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Yarmouk UniversityPhysics Department
Nuclear Techniques Phys. 649
2nd Semester 2012-2013Dr. Nidal M. Ershaidat
© Dr. Nidal M. Ershaidat
Phys. 649: Phys. 649: Nuclear TechniquesNuclear Techniques
Physics Department
Yarmouk University
Overview
3
Textbooks & References
� Radiation Detection and Measurements, Third Edition,
Author: Glenn F. Knoll,
Publisher: John Wiley and Sons, 1999.
� Introductory Nuclear Physics, 2nd Edition
Author: Kenneth S. Krane,
Publisher: John Wiley and Sons, 1988.
� http://ctaps.yu.edu.jo/Physics/Phys641
4
Calendar
06/05/20138. Principles of Neutron Detection
24/04/20137. Semiconductor Diode Detectors
10/04/20136. Scintillators - Photomultipliers
20/03/20135. Gas-Filled Detectors
04/03/20134. General Properties of Radiation Detectors
25/02/20133. Counting Statistics and Error Predictions
13/02/20132. Radiation Interactions with Matter
06/02/20131. Radiation Sources (Review)
9. Application
2
5
15%Wed. 26/04/2013Project
15%Homework
30%Sat. 07/04/20131 to 5Mid Term Exam
GradeDateChapters
40%To be fixed laterAll chaptersFinal Exam
Assessment6
Proposed Activities
There will be at least 2 extra sessions where a mathematical package will be used to understand the concepts covered in this course.To be discussed during the semester.
7
Course’s websites
�http://ctaps.yu.edu.jo/Physics/Phys649
�http://faculty.yu.edu.jo/NErshaidat
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Chapter 1:
Radiation Sources
Supplements
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Chapter 1: Radiation Sources (Preview)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 2
Overview
1. Radiation – Generalities2. Units and Definitions3. Fast Electron Sources4. Sources of Electromagnetic Radiation5. Neutron Sources
3
RadiationIn physics, radiation is a process in which energetic particles or energetic waves travel through a medium or space.Two types of radiation are commonly differentiated in the way they interact with normal chemical matter: ionizing and non-ionizingradiation.
The word radiation is often colloquially used in
reference to ionizing radiation (i.e., radiation having sufficient energy to ionize an atom), but the term radiation may correctly also refer to non-ionizing
radiation (e.g., radio waves, heat or visible light).
The particles or waves radiate (i.e., travel outward in all directions) from a source. This aspect leads to a system of measurements and physical units that are applicable to all types of radiation.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 4
RadiationBoth ionizing and non-ionizing radiation can be harmful to organisms and can result in changes to the natural environment.
In general, however, ionizing radiation is far more harmful to living organisms per unit of energy deposited than non-ionizing radiation, since the ions that are produced by ionizing radiation, even at low radiation powers, have the potential to cause DNA damage.
By contrast, most non-ionizing radiation is harmful to organisms only in proportion to the thermal energy deposited, and is conventionally considered harmless at low powers which do not produce significant temperature rise.
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 5
RadiationUltraviolet radiation in some aspects occupies a middle ground, in having some features of both ionizing and non-ionizing radiation.
These properties derive from ultraviolet's power to alter chemical bonds, even without having quite enough energy to ionize atoms.
Although nearly all of the ultraviolet spectrum of radiation is non-ionizing, at the same time ultraviolet radiation does far more damage to many molecules in biological systems than is accounted for by heating effects (an example is sunburn).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 6
Harm to Biological SystemsThe question of harm to biological systems due to low-power ionizing and non-ionization radiation is not settled.
Controversy continues about possible non-heating effects of low-power non-ionizing radiation, such as non-heating microwave and radio wave exposure.
Non-ionizing radiation is usually considered to have no completely safe lower limit, although at some energy levels, new exposures do not add appreciably to background radiation. The evidence that small amounts of some types of ionizing radiation might confer a net health benefit in some situations is called radiation hormesis.
Radiation from Atomic and
Nuclear Processes
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 8
Radiation - CategoriesWe are concerned in this course by two categories of radiation
Charged particle radiationFast electronsHeavy charged particles
Electromagnetic radiationNeutrons
Uncharged radiation
{
{
The sourcessources of these radiations will be discussed in this chapter.
3
Units
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 10
MKS is no more appropriate when studying physics at the microscopic level (atom, nucleus, nucleons, quarks, etc …). The MLT dimensions are replaced by a new "scale" based on energy and distance,
whose basic units are the electron-volt (eV) and
the Fermi (F or fm)).
New "Appropriate" Units
FMeVnmeV
meVmJhc
⋅⋅⋅⋅====⋅⋅⋅⋅====
⋅⋅⋅⋅××××====⋅⋅⋅⋅××××==== −−−−−−−−
5.12415.1241
102415.1109864457.1 525
mFJeV 1519 101&106.11 −−−−−−−− ====××××====
In this new system we have:
and the useful approximate value:
FMeVc ⋅⋅⋅⋅≈≈≈≈ 200h
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources
11
The eV/c2 Unit وحدة الكتلة :eV/c2
E0 =Energy corresponding
to m0
Electron 9.092933 ×××× 10-31e 8.181 10-14 J ≈≈≈≈ 511003 eV
≈≈≈≈ 511 KeV = 0.511 0.511 MeVMeV
Proton 1.672649 ×××× 10-27p 1.5033168 ×××× 10-10 J ≈≈≈≈ 9.3828××××108 eV
Neutron 1.674955 ×××× 10-27n 1.5074959 ×××× 10-10 J ≈≈≈≈ 9.4218 ×××× 108 eV
Alpha 6.644766 ×××× 10-27αααα
Particle Mass at rest
m0(Kg)
Symbol
3727.409 3727.409 MeVMeV
Table 1: Rest mass for some fundamental particles
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources
12
The eV/c2 Unit وحدة الكتلة :eV/c2
0.5110.511
m0 (in MeV/c2)
939.573939.573
938.28938.28
3727.4093727.409
Table 1: Rest mass for some fundamental particles
Electron 9.092933 ×××× 10-31e
Proton 1.672649 ×××× 10-27p
Neutron 1.674955 ×××× 10-27n
Alpha 6.644766 ×××× 10-27αααα
Particle Mass at rest
m0(Kg)
Symbol
4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 13
me = 0.511 MeV/c2ElectronElectron
mp = 938.3 MeV/c2ProtonProton
For the masses of fundamental particles and using the mass-energy equivalence we have:
Mass and Energy
mαααα = 3727.4 MeV/c2 = 3.7274 GeV/c2Alpha Alpha
mn = 939.573 MeV/c2NeutronNeutron
Appendix 1
Relativity
Appendix 2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources
15
Example – de Broglie Wavelength
The de Broglie wavelength for a particle of
momentum p is:
p
h====λλλλ
For a non-relativistic particle we have:
(((( ))))nmeVcmcm
ch
cvm
ch
e ββββ====
ββββ========λλλλ
)(
12402
02
0
For a relativistic particle (m = γγγγm0) we have:
ββββγγγγ====
γγγγ====λλλλ
200 cm
ch
cvm
ch
See Homework 1
See Homework 1
Spectroscopy
Spectrometry
5
Radiation Sources
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources
22
RadiationFast electrons:• Beta Decay (ββββ- ββββ+ εεεε)• Internal Conversion• Auger Electrons
Electromagnetic radiation:• Gamma-rays and X-rays
• Annihilation Radiation• Bremsstrahlung• Synchrotron Radiation
Heavy charged particles:• alpha particles• Fission products
Neutrons:• Fission
• Radioisotope ( α α α α,n) Sources• Photoneutron Sources• Reactions from Accelerated Charged Particles
Radioactivity
Appendix 3
Fast Electron Sources
6
- Beta Decay
- Internal Conversion
- Auger Electrons
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 27
Main Sourceββββ Decay is the main source!
� Natural beta decay sources:
� Beta decay of artificially produced isotopes
These isotopes are produced by bombarding stable material by a neutrons flux from a reactor.
For these isotopes, beta decay populates, generally, an excited state in the daughter nucleus.
The de-excitation of this exited state produces a gamma ray. It is emitted with the electron from the original beta decay in many common beta sources.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 28
Beta Decay - Decay Scheme
Natural beta decay sources:
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 29
Pure Beta EmittersSome Isotopes beta decay to the daughter's ground state. Table 2 shows some examples:
Nuclide Half-life Endpoint Energy (MeV)
3H 12.26 y 0.0186
14C 5730 y 0.156
32P 14.28 d 0.248
32S 87.9 d 0.167
99Tc 2.12 × × × ×105 y 0.292
Table 2
7
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 30
Beta Decay SpectrumFig. 2 shows the decay scheme for the beta decay of 36Cl. Fig. 3 shows the corresponding electron spectrum
This value is the maximum kinetic energy the electron can have in this decay (in this case the energy of the neutrino is zero).
Figure 3Figure 2
The beta decay spectrum is continuous and has a cut-off value called the endpoint energy.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 31
CalibrationThe beta decay spectrum is continuous and has a cut-off value called the endpoint energy.
This value is the maximum kinetic energy the electron can have in this decay (in this case the energy of the neutrino is zero).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 32
Internal ConversionThis is another source of fast electrons.
In this process, the de-excitation of an excited nuclear state does not occur by gamma emission. The available energy (The difference between the initial excited state and that of the final state (which also could be an excited state) is "transferred" to one of the atom's electron)
If Eex is the nuclear excitation energy and Eb is the
binding energy of the electron in the original shell, then the energy of the emitted electron is:
bexeEEE −−−−====−−−−
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 33
Internal Conversion - Example
Fig. 4-a shows the internal conversion (decay
scheme) of the isomeric level at 393 keV in 113mIn. Fig. 4-b shows the conversion electron spectrum
expected from this decay.
393 keV
0
(100 min) t1/2113mIn
113In
Internal
conversion
Figure 4-bFigure 4-a
8
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 34
Internal Conversion – Common Sources
Table 3 shows some common conversion
electron sources.
Table 3© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 35
Auger Electrons
This is the equivalent process of the internal
conversion but at the level of the atom.
The de-excitation of an excited atom (in general an
atom with a vacancy in a shell) is achieved when
an electron from higher shells "fills" the vacancy.
The energy difference (excitation energy) is
emitted as a characteristic X-ray.
Another competitive process exists, the so-called
Auger effect. Here the energy is (re) transferred to
an electron from the atom itself. The resulting
electrons are called the Auger electronsAuger electrons.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 36
Auger Electrons
The process is schematized in Fig. 5
Figure 5: Auger Effect
Source: The Graz University group Austria.http://surface-science.uni-graz.at/main_frame/techniques/aes.htm
Heavy Charged
Particle Sources
9
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 38
Alpha ParticlesAlpha decay
Figure 6: 238Pu → → → →234U + αααα
Fig. 6 shows the decay
scheme for the alpha
decay of 238Pu and the
corresponding energy spectrum.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 39
Energy and Half-Life
Alpha particles are monoenergetic. As can be seen in Fig. 6 the decay may involve more than
one transition energy.
There is a strong correlation between the energy of an alpha particle and the half-life of its parent. See Supplement 2 Part2(Systematics of alpha decay)
Table 4 shows the common alpha-emitting
radioisotope sources
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 40
Common Alpha-Emitting Sources1
Table 4-a© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 41
Common Alpha-Emitting Sources2
Table 4-b
10
42
Common Alpha-Emitting Sources3
Table 4-c© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 43
Spontaneous FissionFission is the only spontaneous source of energetic
heavy charged particles with mass greater than mαααα.
Fission Fragments are therefore widely used in the calibration and testing of general application to heavy ion measurements.
The most widely used example is 252Cf, which
undergoes spontaneous fission with a half-life of 85 years. However, alpha decay is a competitive process
to fission and for 252Cf the probability of alpha decay is even higher that that of spontaneous fission.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 44
252Cf
Therefore the actual halfTherefore the actual half--life for life for 252252CfCf is is 2.65 2.65 yearsyears and a and a
sample of sample of 1 1 µµµµµµµµgg of this isotope will emit of this isotope will emit 1.921.92××××××××101077 alpha alpha
particle and undergo particle and undergo 6.146.14××××××××101055 spontaneous fissions per spontaneous fissions per
second! A number of fast neutrons is liberated during second! A number of fast neutrons is liberated during
the fission process.the fission process.
Each fission gives rise to two fission fragments, Each fission gives rise to two fission fragments,
which by the conservation of momentum are which by the conservation of momentum are
emitted in opposite directions. Because the normal emitted in opposite directions. Because the normal
physical form for a spontaneous fission source is a physical form for a spontaneous fission source is a
thin deposit on a flat backing, only one fragment thin deposit on a flat backing, only one fragment
can escape from the surface, whereas the other can escape from the surface, whereas the other
fragment is lost by absorption within the backingfragment is lost by absorption within the backing
Homework: How is Homework: How is 224224CfCf obtained?obtained?
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 45
Mass Distribution of Fission Fragments
Fig. Fig. 77 shows the mass distribution of the fission shows the mass distribution of the fission
fragments which are mediumfragments which are medium--weight positive ions.weight positive ions.
Figure 7: Mass distribution of fission fragments
11
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 46
Mass Distribution of Fission Fragments
The fission is predominantly asymmetric so The fission is predominantly asymmetric so
that the fragments are "clustered" into a "that the fragments are "clustered" into a "light light
groupgroup" and a "" and a "heavy groupheavy group" with average mass " with average mass
numbers of numbers of 108108 and and 143143..
The fragments appear initially as positive ions The fragments appear initially as positive ions
for which the net charge approaches the for which the net charge approaches the
atomic mass number (atomic mass number (ZZ) of the fragments. As ) of the fragments. As
the fragment slows down by interacting with the fragment slows down by interacting with
the matter through which it passes, additional the matter through which it passes, additional
electrons are picked up by the positive ion, electrons are picked up by the positive ion,
thus reducing its effective charge.thus reducing its effective charge.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 47
Energy Distribution of Fission Fragments
Fig. Fig. 88 shows the energy shared by the two fragments. shows the energy shared by the two fragments.
It averages about It averages about 185 185 MeVMeV..
Figure 8: Mass distribution of fission fragments
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 48
Energy Spectrum is Asymmetric
The energy spectrum is asymmetric as the The energy spectrum is asymmetric as the
mass distribution, the light fragment mass distribution, the light fragment
receiving a greater fraction of the available receiving a greater fraction of the available
energy.energy.
Fig. 9Fig. 9 shows the initial energy distribution. shows the initial energy distribution.
One should take into account energy loss One should take into account energy loss
and self absorption unless the source and self absorption unless the source
(parent) is prepared in a very thin layer (of (parent) is prepared in a very thin layer (of
material).material).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 49
Degradation of a fission fragment Energy
Figure 9: Mass distribution of fission fragments
12
Sources of
Electromagnetic RadiationSources of Gamma - Rays
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 52
A) Gamma Rays Following Beta Decay
Gamma decay
Excited nuclear states are created in the decay of a parent radionuclide.
Gamma radiation is emitted by excited nuclei in their transition to lower-lying nuclear levels.
Fig. 10 shows the four most common sources of
γγγγ-rays used for calibration in the lab.
In each case, a beta decay (ββββ- ββββ+ or electron capture εεεε) is the cause of populating the excited state in the daughter nucleus.
53
Common Sources
The beta decay in these four cases is rather a relatively
slow process (half-life > 100 days) whereas the excited states in the daughter have a much shorter average lifetime (of the order of picoseconds or less)
t½ = 2.602 y
t½ = 271.8 d
t½ = 5.3 y
t½ =30.17 y
Figure 10: Common Sources
13
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 54
Gamma Rays and Calibration
The nuclear states have very well-defined energies. Thus the energies of gamma rays are nearly monoenergetic. The inherent line width of the photon energy distribution is nearly always compared with the energy resolution of the detectors we shall see in this course.
Broadening of spectral lines© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 55
"High Energy" Calibration Sources
The fours sources we have seen provide γγγγphotons with energies below 2.8 MeV. For higher energies other sources are used.
56Co (t½ = 77 days, Eγγγγ = 3.55 MeV) is used for higher energies calibrations but its use is limited because we need access to accelerators where this isotope is created (through the reaction 56Fe(p,n)56Co). 16N is another useful source for high-energy
calibration (Two gamma's Eγγγγ = 6.13 MeV and Eγγγγ =
7.11 MeV) but again and due to its very short
half-life (t½ = 7 s) its use should be near the production facility!
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 56
"High Energy" Calibration Sourcescont'd
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 57
B) Annihilation Radiation
• The positron loses simply its energy in the "encapsulating" material of the parent sample,
In the case of ββββ+ emitters, the resulting positron which generally travels only a few millimeters may suffer one of the following processes:
•The positron could be "trapped" by an electron and they both form a quasistable combination, the "positronium" which lives for a few nanoseconds.
•A very low energetic positron combine with an electron and they annihilate producing two oppositely directed gamma rays each with energy
0.511 MeV.
14
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 58
C) γ-Rays Following Nuclear Reactions
An example is the production of a 4.4 MeV gamma ray
from the excited state of 12C* produced in the nuclear
reaction:
Those gamma-rays are needed for high energy calibration processes. Again this is restrained by the short life-times of the excited states of the reaction's products
The average lifetime of this state is so short (61 fs) that the recoil carbon atom does not have time to come to rest before the emission of the gamma photon.
nCBe 10
*126
94
42 ++++→→→→++++αααα
This produce a "Doppler effect" that should be taken into account when calibrating detectors! An inherent
spread in the photon's energy of 1% is thus observed.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 59
The 16O example
Energy of the excited state is 6.130 MeV above the
ground state and the lifetime is ~ 2××××10-11 s. This lifetime
is sufficiently long to eliminate all Doppler effects
possible and the resulting 6.130 MeV photons are
essentially monoenergetic.
Another example is 16O produced in the reaction:
The previous examples are also used in neutron production.
nOC 10
*168
136
42 ++++→→→→++++αααα
We shall come back to this in the following section (neutron sources).
See also the neutron capture gamma rays": Knoll p13.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 60
D) Bremsstrahlung – Braking Radiation
In the (classical) electromagnetic theory part of the energy of an accelerated particle is converted into electromagnetic radiation.
The consequent energy loss has a "decelerating" effect on the particle.
The power radiated (energy loss per time unit) is given (for two limiting cases) by:
(((( ))))620
30
622
30
622
||
66 cmc
Eaq
c
aqP va
εεεεππππ====
εεεεππππ
γγγγ====
(((( ))))420
30
422
30
422
66 cmc
Eaq
c
aqP va
εεεεππππ====
εεεεππππ
γγγγ====⊥⊥⊥⊥
q is the charge of the particle and γγγγ is its Lorentz
factor. a being the acceleration
Acceleration // velocity
Acceleration ⊥ ⊥ ⊥ ⊥ velocity
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 61
Bremsstrahlung2
The em radiation seems to "brake" the particle. We call this radiation braking radiation and we use the German word "Bremsstrahlung"
Usually we reserve the term bremsstrahlung for the radiation resulting from the energy loss by electrons into matter. In an X-ray tube, the bremsstrahlung part of the spectrum is very important.
In the case where the acceleration is perpendicular to the velocity of the charged particles, the resulting radiation is called "synchrotron radiation"
The bremsstrahlung spectrum is a continuous one, thus not valid for calibration purposes. It can be altered, as we also we do with X-ray spectra, using the appropriate filters or absorbers to produce "relatively" monoenergetic photons for calibration – See next chapters.
15
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 62
Bremsstrahlung Spectrum Fig. 11 shows the bremsstrahlung spectrum of electrons
of 5.3 MeV incident on a Au-W target.
A 7.22 g/cm3 aluminum filter
was also present.© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 63
E) Characteristic X-Ray
Other sources of bremsstrahlung are beta decay electrons.
A characteristic X-ray spectrum is also emitted when electrons pass through matter
The deceleration could be a serious problem when accelerating charged particles, because of the permanent and continuous energy loss by em radiation.
Nevertheless, this em radiation is used to produce X-rays (The multiple uses of X-rays are (should be) well-known).
As for the synchrotron radiation, we now build electron accelerators (cyclotrons) in order to produce intense beams of photons of major interest in applied and fundamental physics – See later)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 64
Characteristic X-Ray
We leave section E (Knoll pages: 15-19) for self reading.
Revise the structure of an X-ray tube. Be sure you understand a characteristic X-ray spectrum produced in such a tube.
Spectrum of the X-rays emitted by an X-ray tube with a
rhodium target, operated at 60 kV. The continuous curve is due to bremsstrahlung, and the spikes are characteristic K lines for rhodium.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 65
F) Synchrotron RadiationWhen a beam of charged particles is bent into a circular orbit, as in a cyclotron, a small fraction of the beam energy is radiated away during each cycle of the beam. This emitted electromagnetic radiation is called synchrotron radiation.
The term synchrotron radiation is almost exclusively used nowadays for electron accelerators.This radiation is considered as a nuisance for designers of high-energy accelerators (LEP is the most famous one).
Special cyclotrons are built in order to produce this radiation and use it in many important field of physics. (See SESAME, the synchrotron facility now being installed in Jordan. www.sesame.org.jo)
16
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 66
Extra Assignment
1) X-ray Tubes
3) Electron Accelerators – LEP a case study
Due by wed. march 11, 2013
2) SESAME
Write a concise essay (1 to 2 pages) on the following topics:
For this essay and all others that will follow:1)Wikipedia is a starting point. It is not a source.2) Use at least one academic textbook (except Knoll).
Neutron Sources
© Dr. Nidal M. Ershaidat
Next Lecture
Chapter 2Radiation Interactions with
Matter
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Chapter 1: Radiation Sources (Preview)
Neutron Sources
3
Neutron Sources
http://nuclear.dababneh.com/Experimental/Notes/2008/NeutronSourcesByStudents.pdf © Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 4
Neutron SourcesA)Spontaneous Fission
B)Radioisotopes (αααα,n) SourcesC)Photoneutron SourcesD)Reactions from Accelerated Charged Particles
2
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 5
A) Spontaneous FissionMany of the transuranic heavy nuclides have an appreciable spontaneous fission decay probability.
Several fast neutrons are promptly emitted in each fission event, so a sample of such a radionuclide can be a simple and convenient isotopic neutron source.Other products of the fission process are:-The heavy fission products described earlier- Prompt fission gamma rays and- Beta and gamma activity of the fission products accumulated within the sample.
When used as a neutron source, the isotope is generally encapsulated in a sufficiently thick container so that only the fast neutrons and gamma rays emerge from the source.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 6
αααα Particles being available from the direct decay of a number of convenient radionuclides, small self-contained neutron sources are fabricated by mixing an alpha-emitter isotope with a suitable target material.
B) Radioisotope (α,n) Sources
The basic reaction is:
This reaction has a Q-value of 5.71 MeV
α α α α α α α α + + 99BeBe→→→→→→→→ 1212CC + + nn
The best of all targets is Be which gives a
maximum neutron yield.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 7
Q-value of α + 9Be→ 12C + ns
(((( ))))(((( ))))(((( ))))(((( )))) MeVnm
umaCm
umaBem
umaHem
573.939
..000000.12
..012182.9
..002603.4
10
126
94
42
====
====
====
====
(((( )))) (((( )))) (((( )))) (((( ))))[[[[ ]]]](((( ))))
MeV
cnmCmBemHemQ
6992275.5573.9392722275.945
573.9395.931000000.12012182.9002603.4
210
126
94
42
====−−−−====
−−−−××××−−−−++++====
−−−−−−−−++++====
1 a.m.u = 931.5 MeV/c2
8
Neutrons are not monoenergetic
s
http://nuclear.dababneh.com/Experimental/Notes/2008/NeutronSourcesByStudents.pdf
3
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 9
Neutron Yield (α + 9Be→ 12C + n)
sFig. 11 shows the neutron yield from the reaction 9Be(αααα,n)12C.
Fig. 11: Thick Target yield of n from a particles on Be
(Anderson and Hertz – See Knoll reference 22)© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 10
Best AlloyssActinides elements are alpha emitters of practical interest. A stable alloy can be formed between the
actinides and Beryllium of the form MBe13, where Mrepresents the actinide metal. Each alpha particle emitted by the actinide has an opportunity to interact with beryllium nuclei without any intermediate energy loss.The following table shows some of the 9Be(αααα,n)12C
sources and the properties of the produced neutrons.
Practical compromises determine the best alloy/source to be used depending on the application it is intended for.
See Knoll – page 21 for discussion of the Ra-Be and Ac-Besources
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 11
Characteristics of Be (α n) Neutron Sources
s
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 12
Choice CriteriasThe choice is essentially governed by three factors: availability, cost and half-life.
The alloy is sealed within two welded stainless steel cylinders.
Some expansion space is allowed in order to contain the helium gas which may result from the neutralization of some alphas which do
interact with Be.
Fig. 12: A typical construction for a neutron source
4
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 13
Alternative Sourcess
See discussion in Knoll page 24
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 14
Safety ConsiderationssThe actinide isotope in the source is generally very active. A standard container of the neutron source is shown in Fig. 14.
Homework
1) How is 244Cm obtained?
2) How are Be(αααα,n) sources fabricated?
3) How isSeaborgium obtained?
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 15
γγγγ Rays can also be used to produced neutrons.
C) Photoneutron (γ,n) Sources
This implies the use of relatively high energy gammas.
Typical targets are 9Be and 2H (Deuterium).
The reactions are written as:
nBehBe 10
84
94 ++++→→→→νννν++++ --1.666 1.666 MeVMeV
nHhH10
11
21 ++++→→→→νννν++++ --2.226 2.226 MeVMeV
The minus sign means that the photon should have
at least the negative of the Q-value of the reaction
to make the reaction energetically possible.
16
M = mass of recoil nucleus ×××× c2
If the energy of the gamma ray used exceeds this minimum, the corresponding neutron energy can be calculated (simple kinematics) and we have:
Neutron Energy
(((( )))) (((( ))))(((( ))))(((( ))))[[[[ ]]]](((( ))))
44444444 344444444 2143421
21
cos2
2
21
T
n
nn
T
nn
Mm
QEMmMmE
Mm
QEME θθθθ
++++
++++++++++++
++++
++++≅≅≅≅
γγγγγγγγγγγγ
Exercise: Find this equationExercise: Find this equation
θθθθ = angle between the gamma photon and neutron direction,
where
Eγγγγ = gamma energy (assumed << 938 MeV)
mn= mass of neutron ×××× c2 = 939.573 MeV
5
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 18
•If the gamma rays are monoenergetic then the neutrons are nearly monoenergetic
Advantages and Disadvantages
•The relatively small kinematic spread by letting θθθθ
vary from 0 to ππππ broaden the neutron energy
spectrum by only a few percent.
•For large source, a degradation of the spectrum is observed because of the scattering of some neutrons within the source before their escape.
Advantages
Disadvantages
•Main disadvantage is the need of very large activities of the gamma rays source in order to produce neutron sources of attractive intensities.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 19
Photoneutron Source Container
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 20
Neutron Spectra
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 21
Characteristics of Photoneutron Sources
•The half-lives of the gamma emitters are short enough to require their "reactivation" in a nuclear reactor between uses.
6
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 22
Here the projectiles are protons, deuteron, etc …
D) Reactions from Accelerated Charged Particles
Need for accelerators
The most common reactions are:
nHeHH 10
32
21
21 ++++→→→→++++ +3.26 +3.26 MeVMeV
nHeHH 10
42
31
21 ++++→→→→++++ +17.6 +17.6 MeVMeV
DD--D reactionD reaction
DD--T reactionT reaction
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 23
The Coulomb barrier between the incident deuteron and the light target is relatively small, the deuterons need not to be accelerated to very high energy in order to create a significant neutron yield.
Kinematics of the D-D and D-T Reactions
In "Neutron generators", deuterons are
accelerated to 100-300 keV.
The incident deuteron energy is small compared to
the Q-value of the D-D or D-T reactions and hence
the neutrons produced have about the same
energy (~3 MeV for the D-D reaction and 14 MeV for
the D-T reaction).
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources 24
Neutron Yield
Other reactions using Charged Particles
© Dr. Nidal M. Ershaidat
Next Lecture
Chapter 2Radiation Interactions with
Matter
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Supplement 1: Beta Decay
Introduction
Energy Release in ββββ Decay
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
3
IntroductionElectron emission, positron emission (1934, I. & F.
Joliot-Curie) and (orbital) electron capture (1938,
Alvarez) are all known as beta decay processes
eepn νννν++++++++→→→→ −−−− −−−−ββββ
νννν++++++++→→→→ ++++enp ++++ββββνννν++++→→→→++++ −−−− nep εεεε
The electron resulting from a ββββ- decay process is
“created” thanks to the energy available.
(Electrons do not preexist inside a nucleus).
The processes involving protons occur only for
bound protons in nuclei (The presence of the
“nucleus field” is a sine qua non condition)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
4
What really happens! We know now that the weak interaction is
responsible for beta decay.
eepn νννν++++++++→→→→ −−−−
In ββββ-, The W- mediates the interaction. One of the d
quarks of the neutron transform into a u quark.
The neutron, thus, becomes a proton. An electron
and its anti-neutrino are emitted in the process.
Fig 1: ββββ- decay
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
5
7.2 s
2.1××××105 y
38 s
Typical Beta Decay Processes
23Ne →→→→ 23Na + e-+e
ν ββββ- 4.38
Decay Type Q(MeV) t1/2
3.2625Al →→→→ 25Mg + e+ + ννννe ββββ+
4.2 d2.14124I →→→→124Te + e+ + ννννe ββββ+
1.22 s15O + e- →→→→ 15N + ννννe 2.75εεεε
1.0××××105 y0.4341Ca + e- →→→→ 41K + ννννe εεεε
0.2999Tc →→→→ 99Ru + e- + ββββ-e
ν
Exercise: Check the Exercise: Check the QQ--valuevalue for all these reactions.for all these reactions.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
6
We expect to have mono-energetic electrons as we observe
the mono-energetic alpha’s in a decay. But instead we have
for the electrons resulting from a ββββ decay a continuous
spectrum starting at 0 and ending at Emax (Endpoint energy)
which is the energy an electron should have in this decay!
Energy Continuous Spectrum in β Decay
Fig 1: ββββ- decay
spectrum, i.e. Energy Distribution of the
electrons.
Energy Release in ββββ Decay
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
8
Energy Release in β Decay
Beta decay of 210Bi
−−−−++++→→→→ ePoBi 12621084127
21083
Q = ( 209.984095 - 209.982848) ×××× 931.5002 - 0.511
= 0.650 MeV
Neglecting the recoil energy of the daughter,
the maximum energy an electron can have is :
Emax = 0.650 + 0.511 = 1.161 MeV
3
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
9
The Neutrino
Experiments showed that the shape of the
spectrum of electrons emitted in ββββ decay is
characteristic of the electron themselves.
In 1931 Pauli, suggested the presence of a
second particle emitted in the decay which can
carry a part of the available energy and linear
momentum. This particle should have a zero
mass, be neutral and interacts so weakly with
matter that detectors do not “see” it! Pauli
called it The ghost particle and Fermi gave it
the name neutrino (small neutron in Italian).
The neutrino was discovered in 1957!
Kinematics
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
11
Kinematics
Q = (mn - mp - me - mνννν ) c2
eepn νννν++++++++→→→→ −−−− −−−−ββββ
Consider the ββββ- decay of a free neutron (t1/2 = 10 min)
eTTTQ
ep νννν++++++++==== −−−−
In a frame attached to the decaying neutron, the
available energy will be shared by the 3 resulting
particles:
Neglecting the proton’s recoil energy Tp, which is
measured to be 0.3 keV, Q is essentially shared by
the electron and the antineutrino.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
12
The β- Decay Electron is Relativistic
The energy carried by the electron in ββββ- is of
the order of its rest mass energy Te/ me c2 > 0.1,
while the recoil energy is low and can be
taken non relativistically.
4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
13
The electron-(anti)neutrino is massless
The measured value is Q = 0.782 ±±±± 0.013 MeV. This
suggests that the mass of the antineutrino = 0
within the experimental error (13 keV). New
experiments give very much lower limits (few eV's)
2
2
2222
782.0
511.0280.938573.939
cmMeV
cm
cmcmcmcmQ
e
e
eepn
νννν
νννν
νννν
−−−−====
−−−−−−−−−−−−====
−−−−−−−−−−−−==== −−−−
Measurements of the linear momentum of the
electron and the proton indicate that a 3rd
particle should be present.© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
14
β- Decay Kinematics
where N indicates the nuclear mass and energy
and considering a massless antineutrino.
(((( )))) (((( )))) 21 ][ cmYmXmQ
e
AZN
AZN −−−−−−−− −−−−−−−−==== ++++ββββ
−−−−νννν++++++++→→→→ −−−−−−−−++++ eN
AZN
AZ eYX 11
(((( )))) (((( )))) 21 ][ cYmXmQ A
ZAZ ++++ββββ
−−−−====−−−−
Masses here are neutral atomic masses
(tables)e
TTQ e ννννββββ++++====−−−−
Neglecting the electrons binding energies we
have :
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
15
Q is shared between e- and
This is the energy shared between the electron
and the antineutrino.We saw that in the case
of ββββ- Decay of 210Bi,
Q = 1.161 MeV
is in good agreement with the measured value.
−−−−ννννe
This measurement is used to calculate the
mass of the 210Po isotope.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
16
β+ Decay Kinematics
(((( )))) (((( )))) 21 ][ 2 cmYmXmQ
e
AZ
AZ −−−−++++ −−−−−−−−==== −−−−ββββ
eNA
ZNAZ eYX νννν++++++++→→→→ ++++
−−−−−−−− 11
5
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
17
Electron Capture
eNA
ZNAZ YeX νννν++++→→→→++++ ++++−−−−
−−−−11
(((( )))) (((( )))) nA
ZAZ BcYmXmQ −−−−−−−−==== −−−−εεεε
21 ][
Where Bn is the binding energy of the captured
electron from shell n (K, L, M,…)
Note that here the neutrino is mono-energetic. And
if we neglect the recoil energy of AY , Eνννν = Q.
The calculation of Q should take into account
the fact that the daughter nucleus is in an excited state. The resulting X-ray (or X-rays) should have the binding energy of the captured electron Fermi Theory of ββββ Decay
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
19
Fermi Theory of β DecayThe theory to explain ββββ Decay should include the following information :
1) The electron and the neutrino do not preexist in the nucleus
2) The electron and the neutrino are relativistic.
3) The continuous distribution of electron energies
Enrico Fermi proposed, in 1934, a theory of ββββ decay based on Pauli’s neutrino hypothesis. The major idea is that ββββ Decay is the result of a weak interaction (compared to that responsible for the quasi-stationary states). The characteristic times in ββββ decay is of the order of seconds or longer where the nuclear characteristic time is of the
order of 10-20s.© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
20
The barrier potential used in alpha decay (See Suppl2-Alpha decay) does not exist in the case of ββββDecay. And even though if it exists, the
transmission probability is nearly 1.
Fermi Theory - Transition probability
dvψVψV i*ffi ∫∫∫∫====
ρρρρ(Ef) is the final density of final states = dn/dE
dn is the number of final sates per energy interval
dE and Vfi is the matrix element:
(((( ))))ffi EρVπ
λ22
h====
The transition probability is given by :
Fermi’s idea was to consider ββββ decay as a perturbation forcing the quantum system (The parent nucleus) in a transition.
6
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Beta Decay
21
Fermi had no idea about the weak interaction potential so he tried all possible forms consistent
with special relativity, and showed that V can be
replaced by an operator OX, where X gives the form
of the operator O.
Fermi Theory - Perturbation Potential
X = V (vector potential),
X = A (Axial vector)
X = S (Scalar)
X = Pseudoscalar
X = Tensor
Only experiment can help deciding which transformation is the appropriate one. Now we
know that X is the so-called V-A (Vector-Axial transformation).
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Supplement 2: Alpha Decay
Basic αααα Decay Processes
(Introduction and Kinematics)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
3
Seven years after Becquerel’s discovery, Rutherford
(and Mme Curie) identified the naturally emitted αααα
particles as being less penetrating comparatively to
the other emitted ones (β β β β & γγγγ).
Radioactive source
αββββ
γ
Magnetic Field
Introduction
By simply using a deflecting magnetic field Marie Curie
demonstrated that a particles are doubly positively
charged.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
4
Rutherford, using an evacuated closed
chamber with a thin wall accumulated αααα
particles emitted by radium for several days,
proved by atomic spectroscopy that helium gas
is formed and thus that an αααα particle is in fact a 4He nucleus (or a doubly ionized helium atom).
α Particles are Helium 4 Nuclei
This means that an αααα source ejects a cluster of
4 nucleons (2p & 2n).
Why?
One can imagine that it is easier and simpler
for an unstable nucleus to eject a single
nucleon or 2 nucleons.
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
5
Q-Value
The Q-value is simply the difference between
the available initial rest energy and the
resulting final rest energy.
The reaction is energetically possible if and
only if Q > 0.
bBAa ++++→→→→++++In a nuclear reaction
The net energy release called the Q-value of the
reaction is given by:
(((( )))) (((( ))))2222cmcmcmcmQ BbAa ++++−−−−++++====
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
6
Q-Value - Generalization
More generally, in a nuclear reaction involving
s nuclei or nucleons and the result of which is
the creation of t nuclei or nucleons the
(general) definition of the Q-value is:
∑∑∑∑∑∑∑∑========
−−−−====t
ff
s
ii cmcmQ
11
22
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
7
α Decay Kinematics
Example2
422
42 αααα++++→→→→ −−−−
−−−−−−−− N
AZN
AZ YX
αααα++++→→→→ 13422286136
22688 RnRa
MeVE 8.4====αααα
t1/2(226Ra)=1600y
In an αααα decay, an unstable nucleus X emits an αααα
(4He nucleus) and transforms to nucleus Y.
In these conditions the only available energy is the
rest mass energy of AX.
αααα++++++++++++====
ααααTcmTcmcm YYX
222
If we consider that the nucleus AX decays while at rest
(or equivalently we consider a referential attached to
this nucleus) then conservation of energy gives:
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
8
The available energy Q is shared by Y and the αααα
particle inversely proportionally to their respective
masses, i.e. the αααα particle kinetic energy is much
bigger than the nucleus Y’s kinetic energy.
Momentum and Kinetic Energy of α
In the cm referential, X at rest (PX = 0), the
resulting nucleus Y and the αααα particle are
emitted in opposite directions and we have :
12
2
2
2
>>>>========⇒⇒⇒⇒====αααα
αααααααα
αααααααα
m
m
mP
mP
T
TPP Y
YYY
Y
−−−−
AQ
41
Ymm
QT
αααα++++
====αααα1 4>>>>>>>>
→→→→A
αααα++++====−−−−−−−−====
ααααTTcmcmcmQ YYX
222
3
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
9
α Decay is energetically favored
Emitted particle
(mass in MeV/c2) Q (MeV)Decay
n (939.573))))1
10139
23192140
23292 nUU ++++→→→→ - 7.26
p (938.280)0
11140
23191140
23292 HPaU ++++→→→→ - 6.12
2H (1875.628)1
21139
23091140
23292 HPaU ++++→→→→ - 10.70
4He (3728.433)2
4
2138
228
90140
232
92HeThU ++++→→→→ + 5.41
Q=(232.0366 - 231.0357) ×××× 931.502 - 939.573 = - 7.26 MeV
2
1
1
0139
232
92140
232
92
cmmmQnUU
−−−−−−−−====
Let’s look at the possible emission of a
decaying 232U nucleus (mass = 232.0366 u).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
10
Computing Q for “possible” decays of 232U
Decay
Q = (232.0366 - 231.03558)××××931.502 - 938.280= - 6.12 MeV
011140
23191140
23292 pPaU ++++→→→→
Decay
Q = (232.0366 - 230.03457)××××931.502 - 1875.68 = - 10.70 MeV
121139
23091140
23292 HPaU ++++→→→→
Decay
Q = (232.0366 - 228.028715)××××931.502 - 3727.409= + 5.41 MeV
242138
22890140
23292 HeThU ++++→→→→
Decay
Q=(232.0366 - 226.02608- 6.0151)××××931.502 = - 3.79 MeV
363137
22689140
23292 LiAcU ++++→→→→
Decay1
10130
22292131
22392 nUU ++++→→→→
Q = (232.0366 - 231.0357)××××931.502 - 939.573= - 7.26 MeV
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
11
Is Q > 0 the unique condition for an α Decay to occur?
The previous calculations show that only alpha
decay can occur spontaneously (Q > 0). But this is not the only criterion. For example the following
decays are energetically possible(Q > 0)
484136
22288140
23292 BePbU ++++→→→→
6126134
22086140
23292 CRnU ++++→→→→
Exercise : Check that!
However, nuclear spectroscopy shows that such decays have vanishingly small partial decay constants compared to a decay, and thus they are not seen.
In some cases, beta decay is intense enough to mask possible αααα decays.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 1
12
Criteria for an α Decay to occur
1. Q-value should be > 0
2.The partial decay constant should be large
enough This corresponds, according to our
technological limits to half-lives of the order of 1016
years.
3. αααα decay should not be masked by ββββ decay.
Half of the unstable nuclei against alpha decay (A >
190 and many nuclei in the range 150 < A < 190)
verify this criterion.
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Supplement 2:Part 2- Systematics and Theory of αααα emission
Systematics of αααα emission
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
3
1) Emitters with large disintegration energies are
short-lived and vice versa (Geiger & Nuttall, 1911)
Systematics of α Decay
For a factor 2 in energy, the half-lives are 10-24
times different !!
Nuclide t1/2 Qαααα (MeV)
232Th 1.4 x 1010 y 4.08
1.0 x 10-7 s = 100 ns 9.80218Th
(((( ))))(((( )))) 41.2232
218
====αααα
αααα
ThQ
ThQ (((( ))))(((( )))) !10
24
23221
21821 −−−−====
Tht
Tht
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
4
Geiger & Nuttall
Fig. 6.1 : Qαααα vs. t1/2
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
5
2) Adding a neutron to an unstable nucleus
reduces the disintegration energy which
means, according to Geiger-Nuttall findings,
a longer lifetime or more stability.
α Decay & Stability
(See Fig for A > 212). There is a
distinguished discontinuity at (N=126,
A=212). This is conform to the shell-model
and the existence of magic numbers.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
6
α Decay & The Semiempirical Mass Formula
Qαααα = m(Z,A) c2 – m(Z-2, A-4) c2 - mαααα c2
Qαααα = B(4He) + B(Z-2, A-4) – B(Z,A)
∑∑∑∑====
Z
i
eiB
1
m(Z,A) c2 = (Z mpc2 + N mnc2– B(Z,A)) + Z me c
2 +
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
7
α Decay & The Semiempirical Mass Formula
(((( ))))
(((( )))) 432
3132 )1(,
−−−−
−−−−
++++−−−−
−−−−
−−−−−−−−−−−−====
AaA
ZAa
AZZaAaAaAZB
pSym
CSV
(((( )))) (((( )))) (((( ))))
(((( )))) (((( ))))
(((( )))) (((( ))))(((( ))))(((( ))))
(((( )))) 43
2
31
32
4
4
24
4)3(2
444,2
−−−−
−−−−
−−−−++++
−−−−
−−−−−−−−−−−−−−−−
−−−−−−−−−−−−−−−−
−−−−−−−−−−−−====−−−−−−−−
Aa
A
ZAa
AZZa
AaAaAZB
p
Sym
C
SV
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
8
(((( )))) (((( ))))
−−−−++++
++++−−−−
−−−−−−−−====
−−−− 13
413
21
31Z
AZ
ZAZaC (((( ))))(((( )))) (((( )))) 31
3131
14
132−−−−
−−−−−−−− −−−−++++
−−−−−−−−−−−−−−−− AZZa
AAZZa CC
- The volume term: aV (A - 4) - aV A = - 4 aV
B(Z-2,A-4) - B(Z,A)
3132
1
3232
32
3
84
3
2
41
−−−−
>>>>>>>>++++====××××××××
++++
−−−−−−−−
≅≅≅≅ AaA
Aa
AaA
Aa
SSA
SS- The surface term:
- The Coulomb term :
−−−−−−−−
>>>>>>>>>>>>>>>>≅≅≅≅
A
ZAZaC
ZA 3
14 31
11
3
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
10
Qαααα = B(4He) + B(Z-2, A-4) – B(Z,A)
Qα = B (4He) + B(Z-2,A-4) - B(Z,A)
VaQ 43.28 −−−−====αααα
472
32
14 −−−−++++
−−−−−−−− Aa
A
Za PSym
−−−−++++++++ −−−−−−−−
A
ZAZaAa CS
314
3
8 3131
Qαααα = 28.3 - 62 + 7.35 + 36.90 - 3.81 - 0.0077 = 6.75 MeV
For 226Th this gives Q = 6.75 MeV, while the
measured value is 6.45 MeV.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
11
Explaining the Difference
The small difference (0.3 MeV) between the
2 values which comes from the fact that the
semiempirical mass formula’s parameters
are chosen so as to reproduce the maximum
number of nuclei masses, is not significant .
Theory of αααα Emission
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
13
Gamow and Condon & Gurney proposed
independently in 1928 a theory to explain
the alpha decay and its systematics.
Theory of α Decay
The theory is based on the idea that alpha
particles exist in the parent nucleus (or
considered to behave as if they do exist).
The interaction potential between an alpha
particle and the residual (daughter) nucleus
is represented by the following:
4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
14
The Coulomb potential extends inward to a and then suffers
a cut off
Nuclear Potential an α suffers
Qαααα
a b
V(r)
r
rrV
1)( ∝∝∝∝
V0 © Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
15
Three regions of interest appear :
1) A Spherical Potential Well
Qαααα
a b
V(r)
r
rrV
1)( ∝∝∝∝
V0
Region I (r < a) : This is the spherical part
(potential well of depth V0) where the alpha
particle, with a kinetic energy E - V = Qαααα + V0 moves.
The radius a is the sum of the alpha particle and
the residual nucleus.
I
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
16
2) A Barrier potential
Qαααα
a b
V(r)
r
rrV
1)( ∝∝∝∝
V0
I II
Region II (a < r < b) : This is the annular shell
region which plays the role of a barrier potential
because the potential energy here is greater than
the particle’s total energy. Classically the particle
cannot enter this region (Kinetic energy = Q - V < 0).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
17
2) A Classical Accessible Region
Qαααα
a b
V(r)
r
rrV
1)( ∝∝∝∝
V0
I II
Region III (r > b) : This is a classically permitted
region. if the particle succeeds to escape! then it
can move freely in this 3rd region
III
5
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
18
But alpha can escape!Classically the alpha particle rebounds each time it
hits the well at r = a. In quantum mechanics there
is a small probability that it “leaks” through the wall. The wave function “digs” a tunnel and the particle can escape the potential well !!
The escape (tunneling) probability which is related to the decay constant depends on the penetrability
(P) of the barrier region.
The decay constant λλλλ is given by: λλλλ = f Pwhere f represents the frequency with which the
alpha particle presents itself at the barrier and P is
the penetrability, i.e. probability of transmission
from one side to another of the barrier.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
19
3D Barrier Potential - Semi Classical Treatment
(((( ))))(((( ))))[[[[ ]]]] drErVmr
r
212
12
2∫∫∫∫ −−−−====γγγγ
hwith
γγγγ−−−−≈≈≈≈ eP
Here: (((( )))) )(88.2
4
1 2
0
MeVr
Z
r
eZzrV
Fin
′′′′====
′′′′
εεεεππππ====
The transmission probability of a flux of incident
particles on a barrier potential (depth V0 and Length L) can be easily calculated in quantum mechanics.
In a 3D problem with spherical symmetry (V depends
only on r) and zero angular momentum P the
probability to penetrate the complete barrier, also called penetrability, is given by :
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
20
Height of the (Coulomb) Barrier Potential
The height of the barrier part is :
(((( )))) )(88.2
MeVa
ZarB
Fin
======== &&&& B(r = b) = 0
The average width which we shall take as L is : )(2
1ab −−−−
The height of the barrier varies from (B - Q)
above the particle’s energy at r = a to 0 at r = b.
We shall take as a representative height (the
energy difference (E - V0)) the average height :
)(2
1QB −−−−
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
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Parameter bAt a first approximation using the average
height and width the penetrability is :
Typical values for a heavy parent (Z = 90) are :
Q = 6 MeV, a = 7.5 F
(((( ))))(((( )))) (((( ))))abkabkLk eeeP −−−−−−−−−−−−−−−−−−−− ========≈≈≈≈ 222
Q
Z
b
eZzQ b
′′′′====
′′′′
εεεεππππ==== ⇒⇒⇒⇒ 88.2
4
12
0
b can be calculated as being the radius at which
the alpha particle can leave the barrier. At this
point Q = V(b) or
(F)(MeV)
6
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
22
Penetrability
Which gives for Q = 6 MeV : Fb 426
8888.2≅≅≅≅
××××====
(((( )))) 164.1
197
634409.3727 −−−−====−−−−××××
==== Fk
(((( ))))(((( )))) (((( ))))abkabkLk eeeP −−−−−−−−−−−−−−−−−−−− ========≈≈≈≈ 222
(((( )))) MeVarBB 8.335.7
8888.2====
××××============For a = 7.5 F we have :
Thus we have :
(((( ))))
c
QBcm
kh
−−−−
====2
12 2
(((( )))) 255.74264.1105.2
−−−−−−−−××××−−−− ××××====≈≈≈≈ eP© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
23
Estimating f !
f is roughly of the order of v/a, where v is the
relative velocity of the alpha particle.
for a = 7.5 F & Q = 6 MeV and V0 = (B) = 34 MeV:
λλλλ = f P
(((( ))))c
a
cmVQ
af
20 /2
2
1 ++++========
v
1211086.5
5.7
409.3727/402 −−−−++++××××====××××××××
==== scf
λλλλ = 6 ×××× 10+21 ×××× 2.5 ×××× 10-25 = 1.5 ×××× 10-3 s-1 !
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
24
Gamow’s theory
sλ
t 7002ln
21 ≈≈≈≈====
If we Change Q from 6 MeV to 5 MeV then P
becomes 1 ×××× 10-30 and t1/2 = 108 s
The theory, although using a semi-classical
treatment explain remarkably the major
observation by Geiger and Nuttall.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
25
Quantum Theory of α Decay
Y
Y
Y Mm
Mm
mM ++++====µµµµ→→→→++++====
µµµµ αααα
αααα
αααα
111
−−−−
εεεεππππ
µµµµ−−−−==== αααα dr
r
eZzdP Q
2
02 4
122exp
h
Refined calculation is achieved using quantum mechanics principles. First we consider the system
(Daughter, alpha) in the a cm (center of mass)
where the problem reduces to considering a particle with the reduced mass :
Then one divides the Coulomb barrier potential
into spherical shells of radii (r,r+dr). The
transmission probability between r and r + dr is
7
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
26
The Gamow factor G
drr
eZzb
Ra
2
1
2
04
12
1∫∫∫∫ ≡≡≡≡ αααα
−−−−
′′′′
εεεεππππµµµµ==== Q
hG
(((( )))) (((( ))))[[[[ ]]]]xxxeZz
−−−−−−−−′′′′
εεεεππππ====
−−−− 1cos4
1 12
0 vhG
Where G (the Gamow factor) is:
The penetrability is: P = e -2G
Taking x = a/b = Q/B, the integral gives :
Exercise : Check that!
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
27
P in Quantum Mechanics
−−−−
ππππ′′′′
εεεεππππ==== x
eZz2
24
1 2
0 vhG
Thus we have :
−−−−
ππππ××××′′′′××××××××−−−−====
B
Q
Q
ZP 2
2
409.37272
197
4.144exp
−−−−
ππππµµµµ′′′′
εεεεππππ−−−−====
B
Q
Q
c
c
eZzP 2
2
2
4
12exp
22
0 h
For x << 1 ( Q << B or a << b), which is the case
in the case for most decays of interest, G
becomes:
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
28
Half-life in Quantum Mechanics
(((( ))))
−−−−
ππππ′′′′××××++++
++++
µµµµ====
BQZ
QV
c
c
at
12
1
225.25exp
2693.0
0
2
21
with f given by:
(((( ))))c
a
c/mVQ
a
vf
202
2
1 ++++========
Pfλt
2ln2ln21 ========
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
29
Half-life in Quantum Mechanics
220 8.95 10-53.3 x 10-7
8.13222 2.8 x 10-3 6.3 x 10-5
6.45226 1854 6.0 x 101
5.52228 6.0 x 1072.4 x 106
4.77230 2.5 x 10121.0 x 1011
4.08232 4.4 x 10172.6 x 1016
7.31224 1.04 3.3 x 10-2
t1/2 (s)
A Qαααα(MeV) Measured Calculated
For the even-even isotopes of Th (Z =90), the previous calculations give the following table :
8
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
32
The rough and approximate calculations give good
results (discrepancies are within 1 to 2 orders of
magnitude over a range of more than 20 orders of
magnitude).
The Real Situation
• Fermi’s Golden Rule : The initial and final wave functions of the transition,
• The fact that an alpha particle carries an angular momentum,
• The non spherical shape of the majority of nuclei.
• For the highly deformed nuclei (A > 230) , the differences become very significant. We use this fact in the reverse order! Life-times are used to have an approximation of the nuclei radii.
To improve the calculations we should have taken into consideration the following arguments :
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Alpha Decay - 2
33
Application of Gamow’s Theory
• Measurements of the radii of highly
deformed nuclei.
• Calculation and Prediction of heavier nuclei
(12C ) emission
• Calculation and Prediction of single-proton
decay processes.
See Introductory Nuclear Physics
(Kenneth. S. Krane) pages 254-257.
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Supplement 3: Gamma Decay
Introduction
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
3
Introduction
The gamma decay is the emission of an energetic
photon when an excited state of a nucleus decays
to a lower energy state. In general a series of γγγγ
decays is necessary to reach the stable ground
state.
These decays are encountered each time a nucleus
is excited. This excitation could be the result of a
or b decays or a nuclear reaction.
* λλλλ(F) = hc/E = 1240/E(MeV)
The resulting photons are just like the atomic X-
rays in nature, i.e. they are electromagnetic
radiations, but are more energetic (roughly 0.1 to
10 MeV, wavelengths are between 104 F and 100 F*)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
4
γγγγ photons are relatively easy to detect which
makes them a very popular tool for
spectroscopists.
The study of the competitive process to
gamma decay namely, the internal
conversion, is an excellent tool to obtain the
spins and parities of nuclear states!
The Study of excited states is very rich in
information about the nuclei properties.
Importance of γ spectroscopy
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
5
Energetics of γ DecayConsider a nucleus, at rest, in an excited state
Ei decaying to a lower energy state Ef , a γγγγ
photon (Eγγγγ= pγγγγ c) is emitted: A* →→→→ A + γγγγ
Conservation of energy:
Ei = Ef + Eγγγγ + TR (1)
c
E γγγγγγγγ ========⇒⇒⇒⇒ ppR
(2)γγγγ−−−−====⇒⇒⇒⇒ ppR
rr
Conservation of momenta: in the cm we have:
The symbol R stands for "recoil" of the parent.
γγγγ++++==== ppR
rr0
Equation 1 can be written as:
(3)2
2
2 cM
EEEEE fi
γγγγγγγγ −−−−====−−−−====∆∆∆∆
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
6
Eγ ≅ ∆E
∆∆∆∆++++±±±±−−−−====γγγγ
21
2
2 211cM
EcME
Eγγγγ is the solution of the quadratic equation 3, i.e.
(4)
∆∆∆∆E is typically of the order of 1 MeV. The rest
energy Mc2 is of the order of A××××103 MeV, i.e. ∆∆∆∆E <<
Mc2
Expanding the square root we have:
∆∆∆∆++++
∆∆∆∆++++++++−−−−====γγγγ
2
22
2
2
111
cM
E
cM
EcME (5)
From 3 (or 5), if we neglect the term ∆∆∆∆E/Mc2, we
have Eγγγγ = ∆∆∆∆E.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
7
Eγ ≅ ∆E
The correction to Eγγγγ due to the recoil energy
[(∆∆∆∆E)2/2Mc2] is negligible (10-5) and except a
special case in nuclear spectroscopy*, Eγγγγ is
simply taken as equal to ∆∆∆∆E
* Mössbauer Spectroscopy which is dedicated to the use of this correction
(6)(((( ))))2
2
2 cM
EEE
∆∆∆∆−−−−∆∆∆∆====γγγγ
Lifetimes for γγγγ Emission,
Selection Rules
3
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
9
Example: Fig. 1 shows the energy levels of the
(even-even) 72Se (Z=34) isotope
All details are shown in an energy level scheme: spin-parity, energies and/or γγγγ transition energies and half-lives of excited states.
Fig. 1Energies and γγγγ transition
energies are given in keV
Weisskopf Estimates vs. Experiment
1) Evaluation of the partial decay rate for γγγγ emission
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
10
Example: 72Se 3rd excited stateWe’ll have a closer look on the 3rd excited state of 72Se.Spin-parity = 2+
The decay constant λλλλt = ln2/t 1/2 = 8.0××××1010 s-1
Energy = 1317 keV
t1/2 = 8.7x10-12 s
Transition 3rd-1st = 455 keV
Transition 3rd-2nd = 380 keV
Neglecting the (internal) conversion factors, λλλλt is simply the sum of the decay rates of the three transition that depopulate this excited state!, i.e. λλλλt = λλλλg,1317 + λλλλg,445 + λλλλg,380
The measured relative intensities (or branching ratios) are λλλλγγγγ,1317: λλλλγγγγ,445:λλλλγγγγ,380 = 51:39:10
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
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Comparison with Weisskopf Estimates
λλλλγγγγ,1317 = 0.51 * λλλλt = 4.1××××1010 s-1
λλλλγγγγ,455 = 0.39 * λλλλt = 3.1××××1010 s-1
λλλλγγγγ,380 = 0.10 * λλλλt = 0.8××××1010 s-1
The partial decay rates of the 3 transitions
(1317,455,380) are:
The following tables give the calculations of λλλλ(EL)
and λλλλ(ML) for the energies involved in this example
(A=72)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
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λλλλ(EL) (s-1)L
1 λλλλ(E1) = 3.95××××1015
2
3 λλλλ(E3) = 1.20××××106
λλλλ(ML) (s-1)
λλλλ(M1) = 1.28××××1014
λλλλ(M2) = 2.40××××109
λλλλ(M3) = 3.30××××104
λλλλγγγγ,1317 = 4.1××××1010 s-1
λλλλ(E2) = 8.70××××1010
Weisskopf Estimates for E = 455 keV
Weisskopf Estimates for E = 1317 keV
λλλλ(EL) (s-1)L
1 λλλλ(E1) = 1.63××××1014
2
3 λλλλ(E3) = 7.11××××102
λλλλ(ML) (s-1)
λλλλ(M1) = 5.28××××1012
λλλλ(M2) = 1.18××××107
λλλλ(M3) = 1.94××××101
λλλλγγγγ,455 = 3.1××××1010 s-1
λλλλ(E2) = 4.26××××108
4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
13
Weisskopf Estimates for E = 380 keV
λλλλ(EL) (s-1)L
1 λλλλ(E1) = 9.50××××1013
2
3 λλλλ(E3) = 2.02××××102
λλλλ(ML) (s-1)
λλλλ(M1) = 3.07××××1012
λλλλ(M2) = 4.80××××106
λλλλ(M3) = 5.48
λλλλγγγγ,380 = 0.8××××1010 s-1
λλλλ(E2) = 1.73××××108
The previous calculations indicate that the favored
transitions are the E2 ones.
But they also show that the measured values are one
order of magnitude greater than Weisskopf Estimate.
There is a strong evidence for the collective structure
of the nucleus!, since Weisskopf used the shell (a
single individual particle) model to make his
estimations.© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
14
The M4 Transitions (Systematics) Case
This figure shows, in particular, the good agreement
with the expected E-9 dependence.
Fig. 2 represents the experimental data for different
nuclei.
The straight line
represents Weisskopf
estimate
ττττ(M4) = 1.54××××105 A-2 E-9
Fig. 2Fig. 2 : log(ττττ A2) vs. E(in keV)
3 : Selection Rules
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
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Multipoles and Angular Momenta
An em field produced by oscillations of charges and currents produces also angular momentum. In QM this angular momentum carried by the quanta
of energy (photons of energy E = h νννν) is quantized.
The rate at which this angular momentum is radiated, is proportional to the rate at which energy is radiated.
The proportionality is preserved if each emitted photon carries a definite angular momentum.
Conclusion: a multipole of order a multipole of order LL transfers an transfers an
angular momentum of angular momentum of LL per photon.per photon.
A multipole operator of order L includes a spherical
harmonic Ylm(θθθθ,φφφφ), which is associated with an angular momentum L.
5
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
17
Angular Momentum and Parity Selection Rules
Example: For Ii = 3/2 and If = 5/2, the possible values for L are:
1, 2, 3 and 4 and the radiated field would be a mixture of dipole,
quadrupole, octupole and hexadecapole radiation!
Thanks to the rules of addition of angular momenta we know that
L could only have restricted values.
8| Ii – If | ≤ L ≤ Ii + If
The relative parity of the initial and final levels determine the type of the emitted radiation (electric or magnetic). The following table resumes the parities related to em radiations
Consider a γ transition from an initial excited state of angular
momentum Ii and parity pi to a final state (If, πf). Assume Ii ≠ If.
(Spin-parity for these states are and respectively)iπ
iIfπ
fI
Conservation of angular momentum is expressed by:
LII fi
rrr++++==== 7
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
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The following notations are used when studying parity changes –see Table 10
The following table resumes what we know about the parity associated to electric and magnetic transition.
Parity and EM Transitions
9L Electric Transition Magnetic Transition
Even + -Odd - +
Parity
Ii
+
If L
+
∆π
∆π = no- - +
+Even
-
-Odd
+∆π = yes
-
- +
10
The two tables are used to determine the type of the emitted radiation (electric or magnetic).
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
19
In the case ∆π = yes, the transition cannot be magnetic; the
associated parity is (-)1+1 positive. In this case the transition is the electric dipole E1 transition. See Table 18 for the othercases.
ExampleLet’s take again the previous example Ii = 3/2 and If = 5/2.
11
L Transition
∆π = no
2 E2
3 M3
E44
1 M1
L Transition
∆π = yes
2 M2
3 E3
M44
1 E1
For L = 1 and ∆π = no, the transition cannot be electric
because the associated parity is (-)1 negative and this would
give a final parity different from the initial one. In this casethe transition is the magnetic dipole M1 transition.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
20
even electric, odd magnetic
odd electric, even magnetic
Selection RulesThe precedent example suggests restrictions on the possible transitions. These restrictions are called selection rulesselection rules
∆π = no
| Ii – If | ≤≤≤≤ L ≤≤≤≤ Ii + If ; Ii ≠≠≠≠ If.
12
∆π = yes
6
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
21
The Case Ii = If.This exception to the previous selection rules
occurs because it would mean that L = 0 is a
possible value, and there are no monopole transitions in which a single photon is emitted!*
* The magnetic monopole does not exist. For an electric monopole, a spherical distribution of charge (like a single point charge) the Coulomb field is not affected by radial oscillationsand thus no corresponding radiation is produced.
The lowest multipole order allowed in the case
where Ii ≠≠≠≠ If. is L = 1
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
22
Pure Multipole Transition
Another interesting case is when (Ii≠≠≠≠0, If.=0) or
(Ii=0, If ≠≠≠≠ 0)
L is equal to Ii for the first type of transition (or If
for the second one).
For an even Z – even N nucleus (like 72Se) the first
excited state 2+ decays to the ground state 0+
through the emission of a pure E2 (quadrupole)
transition.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
23
Exercise: Pure Multipole Transition
Find the type of transitions for the decay from the fourth state to the ground state in
the case of 72Se
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
24
Internal Conversion
The Ii = If = 0 Case – Internal Conversion
In this process, the available energy is transmitted to an orbital electron which in turn is ejected. Orbital electrons with wave functions penetrating the nucleus field are concerned.
Here the only possible value for L is zero. This case is
not permitted for radiative transitions.
The transition between 2 (excited) states with spin
= 0* occurs through the competitive process we
mentioned before, namely the internal conversion.
* A few even Z – even N nuclei have 0+ first excited states.
Those are forbidden to decay to the 0+ ground state by γγγγemission.
7
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
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Selection Rules and Weisskopf Estimates
In table 18 we showed which transitions are
possible between the 2 states Ii = 3/2 and If =
5/2. Several multipoles are permitted. For
example in the ∆π∆π∆π∆π = no case, M1, E2, M3 and E4are allowed.
To decide which are the ones we observe, we shall make a simple calculation using Weisskopf estimates.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
26
∆π∆π∆π∆π = noλλλλ(σσσσL, A=125, E=1 MeV)
λλλλ(E2) = 1.83××××109 s-1
λλλλ(M3) = 1.00××××104 s-1
λλλλ(E4) = 1.72 s-1
λλλλ(M1) = 5.6××××1013 s-1
λλλλ(σσσσL)/λλλλ(M1)
1
1.4××××10-3
≅≅≅≅2.1××××10-10
1.3××××10-13
Selection Rules and Weisskopf Estimates
Let’s assume a medium-weight nucleus
(A=125)* and E = 1 MeV.
L Transition
2 E2
3 M3
E44
1 M1
* A2/3 = 25, A4/3 = 625
According to Weisskopf, the transition probabilities are as follows:
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
27
λλλλ(σσσσL, A=125,E=1 MeV)
λλλλ(M2) = 4.375××××109 s-1
λλλλ(E3) = 5.31××××104 s-1
λλλλ(M4) = 0.70 s-1
λλλλ(σσσσL)/λλλλ(E1)
1
2.3××××10-7
≅≅≅≅ 2.1××××10-10
2.1××××10-17
Selection Rules and Weisskopf Estimates
The previous results show that the lower orders are dominant. They also indicate that this transition could be composed of M1 radiation with possibly a small mixture of E2.
For the ∆π = yes case, the calculations show that the E1 is
dominant and the other modes are most likely not to occur!
λλλλ(E1) = 1.25××××1016 s-1 L Transition
∆π = yes
2 M2
3 E3
M44
1 E1
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
28
Expectations based on the single-particle Estimates
(((( ))))(((( ))))
(((( ))))(((( ))))
(((( ))))(((( ))))
325 101010'' −−−−−−−− ====××××====
λλλλ
λλλλ××××
λλλλ
λλλλ====
λλλλ
λλλλ
ML
EL
EL
EL
ML
EL
13(((( ))))(((( ))))
(((( ))))(((( ))))
(((( ))))(((( ))))
725 101010'' −−−−−−−−−−−− ====××××====
λλλλ
λλλλ××××
λλλλ
λλλλ====
λλλλ
λλλλ
EL
ML
ML
ML
EL
ML
1) The lowest permitted multipoles usually dominate
3) Emission of multipole L+1 is less probable than
emission of multipole L by a factor of the order of 10-5
2) Electric multipole emission is 2 orders of magnitude more probable than the same magnetic multipole emission.
4) Points 2 and 3 combined give the following relations
(L’ = L +1)
8
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay
29
Important Remark
The precedent calculations are based on a
single particle model with simple
approximations. We observe in the lab.
transitions in which λλλλ(E1) > λλλλ(M1) especially in
transitions between vibrational and rotational
collective states.
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Supplement 4: Weisskopf Estimates
Weisskopf Estimates
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
3
Generalized EM Moments
For a multipole electromagnetic moment of order L the
corresponding em radiation, the moment is said to be an 22LL--
polepole
The electromagnetic theory gives the following properties of
a 2L-pole radiation
Index of the radiation
1 Dipole
2 Quadrupole
3 Octupole
Moment
The γγγγ photons emitted in a nuclear de-excitation are
electromagnetic radiations. This means that an em
interaction is behind this process!
An em field produced by oscillations of charges and currents
produces energy. In QM this energy is emitted as quanta
(photons) of energy E = h νννν
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
4
Properties of a multipole radiation
The classical electromagnetic theory gives the following properties of a 2L-pole radiation
1) The angular distribution of 2L-pole radiation is governed by the
Legendre Polynomials P2L=(cosθθθθ).
For a dipole (L=1) P2 = (3cos2θθθθ – 1) and
For a quadrupole (L=2) P4 = 35 cos4θθθθ – 30 cos2θθθθ +3)
2) The parity of the radiation field is:
(-1)L+1 for an electric multipole and
(-1)L for a magnetic multipole.
For an electric dipole parity is +1 and for a magnetic dipole parity
is -1
3) The radiation power is
(((( ))))(((( ))))(((( ))))[[[[ ]]]]
(((( ))))[[[[ ]]]] 22+2L
0γ Lσm
c
ω
!!1+L2Lε
c1+L2=LσP
1
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
5
Generalized Multipole Moment
In the previous expression, m(σσσσL) is the amplitude
of the time varying electric or magnetic moment.
This moment for L=1 differs from the electric
dipole moment p and the magnetic dipole moment
µµµµ through some relatively unimportant numerical
factors of order unity.
(((( ))))(((( ))))(((( ))))[[[[ ]]]]
(((( ))))[[[[ ]]]] 22+2L
0γ Lσm
c
ω
!!1+L2Lε
c1+L2=LσP
*
* (2L+1)!! is a double factorial and given by (2L-1)!!= (2L+1)x(2L-1) x(2L-3)…3x1
σσσσ = E or M to represent electric or magnetic radiation
respectively
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
6
Multipole moment in Quantum Mechanics
Where we replaced the classical m(σσσσL) by the matrix
element mfi(σσσσL) associated to the (quantum) operator
m(σσσσL), defined by:
The classical properties we have seen are the same in quantum mechanics. We discuss here how the expression of the radiated power (1) is written in QM
(((( ))))(((( )))) (((( ))))
(((( ))))[[[[ ]]]](((( ))))[[[[ ]]]] 2
fi
1+2L
20
Lσmc
ω
!!1+L2Lε
1+L2=
ω
LσP=Lσλ
hh
The decay probability per unit time (the decay
constant) for the emission of photon of energy hhhhωωωωis given by:
(((( )))) (((( )))) vdLmLm
operator
*
∫∫∫∫ ψψψψσσσσψψψψ====σσσσiffi 321
3
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
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Estimation of λλλλ(σσσσL) – Weisskopf estimates
In order to compute λλλλ(σσσσL) we need to know the
initial and final wave functions, i.e. ψψψψi and ψψψψf
- The Electric Transitions Case
Let’s consider that the transition is due to a single
proton which changes from one shell-model state
to another
�For L = 1 (dipole) radiation
The operator m(EL) becomes ez
The multipole operator m(EL) contains a term of
the form e rL Ylm( θ θ θ θ, φ φ φ φ)
�For L = 2 (quadrupole) radiation
The operator m(EL) becomes e(3z2 – r2)© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
8
Radial part of the (electric) transition probability
Approximation of ψψψψi and ψψψψf
If we take the radial parts of ψψψψi and ψψψψf to be
constant for r < R (the nucleus radius) and to be
equal to 0 for r > R then the Radial part of the
transition probability is of the form:
L
3
3L
R
0
2
R
0
L2
R3L
3
R3
1
R3L
1
drr
drrr
++++====
++++====
++++
∫∫∫∫
∫∫∫∫4
Angular part of the (electric) transition probability
We can reasonably take the angular integrals to be
equal to 1 and the EL transition probability is
estimated to be:
(((( ))))(((( ))))
(((( ))))[[[[ ]]]]L2
21L2
0
2
2Rc
3L
3
c
E
c4
e
!!1L2L
1L8LE
++++
πεπεπεπε++++
++++ππππ≅≅≅≅λλλλ
++++
hh5
3
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
9
λλλλ(E1), λλλλ(E2), λλλλ(E3), …Taking R = R0 A1/3 with R0 = 1.2 F, the lower multipole orders are given in the table below (Values are from Krane).
3
0
2
10317197
441
4
−−−−××××.=FMeV
FMeV.=
cπε
e
h
(((( ))))(((( ))))[[[[ ]]]]
(((( )))) 33222323
3
221103
4
3
197
110317
3
161 EA..
!!
πEλ /××××
××××≅≅≅≅ −−−−
6
L λ(EL)
1 λ(E1) = 1.0×1014 A2/3 E3
2 λ(E2) = 7.3×107 A4/3 E5
3 λ(E3) = 34 A2 E7
λλλλ is in s-1 when E is expressed in MeV
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
10
λλλλ(M1), λλλλ(M2), λλλλ(M3), …A similar treatment for the magnetic transitions
gives for λλλλ(ML) the following expression:
(((( ))))(((( ))))
(((( ))))[[[[ ]]]]
22
212
0
22
2
2
2
2
3
41
1
!!12
18
−−−−++++
++++
××××
πεπεπεπε
++++−−−−µµµµ
++++
++++ππππ≅≅≅≅λλλλ
L
L
p
p
RcLc
E
c
e
cm
c
LLL
LLM
h
h
h
7
We replace the factor by 10 where µµµµp is the
nuclear magnetic moment of the proton on which
depends the moment operator and this gives the
following values for λλλλ in s-1 when E is expressed in MeV.
2
1
1
++++−−−−µµµµ
Lp
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
11
Estimation of λλλλ(M1), λλλλ(M2), λλλλ(M3), …
L λλλλ(ML)
1 λλλλ(M1) = 5.6××××1013 E3
2 λλλλ(M2) = 3.5××××107 A2/3 E5
3 λλλλ(M3) = 16 A4/3 E7
4 λλλλ(M4) = 4.5××××10-6 A2 E9
See discussion in Krane p. 332
8
λλλλ is in s-1 when E is expressed in MeV
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
12
λλλλ(EL) and λλλλ(ML)
L λλλλ (EL)
1 λλλλ(E1) = 1.0××××1014 A2/3 E3
2 λλλλ(E2) = 7.3××××107 A4/3 E5
3 λλλλ(E3) = 34 A2 E7
4 λλλλ(E4) = 1.1××××10-5 A8/3 E9
λλλλ(ML)
λλλλ(M1) = 5.6××××1013 E3
λλλλ(M2) = 3.5××××107 A2/3 E5
λλλλ(M3) = 16 A4/3 E7
λλλλ(M4) = 4.5××××10-6 A2 E9
Two major conclusions:
1) The lower multipolarities are dominant. For L > 3 the decay
rates of a γ transition become 10-5 smaller and indicate these transitions due to an L-pole > 4 have infinitesimal probabilities.
2) For a given multipole order, electric transitions are twice as big
as the magnetic transition in medium and heavy nuclei (A > 70)
4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Gamma Decay - Weisskopf Estimates
13
Although Weisskopf estimates are not true theoretical calculations, they provide us with a good tool to compare transition probabilities
Weisskopf estimates vs. Experiment
If the observed transition rates are many order of magnitude smaller than the Weisskopf estimate then we “suspect” that a poor match between the initial and final wave function is slowing the transition.
On the other hand, if the observed transition rates were much greater than the Weisskopf estimate then we might “guess” that more than one nucleon is involved in the transition.
1
© Dr. Nidal M. Ershaidat
Phys. 649: Nuclear Techniques
Physics Department
Yarmouk University
Bremsstrahlung
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
2
D) Bremsstrahlung – Braking Radiation
According to the (classical) electromagnetic theory part of the energy of an accelerated particle is converted into electromagnetic radiation.
The consequent energy loss has a "decelerating" effect on the particle.
The power radiated (energy loss per time unit) is given (for two limiting cases) by:
(((( ))))62
0
3
0
622
3
0
622
||
66 cmc
Eaq
c
aqP va
εεεεππππ
====εεεεππππ
γγγγ====
(((( ))))42
0
3
0
422
3
0
422
66 cmc
Eaq
c
aqP va
εεεεππππ
====εεεεππππ
γγγγ====⊥⊥⊥⊥
q is the charge of the particle and γγγγ is its Lorentz
factor. a being the acceleration
Acceleration // velocity
Acceleration ⊥ ⊥ ⊥ ⊥ velocity
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
3
Bremsstrahlung The em radiation seems to "brake" the particle. We call this radiation braking radiation and we use the German word "Bremsstrahlung"
Usually we reserve the term bremsstrahlung for the radiation resulting from the energy loss by electrons into matter. In an X-ray tube, the bremsstrahlung part of the spectrum is very important.
In the case where the acceleration is perpendicular to the velocity of the charged particles, the resulting radiation is called "synchrotron radiation"
The bremsstrahlung spectrum is a continuous one, thus not valid for calibration purposes. It can be altered, as we also we do with X-ray spectra, using the appropriate filters or absorbers to produce "relatively" monoenergetic photons for calibration – See next chapters.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
4
Bremsstrahlung Spectrum Fig. 11 shows the bremsstrahlung spectrum of
electrons of 5.3 MeV incident on a Au-W target.
A 7.22 g/cm3 aluminum filter
was also present.
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
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E) Characteristic X-Ray
Other sources of bremsstrahlung are beta decay electrons.
A characteristic X-ray spectrum is also emitted when electrons pass through matter
The deceleration could be a serious problem when accelerating charged particles, because of the permanent and continuous energy loss by em radiation.
Nevertheless, this em radiation is used to produce X-rays (The multiple uses of X-rays are (should be) well-known).
As for the synchrotron radiation, we now build electron accelerators (cyclotrons) in order to produce intense beams of photons of major interest in applied and fundamental physics – See later)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
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Characteristic X-Ray
We leave section E (Knoll pages: 15-19) for self reading.
Revise the structure of an X-ray tube. Be sure you understand a characteristic X-ray spectrum produced in such a tube.
Spectrum of the X-rays emitted by an X-ray tube with a
rhodium target, operated at 60 kV. The continuous curve is
due to bremsstrahlung, and the spikes are characteristic Klines for rhodium.
Bremsstrahlung (classical)X-ray peaks
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Bremsstrahlung
7
F) Synchrotron RadiationWhen a beam of charged particles is bent into a circular orbit, as in a cyclotron, a small fraction of the beam energy is radiated away during each cycle of the beam. This emitted electromagnetic radiation is called synchrotron radiation.
The term synchrotron radiation is almost exclusively used nowadays for electron accelerators.This radiation is considered as a nuisance for designers of high-energy accelerators (LEP is the most famous one).
Special cyclotrons are built in order to produce this radiation and use it in many important field of physics. (See SESAME, the synchrotron facility now being installed in Jordan. www.sesame.org.jo)