Transcript Atomic Structure - Nuclear Community

Operator Generic Fundamentals
Nuclear Physics – Atomic Structure
© Copyright 2014
Operator Generic Fundamentals
2
Atomic Structure – Introduction
This Module introduces the student to the building blocks of matter:
•
Properties of atoms and subatomic structure
•
Use of the Chart of Nuclides
•
Mass defect and binding energy
•
Stable and unstable atoms
•
Radiation and interactions with matter
•
Radioactive decay, activity, and half-lives
© Copyright 2014
Introduction
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Terminal Learning Objectives
At the completion of this training session, the trainee will demonstrate
mastery of this topic by passing a written exam with a grade of 80
percent or higher on the following Terminal Learning Objectives (TLOs):
1. Describe 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 Calculate activity levels, halflives, decay constants and radioactive equilibrium.
© Copyright 2014
Introduction
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Atoms – Introduction
TLO 1 – Describe atoms, including components, structure, and
nomenclature.
• John Dalton – determined that elements are made up of distinctly
unique atoms in 1803
– First modern proof for the atomic nature of matter
– Atoms are the smallest component of matter defining an element
– 100 years to prove Dalton’s theories
• Chemical experiments indicated the atom is indivisible
• Electrical and radioactivity experimentation indicated that particles of
matter smaller than the atom do exist
• In 1906, J. J. Thompson won the Nobel Prize for establishing the
existence of electrons
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TLO 1
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Atoms – Introduction
• 1920: Earnest Rutherford named the proton
• 1932: James Chadwick confirmed the existence of the neutron
• 1970’s: the application of the standard model of particle physics
proved the existence of quarks
Figure: Composition and Components of Atoms
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TLO 1
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Enabling Learning Objectives for TLO 1
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.
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TLO 1
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Atomic Structure
ELO 1.1 – Using Bohr's model of an atom, describe the characteristics of
the following atomic particles, including mass, charge, and location within
the atom: proton, neutron, and electron.
• Physicist Ernest Rutherford postulated:
– Positive charge in an atom is at center of the atom
– Electrons orbit around it
• Niels Bohr, from Rutherford's theory and Max Planck’s quantum
theory, proposed orbiting electrons in discrete fixed shells
• An electron in one of these orbits has a specific quantity of energy
(quantum)
• Electron movement between shells results in a photon either emitted
or absorbed
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Neutrons and Protons
• Protons and neutrons are located in
center of atom – called nucleus
• Each element made up of atoms
having a unique number of protons
that defines chemical properties
• Neutrons are electrically neutral – no
electrical charge
• Protons are electrically positive electrical charge of +
– Gives nucleus positive charge
– One proton has a +1, two protons
have +2
• Neutrons and protons essentially
equal in mass
– > 1800 times size of an electron
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ELO 1.1
Figure: Simple Carbon Atom
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Electrons
• Electrons orbit the nucleus
– Orbit in concentric orbits referred to as orbitals or shells
– Mass of 1/1835 the mass of a proton or neutron
• Each electron has -1 electrical charge = in magnitude to one proton
• For the atom to be electrically neutral, number of electrons must
equal protons
• Electrons are bound to the nucleus by electrostatic attraction
(opposite charges attract)
• Atom remains neutral unless some force causes a change in the
number of electrons
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Bohr Model of Atom
• Bohr’s model is shown on the next slide
– An electron is shown to have dropped from the third shell to the
first shell releasing energy
– Energy is released as a photon = hv
o h = Planck's constant - 6.63 x 10-34 J-s (joule-seconds)
o v = frequency of the photon
• Accounts for the quantum energy levels
• Bohr's atomic model is designed specifically to explain the hydrogen
atom – has applicability as first generation model to all atoms
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Bohr Model of Atom
Electron dropped from third
shell to first shell causing
emission of photon
𝑃ℎ𝑜𝑡𝑜𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 = ℎ𝑣
Where:
h = Planck's constant
= 6.63 x 10-34 J-s
v = frequency of photon
Figure: Bohr's Model of the Hydrogen Atom
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ELO 1.1
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Atomic Measuring Units
• Size and mass of atoms are very small
– Use of normal measuring units is inconvenient
• Atomic Mass Unit (amu)
– Unit of measure for mass
– One amu = 1.66 x 10-24 grams
• Electron Volt (eV)
– Unit for energy
– eV is amount of energy acquired by single electron when it falls
through potential difference of one volt
– 1 eV equivalent to 1.602 x 10-19 joules or 1.18 x 10-19 foot-pounds
– Protons eV value is + 1 and Electrons are -1 eV
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ELO 1.1
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Atomic Measuring Units
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
Figure: Nucleus and Orbital Electrons
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Atomic Measuring Units
Knowledge Check
Identify the particles included in the make-up of an atom. (More than
one answer may apply.)
A. neutron
B. electron
C. gamma
D. amu
Correct answers are A and B.
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ELO 1.1
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Atomic Terms
ELO 1.2 – Define the following terms and given the standard notation for
a given nuclide, identify its nucleus and electron makeup: nuclide,
isotope, atomic number, and mass number.
• Atomic number (Z) – # of protons
• Mass number (A) – Total number of neutrons & protons
𝐴=𝑍+𝑁
• Nuclide – Atoms containing a unique combination of protons and
neutrons
– 2500 specific nuclides
• Isotope – Atoms of the same element with the same number of
protons (Z) but different number of neutrons
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ELO 1.2
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Atomic Notation
• Convention for identifying
elements on atomic scale
• Standard notation:
– Element symbol (X)
– Atomic number as subscript
to lower left
o Z = # of protons
– Atomic mass number as
superscript to upper left
Figure: Nomenclature for Identifying Nuclides
o A = # of protons +
neutrons
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ELO 1.2
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Atomic Notation
• Atomic Number – Z
– Total number of protons in
nucleus of atom
– Identifies particular
element
– Each chemical element
has unique atomic number.
– Helium consists of atoms
with only two protons in
nucleus
Figure: Nomenclature for Identifying Nuclides
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Atomic Notation
• Mass Number – A
– Total number of nucleons
(protons and neutrons) in
nucleus
– Atoms of same element
may not always contain
same number of neutrons
and may have different
atomic mass numbers
o Isotopes
Figure: Nomenclature for Identifying Nuclides
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ELO 1.2
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Atomic Notation
• Neutron Number – N
– Number of neutrons in
nucleus
– Does not appear in
standard atomic notation
– Found by: 𝑍– 𝐴 = 𝑁
Figure: Nomenclature for Identifying Nuclides
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Isotopes
• Isotopes – nuclides that have same atomic number and are same
element, but differ in number of neutrons
• Most elements have a few stable isotopes and several unstable,
radioactive isotopes
– Oxygen has three stable isotopes that can be found in nature
(oxygen-16, oxygen-17, and oxygen-18) and eight radioactive
isotopes
– Hydrogen has two stable isotopes (hydrogen-1 and hydrogen-2)
and single radioactive isotope
(hydrogen-3)
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ELO 1.2
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Isotopes of Hydrogen
• Isotopes of hydrogen are unique in that they are each commonly
referred to by unique name instead of common chemical element
name
– Hydrogen-1 – almost always referred to as hydrogen, but also
called protium
– Hydrogen-2 – commonly called deuterium 2
1𝐷
– Hydrogen-3 – commonly called tritium 3
1𝑇
Figure: Isotopes of Hydrogen
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ELO 1.2
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Nuclides
Knowledge Check
• State name of element and number of protons, electrons, and
neutrons in nuclides listed below:
Nuclide
Element
Protons
Nuclide
Nuclide
Element
Element
Protons
Protons Electrons
Electrons
Electrons Neutrons
Neutrons
Neutrons
111𝐻
𝐻
111𝐻
10
10
10
𝐵𝐵
555𝐵
14
14
𝑁
14
777𝑁
𝑁
114
114
48𝐶𝑑
114
48𝐶𝑑
239
48𝐶𝑑
94𝑃𝑢
239
94𝑃𝑢
239
94𝑃𝑢
© Copyright 2014
Hydrogen 1
Hydrogen
1
1
1
0
0
Boron
Boron
Nitrogen
Nitrogen
Cadmium
Cadmium
Plutonium
5
5
5
5
5
Plutonium
5
7
48
94
7
48
94
ELO 1.2
7
48
94
7
48
94
7
7
66
145
66
145
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Forces Acting in the Nucleus
ELO 1.3 – Describe the three forces that act on particles within the
nucleus and how they affect the stability of the nucleus.
• Both protons and neutrons exist in atomic nucleus
– Some attractive force must exist to oppose the repulsive force
between protons
• Forces present in the nucleus are:
– Electrostatic forces between charged particles
– Gravitational forces between any two objects that have mass
• The magnitude of these forces can be calculated based upon
principles from classical physics
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ELO 1.3
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Gravitational Force
• Gravitational force between two bodies is directly proportional to
masses of two bodies and inversely proportional to square of
distance between bodies (Newton)
𝐹𝑔 =
𝐺𝑚2 𝑚2
𝑟2
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)
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Gravitational Force
• Greater gravitational force is because of either
– Larger masses
– Smaller distance between objects
• Distance between nucleons is extremely short
– Makes gravitational force significant
• Gravitational force between two protons separated by 10-20 meters is
about 10-24 newtons
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Electrostatic Force
• Coulomb's Law used to calculate electrostatic force between two
protons
– Electrostatic force is directly proportional to electrical charges of
two particles and inversely proportional to square of distance
between particles
𝐹𝑒 =
𝐾𝑄1 𝑄2
𝑟2
Fe = electrostatic force (newtons)
K = electrostatic constant (9.0 x 109 N-m2/C2)
Q1 = charge of first particle (coulombs)
Q2 = charge of second particle (coulombs)
r = distance between particles (meters)
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Electrostatic Force
• Electrostatic force between two protons separated by distance of 1020 meters is ≈ 1012 newtons
• Results:
– Electrostatic force – 1012 newtons
– Gravitational force – 10-24 newtons
• Gravitational force is so small that it can be neglected
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Nuclear Force
• If only electrostatic and gravitational forces exist stable nuclei
composed of protons and neutrons can’t exist
• Must be some other force at work, specifically nuclear force
• Nuclear Force – strong attractive force that is independent of charge
– Acts equally only between pairs of neutrons, pairs of protons, or a
neutron and a proton
– Has very short range - approximately equal to diameter of nucleus
(10-13 cm)
– Attractive nuclear force drops off with distance much more quickly
than electrostatic force
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Forces Acting in the Nucleus
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
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Nuclear Force
• In stable atoms, attractive and repulsive forces in nucleus balance
• If forces do not balance, the atom is unstable
– Nucleus will emit radiation in an attempt to achieve more stable
configuration
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Forces Acting in the Nucleus
Knowledge Check
Very weak attractive force between all nucleons describes which of the
forces listed below?
A. Electrostatic
B. Nuclear
C. Gravitational
D. Atomic
Correct answer is C.
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TLO 1 Summary
• Atoms consist of three basic subatomic particles:
– Proton – positive charge, same mass as a hydrogen atom, and
exist in the nucleus; +1 charge
– Neutron – no electrical charge, about the same mass as a
hydrogen atom, and exist in the nucleus; no charge
– Electron – negative charge, mass about eighteen hundred times
smaller than the mass of a hydrogen atom, and exist in orbital
shells around the nucleus; -1 charge
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TLO 1
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TLO 1 Summary
• Bohr model – dense nucleus of protons and neutrons surrounded by
electrons traveling in discrete orbits at fixed distances from nucleus
• Nuclides are atoms that contain a particular number of protons and
neutrons
• Isotopes are nuclides that have the same atomic number and are
therefore the same element, but differ in the number of neutrons
• Atomic number – number of protons in the nucleus.
• Mass number of an atom is the total number of nucleons (protons
and neutrons) in the nucleus
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TLO 1
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TLO 1 Summary
• Atomic (standard) notation is used to identify a specific nuclide
– Z = atomic number, # of protons
– A = mass number, equal to the number of nucleons
– X = chemical symbol of the element
– Number of protons = Z
– Number of electrons = Z
– Number of neutrons = A - Z
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TLO 1
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TLO 1 Summary
• Stability of nucleus is determined by different interacting forces:
– Gravitational force – long range, relatively weak attraction
between masses.
o Negligible compared to other nucleus forces
– Electrostatic force – relatively long-range, strong, repulsive force
that acts between the positively charged protons in the nucleus
– Nuclear force – relatively short-range attractive force between all
nucleons
o Balances repulsive electrostatic force in stable nucleus
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TLO 1
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Chart of the Nuclides
TLO 2 – Use the Chart of the Nuclides to obtain information on specific
nuclides.
• The Chart of the Nuclides is a format for presenting a large amount of
scientific information about nuclides in an organized manner
• Important because it provides information about the characteristics of
each elemental isotope
• Used to determined how an unstable atom becomes stable:
– Type and strength of radiation emitted
– Decay chains
– Probabilities of interactions, i.e. absorption or capture
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TLO 2
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Enabling Learning Objectives for TLO 2
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 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
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TLO 2
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Chart of the Nuclides
ELO 2.1 – List information found on the Chart of the Nuclides for isotopes
and describe how stable and radioactive isotopes are identified on the
Chart of the Nuclides.
• Chart of nuclides is a two-dimensional graph plotting
– Number of neutrons on one axis
– Number of protons on the other axis
• Each point plotted on the graph represents a nuclide
• Provides a map of the radioactive behavior of isotopes of the
elements
• Contrasts with a periodic table – maps only chemical behavior
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Chart of the Nuclides
• The Chart of the Nuclides lists stable and unstable nuclides in
addition to pertinent information about each one
• Chart plots box for each nuclide, with number of protons (Z) on
vertical axis and number of neutrons (N = A - Z) on horizontal axis
Figure: Nuclide Chart for Atomic Numbers
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Chart of the Nuclides
• Only 287 isotopes are stable or
naturally occurring
• Stable isotopes are listed in gray
boxes and includes:
– Chemical Symbol
– Number of Nucleons
– % abundance in nature
Figure: Stable Nuclide
– Isotopic mass
– Capture cross sections in
barns
– Indication if it is a fission
product
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Chart of the Nuclides
• 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) (β-,α)
– Beta disintegration energy
in MeV.
– Mass in amu when
available
– Isomeric states
– Indication if it is a fission
product
Figure: Unstable Nuclide
Figure: Line of Stability
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Chart of the Nuclides
Knowledge Check
On the Chart of the Nuclides, a stable isotope is indicated by a
.
A. white square
B. gray square
C. red square
D. black square
Correct answer is B.
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Neutron-Proton Ratio
ELO 2.2 – Describe how an element’s neutron to proton ratio affects its
stability.
• Neutron-Proton ratio (N/Z ratio or nuclear ratio) – ratio of neutrons to
protons making up the nucleus
– As mass numbers increase ratio of neutrons to protons increases
• Light elements up to calcium (Z=20), have stable isotopes with a
neutron/proton ratio of one except:
– 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 N/Z ratio under one
• Uranium-238 has the highest N/Z ratio of any natural isotope at 1.59
(92 protons and 146 neutrons)
• Lead-208 the highest N/Z ratio of any known stable isotope at 1.54
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Neutron-Proton Ratio
• A nuclide existing outside of
the band of stability can
undergo:
– Alpha decay
– Positron emission
– Electron capture, or
– Beta emission to gain
stability.
Figure: Neutron-Proton Plot of the Stable Nuclides
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Neutron-Proton Ratio
• A heavy nucleus splitting
into two fragments, each
fragment would form a
nucleus having
approximately same
neutron-to-proton ratio as
heavy nucleus
• High neutron-to-proton ratio
places fragments below and
to right of stability curve
• Successive beta emissions,
each converting a neutron to
a proton create a more
stable neutron-to-proton
ratio
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Figure: Neutron-Proton Plot of the Stable Nuclides
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Neutron-Proton Ratio
Knowledge Check
Which of the following nuclides has the higher neutron-proton ratio?
A. Cobalt-60
B. Selenium-79
C. Silver-108
D. Cesium-137
Correct answer is D.
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ELO 2.2
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Atomic Quantities
ELO 2.3 – Explain the difference between atom percent, atomic weight
and weight percent; given the atom percentages and atomic masses for
isotopes of a particular element, calculate the atomic weight of the
element.
• Isotopic calculations – determine relative amounts of isotopes in a
given quantity of an element using:
– Atom percent
– Atomic weight
– Weight percent
• Relative abundance of an isotope in nature compared to other
isotopes of same element is relatively constant
– Shown on chart of the nuclides for naturally occurring isotopes
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Atomic Quantities
Atom Percent (a/o)
• Percentage of atoms of an element that are of a particular isotope
• Example:
– If a cup of water contains 8.23 x 1024 atoms of oxygen
– Isotopic abundance of oxygen-18 is 0.20%
– There are 1.65 x 1024 atoms of oxygen-18 in the cup
Atomic Weight
• Average atomic weight of all isotopes of the element
• Calculated by summing products of isotopic abundance with atomic
mass of isotope
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Atomic Weight Calculations
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
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Calculating Atomic Weight of Lithium
Step
Action
Calculation
1.
Determine the abundance of
each isotope present (chart of
the nuclides)
2.
For each isotope determine the Lithium-6, 6.015122 amu ;
atomic mass (chart of the
Lithium-7, 7.016003 amu
nuclides)
3.
For each isotope, multiply
Li6, (.075)(6.015)= .4511 amu ;
abundance times atomic mass Li7, (.925)(7.016)= 6.4898 amu
4.
Sum the products of each
isotope calculation
© Copyright 2014
Lithium-6, 7.5%; Lithium-7,
92.5%
.4511 𝑎𝑚𝑢 + 6.4898 𝑎𝑚𝑢
= 6.9409 𝑎𝑚𝑢
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Weight Percent (w/o)
• Weight Percent (w/o) – percent weight of an element that is a
particular isotope
• Example
– If a sample of material contains 100 kg of uranium that was 28
w/o uranium-235, then 28 kg of uranium-235 is in the sample
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Atomic Quantities
Knowledge Check
Calculate the atomic weight for the element silver with the following
stable isotopes:
Ag-107, abundance 51.84%, Mass 106.905097 amu
Ag-109, abundance 48.16%, Mass 108.904752 amu
A. 107.8886 amu
B. 1078.8860 amu
C. 2907.8109 amu
D. 2.9507 amu
Correct answer is A.
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Enrichment and Depletion
ELO 2.4 – Describe the following terms: enriched uranium and
depleted uranium.
• Natural uranium contains isotopes of:
– 99.2745% of uranium-238
– 0.72% uranium-235
– 0.0055% uranium-234
• All isotopes have similar chemical properties
– But, each isotope has significantly different nuclear properties
• Uranium-235 is the desired material for use in reactors
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Enrichment and Depletion
• Enrichment (separating isotopes from natural quantities) is complex
and expensive.
• For PWRs the Uranium-235 isotope must be “enriched” from natural
uranium 238
– This results in:
o Enriched uranium – uranium having uranium-235 isotope
concentration greater than its natural value
o Depleted uranium – uranium having uranium-235 isotope
concentration less than its natural value (0.72%)
• Depleted uranium is a by-product of the uranium enrichment process
– It does have uses in nuclear field and in commercial and defense
industries
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Enrichment and Depletion
Knowledge Check
Depleted uranium will have ___________ atomic weight than natural
uranium.
A. less
B. the same amount
C. greater
D. much less
Correct answer is C.
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TLO 2 Summary
• 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) provided
• Unstable isotopes – symbol, mode of decay (example, β-,α),
disintegration energy in MeV, mass (amu) when available, and halflife provided
• A high neutron-to-proton ratio places nuclides below and to the right
of the stability curve
• Instability caused by excess neutrons is generally “fixed” by
successive beta emissions (converting a neutron to a proton)
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TLO 2 Summary
• 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 defined as the average atomic weight
of all isotopes of the element
• Weight percent w/o is the percent weight of an element that is a
particular isotope
• Enriched uranium – uranium-235 concentration greater than its
natural value of 0.72%
• Depleted uranium – uranium-235 a concentration less than its natural
value of 0.72%
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TLO 2
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Chart of the Nuclides Group Problem
Correct Answer:
••Using
Np237
decays
through
alpha
with
a ahalf-life
2.14
thedecays
Chart of
the Nuclides,
trace the
decay
chainofofof
Np-237
to the
Which
through
alphaemission,
emission,
with
half-life
46 million
minutes
years
to Pa-233
stable
isotope
of Ti-205.
to EITHER
Ti-209,
•• Which
decays
through
beta-minus
emission,
with
half-life
of 27
decaying
through
beta-minus,
with
a half-life
of a2.2
minutes
to days
Pbto
U-233
209,
OR
• Which decays through alpha emission, with a half-life of 159,200 years
• decaying through beta-minus, to Po-213, which decays with a halfto Th229
life of 4.2 million seconds to Pb-209.
• Which decays through alpha emission, with a half-life of 7430 years to
• Pb-209 decays through beta-minus emission, with a half-life of 3.25
Ra-225
minutes to Bi-209,
• Which decays through beta-minus emission, with a half-life of 15 days
19
• Which
decays
through
alpha
emission,
with
a
half-life
of
1.9
x
10
to Ac-225
Ti-205
• years
Which to
decays
through alpha emission, with a half-life of 10 days to Fr221
Neptunium
– Uranium
– alpha
Protactinium
– Thorium
– Actinium
Radium
• Which decays
through
emission,
with a 5 half-life
of 5 –minutes
to
– Francium
At-217 – Radon – Astatine – Bismuth - Thallium or Polonium –
Lead
- Thallium
• Which
decays through alpha emission, with a half-life of 32.3 seconds
to Bi-213
© Copyright 2014
TLO 2
Operator Generic Fundamentals
59
Mass Defect and Binding Energy
TLO 3 – Describe mass defect and binding energy and their relationship
to one another.
• 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
© Copyright 2014
TLO 3
Operator Generic Fundamentals
60
Enabling Learning Objectives for TLO 3
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 to the atom. Include an explanation for ionization, ionization
energy, nucleus energy and application of the nuclear energy level
diagram.
© Copyright 2014
TLO 3
Operator Generic Fundamentals
61
Mass Defect and Binding Energy
ELO 3.1 – Define mass defect and binding energy.
• Laws of conservation of mass and conservation of energy continue to
hold true on a nuclear level
– However, conversion between mass and energy occurs (E=MC2)
• A single conservation law states that the sum of mass and energy is
conserved
• A mass decrease results in an corresponding energy increase and
vice-versa
– Mass does not magically appear and disappear at random
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
62
Mass Defect and Binding Energy
Mass Defect
• Based on measurement, mass of a particular atom is always slightly
less than sum of masses of individual neutrons, protons, and
electrons
• Mass Defect (∆m) – difference between mass of atom and sum of
masses of its parts
Binding Energy
• Loss in mass, or mass defect, is due to conversion of mass to binding
energy when nucleus is formed
• Binding Energy (BE) – amount of energy that must be supplied to
nucleus to completely separate its nuclear particles (nucleons)
– Also defined as amount of energy that would be released if
nucleus was formed from separate particles
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
63
Mass Defect and Binding Energy
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
Correct answer is B.
© Copyright 2014
ELO 3.1
Operator Generic Fundamentals
64
Calculating Mass Defect and Binding
Energy
ELO 3.2 – 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.
• Mass defect and binding energy are calculated from the following
equations
• Always use the full accuracy of mass measurements because these
differences are very small compared to the atom’s mass
• Rounding off to three or four significant digits prior results in a
calculated mass defect of zero
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
65
Mass Defect Calculations
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 − 𝑚𝑎𝑡𝑜𝑚
∆m = 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 (amu)
Z = atomic number (number of protons)
A = mass number (number of nucleons)
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
66
Mass Defect Calculations
Step
Description
Action
1.
Determine the Z (atomic
number) and A (atomic
mass) of the nuclide.
2.
Determine the mass of the Multiply Z times the mass of a proton and the
mass of an electron: 𝑍(mp + me )
protons and electrons of
the nuclide.
3.
Determine the mass of
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 mass of the
nuclide and the mass of
individual components.
Subtract the mass of the atom of the nuclide:
𝑍 mp + me + 𝐴 − 𝑍 mn − matom
© Copyright 2014
Look up information in the Chart of the
Nuclides.
ELO 3.2
Operator Generic Fundamentals
67
Mass Defect Calculations
Calculate the mass defect for lithium-7 given lithium-7 = 7.016003 amu.
Step
Description
Action
1.
Determine the Z (atomic number) and 𝑍 = 3, 𝐴 = 7
A (atomic mass number) of the
nuclide.
2.
Determine the mass of the protons
and electrons of the nuclide.
3 (1.007826 𝑎𝑚𝑢
+ .000548597 𝑎𝑚𝑢)
= 3.02347979 𝑎𝑚𝑢
3.
Determine the mass of the neutrons.
(7 − 3)(1.008665)
= 4.03466 𝑎𝑚𝑢
4.
Add mass of protons, electrons, and
neutrons.
3.02347979 𝑎𝑚𝑢 + 4.03466 𝑎𝑚𝑢
= 7.058140 𝑎𝑚𝑢
5.
Determine the difference between the 7.058140 𝑎𝑚𝑢 − 7.016003 𝑎𝑚𝑢
= .042135 𝑎𝑚𝑢
atomic mass of the nuclide and the
mass of the individual components.
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
68
Binding Energy Calculation
• Binding energy is energy equivalent of the mass defect
• Calculated using a conversion factor derived from Einstein's Theory
of Relativity
• Einstein's Theory of Relativity is the famous equation relating mass
and energy
𝐸 = 𝑚𝑐2
Where:
c = velocity of light (𝑐 = 2.998 × 108
𝑚
)
sec
m = mass in amu
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
69
Binding Energy
𝐸 = 𝑚𝑐
2
1.6606 × 10−27 𝑘𝑔
= 1 𝑎𝑚𝑢
1 𝑎𝑚𝑢
= 1.4924 × 10−10 𝐽
2.998 ×
108
𝑚
𝑠𝑒𝑐
1𝑁
𝑘𝑔– 𝑚
1
𝑠𝑒𝑐 2
1𝐽
1𝑁 − 𝑚
1 𝑀𝑒𝑉
1.6022 × 10−13 𝐽
= 931.5 𝑀𝑒𝑉
Conversion Factors:
1 𝑎𝑚𝑢 = 1,6606 × 10−27 𝑘𝑔
1 𝑛𝑒𝑤𝑡𝑜𝑛 = 1 𝑘𝑔– 𝑚/sec2
1 𝑗𝑜𝑢𝑙𝑒 = 1 𝑛𝑒𝑤𝑡𝑜𝑛– 𝑚𝑒𝑡𝑒𝑟
1 𝑀𝑒𝑉 = 1.6022 × 10−13 𝑗𝑜𝑢𝑙𝑒𝑠
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
70
Binding Energy
• Since 1 amu is equivalent to 931.5 MeV of energy, binding energy
can be calculated using:
𝐵. 𝐸. = ∆𝑚
© Copyright 2014
931.5 𝑀𝑒𝑉
1 𝑎𝑚𝑢
ELO 3.2
Operator Generic Fundamentals
71
Binding Energy Calculation
Step
1.
2.
3.
Description
Determine the mass
defect of the nuclide
Use the binding
energy equation to
calculate the binding
energy.
Action
Use the equation:
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 −
𝑚𝑎𝑡𝑜𝑚
Use the Equation:
𝐵. 𝐸. = ∆𝑚
931.5 𝑀𝑒𝑉
1 𝑎𝑚𝑢
Calculate the binding Multiply delta mass from step 1 by the
energy
energy conversion to amu
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
72
Binding Energy Calculation
Step
Description
Action
1.
Determine the mass From previous calculation:
defect of the nuclide. .04135 amu.
2.
Use the binding
energy equation to
calculate the binding
energy.
3.
931.5 MeV
Calculate the binding
B. E. = 0.04135 amu
energy.
1 amu
𝐵. 𝐸. = ∆𝑚
931.5 𝑀𝑒𝑉
1 𝑎𝑚𝑢
= 38.5175 𝑀𝑒𝑉
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
73
Binding Energy Example
Calculate the mass defect and binding energy for uranium-235.
– One uranium-235 atom has a mass of 235.043924 amu.
Step 1:
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 − 𝑚𝑎𝑡𝑜𝑚
∆𝑚 = 92 1.007826 𝑎𝑚𝑢) + 235 − 92 1.008665 𝑎𝑚𝑢
− 235.043924 𝑎𝑚𝑢
∆𝑚 = 1.91517 𝑎𝑚𝑢
Step 2:
931.5 𝑀𝑒𝑉
𝐵. 𝐸. = 1.91517 𝑎𝑚𝑢
1 𝑎𝑚𝑢
𝐵. 𝐸. = 1784 𝑀𝑒𝑉
© Copyright 2014
ELO 3.2
Operator Generic Fundamentals
74
Gamma Rays and X-Rays
ELO 3.3 – Explain the difference between an x-ray and a gamma ray and
their effects to the atom. Include an explanation for ionization, ionization
energy, nucleus energy, and application of the nuclear energy level
diagram.
• Defined by their sources – identified by wavelength
– X-rays are emitted by electrons
– gamma rays are emitted from the nucleus – shorter wavelength
• They are both photons and undergo similar interactions
• Electrons that circle the nucleus move in fairly well-defined orbits
• Some of these electrons are more tightly bound in atom than others.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
75
Gamma Rays and X-Rays
• Electrons are attracted to the positive charge of the protons in the
nucleus
– As they orbit further from the nucleus this attraction weakens
– Therefore, less energy is required to remove it
• Removing an electron from its orbit is called ionization
• The energy required for ionization is called ionization energy
– Only 7.38 eV is required to remove outermost electron from lead
atom
– 88,000 eV is required to remove innermost electron
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
76
Energy Levels of Atoms
Ground State
• In a neutral atom (number of electrons = Z) it is possible for electrons
to be in a variety of different orbits, each with a different energy level
• Ground State – state of lowest energy, the atom is normally found in
Exited State
• Atom with more energy than ground state, is in an excited state
• An atom cannot stay in excited state for an indefinite period of time
X-Ray Production
• X-Ray – a discrete bundle of electromagnetic energy emitted when
an excited atom transitions to a lower-energy excited or ground state
• X-ray energy is equal to difference between energy levels of atom
– Range from several eV to 100,000 eV
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
77
Energy Levels of the Nucleus
• Nucleons in nucleus of atom, like electrons that circle nucleus, exist
in shells that correspond to energy states
– Energy shells of nucleus are less defined and less understood
than those of electrons
– There is a state of lowest energy (ground state) and discrete
possible excited states for a nucleus
– Discrete energy states for the electrons of atom are measured in
eV or keV
o Energy levels within the nucleus are considerably greater,
measured in MeV
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
78
Energy Levels of the Nucleus
Gamma Ray Production
• Nucleus in excited state will not remain for an indefinite period
• Nucleons in an excited nucleus transition towards their lowest energy
configuration and emit a discrete bundle of electromagnetic radiation
called a gamma ray
• Differences between x-rays and g-rays are
– Energy levels
– Where they originate from
o X-rays – electron shell
o Y-rays – nucleus
Nuclear Energy Level Diagram
• Nuclear energy-level diagram depicts ground and excited states
• Consists of stack of horizontal bars, one bar for each of excited
states of nucleus
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
79
Energy Levels of the Nucleus
• Vertical distance between bars
represents an excited state
with the bottom bar
representing ground state
• Difference in energy between
ground state and excited state
is called excitation energy of
excited state
• Ground state of nuclide has
zero excitation energy
• Bars for excited states are
labeled with their respective
energy levels
Figure: Energy Level Diagram – Nickel-60
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
80
Gamma Rays and X-Rays
Knowledge Check
In order for Uranium-238 to be "stable", there must be _____ electrons
orbiting the nucleus.
A. 238
B. 146
C. 235
D. 92
Correct answer is D.
© Copyright 2014
ELO 3.3
Operator Generic Fundamentals
81
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 calculated from:
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 − 𝑚𝑎𝑡𝑜𝑚
• Binding energy calculated by multiplying the mass defect by 931.5
MeV per amu from Einstein’s equation
© Copyright 2014
TLO 3
Operator Generic Fundamentals
82
TLO 3 Summary
• The differences between x-rays and gamma-rays are:
– Energy levels
– X-rays are emitted from the electron shell
– Gamma rays are emitted from the nucleus
– Wavelengths
• 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
© Copyright 2014
TLO 3
Operator Generic Fundamentals
83
Crossword Puzzle
• It’s crossword puzzle time!
© Copyright 2014
TLOs 1-3
Operator Generic Fundamentals
84
Nuclear Stability
TLO 4 – Describe the processes by which unstable nuclides achieve
stability.
• Most naturally occurring atoms are stable and do not emit particles or
energy or change state
• Some atoms are not stable
– These atoms emit radiation to achieve more stability
• Important to understand nuclear stability because of effects that
unstable nuclei undergo to become stable
• Unstable nuclides achieve stability by emitting:
– High energy photons
– High energy particles
© Copyright 2014
TLO 4
Operator Generic Fundamentals
85
Enabling Learning Objectives for TLO 4
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.
© Copyright 2014
TLO 4
Operator Generic Fundamentals
86
Conservation Principles
ELO 4.1 – Describe the conservation principles that must be observed
during radioactive decay. Include an explanation of neutrinos.
• For stable nuclides, as the mass number increases the ratio of
neutrons to protons increases
• Non-stable nuclei with an excess of neutrons undergo a
transformation process known as beta (β) decay
• A deficiency of neutrons undergo other processes such as electron
capture or positron emission
– Final nucleus is more stable as a result of the decay processes
• Some naturally occurring heavy elements, such as uranium or
thorium, and their unstable decay chain elements emit radiation
– Uranium and thorium, have an extremely slow rate of decay.
– All naturally occurring nuclides with atomic numbers greater than
82 are radioactive
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
87
Conservation Principles
Principle
Conservation of
Electric Charge
Conservation of
Mass Number
© Copyright 2014
Description
Conservation of electric charge implies that
charges are neither created nor destroyed. Single
positive and negative charges may neutralize each
other. Possible for a neutral particle to produce
one charge of each sign.
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.
ELO 4.1
Operator Generic Fundamentals
88
Conservation Principles
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.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
89
Conservation Principles
Example – Xenon-135
• It’s a radioactive isotope at an excited state
• It decays by emission of beta particle, electron or positron resulting in
a neutron converting to a proton
– Illustrates the conservation of mass and energy
• The beta ejected from the nucleus 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 neutron
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
90
Conservation Principles
Example – Xenon-135
Xenon-135 Decay
134.90720 amu
Resultant Isotope and Energy
Cesium-135
54 protons
134.905977 amu
81 neutrons
55 protons
81 neutrons
∆Mass = .001235 amu
Mass is accounted for in the beta particle
and energy of the gammas emitted.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
91
Conservation Principles
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
Correct answer is B.
© Copyright 2014
ELO 4.1
Operator Generic Fundamentals
92
Decay Processes
ELO 4.2 – Describe the following radioactive decay processes: alpha
decay, beta-minus decay, beta-plus decay, electron capture, gamma ray
emission, internal conversions, isomeric transitions, and neutron
emission.
• 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
– These are explained in this section
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
93
Alpha Decay (α)
• Emission of alpha particles (helium nuclei)
which may be represented as:
4
2𝐻𝑒
or 42𝛼
Figure: Alpha Decay
• When an unstable nucleus ejects an alpha particle, atomic
number is reduced by 2 and mass number decreased by 4
• Uranium-234 decays by ejection of an alpha particle accompanied by
emission of 0.068 MeV gamma
234
92𝑈
© Copyright 2014
→
230
90𝑇ℎ
+ 42𝛼 + 𝛾 + 𝐾𝐸
ELO 4.2
Operator Generic Fundamentals
94
Alpha Decay (α)
• Combined KE of daughter nucleus (Thorium-230) and α particle is
designated as “KE” in equation
• Sum of KE and gamma energy is equal to difference in mass
between original nucleus and final particles
– Equivalent to the binding energy released, since Δm = BE
• Alpha particle will carry off as much as 98% of KE
234
92𝑈
© Copyright 2014
→
230
90𝑇ℎ
+ 42𝛼 + 𝛾 + 𝐾𝐸
ELO 4.2
Operator Generic Fundamentals
95
Beta Decay (β)
• Emission of electrons of nuclear rather than orbital origin
– These particles are electrons that have been expelled by excited
nuclei
– May have a charge of either sign
o β-
o β+
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
96
Beta Minus Decay
• Negative electron emission, converts neutron to proton, increasing
atomic number by one and leaving mass number unchanged
• Common mode of decay for nuclei with excess of neutrons, such as
fission fragments below and to right of neutron-proton stability curve
239
93𝑁𝑝
→
239
94𝑃𝑢
+ −10𝛽 + 00𝑣
• The symbol on the end represents an anti-neutrino
Figure: Beta Minus Decay
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
97
Beta Plus Decay
• Positively charged electrons (beta-plus) are known as positrons
0
+1𝑒
𝑜𝑟 +10𝛽
𝑒 + 𝑜𝑟 𝛽+
– Except for the charge they are nearly identical to their negative
Betas.
• Positron emission decreases the atomic number by one but leaves
the mass number unchanged by changing a proton to a neutron
13
7𝑁
© Copyright 2014
→
13
6𝐶
+ +10𝛽 + 00𝑣
ELO 4.2
Operator Generic Fundamentals
98
Electron Capture (EC, K-capture)
• Nuclei having excess protons may capture an inner orbit electron that
immediately combines with a proton to form a neutron
• The electron is normally captured from the innermost orbit (K-shell),
therefore, this process is also called K-capture
7
4𝐵𝑒
+ −10𝑒 → 73𝐿𝑖 + 00𝑣
• As with beta decays a neutrino is formed, its energy conserving
momentum, photons are given off:
– Atomic mass of the product being appreciably less than the
parent – gamma energy
– Characteristic x-rays are given off when an electron from another
shell fills the vacancy in the K-shell
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
99
Electron Capture
Figure: Electron Capture or K-Capture
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
100
Electron Capture vs. Positron Emission
• Electron capture and positron emission exist as competing processes
– They produce the same daughter product
• For positron emission to occur, the mass of the daughter product
must at least two electrons less than the mass of the parent
– This accounts for the ejected positron and that the daughter has
one less electron than the parent
– If these requirements are not met, then electron capture occurs
and positron emission does not
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
101
Gamma Emission
• High-energy electromagnetic radiation originating in the nucleus
• It is emitted in the form of photons:
– Discrete bundles of energy having both wave and particle
properties
• A daughter nuclide from decay often remains in an excited state
– Resolved by the nucleus dropping to the ground state by the
emission of gamma radiation
• Gamma rays are very penetrating, often requiring several inches of
metal or a couple of feet of concrete to stop (shield)
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
102
Internal Conversion
• Normally an excited nucleus goes from the excited state to the
ground state is by emission of a gamma ray
• In some cases the gamma ray released interacts with one of the
innermost orbital electrons
– Transfers the gamma’s energy to the electron
• This is 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
© Copyright 2014
ELO 4.2
Operator Generic Fundamentals
103
Isomeric Transition
• Nuclear isomer – a nucleus in an excited state, differs in energy and
behavior from other nuclei with identical atomic and mass numbers
• Isomeric transition – when the excited nuclear isomer drops to a
lower energy level
– Commonly occurs immediately after particle emission
– May remain in an excited state for a measurable period of time
before dropping to ground state
• It is also possible for the excited isomer to decay by alternate means
– An example of delayed gamma emission accompanying Beta
emission is illustrated by the decay of nitrogen-16.
16
7𝑁
© Copyright 2014
→
16
8𝑂
16
8𝑂
→
+ 01𝛽 + 00𝛾
16
8𝑂
ELO 4.2
+ 00𝛾
Operator Generic Fundamentals
104
Neutron Emission
• Non-stable nuclei may also emit neutrons (n) in order to become
more stable
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
– Capable of penetrating many materials
• Neutron production and interaction with matter is of great importance
in nuclear physics and will be discussed in greater detail later
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Neutrinos
• Neutrino (𝝂) – emitted with positive electron emission
• Antineutrino (𝝂) – emitted with negative electron (Beta decay)
• Pass through all materials with so few interactions that energy they
possess cannot be recovered
• Neutrinos and antineutrinos carry portion of KE that would otherwise
belong to beta particle
– Considered for energy and momentum to be conserved
• Not significant in context of nuclear reactor applications
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Operator Generic Fundamentals
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Radioactivity Decay
Knowledge Check
Which of the following statements accurately describes alpha 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.
Correct answer is C.
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Operator Generic Fundamentals
107
Stability Curve
ELO 4.3 – 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.
• Radioactive nuclides decay
in a way that results in a
daughter nuclide with a
neutron-proton ratio closer
to the line of stability on the
Chart of the Nuclides
– Helps to predict the type
of decay that a nuclide
undergoes based on its
location relative to the
line of stability
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Figure: Stability Curve
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Predicting Type of Decay
• The line of stability illustrates the method of decay nuclides in
different regions of the chart of nuclides are likely to undergo
– Nuclides below and to the right of the line of stability usually
undergo β- decay
– Nuclides above and to the left of the line of stability usually
undergo either β+ decay or electron capture.
– Nuclides likely to undergo alpha (α) decay are found in the upper
right hand region
– Some exceptions apply especially in the region of heavy nuclides
• Stable isotopes are isotopes that are not radioactive
– They do not decay spontaneously
• Stable isotopes are found on the line of stability
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Predicting Type of Decay
Figure: Types of Radioactive Decay Relative to Line of Stability
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Operator Generic Fundamentals
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Predicting Type of Decay – Examples
• Of the known elements 80 have at least one stable nuclide
– These are the first 82 elements from hydrogen to lead
– Exceptions are technetium-43 and promethium-61
• There are a known total of 254 stable nuclides
• Stable, in this case, means a nuclide that has not been observed to
decay against the natural background
– These elements have half-lives too long to be measured
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Operator Generic Fundamentals
111
Stability Curve
Knowledge Check
Match the 4 areas on the curve with the correct description:
A. Alpha (α) decay
B. Line of Stability
C. β+ decay or electron capture
D. β- decay
Correct answers:
1. C – Beta +
2. D – beta 3. A – alpha
4. B – line of stability
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Operator Generic Fundamentals
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Decay Chains
ELO 4.4 – Given a Chart of the Nuclides, describe the radioactive decay
chain for a nuclide.
• When an unstable nucleus decays the resulting daughter may not be
stable
– Nucleus resulting from decay of parent is often itself unstable, and
will undergo an additional decay(s)
o Common among the larger nuclides
• Steps of an unstable atom can be traced as it goes through multiple
decays trying to achieve stability
• List of original unstable nuclide, nuclides involved as intermediate
steps in decay, and final stable nuclide is known as decay chain
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Decay Chains
• Common method for stating decay chain is to state each nuclide
involved in standard format
• Arrows used between nuclides to indicate where decays occur, with
type of decay indicated above arrow and half-life below arrow
• Example: decay chain of U-238:
– U-238 decays, through alpha-emission, with a half-life of 4.5
billion years to thorium-234
– Decays through beta-emission, with a half-life of 24 days to
protactinium-234
– Decays through beta-emission, with a half-life of 1.2 minutes to
uranium-234
– Decays through alpha-emission, with a half-life of 240 thousand
years to thorium-230
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Decay Chains
Example decay chain of U-238: Cont’d
• Thorium-230 decays, through alpha-emission, with a half-life of 77
thousand years to radium-226
• Decays through alpha-emission, with a half-life of 1.6 thousand years
to radon-222
• Decays through alpha-emission, with a half-life of 3.8 days to
polonium-218
• Decays through alpha-emission, with a half-life of 3.1 minutes to
lead-214
• Decays through beta-emission, with a half-life of 27 minutes to
bismuth-214
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Operator Generic Fundamentals
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Decay Chains
Example decay chain of U-238: Cont’d
• Bismuth-214 decays through beta-emission, with a half-life of 20
minutes to polonium-214
• Decays through alpha-emission, with a half-life of 160 microseconds
to lead-210
• Decays through beta-emission, with a half-life of 22 years to bismuth210
• Decays through beta-emission, with a half-life of 5 days to polonium210
• Decays through alpha-emission, with a half-life of 140 days to lead206, which is a stable nuclide.
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Decay Chains – Example
Use chart of the nuclides and write decay chains for rubidium-91 and
actinium-215
– Continue chains until stable nuclide or nuclide with half-life
greater than 1 x 106 years is reached
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
→
83
81
82
85
0.10 𝑚𝑠
2.14 𝑚𝑖𝑛
4.77 𝑚𝑖𝑛
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TLO 4 Summary
• Conservation principles observed during radioactive decay:
– Electric charge – charges are neither created nor destroyed
– Mass number – no net change in number of nucleons
– Mass and energy – total of KE and BE equivalent of mass in a
system is conserved in all decays and reactions
– Momentum – distribution of 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 has:
– Atomic number 2 less than parent nuclide
– Mass number 4 less than parent nuclide
– Daughter releases its excitation energy by gamma emission
• Beta-minus decay converts neutron to proton with an ejected electron
– Daughter nuclide – atomic number increased by 1, same mass
number
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TLO 4 Summary
• Beta-plus decay converts a proton to a neutron with a positron
ejected
– Daughter nuclide has its atomic number decreased by 1, same
mass number
• 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 electrons of the same atom to eject it from the atom
• An isomeric transition – is 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
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TLO 4 Summary
• Neutrinos - uncharged particles that have very weak interaction with
matter, no mass, and travel at the speed of light
• Many modes of radioactive decay result in a daughter that has
energy level above ground state
– Excitation energy is usually released in the form of a gamma ray
• 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 found by tracing the steps an unstable atom goes
through as it tries to achieve stability
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Emitted Radiation
TLO 5 – Describe how radiation emitted by an unstable nuclide interacts
with matter and materials typically used to shield against this radiation.
• Radiation is comprised of photons (energy waves) and particles that
originate in either the nucleus or electron shells of atoms
• Photons and radiation particles have energy and interact with matter,
transferring their energy
• Radiation reactions with matter are dependent on photon and
particle:
– Mass
– Energy
• It is important to understand the physical properties of the radiation to
understand the potential hazards and methods for protection
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Enabling Learning Objectives for TLO 5
1. Describe the difference between charged and uncharged particle
interaction with matter. Include an explanation of specific ionization.
2. Describe radioactive interactions of the following types with matter:
a. Alpha Particle
b. Beta Particle
c. Gamma
d. Positron
e. 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
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Charged Versus Uncharged Particles
ELO 5.1 – Describe the difference between charged and uncharged
particle interaction with matter. Include an explanation of specific ionization.
• Interactions with matter vary with the different types of radiation
– Large, massive, charged alpha particles have very limited
penetration capabilities
– Neutrinos, the other extreme, have a very low probability of
interacting matter, a large penetrating capability
Charged vs. Uncharged Particles
• Radiation is classified into two groups, charged and uncharged
• Charged particles directly ionize the media through which they pass
• Uncharged particles and photons only cause ionization indirectly or
by secondary radiation
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Charged Versus Uncharged Particles
Charged Particle Interaction
• Charged particles have surrounding electrical fields that interact with
the atomic structure of the material they are traveling through
• Slows the particle and accelerates electrons in the medium’s atoms
• The accelerated electrons acquire enough energy to escape from
their parent atoms causing ionization of the affected atom
Uncharged Particle Interaction
• Uncharged moving particles have no electrical field
– Only lose energy and cause ionization by collisions or scattering
• Photon can lose energy by:
– Photoelectric effect
– Compton Scattering
– Pair production
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Specific Ionization
• Ionizing radiation creates ion-pairs (+ and – charged)
• Specific ionization is defined as:
– Number of ion-pairs formed per centimeter travel in given material
• Specific ionization, the measure of radiation’s ionization power is:
– Roughly proportional to the particle's mass
– Square of its charge
𝑚𝑧 2
𝐼=
𝐾. 𝐸.
Where:
I = ionizing power
m = mass of particle
z = number of unit charges particle carries
K.E. = kinetic energy of particle
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Specific Ionization
• Since mass for α particle is ≈ 7300 times as large as m for β particle,
and z is twice as great (+2) an α will produce much more ionization
per unit path length than β of same energy
– Larger alpha particle moves slower for given energy and thus acts
on given electron for a longer time
𝑚𝑧 2
𝐼=
𝐾. 𝐸.
Where:
I = ionizing power
m = mass of particle
z = number of unit charges particle carries
K.E. = kinetic energy of particle
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Charged Versus Uncharged Particles
Knowledge Check
Charged particles
ionize the media they pass through.
A. directly
B. indirectly
C. never
D. Sometimes
Correct answer is A.
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127
Radioactive Interaction
ELO 5.2 – Describe radioactive interactions of the following types with
matter: alpha particle, beta particle, gamma, positron, and neutron.
• How radiation reacts with matter is dependent on the type of radiation
• The following types of radiation interact with matter in a specific
predictable manner:
– Alpha particle
– Beta particle
– Gamma
– Positron
– Neutron
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Alpha Radiation
Origin
Interaction
• Produced from the radioactive
decay of heavy nuclides and
certain nuclear reactions
• Removes electrons from the
atoms it passes near
• Consists of 2 neutrons and 2
protons, same as a helium atom
• With no electrons, the alpha
particle has a charge of +2
• This strong positive charge
strips electrons from the orbits
of atoms
• Has a high specific ionization
© Copyright 2014
• Removal of electrons requires
energy
• Alpha’s energy is reduced by
each reaction
• Ultimately, the alpha particle
expends its KE, gains 2 electrons,
and becomes a helium atom
• Low penetration power
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Beta Minus Interaction
Origin
• A beta-minus particle is an
electron that has been ejected
at a high velocity from an
unstable nucleus
• Electrons have a small mass
and an electrical charge of -1
• Beta particles cause ionization
by displacing electrons from
atomic orbits
© Copyright 2014
Interaction
• Beta-minus ionization occurs from
collisions with orbiting electrons.
• Each collision removes KE from
the beta particle, causing it to slow
down
• After a few collisions the beta
particle is slowed enough to allow
it to be captured as an orbiting
electron in an atom
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Positron Radiation
Origin
• Positively charged
electrons
• Identical to betaminus particles and
interact with matter
similarly
Interaction
• Positrons are very short-lived and quickly
annihilate via interactions with negatively
charged electrons
• Produces two gammas with energy equal to
the rest mass of the electrons (1.02 MeV)
• These gammas interact with matter via
photoelectric effect, Compton scattering or
pair-production,
0.000549 𝑎𝑚𝑢
2 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
= 1.02 𝑀𝑒𝑉
© Copyright 2014
ELO 5.2
931.5 𝑀𝑒𝑉
𝑎𝑚𝑢
Operator Generic Fundamentals
131
Neutron Interactions
Origin
Interaction
• No electrical charge
• Neutrons are attenuated (reduced
in energy and numbers) by three
major interactions:
• Same mass as a proton
• 1835 times more mass than
electron
̶ Elastic scatter
̶ Inelastic scatter
• 1/4 mass of an alpha particle
̶ Absorption
• Come primarily from nuclear
reactions, such as fission, but
also from decay of radioactive
nuclides
• With no charge, the neutron has
a high penetrating power
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Neutron Interactions
Elastic Scatter
• Neutron collides with a nucleus and bounces off
• Some of the neutron’s KE is transferred to the nucleus
– Results in the neutron slowed and atom gaining KE
• Often referred to as the billiard ball effect
Inelastic Scatter
• Same neutron/nucleus collision occurs as in elastic scatter
• Nucleus receives some internal energy as well as KE
– Slows the neutron and leaves the nucleus in an excited state
• Nucleus decays to original energy level and usually emits a gamma
ray
• The gamma ray goes on to interact with matter via photoelectric
effect, compton scattering or pair-production, as described later
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Neutron Interactions
Absorption
• The neutron is absorbed and captured into the nucleus of an atom
– Atom is left in an excited state
• Called radiative capture if the nucleus emits one or more gamma rays
to reach a stable level
– More probable at lower energy levels
– Gammas rays interact via the photoelectric effect, compton
scattering or pair-production
• May also result in nuclear fission splitting the atom into two smaller
atoms, a couple of neutrons and gamma rays
– Fission fragments may create additional neutrons or gamma
radiation as they decay to stability
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Gamma Interactions
Gamma Radiation
• Electromagnetic radiation - similar to x-ray
o Produced by decay of excited nuclei and by nuclear reactions
• Has no mass and no charge
– Difficult to stop and has very high penetrating power
– Requires several feet of concrete, several meters of water or a
few inches of lead to shield
• Three methods of attenuating gamma rays:
– Photo-electric effect
– Compton scattering
– Pair-production
© Copyright 2014
ELO 5.2
Operator Generic Fundamentals
Gamma Interactions
Photo-Electric Effect
• Occurs when a low energy
gamma strikes an orbital
electron
• Total energy of the gamma is
expended in ejecting the
electron from its orbit
• Atom is ionized and a high
energy electron is ejected
• Rarely occurs with gammas
having energy above 1 MeV
• Energy in excess of BE of
electron is carried off by
electron in form of KE
© Copyright 2014
Figure: Photoelectric Effect
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Gamma Interactions
Compton Scattering
• An elastic collision between an
electron and a photon.
• Photon has more energy than
required to eject the electron from
orbit, or is unable to give up all of its
energy with an electron.
• Since all gamma energy is not
transferred, the photon must scatter.
• Scattered photon has less energy.
• Result is ionization of the atom, a
high energy beta, and a reduced
energy gamma.
Figure: Compton Scattering
• Most predominant with gamma
energy level of 1.0 to 2.0 MeV.
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Gamma Interactions
Pair Production
• At higher energy levels the most likely gamma interaction
• A high energy gamma passes close to a heavy nucleus and
disappears into an electron and positron
• This transformation must take place near a particle, such as a
nucleus, to conserve momentum
• KE of the recoiling nucleus is
very small; all of the photon’s
energy in excess 1.02 MeV
appears as KE of the pair
produced
‒ The original gamma must
have at least 1.02 MeV
Figure: Pair Production
energy
• Electron and positron may
collide and annihilate each other
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Operator Generic Fundamentals
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Radioactive Interactions
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
Correct answer is C.
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Operator Generic Fundamentals
139
Shielding
ELO 5.3 – Describe the type of material that can be used to stop (shield)
the following types of radiation: alpha particle, beta-particle, neutron, and
gamma ray.
• Shielding describes material placed around a radiation source used
to attenuate the radiation level
– Effectiveness is dependent on the material used and the type of
radiation
• Attenuation is the gradual loss in intensity of any kind of radiation flux
through a medium
• Examples:
– Sunlight is attenuated by dark glasses
– X-rays are attenuated by lead
– Neutrons are attenuated by water
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Shielding Properties
Type of Radiation
Shielding Material
• With a strong positive charge and large mass,
the alpha particle deposits a large amount of
energy in a short distance.
Alpha
• Loses energy quickly and therefore has very
limited penetrating power.
• Alpha particles are stopped in a few
centimeters of air or a sheet of paper.
• More penetrating than alpha but still relatively
easy to stop
Beta Particle
• Low power of penetration.
• The most energetic beta radiation is stopped
by thin metal.
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Shielding Properties
Type of Radiation
Shielding Material
• With no charge is difficult to stop – has high
penetrating power.
• Attenuated by:
̶ Elastic scattering
̶ Inelastic scattering
Neutron
̶ Absorption
• Most effective shield is a with similar mass
(neutron) for elastic scattering.
̶ Hydrogenous material such as water
attenuates neutrons effectively.
• 12 inches of water is an effective shield.
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Shielding Properties
Type of Radiation
Shielding Material
• With no mass and no charge, difficult to stop –
very high penetrating power.
Gamma Ray
© Copyright 2014
• Heavy nuclei (i.e. lead) provide large targets
for gamma interactions (3 types).
• Although dependent on gamma energies,
several meters of concrete or water or a few
inches of lead are effective shielding
materials.
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Shielding Properties
Alpha, Beta, and Gamma Shielding
• Shielding thickness are
referred to as 1/2
thickness or 1/10th
thickness.
• These thicknesses are
the amount of material
required to reduce the
original radiation field
strength to 1/2 or 1/10th
respectively
• For example:
– The 1/2 thickness of
lead for gammas is
0.4"
© Copyright 2014
Figure: Effects Various Materials Have on Types of Radiation
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Shielding Properties
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
Correct answer is A.
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145
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 its 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.
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TLO 5 Summary
• Positrons (Beta-plus) interact with matter similar to beta-minus
particles.
– After loosing most of its energy, the positron is annihilated by
interaction with an electron.
– The electron-positron pair disappears and is replaced by two
gammas. Each with the BE equivalent of mass of an electron
(0.51 MeV). Total 1.02 MeV.
• Neutrons interact with matter by:
– Elastic scattering
– Inelastic scattering
– Absorption
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TLO 5 Summary
• Gammas interact with matter:
– Photoelectric effect - interaction with an electron. The entire
energy of the gamma is transferred to the electron, ejecting the
electron.
– Compton scattering - only part of the gamma energy is
transferred to the electron. Electron is ejected from its orbit, and
the gamma is scattered at a lower energy.
– Pair-production, gamma interacts with the electrical field of a
nucleus and is converted into an electron-positron pair. The
gamma must have an energy greater than 1.02 MeV.
• Shielding material:
– Alpha particle – sheet of paper
– Beta (+ or -) – thin sheet of metal
– Neutrons – hydrogenous materials such as water
– Gamma rays – Several meters of concrete or water or a few
inches of lead.
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Radioactive Decay
TLO 6 – Describe radioactive decay terms and calculate activity levels,
half-lives, decay constants and radioactive equilibrium.
• A decay rate of sample of radioactive material is not constant
• As individual atoms of the material decay, there are fewer of them
remaining
• Rate of decay is directly proportional to the number of atoms
– The rate of decay decreases as the number of atoms decreases
• Knowledge of radioactive decay is important for calculating reactivity
poisons in the reactor as well as understanding personnel hazards
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Enabling Learning Objectives for TLO 6
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
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Operator Generic Fundamentals
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Radioactive Decay Terms
ELO 6.1 – Describe the following radioactive terms: radioactivity,
radioactive decay constant, activity, Curie, Becquerel, and radioactive
half-life.
• To understand radioactive decay, knowledge of the terms used to
describing decay rate relationships is important
• The following terms are described:
a. Radioactivity
b. Radioactive decay constant
c.
Activity
d. Curie
e. Becquerel
f.
© Copyright 2014
Radioactive half-life
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Radioactive Decay Terms
Radioactivity
• The process of certain nuclides spontaneously emitting particles or
gamma radiation
• Occurs randomly – cannot be predicted
• However, the average behavior of a large sample can be accurately
determined using statistical methods
Radioactive Decay Constant (λ)
• In a given time interval a specific fraction of a given nuclei in a
sample will decay
• Probability per unit time that an atom of a specific nuclide will decay
is - the radioactive decay constant, λ (lambda)
• Units are inverse times such as 1/second, 1/minute, 1/hour, or 1/year
– Expressed as second-1, minute-1, hour-1, and year-1
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Radioactive Decay Terms
Activity
• Activity (A) is the rate of decay of a sample
• Measured by the number of disintegrations occurring per second
• A sample containing millions of atoms, activity is the product of the
decay constant (λ) and number of atoms present in the sample (N)
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
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Radioactive Decay Terms
Radioactive Half-Life
• Commonly used term
• Estimates how quickly a nuclide is decaying
• Defined as the amount of time required for activity to decrease to
one-half of its original value
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Units of Measurement for Radioactivity
Two common units to measure activity are:
Curie
• Ci, US Measurement
• Measures the rate of radioactive decay - activity
• Equal to 3.7 x 1010 disintegrations per second
• Approximately equivalent to the number of disintegrations that one
gram of radium-226 will undergo in one second
Becquerel
• Bq, Metric System
• A Becquerel is equal to one (1) disintegration per second.
• 1 𝐶𝑢𝑟𝑖𝑒 = 3.7 × 1010 𝐵𝑒𝑐𝑞𝑢𝑒𝑟𝑒𝑙𝑠
© Copyright 2014
ELO 6.1
Operator Generic Fundamentals
155
Radioactive Decay Terms
Knowledge Check
Match the following:
1. The decay of unstable atoms by the emission of particles and
electromagnetic radiation.
2. Unit of radioactivity equal to 3.7 x 1010 disintegrations per second.
3. Unit of radioactivity equal to 1 disintegration per second.
4. Probability per unit time that an atom will decay.
A.
B.
C.
D.
© Copyright 2014
Curie
Radioactivity
Becquerel
Radioactive Decay Constant
ELO 6.1
Correct answers:
1. B – Radioactivity
2. A – Curie
3. C – Becquerel
4. D – Radioactive decay
constant
Operator Generic Fundamentals
156
Convert Between Half-life and Decay
Constant
ELO 6.2 – Convert between the half-life and decay constant for a nuclide.
• With the decay constant or half-life known, calculations can be
performed to determine such things as number of atoms and activity
level.
• The relationship between half-life and the decay constant is
developed from the equation:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
• Half-life is calculated by solving the equation for time, t, when the
current activity, A, equals one-half the initial activity Ao.
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
157
Convert Between Half-life and Decay
Constant
• A relationship between
half-life and decay
constant can be
developed:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
• If A is equal to one-half
of Ao, then A/Ao is equal
to one-half
• Substituting this in the
equation yields an
expression for t1/2
𝐴
= 𝑒 −𝜆𝑡
𝐴𝑜
𝑡1 =
2
𝐴
ln
= −𝜆𝑡
𝐴𝑜
− ln
𝑡=
© Copyright 2014
− ln
1
2
𝜆
ln 2
𝑡1 =
𝜆
2
𝐴
𝐴𝑜
0.693
𝑡1 =
𝜆
2
𝜆
ELO 6.2
Operator Generic Fundamentals
158
Convert Between Half-life and Decay
Constant
Step
1.
Action
Solution
Determine the half-life if decay Use equation:
constant is known.
0.693
𝑡1 =
𝜆
2
2.
Determine the decay constant
if half- life is known
Use equation:
0.693
𝜆=
𝑡1
2
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
159
Example: Determine ‫ ג‬of Cesium-136 –
t1/2 = 13.16 days
Step
Action
1.
Determine the halflife if decay constant
is known.
2.
Method
Calculation
Use equation:
Half life given as
13.16 days.
0.693
𝑡1 =
𝜆
2
Determine the decay Use equation:
constant if half-life is
0.693
known
𝜆=
𝑡1
𝜆=
0.693
13.16 𝑑𝑎𝑦𝑠
𝜆 = 0.0527−1 𝑑𝑎𝑦𝑠
2
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
160
Example: Determine half-life of Potassium-44
‫ = ג‬.0313 -min
Step
Action
1.
Determine the halflife if decay constant
is known.
2.
Method
Use equation:
0.693
𝑡1 =
𝜆
2
Determine the decay Use equation:
constant if half-life is
known
0.693
𝜆=
Calculation
0.693
𝑡1 =
0.03131−𝑚𝑖𝑛
2
𝑡1 = 22.13 𝑚𝑖𝑛
2
Given
𝑡1
2
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
161
Radioactive Half-Life
• Initial number of atoms
No, population
• Activity decrease by
one-half per unit time
(half-life)
• Additional half of
decreases occur
whenever one half-life
time elapses
Figure: Radioactive Decay as a Function of
Time in Units of Half-Life
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
162
Radioactive Half-Life
• After five half-lives, only
1/32, or 3.1%, of original
number of atoms (or
activity) remains
• After seven half-lives, only
1/128, or 0.78%, remains
• Number of atoms existing
after 5 to 7 half-lives is
usually assumed to be
negligible
Figure: Radioactive Decay as a Function of
Time in Units of Half-Life
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
163
Half-life and Decay Constants
Knowledge Check
What is the decay constant for Plutonium-239, which has a half-life of
24,110 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
Correct answer is A.
© Copyright 2014
ELO 6.2
Operator Generic Fundamentals
164
Calculating Activity Over Time
ELO 6.3 – Given the nuclide, number of atoms, half-life or decay
constant, determine current and future activity levels.
• The relationship between activity (N), the number of atoms present
and the decay constant is necessary to understanding the behavior of
radioactive decay
• This section explains how the amount of radiation that a sample of
material has (potential) following a specified time period is estimated
• Decay rate for a given decay constant in a radionuclide sample is
stated in this equation:
𝐴 = 𝜆𝑁
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
165
Calculating Activity Over Time
• The following expressions (derived) are used to calculate the change
in 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
N0 = number of atoms initially present
λ = decay constant (time-1)
t = time
• Since activity and number of atoms are always proportional, they may
be used interchangeably to describe any given radionuclide
population:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
166
Calculating Activity
Step
1.
2.
Action
Solution
Determine the number
of atoms present in the
mass of the isotope
Use the following equation:
Determine the decay
constant
Use the following equation:
𝑁 = 𝑚𝑎𝑠𝑠
1 𝑚𝑜𝑙𝑒
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠
𝜆=
𝑁𝐴
1 𝑚𝑜𝑙𝑒
0.693
𝑡1
2
3.
© Copyright 2014
Determine the activity
Use the following equation:
𝐴 = 𝜆𝑁
ELO 6.3
Operator Generic Fundamentals
167
Activity Calculation Example
A sample of material contains 20 micrograms of californium-252. Halflife of 2.638 years.
• Calculate:
– The number of californium-252 atoms initially present
– The activity of the californium-252 in curies
Step
Action
1.
Determine
the number
of atoms
present in
the
isotope’s
mass
© Copyright 2014
Method
Calculation
Use the equation:
𝑁
= 𝑚𝑎𝑠𝑠
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
1 𝑚𝑜𝑙𝑒
6.022 × 1023 𝑎𝑡𝑜𝑚𝑠
= 20 × 10−6 𝑔
252.08 𝑔
1 𝑚𝑜𝑙𝑒
16
= 4.78 × 10 𝑎𝑡𝑜𝑚𝑠
𝑁𝛼−252 = 𝑚𝑎𝑠𝑠
1 𝑚𝑜𝑙𝑒
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠
𝑁𝐴
1 𝑚𝑜𝑙𝑒
ELO 6.3
Operator Generic Fundamentals
168
Activity Calculation Example
Step
Action
2.
Determine
the decay
constant
3.
Method
Calculation
Use the following
equation:
𝑡1 =
2
=
0.693
𝑡1 =
𝜆
2
0.693
𝜆
0.693
2.638 𝑦𝑒𝑎𝑟𝑠
= 0.263 𝑦𝑒𝑎𝑟 −1
Determine Use the equation: 𝐴 = 𝜆𝑁
the activity 𝐴 = 𝜆𝑁
= 0.263 𝑦𝑒𝑎𝑟 −1 4.78 × 1016 𝑎𝑡𝑜𝑚𝑠
= 3.98 × 108
𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
1 𝑦𝑒𝑎𝑟
365.25 𝑑𝑎𝑦𝑠
3.7 × 1010
1 𝑑𝑎𝑦
24 ℎ𝑜𝑢𝑟𝑠
1 ℎ𝑜𝑢𝑟
3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
1 𝑐𝑢𝑟𝑖𝑒
𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠
𝑠𝑒𝑐𝑜𝑛𝑑
= 0.0108 𝑐𝑢𝑟𝑖𝑒𝑠
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
169
Variation of Radioactivity Over Time
Use the following formula (or derivations) to predict the activity level of
a quantity of an isotope:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
Where:
A = Activity at time t
Ao = Activity initially present
λ = decay constant
t = time
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
170
Variation of Radioactivity Over Time
Step
1.
2.
Action
Equation
If the initial activity is not
Use this equation:
known determine the
1 𝑚𝑜𝑙𝑒
number of atoms present in 𝑁 = 𝑚𝑎𝑠𝑠
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠
the mass of the isotope
Determine the decay
constant if necessary
Use this equation:
𝜆=
𝑁𝐴
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
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
171
Variation of Activity Over Time Calculation
Example
A sample of material contains 20 micrograms of californium-252 with an
activity of 0.0108 curies. Half-life of 2.638 years.
• Calculate:
– The number of californium-252 atoms that will remain in 12 years.
Step
1.
© Copyright 2014
Action
Equation
If the initial activity is
Use the following equation:
unknown, determine the
number of atoms
present in the mass of
1 𝑚𝑜𝑙𝑒
𝑁
=
𝑚𝑎𝑠𝑠
the isotope.
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠
ELO 6.3
𝑁𝐴
1 𝑚𝑜𝑙𝑒
Operator Generic Fundamentals
172
Variation of Activity Over Time Calculation
Step
2.
Action
Equation
Determine the
Use this equation:
decay constant if
0.693
necessary
𝜆=
𝑡1
2
3.
4.
Determine the
initial activity
Use the following
equation:
𝐴 = 𝜆𝑁
Determine the
new activity
Use the following
equation:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
© Copyright 2014
ELO 6.3
Solution
𝜆=
0.693
𝑡1
2
=
0.693
2.638 𝑦𝑒𝑎𝑟𝑠
= 0.263 𝑦𝑒𝑎𝑟 −1
Given: 0.0108 curies
𝐴 = 0.0108 𝑒
−
0.263
12 𝑦𝑟
𝑦𝑟
= 0.00046 𝑐𝑢𝑟𝑖𝑒𝑠
Operator Generic Fundamentals
173
Plotting Radioactive Decay
Step
Action
Method
1.
Calculate the decay
constant of the isotope
Use the equation:
2.
Use the decay constant to
calculate the activity at
various times
Use the equation:
3.
Develop a table of values
from the calculations
performed above
Use above equations
4.
Plot the points from the table Using the correct graph paper, plot
on linear and semi log
the points.
scales
© Copyright 2014
0.693
𝑡1 =
𝜆
2
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
ELO 6.3
Operator Generic Fundamentals
174
Plotting Radioactive Decay
• Useful to plot activity of nuclide as it changes over time
– Used to determine when activity will fall below certain level
– Usually done showing activity on either linear or logarithmic scale
• Decay of activity of single nuclide on logarithmic scale will plot as
straight line because decay is exponential
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
175
Plotting Radioactive Decay – Example
Demonstration
• Plot radioactive decay curve for nitrogen-16 over a period of 100
seconds
• Initial activity is 142 curies and half-life of nitrogen-16 is 7.13 seconds
• Plot curve on both linear rectangular coordinates and on semi-log
scale
0.693
𝑡1 =
𝜆
2
First, calculate λ
0.693
corresponding
𝜆=
𝑡1
to half-life of
2
7.13 seconds
0.693
𝜆=
7.13 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝜆 = 0.0972 𝑠𝑒𝑐𝑜𝑛𝑑 −1
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
176
Plotting Radioactive Decay – Example
Time
Activity
0 seconds
142.0 Ci
20 seconds
20.3 Ci
40 seconds
2.91 Ci
60 seconds
0.416 Ci
80 seconds
0.0596 Ci
100 seconds
0.00853 Ci
© Copyright 2014
• Use the above determined
decay constant to
calculate the activity at
various times using:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
ELO 6.3
Operator Generic Fundamentals
177
Plotting Radioactive Decay – Example
Figure: Linear and Semi-Log Plots of Nitrogen-16 Decay
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
178
Plotting Radioactive Decay
• If a substance contains more than one radioactive nuclide, total
activity is sum of individual activities of each nuclide
• Consider sample of material that contains:
– 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 1925 days
(λ = 4.17 x 10-9 sec-1)
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
179
Plotting Radioactive Decay
Initial activity of each of the nuclides is the product of number of atoms
and 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
𝐴𝐶𝑜–60 = 𝑁𝐶𝑜–60 𝜆𝐶𝑜–60
𝐴𝐶𝑜–60 = (1 × 106 𝑎𝑡𝑜𝑚𝑠)(4.17 × 10−9 𝑠𝑒𝑐 −1 )
𝐴𝐶𝑜–60 = 0.00417 𝐶𝑖
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
180
Plotting Radioactive Decay
• Plotting decay activities for each nuclide illustrates the relative
activities of the nuclides in the sample and combined total over time
– 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
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
181
Calculating Activities
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 Xenon135 and sample activity.
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
Correct answer is A.
Correct answer is A.
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
182
Half-Life and Decay Constants
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
Correct answer is A.
© Copyright 2014
ELO 6.3
Operator Generic Fundamentals
183
Radioactive Equilibrium
ELO 6.4 – Describe the following: radioactive equilibrium, transient
radioactive equilibrium, and secular radioactive equilibrium.
• Describes the combined characteristics of parent and daughter
nuclides as they reach stability
• Important for predicting effects of important nuclides such as Iodine
and Xenon on reactor operation
• There are three terms that describe equilibrium:
– Radioactive equilibrium
– Transient radioactive equilibrium
– Secular equilibrium
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
184
Radioactive Equilibrium
• Radioactive equilibrium
– When radioactive nuclide decay and production rates are equal.
End result equilibrium # atoms
• Transient radioactive equilibrium
– Parent and daughter nuclides decaying at essentially the same
rate
– Half-life of the daughter is shorter than the parent
• Secular equilibrium
– Parent has an extremely long half-life
– Equilibrium activities are set by the half-life of the original parent
– Only exception is the final stable element at the end of the chain
o Its number of atoms are constantly increasing
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
185
Radioactive Equilibrium Example
• Concentration of sodium-24 circulating through a sodium-cooled
reactor
– Assume sodium-24 produced at rate of 1 x 106 atoms per second
– If sodium-24 did not decay, amount of sodium-24 present after
some period of time could be calculated by multiplying production
rate by amount of time
Figure: Cumulative Production of Sodium-24 Over Time
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
186
Radioactive Equilibrium Example
• However, sodium-24 is not stable, and it decays at a half-life of 14.96
hours
• Assume no sodium-24 is present initially and production starts at a
rate of 1 x 106 atoms per second, the decay rate initially starts at zero
because there is no sodium-24 present to decay
• The rate of decay will increase as the amount of sodium-24 increases
• Amount of sodium-24 present initially increases rapidly, then
increases at continually decreasing rate until rate of decay is equal to
rate of production
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
187
Radioactive Equilibrium Example
• The amount of sodium-24 present at equilibrium is calculated by
setting the production rate (R) equal to the decay rate (λ N).
𝑅 =𝜆𝑁
𝑅
𝑁=
𝜆
Where:
R = production rate (atoms/second)
λ = decay constant (sec-1)
N = number of atoms
𝜆=
0.693
𝑡1
𝑁=
2
0.693
1 ℎ𝑜𝑢𝑟
=
14.96 ℎ𝑜𝑢𝑟𝑠 3,600 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
= 1.287 ×
© Copyright 2014
10−5
𝑅
𝜆
𝑎𝑡𝑜𝑚𝑠
𝑠𝑒𝑐
=
1.287 × 105 𝑠𝑒𝑐𝑜𝑛𝑑 −1
1 × 106
= 7.77 × 1010 𝑎𝑡𝑜𝑚𝑠
𝑠𝑒𝑐𝑜𝑛𝑑 −1
ELO 6.4
Operator Generic Fundamentals
188
Radioactive Equilibrium
• This equation is used to calculate the values of the amount of sodium24 present at different times
𝑅
𝑁=
1 − 𝑒 −𝜆𝑡
𝜆
• As the time increases, the exponential term approaches zero, and the
number of atoms present approaches R/λ
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
189
Radioactive Equilibrium Example
Figure: Approach of Sodium-24 to Equilibrium
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
190
Transient Radioactive Equilibrium
Transient Radioactive Equilibrium
• Occurs when parent nuclide and daughter nuclide decay at same rate
(decay constants are not the same)
– Parent must have long half-life when compared to daughter
– Example: Barium-140 decays by beta emission to lanthanum-140,
which in turn decays by beta emission to stable cerium-140
– Decay constant for barium-140 is considerably smaller than
decay constant for lanthanum-140
𝛽−
𝛽−
140
140
140
→
→
𝐵𝑎
𝐿𝑎
𝐶𝑒
56
57
58
12.75 𝑑𝑎𝑦𝑠
1.678 𝑑𝑎𝑦𝑠
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
191
Transient Radioactive Equilibrium
• Decay constant for barium-140 is smaller, but actual rate of decay
(λN) is initially larger than lanthanum-140 because of difference in
initial concentrations
• As concentration of daughter increases
– Rate of decay of daughter will approach and eventually match
decay rate of parent
• When this occurs, they are said to be in transient equilibrium
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
192
Transient Radioactive Equilibrium
Figure: Transient Equilibrium in the Decay of Barium-14
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
193
Secular Equilibrium
Secular Radioactive Equilibrium
• Occurs when parent has extremely long half-life
• In long decay chain for a naturally radioactive element, such as
thorium-232, each descendant builds to an equilibrium amount
(constant # of atoms)
– all decay at a rate set by original parent.
• Only exception is final stable element on end of chain
– Number of atoms is constantly increasing because it is not
decaying
© Copyright 2014
ELO 6.4
Operator Generic Fundamentals
194
Radioactive Equilibrium
Knowledge Check
Parent nuclide has an extremely long half-life is a description of
_____________ .
A. transient equilibrium
B. secular equilibrium
C. stable equilibrium
D. unstable equilibrium
Correct answer is B.
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ELO 6.4
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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 1 disintegration per
second
• Radioactive decay constant (λ) – probability per unit time that an
atom will decay
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TLO 6 Summary
• Radioactive half-life – amount of time required for the activity to
decrease by one-half of its original value
• Activity (A) – rate of decay of the sample
• Activity is calculated from the number of atom and the decay
constant:
𝐴 = 𝜆𝑁
• The amount of activity after a particular time:
𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
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TLO 6 Summary
• The relationship between the decay constant and half-life is:
𝑡1 =
2
0.693
𝜆
• Radioactive equilibrium – when production rate equals removal rate
• Transient radioactive equilibrium – when parent and daughter nuclide
decay at essentially the same rate
– Parent has a long half-life compared to the daughter
• Secular equilibrium
– When the parent has an extremely long half-life
– The decay chain descendants builds to an equilibrium amount
– Except that, the final stable element is constantly increasing
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Operator Generic Fundamentals
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Atomic Structure Group Problem
You are in Texas in an Uranium pit of pure Natural Uranium. If you
Answer:
want to extract enough of this natural element to have 10 curies of
Uranium-235, how many
pounds of natural uranium do you need to
1 curie = 4.62 x 105 grams
remove from the pit?
10 curies = 4.62 x 106 grams
• Uranium-235 – Atom percent a/o = 0.72%
4.62 x 106 grams = 1.02 x 104 lbs of U-235
• Uranium-235 - Weight percent w/o = 0.711%
If •w/oUranium-235
.711% of natural
uranium
U-235 and we need 1.02 x
- Activity
% = is
2.2%
10• 4 lbs
of U-235
– that
is 61.43
x 106 lbm of natural uranium.
Half-life
- 703.8
x 10
years
• U-235 decay constant = 3.12 x 10-17/seconds
• 1 Curie = 3.7 x 1010 Becquerels
• U-235 Activity = 80,011 Bq/g
• Activity U-235 within 1 gram of natural Uranium - 568 Bq
• 1 lb = 453.5924 grams
• 1 kilogram = 2.204623 pounds
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TLO 6
Operator Generic Fundamentals
199
Crossword Puzzle
• It’s crossword puzzle time!
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TLOs 4-6
Operator Generic Fundamentals
200
Atomic Structure Summary
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 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 the materials typically used to shield against this
radiation.
6. Describe radioactive decay terms and calculate activity levels, halflives, decay constants, and radioactive equilibrium.
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Module Summary
Operator Generic Fundamentals