Modes of Radioactive Decay
GE-PP-22502
Author: Ken Jenkins
Approved: Michael J. Kurtzman
Date: 06/14/2003
Revision: 00
Nuclear Stability
Forces Acting Within the Nucleus
FORCE
INTERACTION
RANGE
Gravitational Very weak attractive Relatively long
between nucleons
Electrostatic
Strong repulsive
between protons
Relatively long
Nuclear
Strong attractive
between nucleons
Extremely short
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2
Nuclear Stability
The repulsive electrostatic forces between the
protons have an impact on nuclear stability
The number of neutrons must increase more
rapidly than the number of protons to provide
‘dilution’ and to add additional nuclear forces
If the nuclear (attractive) and electrostatic
(repulsive) forces do not balance, the atom will
not be stable
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Nuclear Stability
An unstable nucleus will eventually
achieve stability by changing its nuclear
configuration
This includes changing neutrons to
protons, or vice versa, and then ejecting the
surplus mass or energy from the nucleus
This emitted mass or energy is called
radiation
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Nuclear Stability
When an atom transforms to become more
stable it is said to disintegrate or decay
The time required for half of a sample of
atoms to decay is known as the half-life
The property of certain nuclides to
spontaneously disintegrate and emit radiation
is called radioactivity
The atom before the decay is the parent and
the resulting atom is called the daughter
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Neutron / Proton Ratio
NUMBER OF PROTONS (Z)
100
N
1
Z
80
60
LINE OF
STABILITY
40
20
0
0
20
40
60
80
100
NUMBER OF NEUTRONS
(N=A-Z)
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120
140
6
Beta Decay
Betas are physically the same as electrons,
but may be positively or negatively charged
Negative beta is a beta minus or negatron
Positive beta is a beta plus or positron
Betas are ejected from the nucleus, not from
the electron orbitals
In all beta decays the atomic number
changes by one while the atomic mass is
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unchanged
Beta (β-) Minus Decay
Occurs in neutron-rich nuclides
The nucleus converts a neutron into a proton
and a beta minus (which is ejected from the
nucleus with an anti-neutrino)
Mass and charge are conserved
1
0
n p e 
1
1
0
1
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Beta (β-) Minus Decay
For beta minus decays,
A
Z
X
90
38
Y   
A
Z 1
0
1
Sr  Y   
90
39
0
1
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Beta (β-) Minus Decay

Anti-neutrino
Daughter
Ca-40
Parent
K-40
0
1

Beta Particle
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Beta (β-) Minus Decay
During radioactive decay energy is released
Source of this energy is from the conversion
of mass
Since energy is conserved, energy
equivalent of the parent must equal energy
equivalent of daughter, particles, and any
energy released
Energy is released as kinetic energy of beta
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minus particle and an
anti-neutrino
Beta (β-) Minus Decay
For beta minus, energy of decay reaction (Q)
is,
Q  (M A X
Z
MeV
 M AY )(931.5
)
Z 1
amu
Mass of beta minus particle is not included
since an additional electron is gained due to
increase of Z
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Beta (β-) Minus Decay
Calculate Q for β- decay of Co60.
60
60
0
27
Co28 Ni 1 
Mass of Co-60 is 59.933813 amu
Mass of Ni-60 is 59.930787 amu
MeV
Q  (59.933813amu - 59.930787amu)(931.5
)
amu
Q  2.819 MeV
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Beta (β-) Minus Decay
The Q value for beta minus decay of Co-60,
for example, is always the same
However, negatrons rarely are emitted with
the same energies
Their energies can range from 0 MeV to the
calculated maximum, Emax
The anti-neutrino carries energy difference
between actual and calculated values
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Beta (β-) Minus Decay
1
E Max
3
EMax
Energy
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Co
99+%
0.83
60
27

1.173


1.332
0.013%
0.12%
Q
Co-60 Decay Scheme
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60
28
Ni
16
Beta (β+) Plus Decay
Occurs in proton-rich nuclides
The nucleus converts a proton into a neutron
and a beta plus (which is ejected from the
nucleus with a neutrino)
As with negatrons, the positron can have a
range of energies from 0 to EMax MeV
Positron is the negatron’s anti-particle
A positron and a negatron will annihilate one
another and release two 0.511 MeV photons
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Beta (β+) Plus Decay
For beta plus decays,
1
1
p n e  
A
Z
X  Y   
1
0
13
7
0
1
A
Z 1
0
1
N  C   
13
6
0
1
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Beta (β+) Plus Decay

Neutrino
Daughter
O-18
Parent
F-18
0
1

Beta Particle
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Beta (β+) Plus Decay
For beta plus, energy of decay reaction (Q)
is,
MeV
Q  [( M A X )  ( M
Z
A
Z 1Y
 2M
0
1 e
)](931.5
)
amu
Since the energy equivalent of two electron
masses is 1.022 MeV, the equation can be
rewritten as,
MeV
Q  [( M A X  M AY )(931.5
)]  1.022 MeV
Z
Z 1
amu
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Beta
+
(β )
Plus Decay
e-
•
• • • •
• • •
13
7
•
•
•
•
• •
13
6
C
N
•

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+
21
Beta (β+) Plus Decay
Calculate Q for β+ decay of F18.
18
18
0
9
F  8 O  1   
Mass of F-18 is 18.000937 amu
Mass of O-18 is 17.999160 amu
MeV
Q  [(18.000937amu - 17.999160amu)(931.5
)]  1.022MeV
amu
Q  0.633 MeV
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Electron Capture
Proton-rich nuclides may also decay via
orbital electron capture (EC)
Usually an innermost K shell electron is
captured and often referred to as K-capture
The electron and a proton are converted into a
neutron and a neutrino is emitted
Electrons from higher orbitals will fill
vacancy and usually emit characteristic x-rays
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Electron Capture
For electron capture decays,
1
1
p  e n  
A
Z
X  e Y 
53
25
Mn  Cr  
0
1
0
1
1
0
A
Z 1
EC
53
24
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Electron Capture
For electron capture, energy of decay
reaction (Q) is,
Q  (M A X
Z
MeV
 M AY )(931.5
)
Z 1
amu
Since the electron was absorbed into the
nucleus and not removed, there is no need to
account for electron mass
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Auger Electrons
When electrons change shells, x-rays are
usually emitted
In some instances, the excess energy is
transferred to another orbital electron, which
is then ejected from the atom
This ejected electron is known as an Auger
electron
Another orbital vacancy now exists and xrays may be emitted if they are filled
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Auger Electrons
•
• • •
• • •
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Beta Interactions
Excitation
The beta, via coulombic interaction, transfers
enough energy to an orbital electron to move it to
a higher energy level, but not to remove it from
the atom
The atom remains electrically neutral
The excited electron will then return to its
ground state and emit the excess energy as x-rays
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Excitation
•
-
• •
•
•
•
•
x-ray
•
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Beta Interactions
Ionization
The beta, via coulombic interaction, transfers
enough energy to an orbital electron to overcome
its binding energy and remove it from the atom
With the loss of the negative electron, the
remaining atom is now a positive ion
If the vacancy is filled, an x-ray will be emitted
The formation of each ion pair in air (gas)
requires about 34 eV of energy from the beta
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e-
•
Ionization
•
-
•
•
•
•
•
•
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Beta Interactions
Bremsstrahlung
German for ‘braking radiation’
Occurs when beta is deflected by the positively
charged nucleus
The kinetic energy lost by the beta is emitted as
a photon (x-ray)
Bremsstrahlung increases with higher Z
materials
For example, a lead blanket may shield betas,
but generate higherGE-PP-22502-00
levels of Bremsstrahlung (x- 32
Bremsstrahlung
•
• • •
•
•
•
x-ray
-
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•
33
Beta Interactions
Betas travel in zig-zag or tortuous paths
Collisions and deflections
Coulombic interactions
Not mono-energetic
Because of this, betas have a definite,
predictable range (given in mg/cm2)
Basic thumb rule is that a 1.0 MeV beta will
travel approximately 12 feet in air
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Beta Interactions
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Beta Interactions
All betas can be stopped, but Bremsstrahlung
photons can be produced
Intensity is proportional to number of betas,
their energy, and Z of the absorber
Shielding is designed to minimize and/or
shield Bremsstrahlung
Low Z materials such as plastic
(hydrocarbons) or aluminum are common
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Beta Interactions
The fraction of beta energy that appears as
photon energy (Bremsstrahlung) can be
estimated with the following equation:
f = E x Z x 10-3
E = beta energy in MeV
Z = atomic number of target (shield) material
Average energy of the Bremsstrahlung photons is
about 1/4 Emax
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Alpha Decay
Alphas are large particles ejected by the
heavier nuclides
Alpha decay is primarily limited to nuclides
with Z > 82
Source is mainly from fuel-related materials
Alpha contains two protons and two neutrons
(no electrons) and is, in effect, a helium nucleus
Thus, the atomic number decreases by two
and the mass number decreases by four
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Alpha Decay
For alpha decays,
A
Z
X
210
84
A 4
Z 2
Y  He
4
2
2
Po Pb He
206
82
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4
2
2
39
Alpha Decay
Daughter
Th-231
Parent
U-235
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4
2
He
2
40
Alpha Decay
Since nothing else is emitted, all energy of
decay goes to the alpha particle (except for a
small amount towards recoil of nucleus)
Alphas, therefore, are mono-energetic
For alpha, energy of decay reaction (Q) is,
Q  [ M A X  ( M A4Y
Z
Z 2
MeV
 M 4 He )](931.5
)
2
amu
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Alpha Decay
Calculate Q for the  decay of Rn222.222
218
4
2
86
Rn  84 Po 2 He
Mass of Rn-222 is 222.017610 amu
Mass of Po-218 is 218.009009 amu
Q  [222.017610amu - (218.009009amu  4.002603amu)](931.5
MeV
)
amu
Q  5.6 MeV
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Alpha Interactions
Alphas interact primarily through
Coulombic interactions due to their +2 charge
Energy transfer occurs through excitation
and ionization
Orbital electrons may receive enough
energy to allow them to cause secondary
ionizations of other atoms
Bremsstrahlung does not occur since the
large alphas are not easily deflected
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Alpha Interactions
Because of their mass and charge, alphas
travel in relatively straight paths over short
distances (higher Z of absorber, less distance)
A 7 MeV alpha travels only about 0.0002
cm in lead
Alphas are considered internal hazards only
When an alpha slows enough, it captures
two free electrons and converts to a helium
atom
44
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Alpha Interactions
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Nuclear De-excitation
Daughter nuclei from radioactive decays are
often ‘born’ with excess energy
Occasionally the excited nucleus will emit
additional alphas or betas
Usually the excited nucleus reaches ground
state via nuclear de-excitation
The excited nucleus and the final ground
state nucleus have the same Z and A and are
called isomers
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Nuclear De-excitation
If the excited state has a half-life >1 sec, it
is said to be a metastable state
The metastable state is denoted by the use of
a lowercase ‘m’, such as Ba-137m
The longest known excited state is Bi-210m
with a half-life of 3.5 x 106 years
During de-excitation no nuclear
transformation occurs, so no ‘new’ element is
formed
47
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Nuclear De-excitation
Internal Conversion
The excess nuclear energy is transferred to an
inner orbital (usually K or L) electron
This electron is then ejected from the atom with
a distinct energy
X-ray emission may follow as electrons shift
orbitals to fill vacancies
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Nuclear De-excitation
Gamma emission
Most frequently the excess energy is relieved
via the emission of one or more gamma rays
Gammas have no mass or electric charge
If gammas are emitted by an isomer in the
metastable state, the emission is known as an
isomeric transition (IT)
Photon Energy (E) = hf
where h is Planck’s Constant (4.14 x 10-15 eV-sec)
f is frequency (sec-1)
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Gamma Ray Radiation
0
1

Gamma Rays
Parent
Co-60
Daughter
Ni-60

Anti-neutrino
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50
Co
99+%
0.83
60
27

1.173


1.332
0.013%
0.12%
Q
Co-60 Decay Scheme
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60
28
Ni
51
Gamma Interactions
Alphas and betas (charged particles)
interact multiple times along their paths
Gammas usually have only one or two
interactions and all of their energy is
transferred
Gammas interact with matter typically
through three processes: photoelectric effect,
Compton effect, and pair production
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Photoelectric Effect
The gamma ray photon transfers all of its
energy to an orbital electron (usually K shell)
The electron is then ejected from the atom
(photoelectron)
Probability of the photoelectric effect
increases with increasing Z of the absorber
Probability of the photoelectric effect
increases with decreasing gamma energy (<1
MeV)
53
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e- •
Photoelectric Effect
•
•
•
•
•
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•

54
Compton Effect (Scattering)
The gamma ray photon transfers some of
its energy to an orbital electron
The electron is ejected (recoil electron)
and the photon is scattered with a lower
energy
Probability of the Compton effect
decreases with increasing gamma energy
(200 keV  5 MeV)
Compton effect is more common with
55
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absorbers of intermediate
Z
Compton Effect
•
e- •
•
•
•
•
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•

56
Pair Production
Photon travels in the vicinity of the nucleus
Photon spontaneously converts into a pair of
particles - an electron and a positron
Since the rest mass energy of an electron is
0.511 MeV (from E = mc2), the photon must
have an initial energy of at least 1.022 MeV
All photon energy in excess of 1.022 MeV is
shared as kinetic energy between the particles
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Pair Production
e- •
•
•
•
•
• •
e+ •
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Pair Production
The electron and positron will lose energy
through excitation, ionization, and
Bremsstrahlung interactions
When the positron slows sufficiently it will
be attracted to an electron and the two will
annihilate one another (anti-particles)
resulting in the formation of two 0.511 MeV
photons
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Pair Production
Though pair production is possible at 1.022
MeV, the process rarely occurs until
approximately 5 MeV photon energy
The likelihood of pair production also
increases proportionally with increasing Z of
the absorber
Few isotopes at Vogtle have sufficient
energies for pair production to occur
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Gamma Ray Attenuation
Gammas interact within an absorber via
photoelectric effect, Compton scattering, and
pair production
Compton scattering and pair production
events result in the emission of photons
The average probability of an event must be
considered for shielding
Theoretically no amount of shielding can
reduce the gamma dose rate to zero
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Gamma Ray Attenuation
Gamma ray intensity is reduced
exponentially with a linear increase in
absorber thickness
 x
I  I 0e
Where:
I = emerging gamma intensity
I0 = incident gamma intensity
x = thickness of absorber (cm)
 = linear attenuation coefficient (cm-1)
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Gamma Ray Attenuation
Pair Production
(annihilation
photons)
No
Interaction
Photoelectric
Effect
Compton
Scatter
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63
Gamma Ray Attenuation
A more simplified shielding calculation
uses the concepts of Half Value Layers
(HVL) and Tenth Value Layers (TVL)
HVL is the thickness of an absorber
necessary to decrease the gamma radiation
to one half of the incident value
TVL is the thickness of an absorber
necessary to decrease the gamma radiation
to one tenth of the incident value
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Gamma Ray Attenuation
To perform calculations with these concepts
use the following equations:
1 n
I  I0 ( )
2
1 n
I  I0 ( )
10
Where n = the number of HVLs or TVLs respectively
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Decay Schemes
Vertical lines represent energy
Horizontal lines indicate atomic number (Z)
Beta minus points down to the right
Alpha and EC point down to the left
Beta plus points down to the left with a
1.022 MeV offset
Parent half-lives are shown
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Decay Schemes
Ground states are bold horizontal lines
Excited states are light horizontal lines
Isomeric states are medium horizontal lines
Total amount of energy for the reaction is
shown (Q)
Abundances (probabilities) of transitions are
shown
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Decay Schemes
What is the half-life of Ar-41?
1.83 hours
What percentage of the Zr-95 beta minus
decays result in an isomer of Nb-95?
2%
Cr-51 decays by what method(s)?
Electron capture (EC)
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Decay Schemes
For every 100 decays of Rb-86, about how
many 1.078 MeV gammas will be produced?
9 (100 x 8.8%)
For every 100 decays of Mn-56, about how
many 1.811 MeV gammas will be produced?
29 (100 x 30% x 97.8%)
What will be the most abundant gamma
energy produced during the decay of Fe-59?
1.095 MeV
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Chart of the Nuclides
Half-life (color
indicates 1-10
days)
I131
Isotope
8.040 d
Beta decay with
most prominent
energy (MeV)
- 0.606, . . .
 364.5, . . .
 0.7, 8
E 0.971
Thermal neutron
Beta disintegration energy (MeV)
cross-section (barns)
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Most prominent
gamma energy
(keV)
Fission
product
70
Chart of the Nuclides
White backgrounds are artificially radioactive
Gray backgrounds are stable and include
percent abundances
Lower half colors represent neutron
absorption properties
Black bar across top of box indicates a
naturally-occurring radioactive isotope
Heavy black outlines indicate ‘magic’
numbers
71
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