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.•
Is there Electron inside of Nucleus?
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.
Is there Electron inside of Nucleus?
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
• <
• >
Is there Electron inside of Nucleus?
- 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.
Is there Electron inside of Nucleus?
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
Composition of the Nucleus
Composition of the Nucleus
27 126mass unit 1.66054 10 kg mass of is exactly 12
Energy equivalent of mass unit 931.49 MeVu C u−→ = × →
→
Composition of the Nucleus
Nuclear Radii
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
Liquid – Drop Model
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 = → ==
Liquid – Drop Model
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
Liquid – Drop Model
( )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
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
∆ = ∆∆ =
= ∆ = = = =∆
Nuclear Exchange Force
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±
+
≈
Nuclear Exchange Force
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 satu- ration 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
→∞ →∞
= − + = −∇ = −
→∞ →