Lecture 7
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
Chapter 3:
Nuclear Models (Review)
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-1
This Chapter
1. Introduction – The Strong Force
2. Nucleon-Nucleon Interaction
3. Shell Model - Choice of a Potential
4. Spin-Orbit Interaction
5. Nuclear Electromagnetic Moments
6. Valence Nucleons
7. Even-Even Nuclei & Collective Structure
8. Collective Models:
The Vibrational and Rotational Models
Reference: Krane Chapter 3
1
Introduction
4
The Nuclear (Strong) Force
We concluded that nucleons are bound
together in a nucleus because there is a
“strong” force between them.
This nucleon-nucleon force must be stronger
than the Coulomb force (which exists between
protons)
We can also understand that this force has a
short range, i.e. in the very close neighborhood
of the nucleus.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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5
Electrons are insensitive to the strong force
In addition to that the atomic electrons do
not respond to this force.
We know that the interaction of nuclei in a
molecule
are
purely
electromagnetic
(Coulomb force).
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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c- Criteria for the validity of a Model
Criteria that should be satisfied by any
model:
1)The model should explain experimental
results
2)The
model
should
predict
additional
properties to be verified by new experiments
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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7
Types of Nuclear Models
Nuclear models are roughly of (divided into)
2 types:
1)Single Particle Models.
2)Collective Structure Models.
semiclassical Quantum Mechanics
Independent Particles
Collective
Fermi Gas
Shell
Liquid Drop
Rotational
Vibrational
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Atomic Physics vs. Nuclear Physics
The major difference between the atomic
physics theory and nuclear physics comes
essentially from the fact that in an atom the
electromagnetic
potential
(Coulomb)
is
between the electrons and the nucleus (that
we treat as if it is an external potential). While
for the nucleus, the nuclear potential itself is
the result of the interaction of nucleons
together (an internal potential somehow)
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Nucleon-Nucleon
Interaction
10
3-2 The nucleon-nucleon interaction
a- Main results from the study of Deuteron
(2-nucleon system)
Chapter 4 and Krane Paragraph 44-4
Experimental results of the nucleon-nucleon
interaction lead to the following conclusions:
1) The nucleon-nucleon force (interaction) is
attractive and only dependent on the distance
between the 2 nucleons (V = V(r)),
2) The nucleon-nucleon force is strongly spin
dependent,
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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11
The nucleon-nucleon interaction
(cont’d)
3) The nucleon-nucleon force is invariant with
respect to parity and time reversal operations
(transformations) (10-7,10-3),
4) The nucleon-nucleon potential includes a tensor
potential. Including a tensor potential means
including a non central term,
5) The nucleon-nucleon force is charge symmetric.
The word charge here refers to the nucleon. The
nucleon-nucleon force sees the proton exactly as it
sees the neutron. The pp potential is identical to the nn
potential (after a small correction for the pp Coulomb
potential),
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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The nucleon-nucleon interaction
(cont’d)
6) The nucleon-nucleon force is nearly charge
independent:
nn, pp and np interactions are identical (after
correction for the Coulomb potential),
7) The nucleon-nucleon force becomes repulsive at
short distances,
8) The nucleon-nucleon force may depend on the
relative motion (momentum or velocity) of the
nucleons.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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14
Why a Shell Model ?
Although theoretical calculations for a
nucleon-nucleon system fairly reproduce
the experimental results, these calculations
encounter a huge mathematical difficulty
when adding a 3rd or more nucleons to the
2-system potential (The many-body problem in
physics).
It is difficult (impossible) to construct a
frame (theory) to account for all the nuclei
properties.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Shell Model in Atomic Physics
Shells in an atom are filled with electrons in order
of increasing energy. The Pauli Exclusion Principle
is being fully respected.
The atomic properties are determined in
considering an atom as:
• an inert core composed of shells (inner ones)
which are filled
•& (+) one or more valence electrons.
These
valence
electrons
determine
the
chemical properties of the atom.
The temptation to introduce shells at the nuclear
level was supported by experimental evidence(s).
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Atomic Radius vs. Z (of Elements)
Atomic radius nm
)
(
Z
Fig 3.1-a
- Smooth variations are observed when filling the same subshell.
- Jumps are observed when jumping from a shell to another.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Ionization Energy (eV)
Ionization Energy vs. Z (of Elements)
Z
Fig 3.1-b
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Experimental evidence supporting the existence of nuclear shells
S2p for a series of isotones (same N) &
S2n for a series of isotopes
Fig 3.2 : Measurements of S2p for a series of isotones (same
N) and S2n for a series of isotopes. The values plotted are
the differences between the measured values of (S2p and
S2n) and the predictions of the mass semiempirical formula.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Nuclear Shells - Experimental Evidence
Fig 3.2 :
- Like the ionization energy in fig. 3.1(b),
S2p and S2n increase gradually with Z or N
and sharp discontinuities (jumps) are
observed when reaching the same proton
and neutron numbers.
These numbers (2, 8, 20,
20 28,
28 50,
50 82 and
126)
126 are called the magic numbers
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Nuclear Shells – Magic Numbers
Using the shell “notion” one is led to assume
that nucleons in a nucleus appear somehow to
be ranged into shells and a magic number
appears to correspond to a major shell.
This is reinforced by considering that one
nucleon feels the effect of an average
“nuclear” potential due to the other nucleons
in the nucleus.
This leads to the assumption that the nucleons
are allowed to occupy the energy levels, of a
potential well, in a series of subshells in a
similar way as electrons in an atom do.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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- Spatial orbits
Another important feature of the atomic
shell theory is the existence of spatial
orbits. The electrons can move in those
orbits relatively freely of collisions with
other electrons.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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- Spatial Orbits in the Nucleus
Assuming the existence of shells in a nucleus
implies that the nucleons (as the atomic electrons)
move in orbits and collisions between these
nucleons, the most likely, do not exist.
To see this let us imagine a collision between 2
nucleons in the innermost shell.
The collision energy would allow one or both
nucleons to jump from their shell to an upper shell.
But the fact that the most immediate neighbor shells
are supposed to be filled would mean that the
colliding nucleons could only escape to the valence
shell.
Such a transition needs an energy far above the
probable collision energy. We conclude that there is
no collision between nucleons.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Additional experimental evidence
α decay
Unstable heavy nuclei emit α particles (4He).
The decay is represented by:
A
Z
XN →
A−4
Z−2
Y N − 2 + 42 α 2
The α particle is emitted with a certain kinetic
energy in the order of the MeV
- The nucleus X is called the father and the
nucleus Y is called the daughter. If Y happens to
decay to a daughter Y’, then Y’ is called the granddaughter of X.
- X is the grand-father of Y’
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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25
The magic number 126
Fig 4.3 :Energy of α particles emitted by isotopes
of radon (Rn). A sudden increase is observed when
the daughter has N=126 (magic number).
See other examples in Krane p.120
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Shell Model
Choice of a Potential
See Supplement 4
12
27
An intermediate form
An intermediate form is chosen:
V (r ) =
− V0
1 + exp[(r − R) a ]
R (the radius) and a (the skin thickness) are
chosen compatible with experimental results.
R = 1.25 A1/3 & a = 0.524 F
4.4:
4.4:
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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- Energy levels
• The first modification to the SHO energy
levels is to remove degeneracies in .
• But
:
The
separation
between
degenerate levels becomes more and
more important. At higher levels it
becomes larger than the spacing between
the energy levels themselves!
• We have the first 3 magic numbers! only
• Fig. 4-5 : Energy levels for the intermediate form
(Figures at the right are the number of nucleons in a shell)
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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What is missing?
We, somehow, have to add a degeneracy of :
• 8 to the 3rd major shell to get 28
• 10 to the shell 40 to get the magic number 50
• 24 to the shell 58 to get the magic number 82.
• 34 to the shell 92 to get the magic number 126 etc …
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Lecture 8
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
3-4 Spin-Orbit
Interaction
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-2
14
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The Shell Model
We have to improve the previous potential in
order to reproduce the magic numbers. We
should not want to make radical changes
because this would destroy the physical
content of this potential
In 1949, Mayer, Haxel, Suess and Jensen
suggested to add a spin-orbit term to V which
solved the problem of finding the magic
numbers and gave the suitable separation
between the shells.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Mayer, Haxel, Suess and Jensen
Source: Maria Goeppert
Mayer’s Nobel Lecture
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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33
Spin-Orbit potential in Atomic Physics
Atomic physicists found that the only way
to explain the fine (and the hyperfine)
structure of the atom is to take into account
the possible interaction between the em
moment of the electron and the magnetic
field it generates in its motion around the
nucleus!.
The idea is to add an interaction term to the
→ →
central potential of the form l • s
34
Spin-Orbit potential in Nuclear Physics
The spin-orbit effect is very small (10-5). In
nuclear physics we consider the same type
of interaction but keeping in mind that an
atomwise electromagnetic effect is certainly
not sufficient to modify the shells and
reproduce the magic numbers
16
35
VSO
The spin-orbit interaction term is written as :
→
VSO l • →
s
As in atomic physics it is better to use the total
angular momentum j
 ± 1

2
j=
1
 2
for > 0
for = 0
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
→ →
The expectation value of
→
36
l •s
→ →
j = l •s
2
→2
2
2
→ →
→
→
→ →


j = l + s  = l + s + 2 l • s


The expectation value of the operator
given by:
→
[
→ →
l • s is
]
1
< j 2 > − < l 2 > − < s2 >
2
→ →
1
< l • s > = [ j ( j + 1) − l (l + 1) − s (s + 1)] 2
2
→
< l • s >=
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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j degeneracy
The effect of adding this spin-orbit term is to split
the subshells according to the j degeneracy. Each
subshell now can contain up to 2j + 1 protons or
neutrons
l
s
j
notation 2j + 1
0
1
2
3
+1/2
-1/2
+1/2
1/2
1/2
3/2
s1/2
p1/2
p3/2
-1/2
+1/2
-1/2
+1/2
3/2
5/2
5/2
7/2
d3/2
d5/2
f5/2
f7/2
2
2
4 6
4
6 10
6
8 14
m & ms become no more good quantum numbers
when introducing spin-orbit interaction.
38
Spin-Orbit l Doublets
Introducing the spin-orbit term adds energy to
the “basic energy” of a shell (coming from the
solution of SWE using the intermediate term).
The energy difference between any pair of 2
states ( > 0) ( (1p1/2, 1p3/2), (1d3/2, 1d5/2) for
example) is given by :
→
→
→
< l•s>
j=+
1
2
→
−< l•s>
j=−
1
2
=
1
(2 + 1) 2
2
This difference is 1.5, 2.5, 3.5 in 2 units for
= 1, 2, 3 respectively. It increases with
increasing .
18
39
Justification of the 1963 Nobel Prize
The idea of Mayer & al. is to make VSO
negative which has for effect to lower the
member of the l doublet with the higher j to
be pushed downward making its level under
the level of the second member
40
Magic ! Numbers (28)
The p and d splitting does not introduce major
changes.
Splitting the shell f into f5/2 & f7/2 with pushing
downward the latter allows us to find the (4th) magic
number 28!!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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41
One More Magic Number (50)
Splitting the shell g into g7/2 & g9/2 with pushing
downward the latter allows us to find the (5th)
magic number 50!!
42
The 82 Magic Number
Splitting the shell h into h9/2 & h11/2 with pushing
downward the latter allows us to find the (6th)
magic number 82!!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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43
Application :
15O
Ground State spin-parity
According to the shell model the filling of
levels of 15O is as follows :
8 protons
The spin parity of
7 neutrons
15O
would be that of the unpaired
−
1
and this is also the
neutron in the 1p1/2 subshell, i.e.
2
value obtained experimentally.
This assumes that the protons filling 2 major
shells do not contribute to the spin-parity.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Application :
17O
Ground State spin-parity
According to the shell model the filling of levels
of 17O is as follows :
8 protons
The spin parity of
9 neutrons
17O
would be that of the
5+
unpaired neutron in the 1d5/2 subshell, i.e.
and
2
this is also the value obtained experimentally.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Ground State spin-parity
The shell model, a single particle type
model, gives in general the same good
predictions in the case of odd-A nuclei (
A<150 and 190< A < 220)
This was a big success for this model. The
measurements of em moments proved to be
consistent with the shell model too. Not
only the shell model explained experimental
results but it predicted other results which
were proved to be valid.!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Lecture 9
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
3-5 Nuclear Electromagnetic
Moments
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-3
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Magnetic Dipole Moment & the Shell Model
We have seen that the mdm is given by
µ = µ N [g l l z + g s s z
]
This was a big success for this model. The
measurements of em moments proved to be
consistent with the shell model too.
µµ==µµNN [gll (jjzz +−(sgz s) +− gsl )sszz ]] µ = µ N [gl jz + ( gs − gl ) sz ] j
[ j ( j + 1) − l (l + 1) + s (s + 1)]
sz =
2 j ( j + 1)
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Magnetic Dipole Moment & the Shell Model
Taking the expectation value when jz = j jz = j ⇒ µ = µ N [gl j + ( gs − gl ) sz
sz =
]
j
[ j ( j +1) − l (l +1) + s (s +1)]
(
)
2 j j +1
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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49
< sz> : Case j = l + ½
j=l+
1
2
sz =
l + 12
(
l + 12 )(l + 12 + 1 ) − l (l + 1) +
1
1
2 (l + 2 )(l + 2 + 1)
[
=
[(
) (
(
1 1
2 2
]
+ 1) ) ]
1
l 2 + 2 l + 43 − l 2 + l + 34 3
2 (l + 2 )
2
=+

1
1 
µ =  g l  j −  + g s  µ N
2 2 
 
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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< sz> : Case j = l - ½
j=l−
sz =
1
2
2 (l −
=
l − 21
1
2
)(l −
1
2
(
[
l − 21 )(l − 21 + 1) − l (l + 1) + 21 (21 + 1)]
+ 1)
1
(
l 2 − 41 ) − (l 2 + l ) +
1
2 (l + 2 )
[
= − j 2( j + 1 )
3
4
]
=−
3
 

j
+




2 1 j
µ =  gl 
−
gs  µN
( j + 1) 2 j + 1 



1
[l − 21 ]
1
2 (l + 2 )
See Eq. 5.9 page
126 in Krane
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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The neutron case (gl = 0)
j=l+
1
2
µ
j=l−
1
2
µ
neutron
=+
1
gs µ N
2
neutron
=−
1 j
gs µ N
2 j+1
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
52
Comparison with Experimental Results
j=l−
1
2
Schmidt Lines:
Solid
for
gs=(gs)
free
dashed
for gs=0.6 (gs) free
j=l +
1
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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53
Comparison with Experimental Results
The model is an oversimplification again,
but it “contains” well the behavior of the
magnetic dipole moment for many nuclei.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Lecture 10
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
3-6 Shell Model
Valence Nucleons
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-4
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The Extreme Independent Particle Model (EIPS)
The main assumption is that nuclear properties
arise from the motion of a single unpaired nucleon.
This assumption would mean that the nuclear
properties (spin-parity for example) of 15O are
obtained from those of the unpaired neutron
(number 7) in the 1p1/2 shell (which is true)
The shell structure for
15O
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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EIPS : The
43Ca
example
Equivalently the nuclear properties of 43Ca are
those of the 23rd neutron in the 1f7/2 shell,
which appears to be not completely true.
The shell structure for
43Ca
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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The Valence Nucleons
The study of the excited states of nuclei indicate
that the simple assumption of the EIPS is no more
valid.
The
EIPS
model
appears
to
be
an
oversimplification and one should take into
account all nucleons in an unfulfilled shell. i.e. we
should consider the 3 neutrons in the 1f7/2 shell in
the case of 43Ca .
In atomic physics we do the same thing. The
valence electrons are those of the last unfulfilled
shell. The valence of an ion is the number of
electrons or holes in the last shell
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
60
Mirror Nuclei
Two nuclei X & Y are said to be mirror
nuclei if they have the same A, and Z(X) =
N(Y). The special mirror converts a proton
into a neutron and vice-versa.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Mirror Nuclei
Consider the 2 mirror nuclei
(Z=9).
17O(
Z=8) and
17F
According to the shell model these nuclei are
composed of a “full” core of 8 nucleons and
one single unpaired nucleon in the 1d5/2 shell*.
The spin parity of the ground state of these 2
nuclei is that of the unpaired nucleon in the
+
5
1d5/2 shell, i.e
2
*A proton in the case of
17F,
a neutron in the case of
17O
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Excited States of the 17O and
17F
Nuclei
Measurement of the energies and spin-parity for
the excited states of the mirror nuclei 17O and 17F:
Excited state
Energy(MeV)
1st
0.90
2nd
3.05
3rd
3.85
4th
4.50
5th
5.10
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
SpinSpin-Parity
1
2
1
2
+
5
2
−
3
2
−
3
2
+
−
29
63
Excited States of Nuclei vs. Shell Model
According to the shell model, the unpaired
nucleon would gain or absorb
the
excitation energy and jump to a higher
level.
This assumption explains the spin parity of
1+
the 1st excited state ( ) and the 5th excited
2
3+
state (
)
2
But how to explain the other 3 (negative
parity) excited states?
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
64
17O
(17F
)
1st
excited state
The unpaired nucleon in the d5/2 shell absorbs the excitation
energy and moves to the nearest shell 2s1/2 .
1+
The spin parity of this 1st excited state is
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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17O
(17F ) 2nd excited state
One of the pair of nucleons in the p1/2 shell absorbs the
excitation energy and moves to the 1d5/2 shell forming a pair
in the 1d5/2 and leaving an unpaired nucleon in the p1/2 .
The pairing energy increases with l. It is energetically
favorable to break a pair with l =1 to from a pair with l =2
−
The spin parity of this excited state is 1
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
66
17O
(17F
)
4th
excited state
A paired nucleon in the p3/2 shell absorbs the
excitation energy and moves to the 2s1/2 .
Energetically,
The spin parity of this excited
−
state is
3
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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67
17O
(17F ) 5th excited state
The unpaired nucleon in the d5/2 shell absorbs the
excitation energy and moves to the 1d3/2.
+
The spin parity of this excited state is 3
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
68
The Negative Parity State 5/2
We successfully accounted for 4 of the mentioned
excited states. We still have some difficulty −in
explaining the occurrence of the excited state
5
2
A possibility for this state to occur is :
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
32
69
Valence Nucleons
Measured spinexcited state
spin-parity suggests that the
2
results from a combination of the spinspin-parity of the nucleons
involved in such a case,
5−
i.e. we have to take all 3 nucleons into consideration when
computing the spinspin-parity of the state resulting from the
migration of an “interne”
interne” nucleon to an upper level
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
70
Valence Nucleons
43Sc(
Z=21) &
43Ti(
Z=22) are another 2-mirror
nuclei example which illustrates the fact
that the EIPS model is not sufficient to
explain experimental measurements of
spin-parity of the excited states of nuclei.
Exercise: See Krane Fig. 5.12 (page 132).
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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71
The “Deep” Shells Nucleons
All properties we have seen until now concern the
outer nucleons, i.e. those near the surface. What
about the nucleons in the inner shells (1s1/2, 1p3/2
for example)?
A good idea to “measure” how valid is the single
particle model in the inner shells is to probe their
wave functions. We know that only the s ( = 0)
wave function can penetrate in the interior of a
nucleus. For the other wave functions ψ → 0 when r
→0
This means that we need to probe a nuclear
property of the interior of the nucleus with a probe
that can penetrate there!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
72
“Deep” Shells Nucleons vs. Shell Model
The distribution of charge is a good “internal”
internal” property and high
energy electrons are used as a probe (high energy electron
scattering).
The contribution of the shell s to the charge density is studied
using 2 nuclei (isotones=
isotones=same N)
N) :
205Tl (Z=81)
206
81) & Pb (Z=82)
82) , N = 124
The 205Tl (Z=81)
81) protons fill all shells except 3s1/2 where it lacks a
proton, whereas 206Pb (Z=82)
82) has all its shells until the magic
number 82 filled with protons
According to the shell model, the charge density difference
between these 2 nuclei is (must be)
be) due to the 206Pb extra proton
in the s1/2 shell.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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73
Probing the “Deep” Shells Nucleons
Theory means simply
the
square
of
a
harmonic oscillator 3s
wave function |Ψ
Ψ3s|2
Charge Density difference
(e/F3)
Fig. 4-7 shows the difference in charge density
between 205Tl and 206Pb.
Theory reproduces rather
well
the
experimental
results
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Radius (F)
74
Shell Model is valid…
Comparing the experimental results with those
expected (actually the probability density of
the 3s wave function of a SHO) gives satisfying
results and tends to confirm us in our
conviction that the shell model is appropriate
for explaining the difference in charge density
between 205Tl and 206Pb.
The Shell model is also valid when explaining
the nuclear properties such as spinspin-parity, em
moments, excited states and even transition
probabilities in nuclear reactions or decay
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
35
Lecture 11
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
3-7 Even-Even Nuclei
& Collective Structure
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-5
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Even-Even Nuclei
The study of the excited states of even-even
nuclei (even Z, even N) reveal a collective
behavior of these nuclei.
A good example is that of
130Sn
(Z=50, N=80).
According to the shell model the 50 protons
fill all shells until the magic number 50,
while neutrons lack the magic shell 82 by 2.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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77
130Sn
Energy Levels
neutrons
50
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
78
130Sn
Excited States
Fig. 3-6 : Energy levels for the excited states of
130Sn
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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79
130Sn
The Odd Parity Excited States
If we suppose that the neutron pair in the 3s1/2
subshell is broken and one of them moves to the
subshell 1h11/2,
We then expect that the resulting spin, according
to
angular
momenta
addition
in
quantum
mechanics will have a value I between |j1 - j2| (11/21/2 = 5 here) and j1 + j2 (6) by steps of 1, i.e. 5 < I < 6!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
80
130Sn
The Odd Parity Excited States
Another possibility would be that one of the
pairs in the 3d3/2 subshell is broken and a
neutron moves to the subshell 1h11/2, then we
expect that the resulting spin I between |j1 - j2|
(11/2-3/2 = 4 here) and j1 + j2 (7) by steps of 1, i.e.
4 < I < 7!
The transition involving an even-parity state
(s) and an odd one (h) the resulting parity is
odd and this can explain the spin-parity of
some of the excited states
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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81
Even-Even Nuclei vs. Shell Model
(130Sn)
It is also possible that the pair in the subshell h11/2 is broken
and the 2 neutrons re-couple again to form a pair with a total
spin (total angular momentum) different than 0.
In this case the resulting spin I will be between 11 and 0!, i.e 0
< I < 11
This excitation involves 2 identical particles which means that
the wave function of such an excited state should be
symmetrized, i.e its parity should be even which leaves us with
the possible spin-parity states 0+, 2+, 4+, 6+, 8+ and 10+.
These possible states are seen (Energy around 2 MeV) and this
leads us to believe that the shell model is valid.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
82
An exception: 2+ excited state at 1.2 MeV
For this exception we can imagine several
possibilities using the shell model, but obviously
the fact that the excitation energy of such a state
is below 2 MeV (which is somehow the necessary
energy for unpairing the neutrons in the different
subshells) indicates that this excited state has a
different origin!
We can always try to sophisticate the shell model
description to explain this state but studying other
even-even nuclei indicate that an anomalous 2+
excited state always appear at or below the energy
needed to form a pair. (Fig 3-7).
A reasonable explanation is possible if we consider
a collective property of these nuclei!
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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83
An exception:
2+
excited state at 1.2 MeV
Fig 5-15a (Krane)
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
84
Collective properties: Vibrations & Rotations
Experiments show that nuclei with A<150
should be treated using a model which
stipulates “vibrations” of the nucleus (as a
whole) about a spherical equilibrium shape.
For nuclei with 150<A<190, they seem to show
a different collective behavior: The nucleus
acts like a nonspherical system which
“rotates”.
Thus nuclear physicists
models: the vibrational
rotational model.
use 2
model
collective
and the
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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85
The Liquid Drop Model
The collective nuclear model in its classical
version is called the “liquid drop” model.
In this model, vibrations and rotations of a
nucleus look like those of a suspended drop of
liquid.
The mathematical frame, used in the treatment
of such a liquid drop, hydrodynamics, is
applied to the nucleus.
Meitner and Frisch were the first to use this
model to explain the fission of uranium
discovered by Hann and Strassmann in 1939.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Lecture 12
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
Nuclear Collective Models
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-6
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87
Nuclear Vibrations
Figure 4-8 shows a vibrating nucleus with a spherical
equilibrium shape.
Figure 3-8
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Representing the Vibrations of a liquid drop
The basic representation of a vibrating nucleus, which
has an average spherical shape is to give the
instantaneous coordinate R(t) of a point on the surface
θ,φ
φ). in terms of the spherical harmonics Yλµ (θ
θ,φ
φ).
at (θ
R(t) would be a linear combination of Yλµ (θ
θ,φ
φ) as
+λ
follows:
R (t ) = R av +
∑ ∑α
λµ ( t ) Y λµ
(θ , φ )
λ ≥ 1 µ = −λ
1) The constant term (Y00) is incorporated in Rav (R0A1/3).
2) The coefficients αλµ are not completely arbitrary.
Two kinds of restrictions are imposed on the α’s:
a) The reflection symmetry which requires that αλµ = αλ-µµ
b)The assumption that the nuclear fluid is incompressible
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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89
Rav
R(t) at θ,φ
φ
The average spherical shape
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Vibrational Modes, Phonons
Case λ = 1 : Dipole vibration (Fig 3-9-1)
This vibration mode cannot exist because its result is a
net displacement (in space) of the center of the
nucleus. Such effect (displacement) should come from
an external force other than the internal nuclear
forces.
Figure 3-9-1
Case λ = 2 : Quadrupole vibration(
vibration(Fig 3-9-2)
As in Quantum Electrodynamics a quantum of energy
is called a photon, in quantum hydrodynamics this
quantum (unit) of energy is called a phonon.
phonon.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Figure 3-9-2
43
91
Phonons
Producing mechanical vibrations means producing
phonons
(kind
of
“carriers
of
vibrational
mechanical energy”)
A single unit of λ = 2 nuclear vibration is thus a
quadrupole phonon.
phonon
A single unit of λ = 3 nuclear vibration would correspond to
the introduction of an octupole
upole phonon.
phonon
Figure 3-9-3
Equivalently we say that a quadrupole phonon carries 2 units of
angular momentum.
momentum. An octupole carries 3 units of angular
momentum, etc ...
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Explaining the
2+
excited state
92
What happens if we add a quadrupole phonon to a 0+ ground state
of an eveneven-even nucleus?
In this case a term containing Y2µµ (5 possible values for m)
m) is added
to the ground state’
state’s wave function.
This 0+ ground state “gains”
gains” an angular momentum l = 2 and the
spin--parity
resultant excited state has positive parity. The resulting spin
+
is 2 .
The phonon energy here appears as an adjustable parameter. It
cannot be predicted by this model*!
off the
This simple operation based on vibrations gives an explanation o
exception 2+ observed in the eveneven-even nuclei.
*This procedure is common in physics.
physics. A model is built with 1 or several free parameters
and experiment gives the best fit value for this or these parameters.
parameters. The most famous
example is the Higgs Boson in the Standard Model
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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93
Two phonons
If we add 2 quadrupole phonons to the 0+
ground state 2 terms Y2µµ are added to the
ground state’s wave function.
There are 25 possible combinations. This
number is reduced because of the fact that
phonons (as photons) have integer spins
(bosons) and this means that they should
have symmetric wave functions
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
94
Possible Excited States
The following table shows all possible 25 values
resulting from the combination of 2 phonons
µ1
-2
-1
0
+1
+2
-2
-1
-4
-3
-2
-1
0
-3
-2
-1
0
+1
+1
µ2
0
+1
+1
+2
+2
-2
-1
0
+1
+1
+2
+2
-1
0
+1
+1
+2
+2
+3
+3
0
+1
+1
+2
+2
+3
+3
+4
+4
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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95
Vibrational Excited States
µ = µ 1 +µ 2
can take one of the values :
-4, -3, -2, -1, 0, +1, +2, +3, +4
There is one possibility to have +4 (1
(1)
There is one possibility to have -4 (1
(1)
There are two possibilities to have +3
Combining (+1
(+1,+2) and (+2
(+2,+1), but because of the
symmetrization of the phonon wave function (The two
phonons are identical), the allowed combination is a
mixture of these 2 states.
states.
This leaves us with one allowed combination for µ = +3
+3
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
96
Vibrational Excited States
Three combinations give µ = +2
+2.
a) (1,1)
1,1) which is already symmetric and
b) (0,+2) and (+2,0
(+2,0)
2,0) which we have to symmetrize.
symmetrize.
A mixture of these 2 combinations is needed to give a
symmetric state.
state. This leaves us with 2 allowed combinations.
combinations.
µ
+4
+3
+2
+1
0
Allowed comb.
1
1
2
2
3
µ
-4
-3
-2
-1
Allowed comb.
1
1
2
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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Vibrational Excited States
Only 15 combinations are possible.
possible.
l=4
l=2
l=0
= +4
+4, +3, +2, +1 , 0, -1, -2, -3, -4
= +2
+2, +1 , 0, -1, -2
=0
Thus we expect a triplet state with spinspin-parity 0+, 2+
and 4+ with an energy as twice as that of the
previous 2+. Two phonons carry twice energy as
much as one
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
98
(An) Experimental Evidence
The 3- state is due to the octupole vibration.
3-phonon quintuplet
2-phonon triplet
Single phonon singlet
Figure 3-10: (low) Excited States of
120Te
47
99
Nuclear Rotations
In the range 150<
150<A<190 and A>
A>220 nuclear vibrations are no
more valid to explain the excited states.
Rotational motion is observed with nonspherical equilibrium
shapes. These “deformed nuclei”
nuclei” can have substantial
distortions from spherical shapes.
Using rotational notions from fluid mechanics, a deformed
nucleus is represented by an ellipsoid of revolution. The
surface of such a shape is represented by :
R(t ) = R av (1+ β Y 20 (θ, φ))
β=
4 π ∆R
3 5 R av
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
100
Equilibrium Shapes of Deformed Nuclei
The deformation parameter is related to the
eccentricity of the ellipse ∆R which is the
difference between the semi-major and the
semi-minor axes of the ellipse.
Using quadrupole moments two deformed
shapes appear:
The prolate shape corresponding to β > 0 and
the oblate shape corresponding to β > 0.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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101
Equilibrium Shapes of Deformed
Nuclei
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
102
The Prolate and Oblate Shapes
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
49
103
Rotations vs. Electric Quadrupole Moments
A large electric quadrupole moment would
indicate a stable deformation of a nucleus.
The relation between β and the quadrupole moment
Q is:
3
2
Q0 =
5π
Rav Z β (1 + 0.16 β )
The subscript 0 refers to a frame attached to the
nucleus. The measured value Q in the laboratory
is different because the nucleus is rotating!
See Appendix 2* for a detailed explanation of the
relationships between Q0, Q and β.
* See also Krane pages 143143-149.
149.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Lecture 12
Phys. 641: Nuclear Physics
Physics Department
1
Yarmouk University 21163 Irbid Jordan
Chapter 3 - Problems
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys641/Lec3-6
50
105
Problems
Problem 1 :
Expected shell model Spin-parity of some
nuclei
a) 7Li
:Z=4,N=3
b)
11Be
:Z=5,N=6
c)
15C
:Z=6,N=9
d)
31P
: Z = 15 , N = 16
e)
141Pr
: Z = 59 , N = 82
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
106
Problem 1: spin parity of 7Li
a) 7Li : Z = 3 , N = 4
protons
neutrons
The 4 paired neutrons do not contribute to
the spin-parity of the nucleus
The spin-parity of 7Li is that of the 3rd
unpaired proton in the 1p3/2 shell, thus
3−
π 7
I Li =
2
( )
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
51
107
Problem 1: spin parity of
a)
11Be
11Be
:Z=5,N=6
protons
neutrons
The 6 paired neutrons do not contribute to the
spin-parity of the nucleus.
The spin-parity of 11Be is that of the 5th
unpaired proton in the 1p3/2 shell, thus
3−
I ( Be) =
2
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
π 11
108
Problem 1: spin parity of
c)
15C
15C
:Z=6,N=9
protons
neutrons
The spin-parity is that of the 9th unpaired neutron
in the 1d5/2 shell, thus
I
+
5+
C =
2
( )
π 15
1+
The measured value is
. This indicates that
2
the 9th neutron is rather in the 2s1/2 shell
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
52
109
Problem 1: spin parity of
d)
31P
31P
: Z = 15 , N = 16
The spin-parity is that of the unpaired neutron in
the 2s1/2 shell, thus
I
+
1+
P =
2
( )
π 31
The shell model prediction agrees well with the
measured value.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
110
Problem 1: spin parity of
d)
141Pr
141Pr
: Z = 59 , N = 82
The neutrons fill all shells until the magic number 82.
The spin-parity is that of the unpaired proton in
the 2d5/2 shell, thus
I
(
π 141
5+
Pr ) =
2
The shell model prediction agrees well with the
measured value.
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
53
111
Problems
Problem 2 :
Expected magnetic dipole moments of
some nuclei
a) 71Ge
: Z = 32 , N = 39
c) 47Sc
: Z = 21 , N = 26
b) 75Ge
: Z = 32 , N = 43
© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
112
Magnetic dipole moment vs. j
for j = + 1
2
<µ> = µN [g (j -
for j = - 1
2
1
1
) + gs ]
2
2


 j+ 3
j




2 1 j

< µ > =  gl
−
gs  µ N
j+1
2 j+1 



© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
54
113
m.d.m of
b)
71Ge
71Ge
: Z = 32 , N = 39
The m.d.m of 71Ge is that of the unpaired neutron,
number 39, (gl =0, gs= -3.826837) in the 2p1/2 (j = 1 - 1/2)
shell.
 1 12

< µ > = − ×
( − 3.8260837 ) µ N = + 0.463 µ N
 2 3 2

© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
114
m.d.m of
b)
75Ge
75Ge
: Z = 32 , N = 43
We have seen that the m.d.m for a nucleon in the level
j=+
1
2
is given by :

< µ > =  gl

1 1 

 j −  + gs  µN
2 2 

for the unpaired neutron (g =0, gs= -3.826837) in the
g9/2 shell :
1
< µ > =  + ( − 3.8260837 ) µ N = − 1.913 µ N
 2

© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
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115
m.d.m of
47Sc
c) 47Sc : Z = 21 , N = 26
The m.d.m of 47Sc is that of the unpaired proton in the
1f7/2 shell (j=7/2, = 3)
We have seen that the m.d.m for a nucleon in the
level
for j = +1
2
1
1
<µ
µ> = [g (j - ) + gs ] µN
2
2


7 1
1
< µ > =  1 ×  −  + ( 5.5856912 ) µ N = 8.793 µ N
  2 2 2

© Dr. N. Ershaidat Phys. 641 Chapter 3: Nuclear Models (Review)
Next Lecture
Chapter 44
Chapter
Nuclear Reactions
Reactions
Nuclear
End of Lecture 12
56