nuclear physics intro

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).

1

• 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.


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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.


3

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:

[email protected]


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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.


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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


6

Physics (to answer Q1).

Constantshttp://physics.nist.gov/cuu/Constants/index.html


7

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


9

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


10

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

(Saed Dababneh).12

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.


13

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


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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


17

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?


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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.


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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


20

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


24

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


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Nuclear Binding Energyg gy

Coulomb effectSurface effect

~200 MeV


HWcHWc 44Thi k f t tThink of a computer program to

reproduce this graph.


27


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.


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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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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


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


36MASS NUMBER


37

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.…..


38

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


39

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


40

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


41

Remember HW 8 … ?!



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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.


44

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

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Effect on:• Fission

⎪⎪⎩

⎪⎪⎨−

=− oddZoddNAa

oddA

p4

30δ

Fission.• Magnetic moment.Effect of high angular momentum


45

⎪⎩ 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.


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• 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.


47


)(),(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)…..


48

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


49

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


50

γ2min∂Z A



51


Double β decay! Both Sides!


52



53

Mass Parabolas and StabilityyVertical spacing b b hbetween both parabolas ?

• Determine constants from atomic masses.

Odd-Odd

Even-Even


54



55

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


56

• 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-


58

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

(Saed Dababneh).60

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


61

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)


62

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.


63

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.


64

…………

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.


65

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.


66

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…..!


68

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.


69

• 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


70

Deuteron: E1 = … MeV r1 = …x10-15 m

Nuclear Force


71

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?


72

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.


73

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.


74

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

(Saed Dababneh).75

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.


76

Nuclear ForceMirror Nuclei


77

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

(Saed Dababneh).78

_ν++→ epnBoson?

Nuclear Force


79

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.


80

• 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..


81

The Deuteron


82

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


83

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


84

bit weaker…?

The Deuteron


85

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


86

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?


87

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.


88

Nuclear Force


89

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 …..


90

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).


91

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.


92

Shell model


93

Shell model


94

Shell model3) Natural abundances.


95

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


96

Shell model5) Binding energy of the last neutron

(S ti E )(Separation Energy).(The measured values are plotted relative to the calculations without η).


97

Shell model6) Excited states.

Pb (even-A) isotopes.


98

Shell model


99

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.


100

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=


101

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


102

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.


103

Shell model


104

Shell model

ZN −HW HW 2323

)(27 MeVA

ZNVas ±=∆

Coulomb repulsion? V (r) = ??Nuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).105

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.


106

Shell modelcentrifugal potential


107

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.


108

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


109

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


110

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


111

21,)1( + slsjmsslsjmS jj h

Shell modelmodel

2j+1

2(2x3 + 1) = 142(2x3 + 1) 14

1f7/2l = 3

7/2

First time

jtime


112

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


113

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.


114

Shell modelFermi Gas

EF ∝ n2/3


115

Shell modelNuclear

ti ?reactions?

Transition probability?


116

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.


117

A < 150190 < A < 220

Shell model

Near No spin-

orbit

Harmonic oscillatorNear drip

line


118

valley of βstability

orbit coupling


119

Shell model


120

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


121

Shell modelExcited states:


122

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


123

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

(Saed Dababneh).124

“valence” nucleons.

Shell model

HWHW 2626 Discuss the energy levels of nuclei with HW HW 2626odd number of nucleons in the 1f7/2 shell.


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 ?


126

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

KNuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).127

Krane


128

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

209Bi 0 37Nuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).129

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

(Saed Dababneh).130

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.


131

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


132

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


133

)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

d t ti f l fNuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).134

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.


135

164Er

Collective modelOdd-A

nucleonEE += 2gω212


136

Collective modelAverageshapeshape

Instantaneousshapeshape


137

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

.


138

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.


139

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.

O t l hNuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).140

• 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


141

Collective modelTwo-phonon triplet at twice the HW HW 3030energy of the single phonon state. Krane 5.10


142

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

Y + bNuclear and Radiation Physics, BAU, First Semester, 2007-2008

(Saed Dababneh).143

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.


144

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?).


145

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.


146

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


147

Nuclear Reactions

Sample


148

Nuclear Reactions


149

Nuclear Reactions


150

Nuclear Reactions


151

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


152

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.


153

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

?


154

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 ….


155

• 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


156

• 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


157

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


158Energy state in “Y”

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


159

0=aT

No dependence on φ

Coulomb Scatteringg


160

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


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


162

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.


163

p g p p g

Coulomb Scatteringg


164

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,


165

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


166

heavy target.

Compound Nucleus Reactionsp


167

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.


168

Direct Reactions

• Random collisions nearly isotropic angular distribution.• Direct reaction component strong angular dependence.

See also Fig. 11.20 in Krane.


169

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.


170

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.


171

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


173


174


nn TOFTOFnn--TOFTOFCERNCERNCERNCERN


175


176


n TOFn TOFn_TOFn_TOFCERNCERNCERNCERN


177



178

Charged Particle Reactionsg

Nuclear

What is the Gamow Peak?


179

NuclearRadius


Electron ScreeningElectron Screening


180

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


181


πη2)(E πησ 2)( −∝ eE

EE 1)( 2 ∝∝ Dπσ

E1 2 )(1)( 2 ESeE

E πησ −=

Nuclear (or astrophysical) S-factor


182


EC = ??EC ??


183

Resonance Reactions

∆E ∆tCN particle emission ∆E ∆E > spacing between p p gvirtual states continuum. (Lower part larger spacing isolated resonances).


184

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?


185

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


186

)()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


188

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


189

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


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?


191

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*


192

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?)


193

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


194

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


195

Experimental upper limit < 1.7 µeV

Neutron Resonance Reactions


196

Neutron Activation Analysisy(Z,A) + n (Z, A+1)( , ) ( , )

β-

( )

γ (β-delayed γ-ray)

(Z+1, A+1)


197

Neutron Attenuation

Neutrons

TargetThickness “x”

ndxIdI

tσ−=nx

oteII σ−=

ISimilar to γ-attenuation. Why?


198

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


199

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.


200

Nuclear Fission


~200 MeV



201

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


202

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.


203

Nuclear Fission

Liquid Drop

MeV

)

Shell

Ene

rgy

(va

tion

EA

cti


204

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.


207

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


208

the process.

Nuclear Fission

The average number of neutrons isneutrons is different, but the distribution is G iGaussian.


209

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.


210

Nuclear Fission


211

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.


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 FissionWhy not use it?Why not use it?

σf,Th 584 2.7x10-6 700 0.019 b


214

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


215

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.


217

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.


221

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


222

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.


225

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


226

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?


228

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)( −=


229

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


230

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).


231

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


232

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


233

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


234

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


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?


236

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.


237

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--


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


239

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.


240

• Only 7 MeV from β-decay appears as heat. Why?Why?

Poisoning


241

Poisoning


242

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


243

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.


244

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


245

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 −+− −

+−

∑+


246

Poisoning

)(∞I )(I)(∞Xe

• Now, we know Xe(t)

eratorcladfuel

Xea

eratorcladfuel

poisona tXet

modmod

)()(∑+∑+∑

−=∑+∑+∑

∑−=∆

σρ


247

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 −=∂


248

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.


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


250

(in this case) for Xe to decay down.

Poisoning


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.


252

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


253

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 ∑∑


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


255

Radioactive Fission Products[ ] sMeVTttxtP /)(101.4)( 2.02.011 −− +−= [ ])()(

per watt of original operating power.T = time of operation.


256

Radioactive Fission Products


257

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.


258

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 =


259

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


260

Fusion Reactions


261

nuclear physics intro

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