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Tony Weidberg Nuclear Physics Lectures 1

Today’s Menu• Why study nuclear physics• Why nuclear physics is difficult• Course synopsis and textbooks• Notation & Units


What is the use of lectures• Definition of a lecture: a process

whereby notes are transferred from the pages of a lecturer to the pages of the student without passing through the head of either.


The use of lectures• Lectures provide:

– best introduction to the subject.– guide to the main concepts, derivations and

applications.– Enough material to start to tackle problems.

• During a lecture you need to:– Add your comments to the notes (the notes are very

terse and are only useful if you do this).– Ask questions if you do not understand something ( if

you don’t understand something, chances are that >50% of the audience also doesn’t understand).

• After a lecture you need to:– go over the notes, make sure you understand them

and make your own notes.– Read material from textbook(s).– Do the problem sets.


Why Study Nuclear Physics?• Understand origin of different nuclei

– Big bang: H, He and Li– Stars: elements up to Fe– Supernova: heavy elements

• We are all made of stardust• Need to know nuclear cross sections

experimental nuclear astrophysics is a hot topic.


Practical Applications• Nuclear fission for energy generation.

– No greenhouse gasses– Safety and storage of radioactive material.

• Nuclear fusion– No safety issue (not a bomb)– Less radioactive material but still some.

• Nuclear transmutation of radioactive waste with neutrons.– Turn long lived isotopes stable or short lived.

• Every physicist should have an informed opinion on these important issues!


Medical Applications• Radiotherapy for cancer

– Kill cancer cells.– Used for 100 years but can be improved by better

delivery and dosimetery– Heavy ion beams can give more localised energy

deposition.• Medical Imaging

– MRI (Nuclear magnetic resonance)– X-rays (better detectors lower doses)– PET– Many others…see Medical & Environmental short

option.


Other Applications• Radioactive Dating

– C14/C12 gives ages for dead plants/animals/people.

– Rb/Sr gives age of earth as 4.5 Gyr.• Element analysis

– Forenesic (eg date As in hair).– Biology (eg elements in blood cells)– Archaeology (eg provenance via isotope

ratios).


Why is Nuclear Physics Hard?

• QCD theory of strong interactions just solve the equations …

• At short distance/large Q coupling constant small perturbation theory ok but long distance/small Q, q large

)(2

).(16

1][

xAAqAAF

AqFFmiL

Not on syllabus !


Nuclear Physics Models• Progress with understanding nuclear physics

from QCD=0 use simple, approximate, phenomenological models.

• Liquid Drop Model: phenomenology + QM + EM.

• Shell Model: look at quantum states of individual nucleons understand spin/parity magnetic moments and deviations from SEMF for binding energy.


Corrections• To err is human … and this is a new

course lots of mistakes.• Please tell me about any mistakes you

find in the notes (I will donate a bottle of wine to the person who finds the most mistakes!).


The Minister of Science• This is a true story honest.• Once upon a time the government science

minister visited the Rutherford Lab (UK national lab) and after a days visit of the lab was discussing his visit with the lab director and he said …

• I hope that you all have a slightly better grasp of the subject by the end!


Course Synopsis - 1• Nuclear sizes and liquid drop model, S.E.M.F.

Applications of S.E.M.F.– Williams, Nuclear & Particle Physics, OUP,

chapters 4 & 5.– Krane, Introductory Nuclear Physics, Wiley,

chapters 3, 8 & 9.• Cross sections and Breit Wigner Resonances

– Cottingham & Greenwood, An Introduction to Nuclear Physics, CUP, chapter 8 and appendices A & D.

• Radioactive decays– Alpha decay, Williams, chapter 6, Krane, chapter 8– Beta decay, Williams, chapter 12, Krane chapter 9


Course Synopsis - 2• Interactions of particles with matter

– Textbook ???• Applications of Nuclear Physics

– Particle detectors, textbook ?– Fission reactors and bombs, Krane

chapter 13– Fusion reactors, Krane, chapter 14.– Radioactive dating, Krane, chapter 6.


Notation• Nuclei are labelled where El is the

chemical symbol of the element, mass number A = number of neutrons N + number of protons Z. eg

• Excited states labelled by * or m if they are metastable (long lived).

ElAZ

Li73


Units• SI units are fine for macroscopic objects like

footballs but are very inconvenient for nuclei and particles use natural units.

• Energy: 1 eV = energy gained by electron in being accelerated by 1V.

– 1 eV= e J.• Mass: MeV/c2 (or GeV/c2)

– 1 eV/c2 = e/c2 kg. – Or use AMU defined by mass of 12C= 12 u

• Momentum: MeV/c (or GeV/c)– 1 eV/c = e/c kg m s-1

• Cross sections: (as big as a barn door) – 1 barn =10-28 m2

• Length: fermi 1 fm = 10-15 m.


Nuclear Masses and Sizes• Masses and binding energies

– Absolute values measured with mass spectrometers.

– Relative values from reactions and decays.• Nuclear Sizes

– Measured with scattering experiments (leave discussion until after we have looked at Rutherford scattering).

– Isotope shifts


Nuclear Mass Measurements• Measure relative masses by energy

released in decays or reactions.– X Y +Z + E – Mass difference between X and Y+Z is E/c2.

• Absolute mass by mass spectrometers (next transparency).

• Mass and Binding energy:• B = [Z MH + N Mn – M(A,Z)]/c2


Mass Spectrometer• Ion Source• Velocity selector

electric and magnetic forces equal and opposite – qE=qvB v=E/B

• Momentum selector, circular orbit satisfies:– Mv=qBr – Measurement r

gives M.

Ion Source

Velocity selector

Detector


Binding Energy vs A• B increases with A up to 56Fe and then

slowly decreases. Why?• Lower values and not smooth at small

A.


Nuclear Sizes & Isotope Shift• Coulomb field modified by finite size of

nucleus. • Assume a uniform charge distribution in the

nucleus. Gauss’s law integrate and apply boundary conditions

• Difference between actual potential and Coulomb

• Use 1st order perturbation theory

32

0)(

4 Rr

rZeE

RZe

RZerrV

03

0

2

83

8)(

)Rr(r4

ZeR8

Ze3R8

Zer)r(V00

30

2

drrrVerrER

)()]()[(4 *

0

2 2/3

00

2/3

0)(2)/exp()(2)(

aZaZr

aZr


Isotope Shifts

2 23

00

2 ( / )5

Ze RE Z a

dr]r4

ZeR8

Ze3R8

Zer)[e()a/Z(4r4E00

30

23R

0

2

5R4drrr4

522

R

0

3R4drr4

32R

0

22R

0R2dr

r1r4

]223

34

104[R

4Ze)a/Z)(e4(E 2

0

3


Isotope Shifts• Isotope shift for optical spectra• Isotope shift for X-ray spectra (bigger

effect because electrons closer to nucleus)

• Isotope shift for X-ray spectra for muonic atoms. Effect greatly enhanced because m~ 207 me and a0~1/m.

• All data consistent with R=R0 A1/3 with R0=1.25fm.


Frequency shift of an optical transition in Hg at =253.7nm for different A relative to A=198.

Data obtained by laser spectroscopy.

The effect is about 1 in 107. (Note the even/odd structure.)

Bonn et al Z Phys A 276, 203 (1976)

A2/3

Isotope Shift in Optical Spectra

E/h

(GH

z)


Data on the isotope shift of K X ray lines in Hg. The effect is about 1 in 106. Again the data show the R2 = A2/3 dependence and the even/odd effect. Lee et al, Phys Rev C 17, 1859 (1978)


Data on Isotope Shift of K Xrays from muonic atoms [in which a muon with m=207me takes the place of the atomic electron].

Because a0 ~ 1/m the effect is ~0.4%, much larger than for an electron.

The large peak is 2p3/2 to 1s1/2. The small peak is 2p1/2 to 1s1/2. The size comes from the 2j+1 statistical weight.

Shera et al Phys Rev C 14, 731 (1976)

58Fe

56Fe

54Fe

Energy (keV)


SEMF• Aim: phenomenological understanding of

nuclear binding energies as function of A & Z.

• Nuclear density constant (see lecture 1).• Model effect of short range attraction due to

strong interaction by liquid drop model.• Coulomb corrections.• Fermi gas model asymmetry term.• QM pairing term.• Compare with experiment: success & failure!


Liquid Drop Model Nucleus• Phenomenological model to understand binding

energies.• Consider a liquid drop

– Ignore gravity and assume no rotation– Intermolecular force repulsive at short distances, attractive

at intermediate distances and negligible at large distances constant density.

E=-n + 4R2T B=n-n2/3

• Analogy with nucleus– Nucleus has constant density– From nucleon nucleon scattering experiments: Nuclear

force has short range repulsion and attractive at intermediate distances.

– Assume charge independence of nuclear force, neutrons and protons have same strong interactions check with experiment!


Mirror Nuclei• Compare binding energies of mirror nuclei

(nuclei n p). Eg 73Li and 7

4Be.• Mass difference due to n/p mass and Coulomb

energy.dQrrQE

R

0 04

)(

323 /3)/()( RZerdQRrZerQ

RZedr

Rr

rZeE

R

0

2

6

5

0 0

2

4)()5/3(

4)(3

3/1

0

2

;2/~;)]2)(1()1([45

3)1,( ARAZZZZZR

eZZEc

3/2)1,( AZZEC


nn and pp interaction same (apart from Coulomb)

“Charge symmetry”


Charge Symmetry and Charge Independence

• Mirror nuclei showed that strong interaction is the same for nn and pp.

• What about np ?• Compare energy levels in “triplets” with

same A, different number of n and p. e.g.

• Same energy levels for the same spin states SI same for np as nn and pp.

MgNaNe 2212

2211

2210


Charge Independence

• Is np force is same as nn and pp?

• Compare energy levels in nuclei with same A.

• Same spin/parity states have same energy.

• np=nn=pp

2311Na 23

12 Mg

2212Mg

2211Na

2210Ne


Charge Independence of Strong Interaction

• If we correct for n/p mass difference and Coulomb interaction, then energy levels same under n p.

• Conclusion: strong interaction same for pp, pn and nn if nucleons are in the same quantum state.

• Beware of Pauli exclusion principle! eg why do we have bound state of pn but not pp or nn?


Asymmetry Term• Neutrons and protons are spin ½ fermions

obey Pauli exclusion principle.• If other factors were equal ground state

would have equal numbers of n & p.

IllustrationNeutron and proton states with same spacing .Crosses represent initially occupied states in ground state.If three protons were turned into neutrons the extra energy required would be 3×3 .In general if there are Z-N excess protons over neutrons the extra energy is ((Z-N)/2)2 . relative to Z=N.


Asymmetry Term• From stat. mech. density of states in 6d phase space = 1/h3

• Integrate to get total number of protons Z, & Fermi Energy (all states filled up to this energy level).

• Change variables p E

3

24hdpVp

dN

3F )h/Vp()3/8(Z

FE

0

2/1

E

0

2/3

2/1 E)5/3(dEE

dEEEAE

dE/dpdp/dNdE/dN

F

F

3/1

0

3/1F A

ZRh)8/3(P

3/2

20

23/2

F AZ

mR2h)8/3(E


Asymmetry Term

• Binomial expansion keep lowest term in y/A

• Correct functional form but too small by factor of 2. Why?

AZNKE

2)(

3/2

20

23/2P

Total AZ

mR2Zh)8/3(

53E

3/53/53/2Total NZ

AKE ZNy

3/53/53/2

3/5

)/1()/1( AyAyAKA

ETotal


Pairing Term• Nuclei with even number of

n or even number of p more tightly bound fig.

• Only 4 stable o-o nuclei cf 153 e-e.

• p and n have different energy levels small overlap of wave functions. Two p(n) in same level with opposite values of jz have AS spin state sym spatial w.f. maximum overlap maximum binding energy because of short range attraction. Neutron number

Neutron separation energy in Ba


Pairing Term• Phenomenological fit to A dependence

• Effect smaller for larger A

2/1 AE

e-e +ivee-o 0o-o -ive


Semi Empirical Mass Formula• Put everything together:

• Fit to measured binding energy. – Fit not too bad (good to <1%).– Deviations are interesting shell effects.– Coulomb term agrees with calculation.– Asymmetry term larger ?– Explain valley of stability.– Explains energetics of radioactive decays, fission

and fusion.

2/13/1

223/2 )(),(

AAZd

AZNcbAaAZNB


The Binding Energy per nucleon of beta-stable (odd A) nuclei.

Fit values in MeV

a 15.56b 17.23

c 23.285

d 0.697

+12 (o-o)

0 (o-e)

-12 (e-e)

A

B/A

(MeV

)

7.5

9.0


Valley of Stability• SEMF allows us

to understand valley of stability.

• Low Z, asymmetry term Z=N

• Higher Z, Coulomb term N>Z.

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