
 Radioactive decay constant (λ): probability per unit time that an
atom decays.
 Radioactive half-life: amount of time required for the activity to
decrease to half its original value.
 The activity (A) of a sample is the rate of decay of that sample.
The activity of a substance is calculated from the number of atoms
and the decay constant based on the equation below:
A = λN
The amount of activity remaining after a particular time is calculated
based on the below equation:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
The relationship between the decay constant and the half-life is based
on the below equation:
𝑡1 =
2
0.693
𝜆

Plots of radioactive decay can be used to describe the variation of
activity over time. When decay is plotted using a semi-log scale, the
plot results in a straight line.
 Radioactive equilibrium exists when the production rate of a
material equals the removal rate.
 Transient radioactive equilibrium exists when the parent nuclide
and the daughter nuclide decay at essentially the same rate, which
occurs only when the parent has a long half-life compared to that of
the daughter.
 Secular equilibrium occurs when the parent has an extremely long
half-life. In the associated decay chain, each of the descendants has
built up to an equilibrium amount and all decay at the rate set by the
original parent. The final stable element on the end of the chain
constantly increases.
Summary
Now that you have completed this lesson, you should be able to do the
following:
1. Describe the following radioactive terms:
a. Radioactivity
b. Radioactive decay constant
c. Activity
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d. Curie
e. Becquerel
f. Radioactive half-life
2. Convert between the half-life and decay constant for a nuclide.
3. Given the nuclide, number of atoms, half-life, or decay constant,
determine current and future activity levels.
4. Describe the following:
a. Radioactive equilibrium
b. Transient radioactive equilibrium
c. Secular radioactive equilibrium
Atomic Structure Summary
This module covered atoms and their composition. We covered how 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. We gained an understanding of these subatomic
interactions that are important to understanding the fission process that
occurs in a nuclear reactor.
Now that you have completed this module, you should be able to
demonstrate mastery of this topic by passing a written exam with a grade of
80 percent or higher on the following TLOs:
1. Describe atoms, including components, structure, and nomenclature.
2. Use the Chart of the Nuclides to obtain information.
3. Describe Mass Defect and Binding Energy and their relationship to
one another.
4. Describe the processes by which unstable nuclides achieve stability.
5. Describe how radiation emitted by an unstable nuclide interacts with
matter and materials typically used to shield against this radiation.
6. Describe radioactive decay terms and perform calculations to
determine activity levels, half-lives and decay constants and
radioactive equilibrium.
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