nuclear physics intro
DESCRIPTION
An introductory course on nuclear physics.TRANSCRIPT
501503742 Nuclear PhysicsNuclear Physics
Instructor: Dr. Saed Dababneh
Nuclear Physics at BAU htt // l b d j /http://nuclear.bau.edu.jo/
This coursehttp://nuclear bau edu jo/nuclear-radiation/
Before we start, let us tackle the following:
http://nuclear.bau.edu.jo/nuclear-radiation/
Before we start, let us tackle the following:• Why nuclear physics?• Why radiation physics?• Why in Jordan?• Interdisciplinary.• Applied
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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• Applied.
General subjects to be covered
This is an introductory course that will cover the following general subjects
• Nuclear properties.• Binding energy and nuclear stability.
• Nuclear models.• Spin and moments.
• Nuclear forces.Th t t f th l• The structure of the nucleus.
• Nuclear reactions: energetics and general cross-section behavior.• Neutron moderation, fission, controlled fission and fusion.
Radioactive decays• Radioactive decays.• Interactions of nuclear radiations (charged particles, gammas, and neutrons) with
matter.
This phenomenological course provides the launch point for other nuclear physics courses that will follow.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Level Test1.Write down whatever you know about nuclear y
spin.2.Describe how can either fission or fusion produce p
energy.3.What is …
1.Compton scattering?2.Energy straggling?gy gg g3.Annihilation?4.Neutron activation?5. Isotopes, Isotones, Isobars, Isomers?6.Parity.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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y
Grading
Mid-term Exam 25%Mid term Exam 25%Project, quizzes and HWs 25%Final Exam 50%Final Exam 50%
• Homeworks are due after one week unless otherwise announced.• Remarks or questions marked in red without being q gannounced as homeworks should be also seriously considered!• Some tasks can (or should) be sent by email:
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Proposed Projectsp j• Experiments to determine nuclear properties.• Nuclear power generation• Nuclear power generation.• Nuclear medicine.• Health physics.
A l t d i t• Accelerator driven systems.• Nucleosynthesis.• Technological applications (e.g. Material Science).• Radioactive ion beams.• Neutrino physics.• Radiological dating.g g• Environmental radioactivity.• ….. (your own selected subject).
Decide on the title of your project within two weeks.Due date (for written version): December 6th.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Presentation: Will be scheduled later.
Scale and ObjectivesjDimensional scale: Order of magnitude of 1 x 10-15 mg≡ 1 femtometer ≡ 1 fm ≡ 1 fermi.Too small for direct investigation.What about time and energy scales?gy
We need to answer …..1 What are the building blocks1. What are the building blocks
of a nucleus?2. How do they move relative to
each other?each other?3. What laws governing them?We need to understand:
N l f (Q2 Q3)• Nuclear forces (Q2, Q3).• Nuclear structure (Q2, Q3).
We also need High Energy
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Physics (to answer Q1).
Constantshttp://physics.nist.gov/cuu/Constants/index.html
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NomenclatureElement vs. Nuclide.94 natural chemical elements total > 10094 natural chemical elements, total > 100.Element Atomic number (Z) chemically identical.~3000 nuclides……? How many are stable?S Z b t diff t t b (N) I t
A X
Same Z but different neutron number (N) Isotopes.Total number of nucleons = Z+N = A mass number.
XA 22Na Na23 Na24NZ X X 1111 Na Na Na
Radioactive Stable Radioactive
Same mass number Isobars chemically dissimilar parallel nuclearredundantSame mass number Isobars chemically dissimilar, parallel nuclear features (Radius …). β decay.Same neutron number Isotones ?????.Same Z and same A Isomers metastable XAmSame Z and same A Isomers metastable.Stable isotope (Isotopic) Abundance.Radioactive isotope Half-life.
X
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Stable Isotopesp
HWcHWc 11HWcHWc 11
Odd A Even A
Isotope N Z N Z
Then plot Z vs NThen plot Z vs. N.Odd A Even A
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Basic Nuclear PropertiesProperties Structure
p
The energy of the nucleon in the nucleus is in the order of 10 MeV.
HWHW 11 Calculate the velocity of a 10 MeV proton and show that it is almostHW HW 11 Calculate the velocity of a 10 MeV proton and show that it is almost 15% of the speed of light. (Perform both classical and relativistic calculations).∴ Relativistic effects are not important in considering the motion of nucleonsin the nucleusin the nucleus.
HW HW 22 Calculate the wavelength of a 10 MeV proton and compare it with the nuclear scale (Perform both classical and relativistic calculations) Is thethe nuclear scale. (Perform both classical and relativistic calculations). Is the nucleus thus a classical or a quantum system?
∴∴HW HW 00 Krane, Ch. 2. HW HW 33 Calculate the wavelength for an electron of the same energy to show that it is much too large to be within the nucleus. (Perform both classical
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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and relativistic calculations). Discuss the proton-electron nuclear hypothesis!
Chadwick, neutron.
Basic Nuclear PropertiespStatic nuclear properties (Time-independent):Electric charge radius mass binding energy angular momentum parityElectric charge, radius, mass, binding energy, angular momentum, parity, magnetic dipole moment, electric quadrupole moment, energies of excited states.Dynamic properties (Time dependent):Dynamic properties (Time-dependent):• Self-induced (Radioactive decay).• Forced (Nuclear reactions) cross sections. .
The key: Interaction between individual nucleons.
Excited states: atomic intervals ~ eV.nuclear intervals ~ 104 – 106 eV.
Decays and reactions: Conservation laws and selection rules.y
HWcHWc 22 Where to find nuclear data???
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Nuclear Mass (Introduction)
• Unified atomic mass unit u based on 12C.R l d b h h i l d h i l b d 16O d l• Replaced both physical and chemical amu based on 16O and natural
oxygen, respectively (Find conversion factors).• 1 u = M(12C)/12 = ……… kg = …………… MeV/c2.• Rest masses
u MeV/c2 kgelectron ………… …………… ………proton ………… …………… ………neutron ………… …………… ………12C 12C 12 …………… ………
• Avogadro’s number .. !! What is the number of atoms in 1 kg of pure 238U?• Mass Stability. E = mc2. Tendency towards lower energy Radioactivity. • Neutron heavier than proton “Free” neutron decays (T = ???):• Neutron heavier than proton Free neutron decays (T½ = ???):_
ν++→ epnNuclear and Radiation Physics, BAU, First Semester, 2007-2008
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Nuclear Mass (Introduction)
• Nuclear masses measured to high accuracy:• mass spectrograph• mass spectrograph.• energy measurement in nuclear reactions.
• Mass decrement = difference between actual mass and mass number:∆ A∆ = m – A
• http://www.eas.asu.edu/~holbert/eee460/massdefect.html• Negative ∆ mass into energy.• Binding Energy?• Stability?• Fission?• Fusion?
• More later ……..
Usually atomic masses are tabulated.Mass of the atom < ZmH + Nmn.
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Mass of the atom ZmH Nmn.
The Valley of Stabilityy y
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Nuclear Size• Different experiments give different results Radius not well defined.
D d b d l t h i• Depends on probe and relevant physics.• Probes should be close to the order of the size of the nucleus ~ 10-14 m.• Visible light? λ much larger• Visible light? λ much larger.
• 1 MeV γ? λ = ?? x 10-12 m. But interacts with orbital electrons.• Suitable probes: p n α e X Charge distribution Mass distribution• Suitable probes: p, n, α, e, X. Charge distribution. Mass distribution.• All experiments agree qualitatively and somehow quantitatively.• Project ….
• R ∝ A⅓ Why? In a while ……• R = r0 A⅓ with r0 dependent on the method• R r0 A with r0 dependent on the method.• Matter distribution ⇔ charge distribution. [Recently some halonuclei, e.g. 11Li, found]. What is that?
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Nuclear SizeHW HW 44 • Experiments show that
t = (2 4 ± 0 3) fm for all nucleit (2.4 ± 0.3) fm for all nuclei
t/R∝ A-1/3
• Is surface effect the same for all• Is surface effect the same for all nuclei?
r 0)( ρρρ0 = nucleon density near the center.
t “ ki ” thi kHWcHWc 33
( ) aRrer /
0
1)( −+=
ρρ
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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t = “skin” thickness. a = thickness parameter.R = Half-density radius.
Compare for A = 4, 40, 120 and 235.
Nuclear Sizeρ0 decreases with A?
YesNo matterech AZ ρρ =arg
High-energy e scattering
YesNo Ag
Light nuclei?~4A Constant R ∝ A⅓
34 3Rπ From some experiments….!
Charge distribution: r0 = 1.07 fm. a = 0.55 fm.R = r0 A⅓
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Matter distribution: r0 = 1.25 fm. a = 0.65 fm.R r0 A
Why?
Nuclear SizeHW HW 55
Nucleus Z/A Charge density
40CaCa ….. …..59Co ….. …..115InIn ….. …..197Au ….. …..
• Charge radius ~ nuclear radius, even though heavy nuclei have more neutrons than protons. Explain…• Density of ordinary atomic matter ~ 103 kg/m3. Density of nuclear mattery y g y~ 1017 kg/m3.• Neutron stars, 3 solar masses, only 10 km across ….. !!!• Surface effect?
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Surface effect?
Nuclear Size
Three conclusions can be drawn:
• Inside the nucleus the density is fairly uniform• Inside the nucleus the density is fairly uniform.• The transitional surface layer is thin.• The central density has a similar value for different nuclei.y
• Saturation?• Get an estimate for nuclear density and thus inter-nucleon distance.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Nuclear SizeNeutronD
1 Ci Pu-Be Neutron Source
Detector
Absorber Beam
nto
to
TeIeII σµ −− ==From Optical Model Di i
2)(2 λπσ += RT
From Optical Model Dimensions
σT
π2T
Differenttargets
31
A
g
How can we get r from the graph?HW HW 66
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APreferably low λ How can we get r0 from the graph?
Nuclear Size
Alpha particle (+2e)( )
Gold nucleus (+79e)d
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Nuclear Size• Closest approach “d”.• Eα = ECoulomb d = 2kZe2/Eα• What about the recoil nucleus?HWHW 77• HW HW 77 Show that
2 2mkZed N=where mN : mass of the nucleus
m : mass of alpha
)( αα mmE N −mα : mass of alpha
What are the values of d for 10, 20, 30 and 40 MeV α on Au?How does this explain … ?
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Nuclear Shapep• Crude Nucleons in the nucleus are confined to an approximatelypp yspherically symmetric structure Nuclear radius.• Deformations…! Consequences….!!• Is there a sharp spherical wall…???!!!Is there a sharp spherical wall…???!!!
• HW HW 88if it is assumed that the charge is uniformly spherically distributed in a g y p ynucleus, show that the electric potential energy of a proton is given by:
eZZ 2)1(3 −R
eZZKE )1(53 −
=
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Nuclear Binding Energyg gyBtot(A,Z) = [ ZmH + Nmn - m(A,Z) ] c2 B mtot( , ) [ H n ( , ) ]Bave(A,Z) = Btot(A,Z) / A HW HW 99 Krane 3.9Atomic masses from: HWHW 1010 Krane 3 12Atomic masses from: HW HW 1010 Krane 3.12http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=&all=all&ascii=ascii&isotype=all
Separation EnergyNeutron separation energy: (BE of last neutron)p gy ( )Sn = [ m(A-1,Z) + mn – m(A,Z) ] c2
= Bt t(A Z) - Bt t(A-1 Z) HWHW 1111 Show that Btot(A,Z) Btot(A 1,Z) HW HW 1111 Show that HW HW 1212 Similarly, find Sp and Sα.HWHW 1313 Krane 3 13 HWHW 1414 Krane 3 14
Magicn mbers
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HW HW 1313 Krane 3.13 HW HW 1414 Krane 3.14 numbers
Nuclear Binding Energyg gyMagic
numbersnumbers
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Nuclear Binding Energyg gyIn generalX Y + aSa(X) = (ma + mY –mX) c2
= BX –BY –BaThe energy needed to remove a nucleon from a nucleus ~ 8 MeV ≅ average binding energy per nucleon (Exceptions???).
Mass spectroscopy B.Nuclear reactions S.Nuclear reactions Q-value
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Nuclear Binding Energyg gy
Coulomb effectSurface effect
~200 MeV
Coulomb effectSurface effect
HWcHWc 44Thi k f t tThink of a computer program to
reproduce this graph.
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Nuclear Binding Energyg gy
HWHW 1515HW HW 1515A typical research reactor has power on the
d f 10 MWorder of 10 MW.
a) Estimate the number of 235U fission events that occur in the reactor per second.p
b) Estimate the fuel-burning rate in g/sb) Estimate the fuel burning rate in g/s.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Nuclear Binding Energyg gy
Is the nucleon bounded equally to everyIs the nucleon bounded equally to everyother nucleon?C ≡ this presumed binding energyC ≡ this presumed binding energy.Btot = C(A-1) × A × ½B = ½ C(A-1) Linear ??!!! Directly proportional ??!!!Bave = ½ C(A-1) Linear ??!!! Directly proportional ??!!!Clearly wrong … ! wrong assumption
finite range of strong forcefinite range of strong force,and force saturation.
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Nuclear Binding Energyg gy
L d i Z 82V)
Lead isotopes Z = 82
For constant ZS (even N) > S (odd N) S
n(M
eV
Sn (even N) > Sn (odd N)For constant NSp (even Z) > Sp (odd Z) E
nerg
y
p ( ) p ( )
Remember HW 14 (Krane 3.14).
arat
ion
208Pb (doubly magic) can then easily remove th “ t ” t i ro
n S
ep
the “extra” neutron in 209Pb. N
eutr
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Neutron Number N
Nuclear Binding Energyg gy
Extra Binding between pairs of “identical” nucleons in the same state (Pauli … !) Stability (e.g. α-particle, N=2, Z=2).( ) y ( g p , , )
Sn (A, Z, even N) – Sn (A-1, Z, N-1)n ( ) n ( )This is the neutron pairing energy.
t bl th dd dd d theven-even more stable than even-odd or odd-even and these are more tightly bound than odd-odd nuclei.
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Abundance SystematicsyOdd N Even N Total
HWHW 11\\Odd ZEven Z
HWcHWc 11\\
Even ZTotal
Compare:• even Z to odd Zeven Z to odd Z.• even N to odd N.• even A to odd A• even A to odd A.• even-even to even-odd to odd-even to odd-odd.
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Neutron ExcessZ Vs N (For Stable Isotopes)
90
70
80
Z = N
Remember HWc 1.
50
60
70 Z = N
40
50
Z
20
30
Odd A
0
10
0 20 40 60 80 100 120 140
Odd A
Even A
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0 20 40 60 80 100 120 140N
Neutron Excess
Remember HWc 1.
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Abundance Systematicsy
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Abundance Systematicsy
UR
E O
NN
CA
PTU
SEC
TIO
EUTR
ON
CR
OSS
Formation processNEUTRON NUMBER
EN
EFormation process
AbundanceN
DA
NC
EAbundanceA
BU
N r s r s
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The Semi-empirical Mass Formulap
• von Weizsäcker in 1935von Weizsäcker in 1935.• Liquid drop. Shell structure.• Main assumptions:Main assumptions:
1. Incompressible matter of the nucleus R ∝ A⅓R ∝ A .
2.Nuclear force saturates.• Binding energy is the sum of terms:Binding energy is the sum of terms:1. Volume term. 4. Asymmetry term.2. Surface term. 5. Pairing term.3. Coulomb term. 6. Closed shell term.…..
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The Semi-empirical Mass FormulapVolume Term Bv = + av A
Bv ∝ volume ∝ R3 ∝ A Bv / A is a constant i.e. number of neighbors of each nucleon is independent of the overall size of the nucleus.
B=
ABV constant
The other terms
AThe other terms are “corrections” to this term
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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this term.
The Semi-empirical Mass FormulapSurface Term Bs = - as A⅔
• Binding energy of inner nucleons is higher than that at the surface.• Light nuclei contain largerLight nuclei contain larger number (per total) at the surface.• At the surface there are:
32
2
322
0 44 ArAr
=ππ Nucleons.
roπ
1
1ABs ∝
31
AA
Remember t/ ∝ A-1/3
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Remember t/R ∝ A /3
The Semi-empirical Mass FormulapCoulomb Term BC = - aC Z(Z-1) / A⅓
• Charge density ρ ∝ Z / R3.• W ∝ ρ2 R5. Why ???ρ y• W ∝ Z2 / R. • Actually: ρπ drr24yW ∝ Z(Z-1) / R. • BC / A =
ρπ drr4
C- aC Z(Z-1) / A4/3
ρπ 3
34 r
Remember HW 8 ?!
3
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Remember HW 8 … ?!
The Semi-empirical Mass Formulap
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The Semi-empirical Mass FormulapQuiz Quiz 11From our information so farso far we can write:
...)1()(),( 31
32
+−++−−−=−AZZaAaAaMMZAMZAM CSVHnn
For A = 125, what value of Z makes M(A,Z) a minimum?, ( , )
Is this reasonable…???
So …..!!!!
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The Semi-empirical Mass FormulapAsymmetry Term Ba = - aa (A-2Z)2 / A
• Light nuclei: N = Z = A/2 (preferable).• Deviation from this “symmetry” less BE and stability.• Neutron excess (N-Z) is necessary for heavier nuclei.• Ba / A = - aa (N-Z)2 / A2.• Back to this when we talk about• Back to this when we talk aboutthe shell model.
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The Semi-empirical Mass FormulapPairing Term Bp = δ
Extra Binding between pairs of identical nucleons in the same state (Pauli !) Stability (e.g. α-particle, N=2, Z=2).
even even more stable than even odd or odd even and these are more tightlyeven-even more stable than even-odd or odd-even and these are more tightly bound than odd-odd nuclei.Remember HWcHWc 11\\ ….?!B t d t d ith A ff t f i d l d ithBp expected to decrease with A; effect of unpaired nucleon decrease with total number of nucleons. But empirical evidence show that:
δ ∝ A-¾δ ∝ A .
⎪⎪⎧+ − evenZevenNAap
43
Effect on:• Fission
⎪⎪⎩
⎪⎪⎨−
=− oddZoddNAa
oddA
p4
30δ
Fission.• Magnetic moment.Effect of high angular momentum
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⎪⎩ p momentum.
The Semi-empirical Mass FormulapClosed Shell Term Bshell = ηshell η
• Extra binding energy for magic numbers f N d Zof N and Z.
• Shell model.• 1 – 2 MeV more binding.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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The Semi-empirical Mass Formulap
• Fitting to experimental data• Fitting to experimental data. • More than one set of constants av, as …..
I h t t t d ?• In what constants does r0 appear?• Accuracy to ~ 1% of experimental values (BE).• Atomic masses 1 part in 104.• Uncertainties at magic numbers.g• Additional term for deformation.
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The Semi-empirical Mass Formulap
)(),(12
−−= MMZAMZAM Hnn
])2()1([ 1231
32
ηδ ++−−−−−− −− AZAaAZZaAaAa aCSV
Variations…….Variations…….Additional physics….Additional physics….Additional physics….Additional physics….Fitting……(Global vs. local)…..Fitting……(Global vs. local)…..
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Work it out …)(),( −−= MMZAMZAM Hnn
)( 2β ZZZAM])2()1([ 123
13
2ηδ ++−−−−−− −− AZAaAZZaAaAa aCSV
?),( 2++= γβα ZZZAM
??=
βα
∂M??=
γβ
?0 min =⇒=∂∂ Z
ZM
?=γ ∂Z A
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Mass Parabolas and Stabilityy
2 HWHW 1616
2
2),( ++= ZZZAM γβα HW HW 1616
32
−−++−= AaAaAaAM aSVn ηδα
31
4)( −−−−−= aAaMM aCHnβ
3114
)(−− += AaAa
aCHn
γ
β
34 += AaAa Caγβ2
0 min −=⇒=∂∂ ZM
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γ2min∂Z A
Mass Parabolas and Stabilityy
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Mass Parabolas and Stabilityy
Double β decay! Both Sides!
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Mass Parabolas and Stabilityy
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Mass Parabolas and StabilityyVertical spacing b b hbetween both parabolas ?
• Determine constants from atomic masses.
Odd-Odd
Even-Even
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Mass Parabolas and Stabilityy
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Nuclear Spinp• Neutrons and protons have s = ½ (ms = ± ½) so they are fermions and obey the Pauli-Exclusion Principleobey the Pauli-Exclusion Principle.•The Pauli-Exclusion Principle applies to neutrons and protons separately (distinguishable from each other) (IsospinIsospin).• Nucleus seen as single entity with intrinsic angular momentum Ι.• Associated with each nuclear spin is a nuclear magnetic momentwhich produces magnetic interactions with its environmentwhich produces magnetic interactions with its environment. •The suggestion that the angular momenta of nucleons tend to form pairs is supported by the fact that all nuclei with even Z and even N h l i 0have nuclear spin Ι=0. • Iron isotopes (even-Z), for even-N (even-A) nuclei Ι=0. • Odd-A contribution of odd neutron half-integer spin.Odd co t but o o odd eut o a tege sp• Cobalt (odd-Z), for even-N contribution of odd proton half-integer spin.
Odd N t o npaired n cleons large integer spin
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• Odd-N two unpaired nucleons large integer spin.
Nuclear Spinp
NaturalZ A Spin NaturalAbundance Half-life Decay
26 54 0 0.059 stable ...
26 55 3/2 ... 2.7y EC26 55 3/2 ... 2.7y EC
26 56 0 0.9172 stable ...
26 57 1/2 0.021 stable ...
26 58 0 0 0028 stable26 58 0 0.0028 stable ...
26 60 0 ... 1.5My β-
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Nuclear Spinp
Z A Spin Natural Abundance Half-life Decay
27 56 4 ... 77.7d β+β
27 57 7/2 ... 271d EC
27 59 7/2 1.00 stable ...
27 60 5 ... 5.272y β-
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Nuclear Magnetic MomentgRemember, for electrons Revise: Torque on a current loop.
Gyromagnetic ratio (g-factor)
Z component ?? Experiment applied magnetic field
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Z component ?? Experiment, applied magnetic field.
Nuclear Magnetic MomentgFor Nuclei
For free protons and neutronsProton: g = 5.5856912 ± 0.0000022 ∼ 3.6 Neutron: g = -3.8260837 ± 0.0000018 ∼ 3.8
The proton g-factor is far from the gS = 2 for the electron, and even the uncharged neutron has a sizable magnetic moment!!!
Internal structure (quarks)Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
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Internal structure (quarks).
Nuclear Magnetic Momentg
Magnetic moment µNuclide Nuclear spin Magnetic moment µ(in µN)
n 1/2 1 9130418n 1/2 -1.9130418
p 1/2 +2.79284562H (D) 1 +0.8574376
17O 5/2 -1.8927957Fe 1/2 +0.0906229357C 7/2 +4 73357Co 7/2 +4.73393Nb 9/2 +6.1705
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Nuclear Parityy
• ψ(r) ψ(-r) Even.ψ(r) ψ( r) Even.• ψ(r) -ψ(-r) odd.• For a nucleon ψ is either of even (π = +) or• For a nucleon ψ is either of even (π = +) or odd (π = -) parity.
F th l• For the nucleus π = π1 π2 π3 … πA.• Practically not possible.• Overall π can be determined experimentally.• Overall Ιπ for a nucleus (nuclear state).Overall Ι for a nucleus (nuclear state).• Transitions and multipolarity of transitions (γ-emission)
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emission).
Electromagnetic momentsg• Electromagnetic interaction information about gnuclear structure.• Charge electric; current magnetic.g ; g• Electromagnetic multipole moments.Field∝1/r2 (zeroth, L=0) electric monopole moment.( , ) p
1/r3 (first, L=1) electric dipole moment.1/r4 (second, L=2) quadrupole moment.( , ) q p………1/r2 magnetic monopole (questionable….!).g p (q )Higher order magnetic moments, we already discussed the magnetic dipole moment.
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g p
Electromagnetic momentsg• Expectation value of the moment. ∫ dvϑψψ *
• Each multipole moment has a parity, determined by ∫ dvϑψψ
the behavior of the multipole operator when r -r.• Parity of ψ does not change the integrand.• Electric moments: parity (-1)L.• Magnetic moments: parity (-1)L+1.• Odd parity vanish.
electric dipole.magnetic quadrupole.electric octupole.
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…………
Electromagnetic momentsg• Electric monopole: net charge Ze.
e e evr
p g• Magnetic dipole: (already discussed). µ = iA
µ = eT
A ; µ = e2π r v
π r2 ; µ = evr2
µ = emvr ; µ = epr ; µ = e Lµ =2m
; µ =2m
; µ =2m
L
• g-factors.
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Electromagnetic momentsg• The nucleus has charge (monopole
t)Classical momentsmoment).
• No dipole moment since it is all positive.B t if th l i t h i ll
moments
• But if the nucleus is not spherically symmetric, it will have a quadrupole
tmoment.
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Electric Quadrupole Momentp• For a point charge e: eQ = e(3z2 - r2).
S h i l t 2 2 2 2/3 Q 0• Spherical symmetry x2 = y2 = z2 = r2/3 Q = 0.• For a proton:
∫I th l Q ⟨ 2⟩
∫ −= dvrzeeQ ψψ )3( 22*
• In the xy-plane: Q ∼ - ⟨r2⟩.• ⟨r2⟩ is the mean square radius of the orbit.
Al Q 2 ⟨ 2⟩• Along z: Q ∼ +2 ⟨r2⟩.• Expected maximum ∼ er0
2A2/3.6 10 30 t 50 10 30 2• 6x10-30 to 50x10-30 em2.
• 0.06 to 0.5 eb.
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Electric Quadrupole MomentpNuclide Q (b)Q ( )2H (D) +0.00288
17O 0 0257817O -0.0257859Co +0.4063Cu -0.209133Cs -0 003
• Closed shell Spherically symmetric core Cs 0.003
161Dy +2.4176L 8 0
symmetric core. • Test for shell model• Strongly deformed nuclei ! 176Lu +8.0
209Bi -0.37
nuclei…..!
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Nuclear Force (Origin of Binding)( g g)
RecallRecall Atomic Binding Energies for hydrogen like atoms:
Dimensionless fine structure constant137
14
,)()(0
2
≈=−
=c
er
ZcrVh
h
πεαα =1
constant.0
NemmZcE =−= µµα1 222
Nen mmn
cE+
== µµα ,2 2
2hwith Bohr radii:
2nc
rn αµh
=
• Coupling constant Strength.• Charge.
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• Mediators (Bosons).
Nuclear ForceThe deuteron: proton-neutron bound state.
!!!!!!!!!1.0
4,)()(
0
2
≈=−
=c
qr
crV SS
S
h
h
πεαα !!!!!!!!!
4 0 cr hπεnp
Sn
mmcE
+=−= µµα ,1
21
222 !!!!!!!!!
np mmn +2 2
2nr h= n
cr
Sn αµ=
Hydrogen: E = eV r = x10-10 mHW HW 1717
Hydrogen: E1 = … eV r1 = …x10 mPositronium: E1 = … eVD t E M V 10 15
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Deuteron: E1 = … MeV r1 = …x10-15 m
Nuclear Force
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Nuclear ForceAttractive but repulsive corerepulsive core. At what separation?
• Saturation?• Saturation?• Get an estimate for nuclear density and thus inter-nucleon distance Ha e o done that?
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nucleon distance. Have you done that?
Nuclear Force
Is the nucleon bounded equally to everyIs the nucleon bounded equally to everyother nucleon?C ≡ this presumed binding energyC ≡ this presumed binding energy.Btot = C(A-1) × A × ½B = ½ C(A-1) Linear ??!!! Directly proportional ??!!!Bave = ½ C(A-1) Linear ??!!! Directly proportional ??!!!Clearly wrong … ! wrong assumption
finite rangefinite range of strong forcefinite range finite range of strong force,and force saturation.
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Nuclear Force• Rate of decay or interaction R ∝ ρ(E).y ρ( )• Coupling constant α. Vertices in the diagrams.• For decays R ∝ 1/T. (T ≡ Lifetime).y ( )• The density of states ρ is a measure of the number of quantum mechanical states per unit q penergy range that are available for the final products. The more states that are available, the p ,higher the transition rate.• The coupling constant α can be interpreted as an p g pintrinsic rate.
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Nuclear Force• Electrostatic and gravitational potential long range (V∝1/r).• Near constancy of nuclear binding energy per nucleon B/A means• Near constancy of nuclear binding energy per nucleon B/A means that each nucleon feels only the effect of a few neighbors. This is called saturation. It implies also that the strong internucleon potential is short range.• Range is of order of the 1.8 fm internucleon separation.• Since volume ∝ A nuclei do not collapse there is a very short range• Since volume ∝ A, nuclei do not collapse, there is a very short range repulsive component. • Exchange.• Some particles are immune. Like what?• Is nuclear physics just quark chemistry?• Charge independenceCharge independence. • Spin dependence. (Deuteron).• Non-central (tensor) component conservation of orbital angular
t ?Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
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momentum….?
Nuclear Force• Spin dependent difference in neutron scattering cross sections of ortho- and para-hydrogen. • Compare n-p to n-n and p-p Charge independence of nuclear force.
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Nuclear ForceMirror Nuclei
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Nuclear Force
If two charges q and q' exchangeWhat about What about
forces betweenforces betweenIf two charges, q and q exchange photons, the Coulomb force occurs between them.
forces between forces between quarks?quarks?Color?Color?
Krane Krane 44..55
If pions are exchanged between two nucleons the strong nuclear forcenucleons, the strong nuclear force occurs. Remember the weak nuclear force…
_Boson?Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
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_ν++→ epnBoson?
Nuclear Force
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Nuclear Force• Only Hadrons.• Typical time: 10-24 s (c 10-15 m)• Typical time: 10-24 s. (c, 10-15 m).• Exchange of light ∼ 140 MeV pions.• ∆t = ħ/∆E = 4.7 x 10-24 s. (Why ∆E?).∆t ħ/∆E 4.7 x 10 s. (Why ∆E?).• Range ∼ ∆t c = ħ/mc = 1.4 x 10-15 m.• Range and time complicated by possibilities of heavier hadron exchange.• Isospin. Conservation of Isospin. Only relevant to hadrons.• Hadron multiplets: Doublet of nucleons and triplet of pions and• Hadron multiplets: Doublet of nucleons and triplet of pions and …• The members of a multiplet have the same strangeness, hypercharge, spin, etc… , but differ in charge and differ slightly in ype c a ge, sp , e c , bu d e c a ge a d d e s g ymass.• Relationship between particle and nuclear physics.
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• Accelerators and large accelerators.
Isospinp• IsospinIsospin Magnitude )1( +TT• T3 can take T, T-1, T-2, ….., -T. • 1,2,3 not x,y,z (Isospin space).
Singlets (T 0) Do blets (T ½) Triplet (T 1) Q artet (??)• Singlets (T = 0), Doublets (T = ½), Triplet (T = 1), Quartet (??).• -T3 for antiparticles.• Isospin addition: for a collection of hadrons (e g in interaction)Isospin addition: for a collection of hadrons (e.g. in interaction)
∑= iTT )(max ∑= iTT )(33 3TT ≥• Example: π+-p scattering, Tmax = 3/2, T3 = 3/2 T can only be 3/2.
i i3
Read Krane Read Krane 1111..33..
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The Deuteron
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The Deuteron• Deuterium (atom).•The only bound state of two nucleons simplest bound state•The only bound state of two nucleons simplest bound state.• Neither di-proton nor di-neutron are stable. Why?
21• Experimentally ∼ 2.224 MeV (Recoil..!).• Also inverse (γ,n) reaction using Bremsstrahlung (Recoil…!).
γ+→+ HHn 21
• Mass spectroscopy mass of D (or deuterium atom).• ∆mc2 = 2.224…??…MeV Very weakly bound.• Mass doublet method all results are in agreement• Mass doublet method all results are in agreement.• Compare 2.224 MeV to 8 MeV (average B/A for nuclei).• Only ground state. (There is an additional virtual state).O y g ou d state ( e e s a add t o a tua state)
HW HW 1818Problems 4 1 4 5 in Krane
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Problems 4.1 - 4.5 in Krane.
The DeuteronV(r) = -V0 r < R
= 0 r > R• Oversimplified.HW HW 1919Assuming Assuming ll = = 00, show , show that Vthat V00 ∼∼ 35 35 MeV.MeV.(Follow Krane Ch.(Follow Krane Ch.4 4 and and Problem Problem 44..66), or similarly ), or similarly any other reference.any other reference.• Really weakly bound.• What if the force were a
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bit weaker…?
The Deuteron
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The Deuteron• Experiment deuteron is in triplet state Ι = 1.• Experiment even parity.
• Ι = l + sn + sp parity = (-1)lp
• Adding spins of proton and neutron gives:s = 0 (antiparallel) or s = 1 (parallel).
• For Ι = 1parallel s-state evenparallel p-state oddantiparallel p-state odd
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parallel d-state even
The Deuteron• Experiment µ = 0.8574376 µN spins are aligned…..But.?• Direct addition 0.8798038 µN.• Direct addition of spin components assumes s-state (no orbital component).• Discrepancy d-state admixture.
ψ = a0ψ0 + a2ψ2
µ = a02µ0 + a2
2µ2HW 20 In solving HW 19 you assumed an s-state. How good was that assumption?
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The Deuteron
• S-state No quadrupole moment.S state No quadrupole moment.• Experiment +0.00288 b.HWHW 2121HW HW 2121Discuss this discrepancy.
• From µ and Q, is it really admixture?• What about other effects?• What about other effects?• Important to know the d-state wavefunction.
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Nuclear Force
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Nuclear Models• Nuclear force is not yet fully understood.• No absolutely satisfying model, but models.• Specific experimental data specific model.• Model success in a certain range.• Some are:
Individual particle model. (No interaction, E. states, static properties, …).
Liquid drop model. (Strong force, B.E., Fission, …).Collective model.α-particle model.Optical modelOptical model.Fermi Gas model.others …..
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others …..
Shell model• Electron configuration….
1 2 2 2 2 6 3 2 3 6 4 2 3d10 4 6Chemistry!
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 ….• AtomicAtomic Electron magic numbers: 2, 10, 18, 36, 54, …
Common center of “external” attraction.Well understood Coulomb force.One kind of particles.Clear meaning for electron orbits.g…
• NuclearNuclear magic numbers: 2 8 20 28 50 82 126• NuclearNuclear magic numbers: 2, 8, 20, 28, 50, 82,126, …(for Z or N).
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Shell model
Evidence:Evidence:1) End of radioactive series:1) End of radioactive series:
thorium series 208Pburanium series 206Pbactinium series 207Pbneptunium series 209Bi
2) A Z d N ’ h l i l l b2) At Z and N mn’s there are relatively large numbers of isotopes and isotones.
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Shell model
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Shell model
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Shell model3) Natural abundances.
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Shell model
4) N t t ti4) Neutron capture cross section.
PTU
RE
TIO
NR
ON
CA
PSS
SEC
TN
EUTR
CR
OS
NEUTRON NUMBER
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Shell model5) Binding energy of the last neutron
(S ti E )(Separation Energy).(The measured values are plotted relative to the calculations without η).
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Shell model6) Excited states.
Pb (even-A) isotopes.
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Shell model
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Shell model
7) Quadrupole moments ?
HWHW 2222
7) Quadrupole moments ….. ?
HW HW 2222
Work out more examples for theWork out more examples for the above evidences. For example, take part of a plot and work on a group of relevant nuclides.relevant nuclides.
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Shell model• Nucleons are in definite states of energy and gyangular momentum.• Nucleon orbit ?? Continuous scattering expected ..!!g p• No vacancy for scattering at low energy levels.• Potential of all other nucleons.• Infinite square well:
⎨⎧ < Rr
V0
• Harmonic oscillator:⎩⎨ =∞
=Rr
V
22
21 rmV ω=
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2
Shell model ?model
???
2(2l 1) ?2(2l + 1)accounts correctlycorrectly for the number ofnumber of nucleons in each each
Infinite spherical well(R=8F)
Harmonic oscillatorlevellevel.But what about ων h)( 2
3+=Eabout magic magic numbersnumbers?
ωνν h)( 2+E
ωh)2( 21−+= lnEnl
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numbersnumbers?
Shell model• More realistic! (Can it solve the problem?)• Finite square well potential:
⎧ ≤− RrV0
⎩⎨⎧
>≤
=RrRrV
V0
0
V)( 0
• Rounded well potential:⎩
Adjusted by the separation energies.
( ) MeVVeVrV aRr 57~
1)( 0/
0−+
−=
• Correction for asymmetry and Coulomb repulsion.
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Shell model
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Shell model
ZN −HW HW 2323
)(27 MeVA
ZNVas ±=∆
Coulomb repulsion? V (r) = ??Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
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Coulomb repulsion? Vc(r) = ??
Shell model• Separation of variables:
F i h i ll t i t ti l V( )
),()()()()(),,( φθφθφθψ mlYrRrRr =ΦΘ=
• For a given spherically symmetric potential V(r),the bound-state energy levels can be calculated f di l ti f ti l bit lfrom radial wave equation for a particular orbital angular momentum l. HW HW 2424• Notice the important centrifugal potential.
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Shell modelcentrifugal potential
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1s 1p 1d 2s 1f 2p 1g 2d 3sml2(2l +1) 2 6 10 2 14 6 18 10 2Total 2 8 18 20 34 40 58 68 70ms
• 2, 8, 20 ok.• What about other magic at about ot e ag cnumbers?• Situation does not improve pwith other potentials.• Something very fundamental g yabout the single-particle interaction picture is missing in p gthe description…..!!!!!• Spin-orbit coupling.
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p p g
Shell model• So far, 2(2l + 1) accounts correctly for the number of nucleons in each level, since we already considered both orbital angular momentum, and spin, but still not for closed shells.
sl ms
mlsl Ymsml χ≡,,, slsl χ,,,
Spherical Harmonics
But this representation does not solve theHarmonics,
Eigenfunctions of L2 and Lz. S mm h 21)1( 22
does not solve the problem.
z
smsmS
sssS
smss
msz
ms
ms
ss
ss
≤≤−=
=+=
χχ
χχ
h
h 21)1( 22
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Shell model
SpinSpin--Orbit CouplingOrbit Coupling• M. G. Mayer and independently Haxel, Jensen, and Suess.• Spin-Orbit term added to the Hamiltonian:
LSrVrVm
pH SO .)()(2
2
++=m2
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Shell model
LSJLSJLS +=−−= 2/)( 222
LL ULL S
LSJLSJLS +== 2/)(.
LLantiparallel
ULparallelJ
)1( 22 +≤≤−+= sljsllsjmjjlsjmJ h ,)1(
≤≤−=
+≤≤+
jmjlsjmmlsjmJ
sljsllsjmjjlsjmJ
jjjj
jj
h
h
,....2,1,0,)1(
,22 =+=
≤≤
llsjmlllsjmL
jmjlsjmmlsjmJ
jj
jjjjz
h
h
21,)1(
,....2,1,0,)1(22 =+=
+
slsjmsslsjmS
llsjmlllsjmL
jj
jj
h
h
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21,)1( + slsjmsslsjmS jj h
Shell modelmodel
2j+1
2(2x3 + 1) = 142(2x3 + 1) 14
1f7/2l = 3
7/2
First time
jtime
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Shell model
2
LSrVrVm
pH SO .)()(2
2
++=m2
2)]1()1()1([21. h+−+−+= sslljjSL2
HW HW 2525 0,)12(21 2 >+∝ llgap h
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2
Shell modelNotes:1 The shell model is most useful when applied to closed shell or1. The shell model is most useful when applied to closed-shell or near closed-shell nuclei. 2. Away from closed-shell nuclei collective models taking into y gaccount the rotation and vibration of the nucleus are more appropriate. 3. Simple versions of the shell model do not take into account pairing forces, the effects of which are to make two like-nucleons combine to give zero orbital angular momentum The pairingcombine to give zero orbital angular momentum. The pairing force increases with l.4. Shell model does not treat distortion effects (deformed nuclei) ( )due to the attraction between one or more outer nucleons and the closed-shell core.
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Shell modelFermi Gas
EF ∝ n2/3
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Shell modelNuclear
ti ?reactions?
Transition probability?
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Shell modelGround state: (near closed shells)( )
1. Angular momentum of odd-A nuclei is determined by the angular momentum of the last nucleon that is odd. g2. Even-even nuclei have zero ground-state spin, because the net angular momentum associated with geven N and even Z is zero, and even parity. 3. In odd-odd nuclei the last neutron couples to the last pproton with their intrinsic spins in parallel orientation.
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A < 150190 < A < 220
Shell model
Near No spin-
orbit
Harmonic oscillatorNear drip
line
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valley of βstability
orbit coupling
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Shell model
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Shell model
• 17 p, 21 n.• p in 1d3/2 l s π = +p in 1d3/2 l s π • n in 1f7/2 l s π = -• Rule 3 sp sn lp ln
total π = -Rule 3 sp sn lp ln
• ½ + ½ + 3 – 2 = 2
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Shell modelExcited states:
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Shell modelExtreme independent particle model!!! Does the core really remain inert?
1d3/2
?1p1/2
l pairing
2s1/2
l pairing
1d5/2
2s1/2
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1d5/2
Shell model
Core 20Core• Extreme independent
20
particle model only 23rd neutron.• More complete shell model all three “Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
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“valence” nucleons.
Shell model
HWHW 2626 Discuss the energy levels of nuclei with HW HW 2626odd number of nucleons in the 1f7/2 shell.
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125and 43Sc, 43Ti.
Shell modelDipole Magnetic Moment
Nj jg µµ =
)1()1()1()1()1()1( +−++++
+++−+ sslljjgsslljjgg
HW HW 2727 Show that
)1(2)1(2 ++
+=
jjg
jjgg lsj
and examineexamine Eqs 5 9 in Krane In additionand examineexamine Eqs. 5.9 in Krane. In addition, work out problem 5.8 in Krane Conclusion?P t (f ) 5 5856912 ? 1 ?Proton: gs(free) = 5.5856912 ? gl = 1 ?Neutron: gs(free) = -3.8260837 ? gl = 0 ?Wh t b t + d ?
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What about π+ and π-?
Shell modelElectric Quadrupole Moment In the xy-plane: Q ∼ - ⟨r2⟩.
Refined QM ⎤⎡ −− 112 223 nj⎥⎦
⎤⎢⎣
⎡−
−+
−=12121
)1(212 32
053
jnAr
jjQ
⎦⎣+ 12)1(2 jjjn 21 ≤≤Extremes
Single particle: n = 1 ive Q Number of protons in a subshell
Single particle: n = 1 - ive QSingle hole: n = 2j +ive Q
E i T bl 5 1 <r2> for a uniformly charged sphere
Examine Table 5.1 and Fig.5.10 in
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Krane
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Shell model
Validity Nuclide Q (b)ValidityA < 150
190 < A < 220
Nuclide Q (b)2H (D) +0.00288
190 < A < 22017O -0.02578
59Co +0.4063Cu -0.209133Cs 0 003133Cs -0.003
161Dy +2.4y176Lu +8.0
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209Bi -0.37
Collective model• Large quadrupole moments nucleus as a collective b d (Li id d d l)body (Liquid drop model).• Interactions between outer nucleons and closed shells
t d f ticause permanent deformation.• Single-particle state calculated in a non-spherical
t ti l li t dpotential complicated.• Spacing between energy levels depends on size of di t tidistortion.• Doubly magic 1st excited state away from GS.
N l i l ti l t t• Near closure single-particle states.• Further away from closure collective motion of the
it d t tNuclear and Radiation Physics, BAU, First Semester, 2007-2008
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core excited states.
Collective model
• A net nuclear potential due to filled core• A net nuclear potential due to filled core shells exists.
• Collective model combines both liquid drop model and shell model.p
• Two major types of collective motion:R t ti R t ti f d f d hRotations: Rotation of a deformed shape.Vibrations: Surface oscillations.
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Collective model• Rotational motion observed for non-spherical nuclei.• Deformed nuclei are mainly 150 < A < 190 and A > 220.• Ellipsoid of surface:
[ ]31
200 ),(1),( YArR += φθβφθDiff
31
0534
ArR∆
=πβ
Difference between
semimajor and053 Ar semimajor and semiminor axes.Deformation
parameter Rparameter.
HW HW 2828 Problems Problems 55..11 11 and and 55..12 12 in Krane.in Krane.Di ff t d l t
Rav
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Discuss effect on quadrupole moment.β > 0 β < 0
Collective modelSymmetry
axis 1E 2axis
2=
=
l
E
g
g 2
ω
ω
)1(22
+
=
IIlE
l g
h
ω
)1(22
+== IIEgg
GS (even-even) 0+GS (even even) 0Symmetry only even I
0)0( =+E
)2/(20)4(
23.152/4.91)2/(6)2(0)0(
2
22 gg
h
hh =⇒==
=
+
+
E
keVkeVEE
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)2/(20)4( 2 gh=+E
Collective model
HW HW 2929 compare measured energies of the states of the ground state rotational band to the calculations.Rigid body or liquid drop? Intermediate Short
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range and saturation of nuclear force.
Collective model
Spin parity Emeasured E/E(2+) I(I + 1)/6HWHW 2929 (continued)(continued) Spin (keV) E/E(2 ) I(I + 1)/6HW HW 29 29 (continued)(continued)
12+
Higher angular momentum centrif gal stretching 12
10+ 1518.00 16.61 18.338+
centrifugal stretching higher moment of
inertia lower6+ 7.004+ 299.44 3.28 3.33
inertia lower energy than expected
additional 2+ 91.4 1.0 1.00+ 0
evidence for lack of rigidity.
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164Er
Collective modelOdd-A
nucleonEE += 2gω212
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Collective modelAverageshapeshape
Instantaneousshapeshape
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Collective modelλ
∑ ∑µ-λ,λµ αα =
φ),(θY(t)αRR(t) λµλ λ µ
λµav ∑ ∑−=
+=µ λ,λµ
Instantaneouscoordinate
Symmetry
Amplitude Sphericalharmonics
r0A1/3coordinate harmonics
λ = 1 λ = 2 λ = 3http://wwwnsg.nuclear.lu.se/basics/excitations.asp?runAnimation=beta10
λ = 0monopole
λ = 1dipole
λ = 2quadrupole
λ = 3octupole
.
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Collective model
R(t) = Ravr +α00 Y00λ = 0
l
)(YαRR(t)1
+= ∑ φθ
R(t) Ravr α00 Y00 monopole
λ = 1
YYYR
),(YαRR(t) 1µ1 µ
1µavr
+++
+= ∑−=
φθ λ = 1dipole
YαR YαYαYα R
1010avr
1- 1,1- 1,10101111avr
+=
+++=
1010avr
Both monopole and dipole excitations requireBoth monopole and dipole excitations require high energy.
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Collective model
λ = 22
)(YRR(t) + ∑ φθ λquadrupole
22222 12 1202021212222avr
2µ2 µ
2µavr
YαYαYαYαYα R
),(YαRR(t)
+++++=
+= ∑−=
φθ
Q ti ti f d l ib ti i ll d2020avr
2-2,2-2,2,-12,-1202021212222avr
YαR +=
• Quantization of quadrupole vibration is called a quadrupole phonon.
A h i t it f l t• A phonon carries two units of angular momentum and even parity (-12).
Thi d i d i t F t l i• This mode is dominant. For most even-even nuclei, a low lying state with Jπ=2+ exists.
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• Octupole phonon.
Collective modell = 4 µ = +4, +3, +2, +1, 0, -1, -2, -3, -4l 2 2 1 0 1 2Triplet l = 2 µ = +2, +1, 0, -1, -2l = 0 µ = 0
Triplet0+, 2+, 4+
-2 -1 0 1 2
-2 -4 -3 -2 -1 0-2 -4 -3 -2 -1 0
-1 -3 -2 -1 0 +1
0 -2 -1 0 +1 +2
1 1 0 +1 +2 +31 -1 0 +1 +2 +3
2 0 +1 +2 +3 +4
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Collective modelTwo-phonon triplet at twice the HW HW 3030energy of the single phonon state. Krane 5.10
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Nuclear Reactions
X(a b)YX(a,b)Y• First in 1919 by Rutherford:
4He + 14N 17O + 1H4He + 14N 17O + 1H14N(α,p)17O
I id t ti l h di ti l• Incident particle may: change direction, lose energy, completely be absorbed by the target……
T t t t il• Target may: transmute, recoil……• b = γ Capture reaction.
If B E it fi i ( bl )• If B.E. permits fission (comparable masses).• Different exit channels a + X Y1 + b1
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Y2 + b2Y3 + b3 …….
Nuclear Reactions• Recoil nucleus Y could be unstable β or γ emission.
• One should think about:Reaction dynamics and conservation laws i.e.
conditions necessary for the reaction to be energetically possible.
Reaction mechanism and theories which explain the reaction.
Reaction cross section i.e. rate or probability.
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Nuclear ReactionsConservation Laws• Charge, Baryon number, total energy, linear momentum, angular momentum, parity, (isospin??) …….
bθa
pb
bQTTcmcm iffi =−=− 22
φa
pa XpY Y
+ve Q-value exoergic reaction.-ve Q-value endoergic reactionY ve Q value endoergic reaction.
aYb TQTT +=++ve Q-value reaction possible if Ta 0.-ve Q-value reaction not possible if Ta 0. (Is Ta > |Q| sufficient?).
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Conservation of momentum ……
Nuclear Reactions• Conservation of momentum.
W ll d t d t t YHW HW 3131
• We usually do not detect Y.Show that:
2
Th th h ld (f T )bY
aaYYbYabaabab mm
TmmQmmmTmmTmmT
+−+++±
=])()[(coscos 2θθ
• The threshold energy (for Ta): (the condition occurs for θ = 0º).bY
ThmmQT +
−=
• +ve Q-value reaction possible if Ta 0.C l b b i !!!
abYTh mmm
Q−+
• Coulomb barriers…….!!!• -ve Q-value reaction possible if Ta > TTh.
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Nuclear Reactions
HWHW 3131 (continued)(continued)
• The double valued situation occurs between TTh and the
HW HW 3131 (continued)(continued)
The double valued situation occurs between TTh and the upper limit Ta
\.YmQT −=\
Double valued in a forward coneaY
a mmQT
−=
• Double-valued in a forward cone.
aaYYbY TmmQmmm ])()[(cos2 −++θaba
aaYYbY
TmmQ )()(cos max −=θ
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Nuclear Reactions
Sample
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Nuclear Reactions
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Nuclear Reactions
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Nuclear Reactions
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Nuclear Reactions• If the reaction reaches excited states of Y
exbexYaXex EQcmEcmcmcmQ −=−+−+= 02222 )(
58Ni(α,p)61Cu
even less ….
less proton energy
Highest proton energy
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See Figures 11.4 in Krane
Nuclear ReactionsNeutron scattering• Inelastic Q = -Eex (=-E*).• Elastic Q = 0.
HW HW 3232Discuss the elastic and inelastic scattering of neutrons using the relations you derived inneutrons using the relations you derived in HW 31.
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Nuclear ReactionsCategorization of Nuclear Reactions• According to: bombarding particle, bombarding energy, target, reaction product, reaction mechanism.
Bombarding particle:• Bombarding particle:Charged particle reactions. [ (p,n) (p,α) (α,γ) heavy ion reactions ].Neutron reactions. [ (n,γ) (n,p) ….. ].Photonuclear reactions. [ (γ,n) (γ,p) … ].Electron induced reactions………….
• Bombarding energy:Bombarding energy:Thermal. Epithermal.Slow Neutrons.?Slow.Fast. Low energy charged particles.Hi h h d ti l
?
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High energy charged particles.
Nuclear Reactions• Targets:
Light nuclei (A < 40).Light nuclei (A 40).Medium weight nuclei (40 < A < 150).Heavy nuclei (A > 150).
• Reaction products:• Reaction products:Scattering. Elastic 14N(p,p)14N
Inelastic 14N(p,p/)14N* Radiative capture.Fission and fusion.Spallation.…..
• Reaction mechanism:Direct reactionsDirect reactions.Compound nucleus reactions.
• More in what follows ….
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• What is a transfer reaction….?????
Reaction Cross Section(s) (Introduction)( )• Probability.
P j til ill b bl hit t t if i l• Projectile a will more probably hit target X if area is larger.• Classically: σ = π(Ra + RX)2.Classical σ = ??? (in b) 1H + 1H 1H + 238U 238U + 238UClassical σ = ??? (in b) H + H, H + U, U + U • Quantum mechanically: σ = π D2.
Xa mm hhD
+HW HW 3333
• Coulomb and centrifugal barriers energy dependence of σ
CMaXaXaaX
Xa
EEmmmm
µ22hh
D ==
Coulomb and centrifugal barriers energy dependence of σ.• Nature of force:
Strong: 15N(p,α)12C σ = 0.5 b at Ep = 2 MeV.g (p, ) pElectromagnetic: 3He(α,γ)7Be σ = 10-6 b at Eα = 2 MeV.Weak: p(p,e+ν)D σ = 10-20 b at Ep = 2 MeV.E i t l h ll t l X ti
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• Experimental challenges to measure low X-sections..
Reaction Cross Section(s) (Introduction)( )Detector for particle “b”dΩ
θ,φIa
p
d“b” particles / s
2θ,φ
NIdRd b=σ
cm2
NIa
Typical nucleus (R=6 fm): geometrical πR2 ≈ 1 b.Typical σ: <µb to >106 b
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Typical σ: <µb to >10 b.
Reaction Cross Section(s) (Introduction)( )Many different quantities are called“ ti ”
Angular distribution“cross section”.Krane Table 11.1 drdRb π
φθ4
),( Ω=Units !
rd φθσπ
),(4
=“Differential” cross sectionσ(θ φ) or σ(θ )
Units … !
NId aπ4Ωσ(θ,φ) or σ(θ )or “cross section” …!!
=Ω ddd φθθsin D bl diff ti l
∫ ∫ ∫=Ω=ddddd σφθθσσ
φθθπ π2
sin d σ2Doubly differential
dσ∫ ∫ ∫ ΩΩ ddφ
0 0 ΩddEbdEσ for all “b” particles
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σt for all b particles.
Coulomb Scatteringg• Elastic or inelastic.• Elastic Rutherford scattering.• At any distance:V = 0
Ta = ½mvo2 r
zZemvmvo
22
212
21
41πε
+=
vovmin
a ol = mvob
ro4πε
b rmin
o
dd4
1 2
=o d
zZeVπε
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0=aT
No dependence on φ
Coulomb Scatteringg
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Coulomb Scatteringg
bnxf π)( 2=b θ
bdbnxdfbnxfπ
π2)(
)(==b θ
< b > θdb dθ bdbnxdf π2)(=
n ≡ target nuclei / cm3
x ≡ target thickness (thin).
db dθ
x target thickness (thin).nx ≡ target nuclei / cm2
HW HW 3434 θd
b
Show thatand hence 2
cot2
θdb =
222 ⎞⎛⎞⎛
24
222
sin1
41
4)(
θπεθσ
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
Ω ao TzZe
dd
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161Rutherford cross section
Coulomb ScatteringgStudy Fig. 11.10 (a,b,c,d) in Kranein Krane
See also Fig 11 11 inSee also Fig. 11.11 in Krane.
HW HW 3535Show that the fraction of incident alpha particles scattered at backward angles from a 2 µm
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g µgold foil is 7.48x10-5.
Coulomb Scatteringg• Elastic Rutherford scattering.• Inelastic Coulomb excitation.
See the corresponding alpha spectrum of Fig. 11.12 in Krane.
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p g p p g
Coulomb Scatteringg
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Nuclear Scatteringg• Elastic or inelastic.• Analogous to diffraction• Analogous to diffraction.• Alternating maxima and minima.• First maximum at λθ ≈
1
ph
=λ Rθ ≈
• Minimum not at zero (sharp edge of the nucleus??)
31
ARR o=
of the nucleus??)• Clear for neutrons.• Protons? High energy, large g gy, gangles. Why?• Inelastic Excited states,
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energy, X-section and spin-parity.
Compound Nucleus Reactionsp
Direct• Time.
CN decays• Energy.
CN decays
• Two-step reaction. CN “forgets” how it was formedE CM • CN “forgets” how it was formed.
• Decay of CN depends on statistical factors that are functions
EaCM
statistical factors that are functions of Ex, J.• Low energy projectile, medium or
QCN
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heavy target.
Compound Nucleus Reactionsp
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Compound Nucleus Reactionsp• Consider p + 63Cu at Ep
CM= 20 MeV.• Calculate E + [m(63Cu) + m(p) m(64Zn)]c2• Calculate Ep + [m(63Cu) + m(p) – m(64Zn)]c2.• Divide by 64 available energy per nucleon << 8 MeV.• Multiple collisions “long”Multiple collisions long time statistical distribution of energy small chance for a nucleon to get enough energy EvaporationEvaporation.
Hi h i id t• Higher incident energy more particles “evaporate”.
See also Fig. 11.21 in Krane.
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Direct Reactions
• Random collisions nearly isotropic angular distribution.• Direct reaction component strong angular dependence.
See also Fig. 11.20 in Krane.
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Direct Reactions• Peripheral collision with surface nucleon.• 1 MeV incident nucleon D ≈ ?? more likely to interact with• 1 MeV incident nucleon D ≈ ?? more likely to interact with the nucleus CN reaction.• 20 MeV incident nucleon D ≈ ?? peripheral collision20 MeV incident nucleon D ?? peripheral collision Direct reaction.• CN and Direct (D) processes can happen at the same incident particle energy. Distinguished by:
D (10-22 s) CN (10-18-10-16 s).[Consider a 20 MeV deuteron on A=50 target nucleus][Consider a 20 MeV deuteron on A=50 target nucleus].
Angular distribution.
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Direct Reactions• (d,n) stripping (transfer) reactions can go through both processes.• (d,p) stripping (transfer) reactions prefer D rather than CN; protons do not easily evaporate (Coulomb) [(p d) is a pickup reaction]protons do not easily evaporate (Coulomb). [(p,d) is a pickup reaction].• What about (α,n) transfer reactions?HW HW 3636 Show that for a (d,p) reaction taking place on the surface of a 90Zr nucleus, and with 5 MeV deuterons, the angular momentum transfer can be approximated by l = 8sin(θ/2), where θi th l th t i t k ith th i id t d tis the angle the outgoing proton makes with the incident deuteron direction. (Derive a general formula first).
l 0 1 2 3Jπ(90Zr ) = 0+ Fig 11 23 in Krane l 0 1 2 3θ 0º 14.4º 29º 44º
Jπ(90Zrgs) = 0J(91Zr) = l ± ½, π = (-1)l
Optical model, DWBA, Shell model, Spectroscopic Factor.
Fig. 11.23 in Krane.
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Optical model, DWBA, Shell model, Spectroscopic Factor.
calcmeas ddS
dd
⎟⎠⎞
⎜⎝⎛Ω
=⎟⎠⎞
⎜⎝⎛Ω
σσ
Neutron-induced Reactions22 nXHCCHbY ++∝ DσX(n,b)Y nXHCCHbY IIIn ++∝ Dσ( , )
Γn(En)Γb(Q+En)2
11∝∝ Γn(En)b(Q n)2vE
)( nln EPvn
∝Probability to penetrate the potential barrier
For thermal neutronsQ >> E
Γb(Q) ≈ constant
potential barrier
Po(Ethermal) = 1P (E ) = 0Q >> En P>o(Ethermal) = 0
E 1)( ∝σNon-resonantNuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).172v
Enn )( ∝σNon-resonant
Neutron-induced Reactions
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Neutron-induced Reactions
nn TOFTOFnn--TOFTOFCERNCERNCERNCERN
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Neutron-induced Reactions
n TOFn TOFn_TOFn_TOFCERNCERNCERNCERN
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Neutron-induced Reactions
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Charged Particle Reactionsg
Nuclear
What is the Gamow Peak?
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NuclearRadius
Charged Particle Reactionsg
Electron ScreeningElectron Screening
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Charged Particle Reactionsge2 = 1.44x10-12 keV.m
2 eZZ 221
HW HW 3737Tunneling probability: πη2−≅ eP vh
21=η
S f ld t
)(uµ
Sommerfeld parameterGamow factor
In numerical units: )(
)(29.312 21 keVEuZZ CM
µπη =
For γ-ray emission: 12)( +=Γ LLL EE γγ α)( LL γγ
31)( γγ α EEDipole =Γ
Multipolarity
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Charged Particle Reactionsg
πη2)(E πησ 2)( −∝ eE
EE 1)( 2 ∝∝ Dπσ
E1 2 )(1)( 2 ESeE
E πησ −=
Nuclear (or astrophysical) S-factor
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Charged Particle Reactionsg
EC = ??EC ??
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Resonance Reactions
∆E ∆tCN particle emission ∆E ∆E > spacing between p p gvirtual states continuum. (Lower part larger spacing isolated resonances).
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D bound states γ-emission ∆E isolated states.
Resonance ReactionsHW HW 3838
In the 19F(p,αγ) reaction:Th Q l i 8 ??? M V• The Q-value is 8.??? MeV.
• The Q-value for the formation of the C.N. is 12.??? MeV.• For a proton resonance at 668 keV in the lab system theFor a proton resonance at 668 keV in the lab system, the corresponding energy level in the C.N. is at 13.??? MeV.• If for this resonance the observed gamma energy is 6.13 MeV, g gywhat is the corresponding alpha particle energy?• If for this resonance there has been no gamma emission b d h t ld th b th l h ti l ?observed, what would then be the alpha particle energy?
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Resonance Reactions
ExcitedExJπ a + X Y + b Q > 0
Entrance Ch l
ExcitedState
a + X Y + b Q > 0b + Y X + a Q < 0
Channela + X
ExitChannelb + YCompound
Inverse ReactionNucleus C*
22 )1()12)(12(
12 XaHCCHbYJJ
JIIIaXaXaX +++
+++
= δπσ D)12)(12( JJ Xa ++
QM StatisticalF ( )
Identicalparticles
• Nature of force(s).• Time reversal invarianceFactor (ω) particles • Time-reversal invariance.
22 )1()12)(12(
12 YbHCCHXaJJ
JIIIbYbYbY +++
+= δπσ D
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)()12)(12( JJ IIIbY
YbbYbY ++
??=bY
aX
σσHW HW 3939
Resonance ReactionsProjectile Projectile
Target
Projectile
TargetgQ-value Q-value
g
Q + ER = Er
N t ResonantEγ = E + Q - Eex
Q R r
Non-resonant Capture
(all energies)
Resonant Capture
(selected energies(all energies) (selected energies with large X-section)
2XaHY +∝σ 22
XaHEEHE +∝σNuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).187
XaHY +∝ γγσ XaHEEHE CNrrf +∝ γγσ
baΓΓ∝γσ
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Resonance Reactionsh
h bbplL ===D
h bbplL
Dlb = Dlb =222
1max )12( Dπππσ +=−= + lbb lll 1max, )(+ lll
7.656)(2 bπ =DHWHW 4040)()(
)(keVEu
b CMµπDHW HW 4040
122 J + )1()12)(12(
122max aX
XaaX JJ
J δπσ +++
+= D
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ω
Resonance ReactionsDamped OscillatorDamped Oscillator Oscillator strength
22 )()( δωω +∝
fresponseg
2 )()( δωω +− o
Damping1 Dampingfactor0
1t
=δ
eigenfrequency
ΓΓ2
22 )()(
)(Γ+−
ΓΓ∝
R
ba
EEEσ
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190h=Γ ot
Resonance Reactions
12 ΓΓJ2
22
2
)()()1(
)12)(12(12)(
Γ+−ΓΓ
+++
+=
R
baaX
XaaX EEJJ
JE δπσ D2 )()())(( RXa
BreitBreit--Wigner formulaWigner formulaba Γ+Γ=Γ
• All quantities in CM system• Only for isolated resonances.
ba
baR
ΓΓ∝ΓΓ∝
σσ Reaction
Elastic scatteringUsually Γa >> Γb.
bR
aae
ΓΓ
=
ΓΓ∝σσ Elastic scattering
HW HW 4141 When does σR take its maximum value?
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ae Γσ
Resonance Reactions
J + JX + l = J ExitCh lEJπJa + JX + l J
(-1)l π(Ja) π(JX) = π(J)Channelb + Y
ExcitedState
ExJ
Entrance
(-1)l = π(J) Natural parity.Compound
Entrance Channela + X
Compound Nucleus C*
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Resonance ReactionsWhat is the “Resonance Strength” …? HWHW 4242What is its significance?In what units is it measured?
ΓΓJ 12
HW HW 4242
ΓΓΓ
+++
+= ba
aXXa JJ
J )1()12)(12(
12 δωγ
tion
Charged particle
oss s
ect Charged particle
radiative capture (a,γ)(What about neutrons?)
Cro
ECωγ ∝ Γa ωγ ∝ Γγ
(What about neutrons?)
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Energy
Resonance Reactions14N(p,γ) HWHW 4343(p,γ)• Q = ??• E = ??
HW HW 4343
• EC = ??• ER = 2.0 MeV
Formation via s-wave protons, Take J = ½, Γp = 0.1 MeV, p , , p ,dipole radiation Eγ = 9.3 MeV, Γγ = 1 eV.Show that ωγ = 0.33 eV.
If b t t 10 k V• If same resonance but at ER = 10 keVΓp = ?? Eγ = ?? Γγ = ??Show that ωγ = 3 3x10-23 eVShow that ωγ = 3.3x10 eV.
Huge challenge to experimentalists
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g g p
α-transfer reactions
Resonance Jπ EstimatedEnergy (keV) ωγ (µeV)
α transfer reactionsAngular distribution
Energy (keV) ωγ (µeV)566 2+ 1.9
3- 0.154+ 0 01
18O(α, γ)22Ne4+ 0.01
470 0+ 0.6 1- 0.2
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Experimental upper limit < 1.7 µeV
Neutron Resonance Reactions
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Neutron Activation Analysisy(Z,A) + n (Z, A+1)( , ) ( , )
β-
( )
γ (β-delayed γ-ray)
(Z+1, A+1)
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Neutron Attenuation
Neutrons
TargetThickness “x”
ndxIdI
tσ−=nx
oteII σ−=
ISimilar to γ-attenuation. Why?
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Neutron ModerationShow that, after elastic scattering the HW HW 4444, gratio between the final neutron energy E\
and its initial energy E is given by:gy g y
2
2\
)1(cos21
+++
=A
AAEE CMθ
For a head-on collision:
)1( +AE2\ 1⎠⎞
⎜⎝⎛ −
=⎞
⎜⎜⎛ AE
After n s-wave collisions:min 1⎠
⎜⎝ +⎠
⎜⎝ AE
ζnEEn −= lnln \
where
ζn
11ln
2)1(1ln
2
\
−−+=⎥⎦
⎤⎢⎣⎡=
AA
AA
EEζ
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12\ +⎥⎦⎢⎣ AAE av
ζ
Neutron ModerationHW HW 44 44 (continued)(continued)
How many collisions are needed to thermalize a 2 MeV neutron if the moderator was:neutron if the moderator was:
1H 2H 4He 12C 238U
Discuss the effect of the thermal motion of the moderator atomsatoms.
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Nuclear Fission
Coulomb effectSurface effect
~200 MeV
Coulomb effectSurface effect
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Nuclear Fission• B.E. per nucleon for 238U (BEU) and 119Pd (BEPd) ?
2 119 BE 238 BE ?? K E f th• 2x119xBEPd – 238xBEU = ?? K.E. of the fragments ≈ 1011 J/g
B i l 105 /• Burning coal 105 J/g• Why not spontaneous?
T 119 d f t j t t hi Th C l b• Two 119Pd fragments just touching The Coulomb barrier is:
)46( 2
MeVMeVfm
fmMeVV 2142502.12
)46(.44.12
>≈=
• Crude …! What if 79Zn and 159Sm? Large neutron l d t h t ti l d !
f
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excess, released neutrons, sharp potential edge…!
Nuclear Fission
• 238U (t½ = 4.5x109 y) for α-decay.• 238U (t½ ≈ 1016 y) for fissionU (t½ ≈ 10 y) for fission.• Heavier nuclei??• Energy absorption from a neutron (for example) couldEnergy absorption from a neutron (for example) could form an intermediate state probably above barrier induced fissioninduced fission.• Height of barrier is called activation energy.
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Nuclear Fission
Liquid Drop
MeV
)
Shell
Ene
rgy
(va
tion
EA
cti
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Nuclear Fission34 Rπ ε+= )1(Ra23 abR =3
24 abπ=
=Rb
abR =
Surface Term B = a A⅔
3abπ ε+1
)1( 22 ++ ε
Volume Term (the same)
Surface Term Bs = - as A⅔
Coulomb Term BC = - aC Z(Z-1) / A⅓
...)1( 52 ++ ε
...)1( 251 +− ε
32
31
52
51 )1( AaAZZa SC >− − fission
47~2
>Z Crude: QM and original shape
could be different from sphericalNuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).205
A could be different from spherical.
Nuclear Fission
)120( 2
48300
)120(=
300Consistent with activation energy curve for A = 300curve for A = 300.
Extrapolation to 47 10-20 sNuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).206
Extrapolation to 47 ≈ 10 20 s.
Nuclear Fission
235U +235U + n
93Rb + 141Cs + 2nRb + Cs + 2nNot unique.
Low-energy fissionfission processes.
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Nuclear FissionZ1 + Z2 = 921 2Z1 ≈ 37, Z2 ≈ 55A1 ≈ 95, A2 ≈ 1401 , 2Large neutron excessMost stable:Z=45 Z=58
16Prompt neutronsPrompt neutrons within 10-16 s.Number ν depends on nature of fragments and on incident particle energy.The average number is characteristic of
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the process.
Nuclear Fission
The average number of neutrons isneutrons is different, but the distribution is G iGaussian.
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Higher than S ?
Delayed neutronsDelayed neutrons
Higher than Sn?
Delayed neutronsDelayed neutrons~ 1 delayed neutron
100 fi i b tper 100 fissions, but essential for control of the reactorof the reactor.
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Nuclear Fission
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Nuclear Fission
1/v Fast neutrons should be
d t dmoderated.
235U thermal cross sectionsσfission ≈ 584 b.σscattering ≈ 9 b.σ ≈ 97 bσradiative capture ≈ 97 b.
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212Fission Barriers
Nuclear Fission
• Q for 235U + n 236U is 6.54478 MeV.• Table 13 1 in Krane: Activation energy E for 236U ≈ 6 2 MeV• Table 13.1 in Krane: Activation energy EA for 236U ≈ 6.2 MeV (Liquid drop + shell) 235U can be fissioned with zero-energy neutrons.
• Q for 238U + n 239U is 4.??? MeV.• EA for 239U ≈ 6.6 MeV MeV neutrons are needed.• Pairing term: δ = ??? (Fig 13 11 in Krane)• Pairing term: δ = ??? (Fig. 13.11 in Krane).• What about 232Pa and 231Pa? (odd Z).• Odd-N nuclei have in general much larger thermal neutron
213
g gcross sections than even-N nuclei (Table 13.1 in Krane).
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
Nuclear FissionWhy not use it?Why not use it?
σf,Th 584 2.7x10-6 700 0.019 b
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Nuclear Fission• 235U + n 93Rb + 141Cs + 2n• Q = ????• What if other fragments?
Diff t b f t• Different number of neutrons.• Take 200 MeV as an average.
66 MeV 98 MeV
Heavy LightHeavyfragments
g tfragments
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miscalibrated
Nuclear Fission• Mean neutron energy ≈ 2 MeVMeV.• ≈ 2.4 neutrons per fission (average) ≈ 5 MeV(average) 5 MeV average kinetic energy carried by prompt neutrons per fission.
HW HW 4545• Show that the average momentum carried by a neutron is only ≈1.5 % that carried by a fragment. • Thus neglecting neutron momenta, show that the ratio between kinetic energies of the two fragments is the inverse of the ratio of their masses ENuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).216
their masses.
1
2
2
1
mm
EE
≈14095
9866
≈
Nuclear Fission
Distribution of fission energyEnge Distribution of fission energyg
Krane sums
them up as β Lost … !
decays.
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Nuclear FissionSegrè Distribution of fission energy
aLost … ! b
c
•• How much is recoverable?How much is recoverable?How much is recoverable?How much is recoverable?•• What about capture gammas? What about capture gammas? (produced by (produced by νν--1 1 neutronsneutrons))•• Why c < (Why c < (a+ba+b) ?) ?
218
Why c < (Why c < (a+ba+b) ?) ?Nuclear Reactors, BAU, 1st Semester, 2007-2008
(Saed Dababneh).Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).
Nuclear Fission• Recoverable energy release ≈ 200 MeV per 235U fission.
Fi i t 2 7 1021 fi i d i MW• Fission rate = 2.7x1021 P fissions per day. P in MW.
• Burnup rateBurnup rate: 1.05 P g/day. P in MW.)(Eσ HWHW 4646• Capture-to-fission ratio:
C iC i 1 05(1 ) P /d)()(
)(EE
Efσ
σα γ= HW HW 4646
• Consumption rateConsumption rate: 1.05(1+α) P g/day.• 1000 MW reactor.• 3 1x1019 fissions per second or 0 012 gram of 235U per second• 3.1x10 fissions per second, or 0.012 gram of U per second.• Two neutrinos are expected immediately from the decay of the two fission products, what is the minimum flux of neutrinos
219Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
pexpected at 1 km from the reactor.
4.8x1012 m-2s-1
Nuclear Fission
• 3 1x1010 fissions per second per W• 3.1x1010 fissions per second per W.• In thermal reactor, majority of fissions occur in thermal energy region φ and Σ are maximumthermal energy region, φ and Σ are maximum.• Total fission rate in a thermal reactor of volume V
• Thermal reactor powerThermal reactor power (quick calculation)(quick calculation)
φfVΣ• Thermal reactor powerThermal reactor power (quick calculation)(quick calculation)
V fφΣ101013 x
VP f
th
φΣ=
220Nuclear Reactors, BAU, 1st Semester, 2007-2008 (Saed Dababneh).
101.3 x220Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).
Controlled Fission• 235U + n X + Y + (~2.4)n Fast second generation neutrons
• Moderation of second generation neutrons Chain reaction.• Net change in number of neutrons from one generation to th t k ( t d ti f t )the next ≡ k∞ (neutron reproduction factor).
• k ≥ 1 Chain reactionInfinite medium (ignoring leakage at the surface).
k∞ ≥ 1 Chain reaction.• Water, D2O or graphite moderator.• k < 1 subcritical system. Chain reacting pileChain reacting piley• k = 1 critical system.• k > 1 supercritical system.
F t d l f ( t d
Chain reacting pileChain reacting pile
• For steady release of energy (steady-state operation) we need k =1.
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Controlled FissionProbability for a thermal neutron to
fi i 235 i235U thermal cross sections
cause fission on 235U isU e c oss sec o s
σfission ≈ 584 b.σscattering ≈ 9 b. σ
=≈1f
σradiative capture ≈ 97 b. ασσ γ ++ 1f
If each fission produces an average of ν neutrons, then the mean number of fission neutrons produced per thermal neutron = ηnumber of fission neutrons produced per thermal neutron = η
νσσ ff
αν
σσσ
νσσ
νηγ +=
+==
1f
f
a
f η <ν
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Controlled Fission235U• Assume Assume natural natural uranium:uranium:
99 2745% 238U 0 7200% 235U99.2745% 238U, 0.7200% 235U.
Thermal σf = 0 b 584 bTh l 2 75 b 97 b
24 RπThermal σγ = 2.75 b 97 b
NN yyxxyx σσ +=Σ+Σ=Σ
4 Rπ
238UΣ / N (0 992745)(0)
Nyyxx )( σγσγ +=
• Σf / N = (0.992745)(0) + (0.0072)(584)
= 4 20 b24 Rπ
Doppler effect?Doppler effect?
4.20 b.• Σγ / N = (0.992745)(2.75) +
(0.0072)(97)
223
= 3.43 b.Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
Using the experimental elastic scattering data the radius of the nucleus can be estimated.
Moderation (to compare x-section)
1H2H 1H(n,n)(n,n)
2H
(n,γ) (n,γ)(n,γ) (n,γ)
• Resonances?224Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).
Controlled Fissionσ
≈ f• Probability for a thermal neutron to cause fissionγσσ +f
• For natural uraniumnatural uranium 55.0433204
20.4=
+=
• If each fission produces an average of ν = 2.4 neutrons, then the mean number of fission neutrons produced per thermal neutron =
43.320.4 +
mean number of fission neutrons produced per thermal neutron η = 2.4 x 0.55 ≈ 1.3
γσσσ
νη+
=f
f
• This is close to 1. If neutrons are still to be lost, there is a danger of losing criticality.
For enriched uraniumenriched uranium (235U = 3%) = ????? (> 1 3)• For enriched uraniumenriched uranium (235U = 3%) η = ????? (> 1.3).• In this case η is further from 1 and allowing for more neutrons to be lost while maintaining criticality.
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be lost while maintaining criticality.
Controlled Fission• N thermalthermal neutrons in one generation have produced so far have produced so far
NN f t tf t tηηNN fast neutrons.fast neutrons.• Some of these fast neutrons can cause 238U fission more fast neutrons fast fission factor = ε (= 1 03 for natural uranium)neutrons fast fission factor = ε (= 1.03 for natural uranium).• Now we have Now we have εηεηNN fast neutrons.fast neutrons.• We need to moderate these fast neutrons use graphite for 2 g pMeV neutrons we need ??? collisions. How many for 1 MeV neutrons?
Th t ill th h th 10 100 V i d i th• The neutron will pass through the 10 - 100 eV region during the moderation process. This energy region has many strong 238Ucapture resonances (up to 1000 b) Can not mix uranium andcapture resonances (up to 1000 b) Can not mix uranium and graphite as powders.• In graphite, an average distance of 19 cm is needed for
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g p gthermalization the resonance escape probability p (≈ 0.9).
Controlled Fission• Now we have Now we have ppεηεηN N thermal neutrons.thermal neutrons.• Graphite must not be too large to capture thermal neutrons;Graphite must not be too large to capture thermal neutrons; when thermalized, neutrons should have reached the fuel.• Graphite thermal cross section = 0.0034 b, but there is a lot of it present.• Capture can also occur in the material encapsulating the fuel l telements.
• The thermal utilization factor f (≈ 0.9) gives the fraction of thermal neutrons that are actually available for the fuelthermal neutrons that are actually available for the fuel.• Now we have Now we have ffppεηεηNN thermal neutronsthermal neutrons, could be > or < Nthus determining the criticality of the reactor.
k fk∞ = fpεη The fourThe four--factor formula.factor formula.
k = fpεη(1-lf t)(1-lth l)Nuclear and Radiation Physics, BAU, First Semester, 2007-2008
(Saed Dababneh).227
k fpεη(1 lfast)(1 lthermal)Fractions lost at surface
Neutron reproduction
f
x x 00..99Th lTh l
factork = 1.000
Thermal Thermal utilization utilization factor “f”factor “f”
x η
x x 00..99Resonance Resonance
escape escape b bilitb bilit
x x 11..0303Fast fissionFast fission
probability probability ”p””p”What is:
• Migration length? Fast fission Fast fission factor “factor “εε””
g g• Critical size?How does the
t ff t thgeometry affect the reproduction factor?
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Controlled FissionTime scale for neutron multiplicationTime scale for neutron multiplication• Time τ includes moderation time (~10-6 s) and diffusion time of thermal neutrons (~10-3 s).
Time Average number of thermal neutronsTime Average number of thermal neutronst N
t + τ kNt τ kNt + 2τ k2N
NkNdN• For a short time dtτ
NkNdtdN −
=
•• Show thatShow that τtkeNtN )1(0)( −=
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0)(
Controlled FissionτtkeNtN )1()( −=
• k = 1 N is constant (Desired).• k < 1 N decays exponentially
eNtN 0)( =• k < 1 N decays exponentially.• k > 1 N grows exponentially with time constant τ / (k-1).• k = 1.01 (slightly supercritical) e(0.01/0.001)t = e10 = 22026 in in 11s. s. ( g y p )• Cd is highly absorptive of thermal neutrons.• Design the reactor to be slightly subcritical for prompt neutrons.• The “few” “delayed” neutrons will be used to achieve criticalitywill be used to achieve criticality, allowing enough time tomanipulate the control
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prods. Cd control rodsCd control rods
Fission ReactorsEssential elements:Essential elements:• Fuel (fissile material).• Moderator (not in reactors using fast neutrons). Core
• Reflector (to reduce leakage and critical size).• Containment vessel (to prevent leakage of waste).• Shielding (for neutrons and γ’s).• Coolant.• Control system.• Emergency systems (to prevent runaway during failure).
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Fission ReactorsTypes of reactors:Types of reactors:Used for what?Used for what?• Power reactors: extract kinetic energy of fragments as heat boil water steam drives turbine electricity.• Research reactors: low power (1-10 MW) to generate neutrons (~1013 n.cm-2.s-1 or higher) for research.• Converters: Convert non-thermally-fissionable material to a thermally-fissionable material.
_239min23239238 νβ ++⎯⎯ →⎯→+ −NpUnU
_2393.2 νβ
β
++⎯⎯→⎯ −Pu
p
dFertile
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βFertileσf,th = 742 b
Fission Reactors_
233min22233232 β +++ −PThTh_
23327
233min22233232
νβ
νβ
++→
++⎯⎯ →⎯→+
−U
PaThnTh
d νβ ++⎯⎯→⎯ UFertile
• If η = 2 Conversion and fissionσf,th = 530 b
• If η = 2 Conversion and fission.• If η > 2 Breeder reactor.• 239Pu: Thermal neutrons (η = 2.1) hard for breeding.(η ) g
Fast neutrons (η = 3) possible breeding fast breeder reactors.
After sufficient time of breeding, fissile material can be easily (chemically) separated from fertile material.C t ti 235U f 238U
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Compare to separating 235U from 238U.
Fission ReactorsWhat neutron energy?What neutron energy?• Thermal, intermediate (eV – keV), fast reactors.• Large, smaller, smaller but more fuel.What fuel?What fuel?• Natural uranium, enriched uranium, 233U, 239Pu., , ,
From converter or breeder reactorHow??? breeder reactor.How???
HW HW 4747
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Fission ReactorsWhat moderator?What moderator?1 Ch d b d t1. Cheap and abundant.2. Chemically stable.3 V l ( 1)3. Very low mass (~1).4. High density.5 Mi i l t t ti5. Minimal neutron capture cross section.• Graphite (1,2,4,5) increase amount to compensate 3.
W t (1 2 3 4) b t d i h d i• Water (1,2,3,4) but n + p → d + γ enriched uranium.• D2O (heavy water) has low capture cross section
t l i b t if t dnatural uranium, but if capture occurs, produces tritium.B d B O b t i
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235
• Be and BeO, but poisonous.
Fission ReactorsWhat assembly?What assembly?
H t d t d f l l d• Heterogeneous: moderator and fuel are lumped. • Homogeneous: moderator and fuel are mixed together.
I h t it i i t l l t d• In homogeneous systems, it is easier to calculate p andf for example, but a homogeneous natural uranium-
hit i t t iti lgraphite mixture can not go critical.What coolant?What coolant?• Coolant prevents meltdown of the core.• It transfers heat in power reactors.• Why pressurized-water reactors.• Why liquid sodium?
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Boiling water reactorBoiling water reactor
Pressurized Pressurized water water ateate
reactorreactor
•• Light water reactors.Light water reactors.•• Both use “light” water asBoth use “light” water as•• Both use light water as Both use light water as coolant and as moderator, coolant and as moderator, thus enriched (thus enriched (22--33%)%)(( ))uranium is used.uranium is used.•• Common in the US.Common in the US.
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CANDU CANDU reactorreactor
•• Canada has DCanada has D22O O and natural uranium.and natural uranium.
GasGasGas Gas cooled cooled reactorreactor
•• Most Most power power
ii reactorreactorreactors in reactors in GB are GB are graphitegraphitegraphite graphite moderated moderated gasgas--
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238
ggcooled.cooled.
•• Liquid sodium cooled, fast breeder reactor.Liquid sodium cooled, fast breeder reactor.
Breeder Breeder reactorreactor
qq•• Blanket contains the fertile Blanket contains the fertile 238238UU..•• Water should not mix with sodium.Water should not mix with sodium.
reactorreactor
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More on Fission Products• β and γemissions fromemissions from radioactive fission products carry partproducts carry part of the fission energy, even after shut down. • On approaching end of the chain, the decay energy decreases and half-life increases Long-lived isotopes constitute the mainand half-life increases. Long-lived isotopes constitute the main hazard.• Can interfere with fission process in the fuel. (poisoning).(poisoning).p (p g)(p g)• Important for research.• β-decay favors high energy 17 MeV compared to 6 MeV for γ.
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• Only 7 MeV from β-decay appears as heat. Why?Why?
Poisoning
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Poisoning
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Poisoning• Not anticipated! Reactor shut down! Time scale:Time scale:
Hours and days.Hours and days.135Xe 149SmXe106 b
Sm105 b
φσmXe
a
XeγIγ
I
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φσ Ia φσ Xe
a
PoisoningHW HW 4848 k
kk reactor). (Infinite use uslet ,1Reactivity −
=≡ ∞ρ
eratorcladfuel
fuelaf
k
mod1 (critical) ∑+∑+∑
∑=
itl df l
fuela
aafa
f d2∑
=
∑+∑+∑
poisona
poisona
eratora
clada
fuela
f mod2
thatShow ∑==∆
∑+∑+∑+∑
ρρρ eratora
clada
fuela
mod12 that Show∑+∑+∑
−=−=∆ ρρρ
Negative reactivity due to poison buildup. It is proportional to the amount of poison.
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p opo t o a to t e a ou t o po so
Poisoning),(),(),(),(),( trtrItrItr
ttrI I
aIfIrrrr
r
φσλφγ −−∑=∂
∂ small
),(),(),(),(),(),( trtrXetrXetrItrt
trXet
XeaXeIfXe
rrrrrr
φσλλφγ −−+∑=∂
∂∂
t∂Initial conditions?Initial conditions?• Clean Core StartupClean Core Startup Assume no spacialClean Core Startup.Clean Core Startup.• Shutdown (later).
Assume no spacial dependence.
)()(ldFuel.Fresh 0)0()0( ==φφ
XeIconstant.)0()( assumeuslet and ==φφ t
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Poisoning
)1()( 0 tfI IetI λ
λφγ −−
∑=The solution is
Iλ
)(∞I )(∞I)(∞Xe
)1()(
)( )(0 0 tXe
fXeI XeaXeetXe φσλ
φλφγγ +−−
∑+=and )()(
)(0
0
ttfI
XeaXe
Xe λφλφγφσλ
+∑+
)( )(
0
0 0 ttXeaIXe
fI IXeaXe ee λφσλ
φσλλφγ −+− −
+−
∑+
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Poisoning
)(∞I )(I)(∞Xe
• Now, we know Xe(t)
eratorcladfuel
Xea
eratorcladfuel
poisona tXet
modmod
)()(∑+∑+∑
−=∑+∑+∑
∑−=∆
σρ
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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aaaaaa ∑+∑+∑∑+∑+∑
Poisoning•• Shutdown. Shutdown. After the reactor has been operating for a “long” time.
)()()()(),( ttItIttrI I rrrrr
φλφ∑∂
)()()()()(),(
),(),(),(),(),(
ttXtXtIttrXe
trtrItrItrt
Xe
IaIfI
rrrrrr
φλλφ
φσλφγ
+∑∂
−−∑=∂
),(),(),(),(),(),( trtrXetrXetrItrt
XeaXeIfXe φσλλφγ −−+∑=
∂
)()0( ∞= II ),(),( trItrII
rr
λ−=∂
0)0()()()0(
==∞=
φφ tXeXe
)()(),(
),(
tXtItrXet I
rrr
λλ∂∂
.0)0()( ==φφ t ),(),(),( trXetrIt XeI λλ −=∂
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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PoisoningThe solution is
)()()()(
)()(
ttIt
tI
IXX
eItI
λλλ
λ
λ
−
∞∞=
)()()()( tt
XeI
It IXeXe eeIeXetXe λλλ
λλλ −−− −−
+∞=
f> 0 ?
Height of the peak depends on I(∞)
dand Xe(∞), i.e. depends on φ.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
249
PoisoningShutdown Xe negative ∆ρ move control rods out try to add positive reactivity need to have
If, the available excess
y p yenough reserve costly to do that.
reactivity can compensate for less than 30 minutes ofthan 30 minutes of poison buildup, can’t startup again after ~30startup again after 30 minutes of shutdown, because you can’t achieve criticality. Need to wait some 40 hours (in this case) for Xe to
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
250
(in this case) for Xe to decay down.
Poisoning
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
251
Poisoning
StrategiesStrategies• If you plan to shut down for “short maintenance” think• If you plan to shut down for short maintenance , think about stepback.• Examine different scenarios using this code from• Examine different scenarios using this code from
http://www.nuceng.ca/• You will get more experience after you finish the• You will get more experience after you finish the computational physics course.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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PoisoningXeXe OscillationsOscillations• φ(r,t) (spacial dependence) flux locally Xeburnup ρ (reactivity) flux further control rods globally in flux elsewhere Xe burnup ρ ….. Xe oscillation but limited by opposite effect due to increase (decrease) of I in the high (low) flux region.• In large reactors (compared to neutron diffusion length) local flux, power and temperature could reach unacceptable values for certain materials safety issues. • Think of one sensor and one control rod feel average flux apparently OK more sensors and
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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control rods to locate and deal with local changes.
PoisoningPermanent PoisonsPermanent Poisons• 149Sm has sizeable but lower cross section than 135Xe.• It does not decay.
C t t)0()0()()(),( tttrSm φφ ∑∑∂ rrrr
r
• Start from fresh fuel:
Constant.)0,()0,(),(),(),(SmfSmfSm rrtrtr
tγφγφγ =∑=∑≈
∂
h time.Linear wit .Constant . )0,()0,(),( ttrrtrSm SmfSm γφγ =∑=rrr
• Thus:h time.Linear wit )0,()0,(),( trrtr fSm
Sma
Sma
rrr φγσ ∑=∑ t e.ea w t)0,()0,(),( tt fSmaa φγσ ∑∑
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
254
Radioactive Fission Products
Read sectionRead section 1313 77 in Kranein KraneRead section Read section 1313..7 7 in Krane.in Krane.
Look at sections Look at sections 1313..8 8 and and 1313..9 9
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Radioactive Fission Products[ ] sMeVTttxtP /)(101.4)( 2.02.011 −− +−= [ ])()(
per watt of original operating power.T = time of operation.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Radioactive Fission Products
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Fusion Reactions• Neutron-induced fission No Coulomb barrier.• Charged particle-induced fusion Coulomb barrier.
kTmv 321 =• Thermonuclear reactions
kTmv 22 =
• At room temperature kT = 0.025 eV.• Practically, keV available energy but much higher y, gy gCoulomb barrier.• What is the temperature required to classically p q yovercome the barrier for a D-D reaction.• Penetration probability much lower temperatures.
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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p y p
Fusion Reactions
• We formulated the cross section when we considered charged particle reactions.
1co s de ed c a ged pa t c e eact o s
)(1)( 2 ESeE πησ −= )()( ESeE
Eσ =
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Fusion Reactions•• Show thatShow that
bb mmQvm
+=2
21
1 Yb
Qmm+
21
1
bYYY mm
Qvm+
=221
1
Ybb mvm=21
221
bYY mvm 221
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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Fusion Reactions
Nuclear and Radiation Physics, BAU, First Semester, 2007-2008 (Saed Dababneh).
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