Transcript Atomic Structure Instructor Guide Rev 2x

Operator Generic Fundamentals
Nuclear Physics – Atomic Structure
© Copyright 2016
Operator Generic Fundamentals
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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,
half-lives, decay constants and radioactive equilibrium.
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TLOs
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Atoms – Introduction
TLO 1 – Describe atoms, including components, structure, and
nomenclature.
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
1.2 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
1.3 Describe the three forces that act on particles within the nucleus and
how they affect the stability of the nucleus.
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Atoms – Introduction
• 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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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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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 the 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 +1
– Gives nucleus positive charge
– One proton has a +1, two
protons have +2
• Neutrons and protons are
essentially equal in mass
– > 1800 times size of an
electron
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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 equal 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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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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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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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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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
Figure: Nomenclature for Identifying Nuclides
– Atomic mass number as
superscript to upper left
o A = # of protons +
neutrons
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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
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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
Figure: Nomenclature for Identifying Nuclides
o Isotopes
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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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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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Nuclides
Knowledge Check
State name of element and number of protons, electrons, and neutrons
in nuclides listed below:
Nuclide
Element
Protons
Nuclide
Element
Protons
Electrons
Neutrons
Nuclide
Element
Protons Electrons
Electrons Neutrons
Neutrons
1
11𝐻
1𝐻
11𝐻
1010𝐵
10
𝐵
5
55𝐵
14
14 7𝑁
147𝑁
7𝑁
114
11448𝐶𝑑
𝐶𝑑
48
114
239
𝐶𝑑
4894
𝑃𝑢
239
94𝑃𝑢
239
94𝑃𝑢
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Hydrogen
Hydrogen
Boron
Boron
Nitrogen
Nitrogen
Cadmium
1
5
5
7
48
Cadmium 94
Plutonium
Plutonium
1
1
5
1
5
7
7
48
94
ELO 1.2
48
94
0
0
5
5
7
7
48
94
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 an atom’s nucleus
– Some attractive force must exist to oppose the repulsive force
between protons
• Forces present in the nucleus are:
– Gravitational forces between any two objects that have mass
– Electrostatic forces between charged particles
– Nuclear forces between nucleons (protons and neutrons)
• The magnitude of gravitational and electrostatic forces can be
calculated based upon principles from classical physics
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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)
𝐺𝑚1 𝑚2
𝐹𝑔 =
𝑟2
• Where:
Fg = gravitational force (newtons)
m1 = mass of first body (kilograms)
m2 = mass of second body (kilograms)
G = gravitational constant (6.67 x 10-11 N-m2/kg2)
r = distance between particles (meters)
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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
𝐾𝑄1 𝑄2
𝐹𝑒 =
𝑟2
– Where:
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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Chart of the Nuclides
TLO 2 – Use the Chart of the Nuclides to obtain information on specific
nuclides.
2.1 Describe the information for stable and radioactive isotopes found
on the Chart of the Nuclides.
2.2 Describe how an element’s neutron to proton ratio affects its
stability.
2.3 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.
2.4 Describe the following terms:
a. Enriched uranium
b. Depleted uranium
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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 the 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
– Isotopic mass
– Capture cross sections in
barns
– Indication if it is a fission
product
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Figure: Stable Nuclide
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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
• Fission fragments have
approximately the same
neutron-to-proton ratio as
heavy nucleus
• Places fragments below and
to right of stability curve
• Beta-minus decays convert a
neutron to a proton
– create a more stable
neutron-to-proton ratio
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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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 Lithium-6, 6.015122 amu ;
the atomic mass (chart of
Lithium-7, 7.016003 amu
the nuclides)
3.
For each isotope, multiply
abundance times atomic
mass
Li6, (.075)(6.015)= .4511
amu ; Li7, (.925)(7.016)=
6.4898 amu
4.
Sum the products of each
isotope calculation
.4511 amu + 6.4898 amu
= 6.9409 amu
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Lithium-6, 7.5%; Lithium-7,
92.5%
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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. 107.8681 amu
B. 1078.6813 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.
• 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 has uses in both 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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Mass Defect and Binding Energy
TLO 3 – Describe mass defect and binding energy and their relationship
to one another.
3.1 Define mass defect and binding energy.
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.
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.
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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 mass decrease results in an corresponding energy increase and
vice-versa
– Mass does not magically appear and disappear at random
– The total mass and corresponding energy remains constant
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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
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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.
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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
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Mass Defect Calculations
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 − 𝑚𝑎𝑡𝑜𝑚
• Where:
∆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)
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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 2016
Look up information in the Chart of the Nuclides.
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Mass Defect Calculations
Calculate the mass defect for lithium-7 given lithium-7 = 7.016003 amu.
Step
Description
1.
Determine the Z (atomic number) and A
(atomic mass number) of the nuclide.
𝑍 = 3, 𝐴 = 7
2.
Determine the mass of the protons and
electrons of the nuclide.
3 (1.007276467𝑎𝑚𝑢
+ .000548597 𝑎𝑚𝑢)
= 3.023475192 𝑎𝑚𝑢
3.
Determine the mass of the neutrons.
(7 − 3)(1.008665) = 4.03466 𝑎𝑚𝑢
4.
Add mass of protons, electrons, and
neutrons.
3.023475192 𝑎𝑚𝑢 + 4.03466 𝑎𝑚𝑢
= 7.0581352 𝑎𝑚𝑢
5.
7.0581352 𝑎𝑚𝑢 − 7.016003 𝑎𝑚𝑢
Determine the difference between the
atomic mass of the nuclide and the mass = .042132 𝑎𝑚𝑢
of the individual components.
© Copyright 2016
Action
ELO 3.2
Operator Generic Fundamentals
58
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 2016
ELO 3.2
Operator Generic Fundamentals
59
Binding Energy
𝐸 = 𝑚𝑐
2
1.6605 × 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 = 1,6605 × 10−27 𝑘𝑔
1 𝑛𝑒𝑤𝑡𝑜𝑛 = 1 𝑘𝑔– 𝑚/sec2
1 𝑗𝑜𝑢𝑙𝑒 = 1 𝑛𝑒𝑤𝑡𝑜𝑛– 𝑚𝑒𝑡𝑒𝑟
1 𝑀𝑒𝑉 = 1.6022 × 10−13 𝑗𝑜𝑢𝑙𝑒𝑠
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
60
Binding Energy
• Since 1 amu is equivalent to 931.5 MeV of energy, binding energy
can be calculated using:
931.5 𝑀𝑒𝑉
𝐵. 𝐸. = ∆𝑚
1 𝑎𝑚𝑢
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
61
Binding Energy Calculation
Step
Description
1.
Determine the
mass defect of the
nuclide
Action
Use the equation:
∆𝑚 = 𝑍 𝑚𝑝 + 𝑚𝑒 + 𝐴 − 𝑍 𝑚𝑛 − 𝑚𝑎𝑡𝑜𝑚
2.
Use the binding
Use the Equation:
energy equation to
931.5 𝑀𝑒𝑉
𝐵.
𝐸.
=
∆𝑚
calculate the
1 𝑎𝑚𝑢
binding energy.
3.
Calculate the
binding energy
© Copyright 2016
Multiply delta mass from step 1 by the
energy conversion to amu
ELO 3.2
Operator Generic Fundamentals
62
Binding Energy Calculation
Step
Description
1.
Determine the
mass defect of the
nuclide
Action
From previous calculation:
.04135 amu.
2.
931.5 𝑀𝑒𝑉
Use the binding
𝐵.
𝐸.
=
∆𝑚
energy equation to
1 𝑎𝑚𝑢
calculate the
binding energy.
3.
Calculate the
binding energy
931.5 MeV
B. E. = 0.04135 amu
1 amu
= 38.5175 𝑀𝑒𝑉
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
63
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:
– 𝐵. 𝐸. = 1.91517 𝑎𝑚𝑢
931.5 𝑀𝑒𝑉
1 𝑎𝑚𝑢
– 𝐵. 𝐸. = 1784 𝑀𝑒𝑉
© Copyright 2016
ELO 3.2
Operator Generic Fundamentals
64
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 2016
ELO 3.3
Operator Generic Fundamentals
65
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 2016
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Operator Generic Fundamentals
66
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
Excited 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
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ELO 3.3
Operator Generic Fundamentals
67
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 2016
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Operator Generic Fundamentals
68
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 (ˠ)
• Differences between x-rays and ˠ-rays are
– Energy levels
– Where they originate from
o X-rays – electron shell
o ˠ-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
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Operator Generic Fundamentals
69
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
© Copyright 2016
Figure: Energy Level Diagram – Nickel-60
ELO 3.3
Operator Generic Fundamentals
70
Gamma Rays and X-Rays
Knowledge Check
Uranium-238 would be "stable" with _____ electrons orbiting the
nucleus.
A. 238
B. 146
C. 235
D. 92
Correct answer is D.
© Copyright 2016
ELO 3.3
Operator Generic Fundamentals
71
Nuclear Stability
TLO 4 – Describe the processes by which unstable nuclides achieve
stability.
4.1 Describe the conservation principles that must be observed during
radioactive decay. Include an explanation of neutrinos.
4.2 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
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.
4.4 Given a Chart of the Nuclides, describe the radioactive decay chain for a
nuclide.
© Copyright 2016
TLO 4
Operator Generic Fundamentals
72
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
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
73
Conservation Principles
Principle
Conservation of
Electric Charge
Conservation of
Mass Number
© Copyright 2016
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
74
Conservation Principles
Principle
Description
Conservation Conservation of mass and energy implies that the
of Mass and total of the kinetic energy and the energy equivalent
Energy
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 mass-equivalent energy must
be constant.
Conservation Conservation of momentum is responsible for the
of Momentum 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 might be distributed differently
among entirely different nuclides and/or particles.
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ELO 4.1
Operator Generic Fundamentals
75
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 2016
ELO 4.1
Operator Generic Fundamentals
76
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
80 neutrons
∆Mass = .001235 amu
Mass is accounted for in the beta particle
and energy of the gammas emitted.
© Copyright 2016
ELO 4.1
Operator Generic Fundamentals
77
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 2016
ELO 4.1
Operator Generic Fundamentals
78
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 2016
ELO 4.2
Operator Generic Fundamentals
79
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 2016
→
230
90𝑇ℎ
+ 42𝛼 + 𝛾 + 𝐾𝐸
ELO 4.2
Operator Generic Fundamentals
80
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 2016
→
230
90𝑇ℎ
+ 42𝛼 + 𝛾 + 𝐾𝐸
ELO 4.2
Operator Generic Fundamentals
81
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 2016
ELO 4.2
Operator Generic Fundamentals
82
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 2016
ELO 4.2
Operator Generic Fundamentals
83
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 2016
13
6𝐶
+ +10𝛽 + 00𝑣
ELO 4.2
Operator Generic Fundamentals
84
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 2016
ELO 4.2
Operator Generic Fundamentals
85
Electron Capture
Figure: Electron Capture or K-Capture
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
86
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 be 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 2016
ELO 4.2
Operator Generic Fundamentals
87
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 2016
ELO 4.2
Operator Generic Fundamentals
88
Internal Conversion
• Normally an excited nucleus goes from the excited state to the
ground state by emission of a gamma ray
• In some cases the gamma ray released interacts with one of the
innermost orbital electrons
– Transfers the gamma’s energy to the electron
– When this occurs the atom is said to be 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 2016
ELO 4.2
Operator Generic Fundamentals
89
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
16
7𝑁 →
8𝑂
16
16
8𝑂 → 8𝑂
© Copyright 2016
+ 01𝛽 + 00𝛾
+ 00𝛾
ELO 4.2
Operator Generic Fundamentals
90
Neutron Emission
• Non-stable nuclei may also emit neutrons (n) in order to become
more stable
𝛽−
𝑛
87
86
→ 87
36𝐾𝑟 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠→ 36𝐾𝑟𝑠𝑡𝑎𝑏𝑙𝑒
35𝐵𝑟
55.9 𝑠𝑒𝑐
• 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
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
91
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
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
92
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.
© Copyright 2016
ELO 4.2
Operator Generic Fundamentals
93
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
© Copyright 2016
Figure: Stability Curve
ELO 4.3
Operator Generic Fundamentals
94
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
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
95
Predicting Type of Decay
Figure: Types of Radioactive Decay Relative to Line of Stability
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
96
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
© Copyright 2016
ELO 4.3
Operator Generic Fundamentals
97
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
© Copyright 2016
4
ELO 4.3
Operator Generic Fundamentals
98
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
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
99
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
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
100
Decay Chains
Example Decay Chain of U-238 Continued:
• 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
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
101
Decay Chains
Example Decay Chain of U-238 Continued:
• 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.
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
102
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
© Copyright 2016
ELO 4.4
Operator Generic Fundamentals
103
Emitted Radiation
TLO 5 – Describe how radiation emitted by an unstable nuclide interacts
with matter and materials typically used to shield against this radiation
5.1 Describe the difference between charged and uncharged particle
interaction with matter. Include an explanation of specific ionization.
5.2 Describe radioactive interactions of the following types with matter:
a. Alpha Particle
b. Beta Particle
c. Gamma
d. Positron
e. Neutron
5.3 Describe the type of material that can be used to stop (shield) the
following types of radiation:
Alpha particle
Beta-particle
Neutron
Gamma ray
© Copyright 2016
TLO 5
Operator Generic Fundamentals
104
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 with 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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Operator Generic Fundamentals
109
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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Operator Generic Fundamentals
110
Alpha Radiation
Origin
Interaction
• Produced from the radioactive
decay of heavy nuclides and
certain nuclear reactions
• Consists of 2 neutrons and 2
protons, same as a helium atom
• With no electrons, the alpha
particle has a charge of +2
• Removes electrons from the
atoms it passes near
• Removal of electrons requires
energy
• Alpha’s energy is reduced by
each reaction
• This strong positive charge
strips electrons from the orbits
of atoms
• Ultimately, the alpha particle
expends its KE, gains 2
electrons, and becomes a
helium atom
• Has a high specific ionization
• Low penetration power
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111
Beta Minus Interaction
Origin
• A beta-minus particle is an
electron that has been ejected
at a high velocity from an
unstable nucleus
Interaction
• Beta-minus ionization occurs
from collisions with orbiting
electrons.
• Electrons have a small mass
and an electrical charge of -1
• Each collision removes KE
from the beta particle, causing
it to slow down
• Beta particles cause ionization
by displacing electrons from
atomic orbits
• 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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Operator Generic Fundamentals
112
Positron Radiation
Origin
• Positively charged
electrons
Interaction
• Positrons are very short-lived and quickly
annihilate via interactions with negatively
charged electrons
• Identical to betaminus particles and • Produces two gammas with energy equal
interact with matter
to the rest mass of the electrons (1.02
similarly
MeV)
• These gammas interact with matter via
photoelectric effect, Compton scattering
or pair-production,
0.000549 𝑎𝑚𝑢 931.5 𝑀𝑒𝑉
2 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
𝑎𝑚𝑢
= 1.02 𝑀𝑒𝑉
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Neutron Interactions
Origin
Interaction
• No electrical charge
• Same mass as a proton
• 1835 times more mass than
electron
• Neutrons are attenuated
(reduced in energy and
numbers) by three major
interactions:
̶ Elastic scatter
• 1/4 mass of an alpha particle
̶ Inelastic scatter
• Come primarily from nuclear
reactions, such as fission,
but also from decay of
radioactive nuclides
̶ Absorption
• 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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Operator Generic Fundamentals
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Gamma Interactions
Gamma Radiation
• Electromagnetic radiation - similar to x-ray
– 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 2016
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 2016
Figure: Photoelectric Effect
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Operator Generic Fundamentals
118
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.
• Most predominant with gamma
energy level of 1.0 to 2.0 MeV.
© Copyright 2016
ELO 5.2
Figure: Compton Scattering
Operator Generic Fundamentals
119
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 energy
• Electron and positron may collide and annihilate each other
Figure: Pair Production
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Operator Generic Fundamentals
120
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
121
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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Operator Generic Fundamentals
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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 for neutrons is a
material with similar mass 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 2016
• 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"
ˠ
Figure: Effects Various Materials Have on Types of Radiation
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126
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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Operator Generic Fundamentals
127
Radioactive Decay
TLO 6 – Describe radioactive decay terms and calculate activity levels,
half-lives, decay constants and radioactive equilibrium.
6.1 Describe the following radioactive terms:
a. Radioactivity
b. Radioactive decay constant
c. Activity
d. Curie
e. Becquerel
f. Radioactive half-life
6.2 Convert between the half-life and decay constant for a nuclide.
6.3 Given the nuclide, number of atoms, half-life or decay constant,
Determine current and future activity levels.
6.4 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.
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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
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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. Curie
B. Radioactivity
C. Becquerel
D. Radioactive
Decay Constant
Correct answers: 1-B, 2-A, 3-C, 4-D.
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134
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.
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Operator Generic Fundamentals
135
Convert Between Half-life and Decay
Constant
• A relationship between half-life
and decay constant can be
developed:
𝑒 −𝜆𝑡
𝐴 = 𝐴𝑜
𝐴
= 𝑒 −𝜆𝑡
𝐴𝑜
𝐴
ln
= −𝜆𝑡
𝐴𝑜
𝐴
− ln
𝐴𝑜
𝑡=
𝜆
© Copyright 2016
• 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
1
− ln
2
𝑡1 =
𝜆
2
ln 2
𝑡1 =
𝜆
2
0.693
𝑡1 =
𝜆
2
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Operator Generic Fundamentals
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Convert Between Half-life and Decay
Constant
Step
1.
Action
Solution
Determine the half-life if
decay constant is known.
Use equation:
𝑡1 =
2
2.
Determine the decay
constant if half- life is
known
0.693
𝜆
Use equation:
0.693
𝜆=
𝑡1
2
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137
Example: Determine ‫ ג‬of Cesium-136 –
t1/2 = 13.16 days
Step
Action
Method
Calculation
1.
Determine the half- Use equation:
life if decay constant
0.693
𝑡1 =
is known.
𝜆
2
Half life given as
13.16 days.
2.
Determine the decay Use equation:
constant if half-life is
0.693
𝜆=
known.
𝑡1
𝜆=
0.693
13.16 𝑑𝑎𝑦𝑠
𝜆 = 0.0527−1 𝑑𝑎𝑦𝑠
2
© Copyright 2016
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Operator Generic Fundamentals
138
Example: Determine half-life of
Potassium-44 ‫ = ג‬.0313 -min
Step
1.
2.
Action
Determine the
half-life if decay
constant is
known.
Method
Use equation:
0.693
𝑡1 =
𝜆
2
Determine the
Use equation:
decay constant if
0.693
half-life is known. 𝜆 = 𝑡1
Calculation
𝑡1 =
2
0.693
0.03131−𝑚𝑖𝑛
𝑡1 = 22.13 𝑚𝑖𝑛
2
Given
2
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Operator Generic Fundamentals
139
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
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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
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Operator Generic Fundamentals
141
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.
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Operator Generic Fundamentals
142
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 (A), the number of atoms present
(N) and the decay constant (𝜆) is necessary in order to understanding
the behavior of radioactive decay
• This section explains how the activity of a sample of material varies
with time.
• Activity for a given material and a given amount of material is
determined by the following equation:
𝐴 = 𝜆𝑁
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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
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Operator Generic Fundamentals
144
Calculating Activity
Step
Action
Solution
1.
Determine the
number of atoms
present in the mass
of the isotope
Use the following equation:
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁 = 𝑚𝑎𝑠𝑠
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
2.
Determine the decay Use the following equation:
constant
0.693
𝜆=
𝑡1
2
3.
© Copyright 2016
Determine the
activity
Use the following equation:
𝐴 = 𝜆𝑁
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Operator Generic Fundamentals
145
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 2016
Method
Calculation
Use the equation:
𝑁 = 𝑚𝑎𝑠𝑠
1 𝑚𝑜𝑙𝑒
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 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
𝑁𝐴
1 𝑚𝑜𝑙𝑒
ELO 6.3
Operator Generic Fundamentals
146
Activity Calculation Example
Step
2.
Action
Determine
the decay
constant
Method
Calculation
Use the following
equation:
𝑡1 =
2
0.693
𝑡1 =
𝜆
2
𝜆=
0.693
𝜆
0.693
2.638 𝑦𝑒𝑎𝑟𝑠
𝜆 = 0.263 𝑦𝑒𝑎𝑟 −1
3.
Determine
the activity
Use the equation: 𝐴 = 𝜆𝑁
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 2016
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Operator Generic Fundamentals
147
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 2016
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Operator Generic Fundamentals
148
Variation of Radioactivity Over Time
Step
1.
2.
Action
Equation
If the initial activity is not
Use this equation:
known determine the
1 𝑚𝑜𝑙𝑒
𝑁
=
𝑚𝑎𝑠𝑠
number of atoms present in
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠
the mass of the isotope
Determine the decay
constant if necessary
𝑁𝐴
1 𝑚𝑜𝑙𝑒
Use this equation:
0.693
𝜆=
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
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ELO 6.3
Operator Generic Fundamentals
149
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 2016
Action
Equation
If the initial activity is
Use the following equation:
unknown, determine the
1 𝑚𝑜𝑙𝑒
𝑁𝐴
𝑁
=
𝑚𝑎𝑠𝑠
number of atoms
𝑖𝑠𝑜𝑡𝑜𝑝𝑖𝑐 𝑚𝑎𝑠𝑠 1 𝑚𝑜𝑙𝑒
present in the mass of
the isotope.
ELO 6.3
Operator Generic Fundamentals
150
Variation of Activity Over Time Calculation
Step
2.
Action
Equation
Solution
Determine the Use this equation: 𝜆 = 0.693
𝑡1
decay constant
0.693
2
𝜆=
𝑡1
if necessary
0.693
2
=
2.638 𝑦𝑒𝑎𝑟𝑠
= 0.263 𝑦𝑒𝑎𝑟 −1
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 2016
ELO 6.3
Given: 0.0108 curies
𝐴 = 0.0108 𝑒
−
0.263
12 𝑦𝑟
𝑦𝑟
= 0.00046 𝑐𝑢𝑟𝑖𝑒𝑠
Operator Generic Fundamentals
151
Plotting Radioactive Decay
Step
1.
Action
Method
0.693
𝜆
Calculate the decay
constant of the isotope
Use the equation: 𝑡1 =
2.
Use the decay constant to
calculate the activity at
various times
Use the equation: 𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
3.
Develop a table of values
from the calculations
performed above
Use above equations
4.
Plot the points from the
table on linear and semi
log scales
Using the correct graph
paper, plot the points.
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2
ELO 6.3
Operator Generic Fundamentals
152
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
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Operator Generic Fundamentals
153
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
• First, calculate 𝜆 corresponding to half-life of 7.13 seconds
0.693
𝑡1 =
𝜆
2
0.693
𝜆=
𝑡1
2
0.693
𝜆=
7.13 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝜆 = 0.0972 𝑠𝑒𝑐𝑜𝑛𝑑 −1
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Operator Generic Fundamentals
154
Plotting Radioactive Decay – Example
• Use the determined decay constant (𝜆 = 0.0972 𝑠𝑒𝑐𝑜𝑛𝑑 −1 )
calculate the activity at various times using: 𝐴 = 𝐴𝑜 𝑒 −𝜆𝑡
• The results are shown below and are then plotted on the decay
charts.
© Copyright 2016
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
ELO 6.3
Operator Generic Fundamentals
155
Plotting Radioactive Decay – Example
Figure: Linear and Semi-Log Plots of Nitrogen-16 Decay
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ELO 6.3
Operator Generic Fundamentals
156
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)
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Operator Generic Fundamentals
157
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
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Operator Generic Fundamentals
158
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 2016
ELO 6.3
Operator Generic Fundamentals
159
Calculating Activities
Knowledge Check
A sample contains 100 grams of Xenon-135. The Half-life of Xenon-135
is 9.14 hours and an atomic mass 134.907 amu. Calculate the decay
constant of Xenon-135 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.
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Operator Generic Fundamentals
160
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.
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Operator Generic Fundamentals
161
Radioactive Equilibrium
ELO 6.4 – Describe the following: 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 two terms that describe equilibrium:
– Radioactive equilibrium
– Secular equilibrium
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ELO 6.4
Operator Generic Fundamentals
162
Radioactive Equilibrium
• Radioactive equilibrium
– When radioactive nuclide decay and production rates are equal.
End result equilibrium # atoms
• 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
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Operator Generic Fundamentals
163
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
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Operator Generic Fundamentals
164
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
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Operator Generic Fundamentals
165
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 × 10
© Copyright 2016
−5
𝑠𝑒𝑐𝑜𝑛𝑑
−1
𝑅
𝜆
𝑎𝑡𝑜𝑚𝑠
𝑠𝑒𝑐
=
1.287 × 105 𝑠𝑒𝑐𝑜𝑛𝑑−1
1 × 106
= 7.77 × 1010 𝑎𝑡𝑜𝑚𝑠
ELO 6.4
Operator Generic Fundamentals
166
Radioactive Equilibrium
• This equation is used to calculate the values of the amount of
sodium-24 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
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ELO 6.4
Operator Generic Fundamentals
167
Radioactive Equilibrium Example
Figure: Approach of Sodium-24 to Equilibrium
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ELO 6.4
Operator Generic Fundamentals
168
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
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Operator Generic Fundamentals
169
Radioactive Equilibrium
Knowledge Check
A parent nuclide that 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.
© Copyright 2016
ELO 6.4
Operator Generic Fundamentals