nuclear techniques phys. 649 -...

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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

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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

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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

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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.

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Course’s websites

�http://ctaps.yu.edu.jo/Physics/Phys649

�http://faculty.yu.edu.jo/NErshaidat


Phys. 649: Nuclear Techniques

Physics Department

Yarmouk University

Chapter 1:

Radiation Sources

Supplements

1



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.


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


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).


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


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


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


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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


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


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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


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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

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- Beta Decay

- Internal Conversion

- Auger Electrons


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.


Beta Decay - Decay Scheme

Natural beta decay sources:


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

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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.


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).


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 −−−−====−−−−


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

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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.


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

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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.


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


Common Alpha-Emitting Sources1

Table 4-a© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources 41


Table 4-b

10

42


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.


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?


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

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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.


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..



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).


Degradation of a fission fragment Energy


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Sources of

Electromagnetic RadiationSources of Gamma - Rays


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

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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!


"High Energy" Calibration Sourcescont'd


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.

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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.


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.


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

γγγγ====⊥⊥⊥⊥

q is the charge of the particle and γγγγ is its Lorentz

factor. a being the acceleration

Acceleration // velocity

Acceleration ⊥ ⊥ ⊥ ⊥ velocity


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.

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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)


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.


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)

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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


Next Lecture

Chapter 2Radiation Interactions with

Matter

1



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.


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.


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


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


Characteristics of Be (α n) Neutron Sources

s


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


Alternative Sourcess

See discussion in Knoll page 24


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?


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


•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.


Photoneutron Source Container


Neutron Spectra


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


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


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).


Neutron Yield

Other reactions using Charged Particles


Next Lecture

Chapter 2Radiation Interactions with

Matter

1



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)


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


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.


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


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


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


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.


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


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 :


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.


16

β+ Decay Kinematics

(((( )))) (((( )))) 21 ][ 2 cmYmXmQ

e

AZ

AZ −−−−++++ −−−−−−−−==== −−−−ββββ

eNA

ZNAZ eYX νννν++++++++→→→→ ++++

−−−−−−−− 11

5


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


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


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



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.


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


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 ++++−−−−++++====


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


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:


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


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).


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


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.


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



Physics Department

Yarmouk University

Supplement 2:Part 2- Systematics and Theory of αααα emission

Systematics of αααα emission


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


4

Geiger & Nuttall

Fig. 6.1 : Qαααα vs. t1/2

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.


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 +


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


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


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.


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


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


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


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).


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


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.


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

′′′′====

′′′′


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 :


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 −−−−


21

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


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 !


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.


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


26

The Gamow factor G

drr

eZzb

Ra

2

1

2

04

12

1∫∫∫∫ ≡≡≡≡ αααα

−−−−

′′′′

εεεεππππµµµµ==== Q

hG

(((( )))) (((( ))))[[[[ ]]]]xxxeZz

−−−−−−−−′′′′


−−−− 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!


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:


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 ========


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


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 :


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



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*)


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


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 −−−−====−−−−====∆∆∆∆


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.


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


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


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


11

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)


12

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


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


13


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


16

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


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


18

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).


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.


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


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


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.


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


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


25

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.


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


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:


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


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


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


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



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


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


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


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


7

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


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


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


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


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


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



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


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.


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.

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5

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)


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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


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)

nuclear techniques phys. 649 -...

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