nuclear structure, nuclear force - mail :: welcome to...

Nuclear Structure, Nuclear Force

Gocha Khelashvili, Ph.D.

Nuclear Structure

• Composition of the Nucleus • Atomic Constituents • Nuclear Shape • Nuclear Stability • The Nuclear Force • Liquid – Drop Model • SEMF • Binding Energy • Nuclear Force Revisited

Nuclear Constituents

- number of protons (charge of the nucleus).

The nucleus has a mass - led to believe that nucleus

has protons

1 The nucleus has a chagre 2

p

Z

A mA

Z A

•

• ≈ ⋅

• ≈

Is there Electron inside of Nucleus?

Before discovery of neutron it was believed that there are electrons inside of nucleus.

The existance of electron inside of nucleus was supported by the observation of radioactive de

A Z

β −

• −

•

cay, in which electrons are ejected by certain radiocative nuclei.

There were significant problems with this model.•


14min

Uncertainity Principle in order for electron to be confined in the region of 10 m, electron should have 100 MeV energy.

Hovewer, energies of electrons in decay are 1-2 MeV.

There

r E

β

−

• →

< ≈

•

• is no evidence of force 50-100 MeV between electron and nuclei.


If 0 - electron would never escape

If 0 - There is no barrier to overcome; All naturally occuring emitters should have dissapeared from the Earth long time ago.

e

e

E

Eβ

• <

• >


- measured magnetic moment of nuclei (order or nuclear magneton). 2

- should be of order or Bohr magneton if electron is inside2

of nuclei.

1 - m2000

Np

Be

N

B

em

em

µ

µ

µµ

• =

• =

• ≈

easurement shows 2000 times smaller magnetic momentum

compared to Bohr magneton.


14

14

- nitrogen nucleus - angular mometum has quantum number 1 (observed in hyperfine structure).

If containes 14 protons and 7 electrons, the resulting angular momentum would have qua

N

N

•

•ntum numbers 1/2, 3/2, 5/2 etc. It would be Fermion and it

would obey Fermi-Dirac statistics.

But it obeys Bose-Einstein statistics.

Rutherford suggested existance of neutral particle, possibly a

•

• bound state of proton and electron and cal nl eed u it tron

Discovery of Neutron • 1930 W. Bothe and H. Becker - Alpha particles incident

on a beryllium foil cause the emission of uncharged radiation capable of penetrating lead.

• 1932 Irene Joliot-Curie and Frederic Joliot - Protons of up to 5.7 MeV are ejected when the radiation strikes a paraffin slab. They assumed that radiation consisted of gamma ray photons and the protons are knocked out of the hydrogen-rich paraffin in Compton collisions.

• Using Compton’s theory they estimated energy of incident gamma rays and found that energies must be at least 55 MeV.

• 1932 James Chadwick - Radiation consists of neutral particles of approximately proton mass. In this case their energies need not exceed 5.7 MeV since in head on collision between particles of the same mass all energy is transferred to target particles – protons.

Composition of the Nucleus

After Chadwick's discovery of neutron the idea of neutron being a tightly bound state of proton and electron was abandoned

•

- number of protons (charge of the nucleus). - number of neutrons. - Mass number of nucleus. A particular nuclear species is called a Nuclides are denote

nuclide.d by the chemical s

ZNA Z N

••• = +••

16 15

158

158 7

ymbol of element with presuperscript giving value of - O, O,.... Sometimes is given as presubscript - O (not necessary - chemical symbol )

Sometimes is given as subscript - O

AZ Z

N

•

•

16 15

13 14

IsotopesIsotone

(not necessary - )

- same (protons) and different (neutrons) - Oand O - same (neutrons) and different (protonssIs

) - Cand N - same (neutrons protoobars ns)

N A Z

Z NN ZA

= −

•

•

• + 14 14 - Cand N


27 126mass unit 1.66054 10 kg mass of is exactly 12

Energy equivalent of mass unit 931.49 MeVu C u−→ = × →

→


Nuclear Radii

Nuclear Radii

( ) ( ) ( )

( )

( )

15 15

2

0

15 15

222

0

1/3 150 0

O N Measure Decay Energy

3 1 Potential energy of charged sphere: 5 4

O and N 1

3 1 15 4

where 1.2 0.2 10 m 1.2 0.2 fm

e Q

qUR

q Ze q Z e

eQ U Z ZR

R R A R

ν

πε

πε

−

−

• → + + →

• =

• = = −

• = ∆ = − −

• = ≈ ± × = ±

Ground-State Properties of Nuclei

After 27 MeV energy of alpha particle experimental curve deviates from Rutherford formula.

Energetic alpha particle penetrates nucleus deep enough to interact directly with protons and neut

•

•

rons with attractive nuclear force.

Thus scattering intensity falls. •

Nuclear Radii

( )

SLAC experiment (1953) - nuclei bombarded with electrons having 200-500 MeV energies.

500 MeV electron 2.5 fm (de Broglie formula)

2.5 fm radius of heavy nuclei

Study a structure of heavy

λ

•

•

• <

•

nuclei by analyzing electron diffraction patteren.

0.61 First minimum - sinRλθ• =

Nuclear Radii

16

16

0

2 2 2

Using the data for 420 MeV electron scattered from O, estimate the radius of O nucleus.

0.61 , where 44sin

de Broglie wavelenght of 420 MeV electron -

Example:

R

hp

p c E

λ θθ

λ

= =

=

= ( ) ( ) ( ) ( )

( )( ) ( )

2 2 2 22

66

0

420 0.511 420 MeV

or1239.8 eV

420 MeV/c 2.95 10 nm 2.95 fm420 MeV 420 10 eV

0.61 2.95 2.59 fmsin 44

mc

nmhcp

R

λ −

− = − ≈

⋅= → = = = × =

×

⋅= =

Nuclear Size and Shape

( )

( )( )( )

( )

1/30

150

126

1/3

27

33 15

17 3

where 1.2 0.2 10 m 1.2 0.2 fm

find the density of the nucleus

1.2 12 fm 2.7 fm

12 1.66 10 /4 4 2.7 103 3

Ex

2.4 10 kg/m

ample:

R R A

R

C

R

u kg um

R mρ

π π

ρ

−

−

−

=

≈ ± × = ±

≈ × =

×= =

×

= × → 4 billion tons per cubic inch!

The Nuclear Force – Range Behavior

Strongly attractive component which acts only over short range At very short distances (<0.5 fm) repulsive component Equilibrium - leads to the saturation of nuclear force. Evidence - approximat

•••• ely constant density of nuclear matter.

The Nuclear Force – Charge Dependence

Nuclear force is charge symmetric: n-n and p-p forces are the same in a given state Evidence: Stability of nucleus ( 40). This would not be a case if n-n

p-p forces are different. Why do

N Z A

•

• ≈ ≤

• we see different potentials on the picture above?

Stability Curve

For 40 for stable nuclei

For 40 for stable nuclei

A N Z

A N Z

• ≤ → ≈

• > → >

Nuclear Stability

Ignore electrostatic repulsion between protons for light nuclei ( 40) Energy is smallest if A/2 are neutrons and A/2 are protons Energy is greatest if there only one type of particle (exclusion p

A• ≤•• rinciple)

The Nuclear Force – Charge Dependence

2

For heavy nuclei ( 40), electrostatic repulsion between protons becomes important. Potential energy becomes dependent. The energy inceased less by adding two neutrons than by adding one neutro

AZ

• >

•• n and one proton. increases for stable nuclei with inceasing .

N Z Z• −

Stable Isotopes (N vs. Z)

There are about 270 stable nuclides and about 100 different elements. 2.7 stable isotopes per element There are larger than average number of stable isotopes with nuclei

with equal 2, 8, 20, Z

•••

28, 50, 82 and 126 (last is theoretical for now) "Magic Numbers" - closed shell structure, very much as "magic atomic

numbers" - 2, 10, 18 and 36 corresponds to closed-electron shell structure.

•

The Nuclear Force – Charge and Spin Dependence

Nuclear force is almost charge idependent same for n-p, n-n and p-p

n-n0

1/ 2 1/2

p-p0

1/ 2 1/2

p-n1/ 2 1/2 0,1 Force is stronger for 1 state.1/ 2 1/2

S

S

S S

• ≈

→ =↑ ↓

→ =↑ ↓

↑ ↓ → = =↑ ↑

Binding Energy

11

21

Mass of H atom 1.007825u Mass of neutron 1.008665u Expected mass of H atom

+ +

21

2.016490u

Measured mass of the H atom is only 2.014102u (less than expected value) 2.016490u 2.014102uThe energy equivalent of missing mass

0.0023is:

88u

m∆ = − =

( )( )21

0.002388u 931.49 MeV/u

Missing mass (energy) corresponds to energy given off when H nucleus is formedor it is energy required to break apart a deuterium nucleus i

2.224

nto se

Me

parate

VE∆ = =

neutron andproton Binding En ergy

Binding Energy per Nucleon

2 2091 83

1

Nuclear binding energies are strikingly high. The range for stable nuclei is from2.2 MeV for H (deuterium) to 1640 MeV for Bi (an isotope of the metal bismuth).

Typical binging energy is - 8 10×

( )

1

4

21

kJ/kg 800 billion kJ/kgBoiling Water - 2260 kJ/kg (heat of vaporization)Burning gasoline - 4.7 10 kJ/kg (17 million times smaller)

Total Binding EnergyBinding Energy per nucleon

2.2 MeVH 12

A

B

−

×

=

= = ( )20983

1640 MeV.1 MeV and for Bi 7.8 MeV209

B = =

Binding Energy Curve

Binding Energy Calculations

4220

(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotope Ca.(b) Find the energy needed to remove a proton from this nucleus(c) Why are these energies

Exam

dif

ple:

fere

( ) ( ) ( ) ( )

( )

41 4220 20

nt?

Ca 40.962278 , Ca 41.958622 n 1.008665 , 1.007276

(a) 40.962278 1.008665 41.970943 41.970943 41.958622 0.012321 0.012321 u

M u M u M u M p u

u u u M u u uB

= = = =

+ = → ∆ = − =

=42 4120 19

931.49 MeV/u

(b) Removing a proton from Ca leavs the potassium isotope K (40.961827 u). A similar calculation give binding energy of 10.27 MeV for the missing proton

11.48 MeV

(c) Diffe

× =

rence is due electrostatic repulsion between protons

Binding Energy Calculations

2 2 2 2

2

- mass of the nucleus, - mass of the atom, - mass of the electron, - mass of the proton, - mass of the neutron,

0 (compared to )

N A

e p n

atom N e A nuclear

nuclear p n

M Mm m m

B M c Zm c M c mc B

B Zm c Nm

= + − = ∆ ≈

= + ( )2 2

2 2 2 2 2 2 2

2 2 2

H A

N p e n N e

ZM c M c

nuclear H n A

c M c Zm c Zm c Nm c M c Zm c

B ZM c Nm c M c

− = + + − +

= + −

Liquid – Drop Model

( ) ( )22/3 1/3 1 1/21 2 3 4 51 2B a A a A a Z Z A a A Z A a A− − −= − − − − − ±

Semiempirical Mass Formula

23.7


3

11/3

0

4 3

Reason why the binding energy per nucleon is approximately constant

volumevolume

E V RE a A

R R A

π = → ==


2/322

1/30

Correction to first term 4

Implies fewer interactions for surface nucleus and thus smaller binding energy Explains shar

surfacesurface E a AE S RR R A

π

• → = −= =

••

p decline in the binding energy per nucleon at low values.A


( )22 1/32 1/3

30

1/30

Correction to first term

3 5 4

Energy of repulsion decreases the binding energy (thus < 0) Explains slow d

coulombcoulomb

Ze E a Z AE Z AR

R R Aπε

−−

• → = −= =

••

ecline in the binding energy per nucleon at large values.A

Semiempirical Mass Formula

SEMF – Asymmetry Energy

( ) ( ) ( )

( ) ( ) ( )2 2

2 14

1

energy incease 1 1number of new neutronsnew neutron 2 2 2

2 28 8 assymetry

E N Z N Z

E N Z A ZE E a A Z A

A

ε

ε ε

ε

−

−

∆ = = − − ∆ = − = − → = ∆ = − −

2N Z A Z Z A Z− = − − = −

SEMF – Pairing Energy

The last term arises from the tendency of proton pairs and neutron pairs to occur. Even-even nuclei are most stable and have higher binding energies than would

otherwise expected Such nuclei as

••

• 4 12 162 6 8 He, C and O appear as peaks on the empirical curve

Odd-Odd nuclei have both unpaired protons and neutrons and have relatively low B.E.•

1/25pairingE a A−= ±

Weizsäcker’s SEMF

23.7

23.7

More Binding Energy Calculations

( )( ) ( )( )

6430

6430

The atomic mass of the zinc isotope Zn is 63.929 u. Compare its binding energy with the prediction of SEMF

The binding energy of Zn is:

30 1.007825 u 34 1.

Example:

So

008665 u

lu

63.929 u

tion:

bE = + − ( )

( )( ) ( )( ) ( )( )( )

( )( )

( )

22/3

1/3

1/2

931.49 MeV/u

Using SEMF:

0.75 MeV 30 23.7 MeV 1615.67 MeV 64 17.23 MeV 64

646412 MeV

559.1 MeV

554.1 MeV 64

559.1 554.8.73 1 or 64 64

MeV 8.66 M

b

b b

E

E EA A

× =

= − − −

+ =

= = = =

less than 0.1% differ

eV

ence

More Binding Energy Calculations

Nuclear Exchange Force


Distribution of charge produces an electric field , and the force fe

Electrostatic Interaction (Classical Pict

lt by another charge

located in the fie

ure

ld i

)

s

E q

the product . Any change in the charge distribution changes , however, the information that change has occured does not appear instantaneously throughoutthe field, but it is propageted outward a

qE E

Electrostatic Interaction (Quantum Mechanical P

t the speed of light

Every charge is continually emitting and absorbing photons, even when it is not moving

ictu

.

re)

Thesephotons are called meaning they are not directly observable. A charge can emmita virtual photon of energy without changing its energy or recoiling. Energy and momentumconservat

,

i

onh

virtual photonsf

laws are not violated provided that photon exists for no longer than / ,where , as required by uncertainty principle.

2 2

t EE hf

c c cR c tE hf f

λπ π

∆ = ∆∆ =

= ∆ = = = =∆


2

28

1935 Hideki Yukawa Nuclear interaction is carried out by virtual particles - Short Range - meson has a mass

1 fm

3.5 10 kg 380 20e

E mc

cR c tE mc

m

e s

m

m son

−

∆ ≥

= ∆ = =∆

≈ × ≈ ≈

20 MeV/c

0

2

Charge Independence - mesons carry , 0, - chargeExperimentaly , are found in 1947

140 MeV/c

e e

mπ

π π±

+

≈

Probability Density of the Exchange Mesons

( )( )( )( )

3424

2 2 2 13

20

Virtual Meson Exists :

1.055 10 J×s5 10

140 MeV/c 1.6 10 J/MeV

Thus, a 10 second time-exposure "snapshot" of a nucleon would show a cloudconsisting more than 10,000 mesons surround

t smc c

−−

−

−

×∆ = = = ×

×

( ) ( )( ) ( ) ( )

2 22 222

22 22 2 2

ing the nucleon!

,1 , ,,

mc E pc r t mcr t r tc tE i p

t

= − ∂ Φ ⇒ ∇ Φ − = Φ ∂ ∂ → → − ∇∂

Probability Density of the Exchange Mesons

( ) ( ) ( )

( ) ( ) ( )

( )

222

2 2

2 /2

2 2 /2

2

,1 , , For stationary solution

Probability Density

r R

r R

r t mcr t r tc t

mc Aer r rr

A er

r

−

−

∂ Φ ∇ Φ − = Φ → → ∂

∇ Φ = Φ → Φ = →

→ Φ =

Probability Density of the Exchange Mesons For 0.5 curve agrees with experimental

results.Breaks down for 0.5 Measured meson density is much lower than

figure would suggest Reason - at 0.5 quark composition

of nuclei and mes

r R

r R

r R

• >

• <•

• <ons becomes important

Decrease in meson density is due to saturation of strong force - called assymtotic freedom of quarks.

•

QCD – Asymptotic Freedom

18

For low , strong force is diluted by gluon self- interactions.

1 0 faster than

For 10 m, quarks move as almost free particles.

Asymtotic Freedom - confirmed by electron deep sca

S

r

r

r

α

−

•

• →

• <

•

ttering experimensts.

QCD – Asymptotic Freedom

Quark Confinement

( ) ( ) ( )

( ) ( )

2

4 4, 3 3

lim , lim const

S SQCD QCD QCD

QCD QCDr r

V r kr F r V r kr r

V r F r

α α

→∞ →∞

= − + = −∇ = −

→∞ →

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