___________________________________________________________
UNIT 1 BASIC PROPERTIES OF NUCLEI
Structure
1.0
Introduction
1.1
Objectives
1.2
Methods for determination of nuclear size and their interpretations
1.2.1 Rutherford alpha scattering method
1.2.2 Mesonic X-rays method
1.2.3 Electron scattering method
1.2.4 Mirror nuclei method
1.3
Binding energy curve for nuclei and its consequences: Numerical
problems
1.4
Nuclear spin, magnetic and quadrupole moments of nuclei
1.5
Semiemprical mass formula and its application to mass parabolas.
1.6
Mirror nuclei and isotopic spin formalism
1.7
Let us sum up
1.8
Check your progress : The Key
___________________________________________________________
1.0 INTRODUCTION
The discovery of the atomic nucleus the central massive core of the atom was
made by Lord Rutherford in 1911 through his  - particle scattering experiments in
which positively charged  - particles (doubly ionized thelium atoms ) were scattred
by thin metallic foils. In a bid to account numerically for large angle scattering of  particles. Lord Rutherford contended that the scattering was due to electrostatic
interaction between the positive charge of the  - particle and a concentrated
positively charged region within the atom. On grounds of symmetry, this positively
charged region within the atom was taken to be spherical and located at the center of
the atom and was named ‘Nucleus’. The radius of the nucleus was established to be of
the order of 10-12 or 10-8 cms.
The atomic nuclei possess following important properties:
1-
All nuclei are positively charged and the magnitude of this charge is an
integer, Z times the charge of an electron and is of opposite sign.
2-
More than 99.99% of the mass of the atom is concentrated within the
small volume of the nucleus.
Properties of Nuclei and Scattering
3-
The existing experimental evidence supports the view that within the
nucleus, the distribution of the positive charge is uniform i.e. atomic
nuclei are spherical in shape.
A number of experimental evidences lead to the conclusion that ‘nuclear volume’ is
substantially proportional to the number of nucleons (protons an neutrons) in a given
nucleus. The number of nucleons in a nucleus is denoted by A and represents the mass
number. In the constant density model of the nucleus, the nuclear radius is given by
R = R0A1/3
Where R0 varies slightly from one nucleus to another but is roughly constant for
A > 20. The present experimental evidence shows that R0 = (1.2± 0.1) × 10-13 cm. The
variation in R0 is from 1.2 to 1.5 fermi (1 fermi= 10-13 cm).
___________________________________________________________
1.1 OBJECTIVES
The main aim of this unit is to study the basic properties of the nuclei. After going
through this unit you should be able to:

Know the basic properties of nuclei Viz., its size, spin, quadrupole moment,
binding energy etc.
 Study various method used to determine the size of the nucleus. How the
Rutherford alpha scattering method provides satisfactory results for nucleus
radius ( r0 ~10-14 m).

Understand the deviation from spherical symmetry by studying the quadrupole
moment of the nuclei.

Derive the semi empirical mass formula which comes by adding the various
correction term in binding energy expression and the mass parabolas for
isobaric nuclei.
___________________________________________________________
1.2 METHODS FOR DETERMINATION OF NUCLEAR SIZE
AND THEIR INTERPRETATIONS
1.2.1 Rutherford alpha scattering method
The theory of  - particle scattering was first developed by Rutherford in 1911. When
 - particles strike matter, some of them scatter to a greater angle while others to a
2
Basic Properties of Nuclei
small angle. Rutherford put forward an explanation of this large angle scattering by
assuming that
(i)
the scattering is due to a single encounter between the  -particle and
the atom,
(ii)
the whole mass of the atom is concentrated in the core called nucleus,
(iii)
the nucleus is a positively charged body of the size of  10-13cm.
(iv)
The  -particle penetrates very close to the nucleus until the repulsive
force on it becomes very large and the particle is scattered to large
angle .
(v)
The  -particle when near the nucleus, is relatively very far from the
negative charges which are spread over a much larger volume so that
the attractive forces exerted by the electrons are negligible.
Let us consider an  -particle moving along PO approaching the heavy nucleus which
is stationary at S which is the external focus for the hyperbola. The particle is
deflected along OP’.
P
Fig. 1.  - particle scattering
P’
Let SN-p, the perpendicular drawn from s on the direction of the incident particle . (p
is also known as impact parameter). Let E,M and V be the chare mass and velocity
(initial) of the  -particle and Ze be the charge on the nucleus –Z. being the atomic
number. The velocity of the  - particle varies with its path and the velocity at A-the
closest point of approach to the nucleus. Is v. The particle is moving perpendicular to
SO when at A. Hence from the principle of conservation of angular momentum we
have
M.V.p=M.v.SA
(1)
And from the law of conservation of energy we have the initial kinetic energy equal to
the sum of potential and kinetic energies at A so that
3
Properties of Nuclei and Scattering
1
1
Ze.E
MV 2  Mv 2 
,
2
2
SA
b 

 V 2 1 

 SA 
v2  V 2 
Where b 
2 Ze.E
SA.M
2 Ze.E
MV 2
From equation (1). we have
p  SA.
v
v2
or p 2  SA2 . 2
V
V
p 2  SA2 .
b 

1 

 SA 
 SA( SA  b).
On solving it we get,
 b  2 p cot
(   )

 2 p tan . ,
2
2

2p

 cot .
b
2
(2)
Hence the number N’ of  -particles striking unit area of the screen at an inclination 
from the incident direction is given by
1 2
b ntQ. cot( / 2). cos ec 2 ( / 2).d
,
N' 4
2r 2 sin .d
Substituting b 

Qntb 2

cos ec 4
2
16r
2
2Ze.E 2 4Z 2e 2 .E 2
,b 
,
MV 2
M 2V 4
N'
Qnt ( Ze) 2 .E 2

cos ec 4 .
2
2 4
4r M V
2
(3)
Equation is known as Rutherford scattering formula which states that the number of
 -particles N’ striking unit area of a fluorescent screen at a distance r from the point
of scattering must be proportional to

2
(1)
cosec4
(2)
The thickness t of the scatterer.
(3)
The square of the nuclear charge (Ze)2.
(4)
Inversely to the square of the initial kinetic energy.
The results of Rutherford scattering of  -particles are important because the
quantitative information about the size of the nucleus was first obtained with its help.
The quantity b 
2 Ze.E
defined above gives the distance of closest approach of the
MV 2
 -particle to the nucleus. Let us calculate the value of b for copper.
Z for copper = 29,E for  -particle =2e=2×4.8×10-10 e.s.u.
4
Basic Properties of Nuclei
M for  -particle = 4×1.67 × 10-24 gm.,V for  -particles from Radon = 1.6×109
cm./sec.
4  29  (4.8 1010 ) 2
b
4 1.67 1024  (1.6 109 ) 2
 1.7 10 12 cm.
This therefore, suggests that the radius of the nuclei is of the order of 10-12 cm.
1.2.2 Mesonic X-rays method
In this method the probing particle is a meson (µ) . The advantage of the µ meson
over the electron is its larger mass (about 207 times the mass of the electron) which
allows the µ-meson to penetrate the nucleus long before it decays or is captured by the
nucleus.Fitch and Rainwater conducted the experiment using a beam of 385 Mev
protons to produce negative pions of kinetic energy 110 Mev and the beam of pions of
required energy and charge was selected by magnetic analyzer. Positive pions decay
as follows having a life time of only -10-8 secs.
     
   e   v  v.
(v represents anti-neutrino)
The beam of positive pions becomes mixed with muons by this decay process. The
mesons can be slowed down by a suitable thickness of the absorber so that the muons
are brought to rest in the target nuciei.
The negative  -mesons are specially interesting. If they are not captured by a
nucleus, they decay into negative muons as fellows:
      v.
The negative muon has the opportunity of being slowed down by ionizing collisions
to a substantially thermal velocity and then of being captured by a nucleus. In 1947
Fermi and Teller suggested the possibility of existence of ‘Mesonic atoms’. The
capture produces is supposed to proceed as follows:
The Bohr-like orbits will be much closer to the nucleus than the electron orbits due to
the much greater mass of the mesons. For a one- electron atom, the radius and the
energy of the stationary orbits is given by.
n2h2
mZe2
mZ 2 e 4
E 2 2
2n h
r
5
Properties of Nuclei and Scattering
As the  -meson mass is 207 times of the electron mass. The muon orbit is 207 times
less than the electron orbit and , on the other hand, the energy of the bound muon is
207 times greater than that of the corresponding bound electron.
It is a well known fact that interaction between muons and the nucleus is Colombian
in character. Magnetic moment measurements of muons confirm that it is a Dirac
particle of spin
1
2
. The de Broglie wavelength of bound muon is much shorter than
the wavelength of a bound electron. Hence it can be localized more sharply than the
electron. The results for some important elements are listed below:
TABLE – 1
S.No.
Element (Z)
Calculated
Observed
R0
energy
energy
(in Fermi)
1.
Ti
22
1.045
0.955
1.17
2.
Cu
29
1.826
1.55
1.21
3.
Sb
51
5.83
3.5
1.22
4.
Pb
82
16.41
6.02
1.17
Thus the radius constant R0 determined from muon X-ray observations comes out to
be 1.20 ± 0.03 fm.
1.2.3 Electron scattering method
In this method elastic scattering of high energy electrons is studied and from the
knowledge of the scattered electrons the radius constant R0 is computed. The reduced
  
de Broglie wavelength  
 of a 200 Mev electron is about 10-13 cm. And
 2 
therefore such particles which interact strongly with electric charges are very useful
for probing the nucleus. The experiment was performed by Hofstadter and others at
Standford University.)
The electron beam from a linear accelerator after momentum analysis is allowed in
the scattering chamber with the help of two deflecting magnets. In the center of the
scattering chamber, the beam passes through a thin gold foil (or any other scattering
material) and the electrons scattered by the electric fields of the nuclei then fall on a
magnetic spectrometer when they again undergo momentum analysis to record only
the elastic events. The spectrometer can be rotated about an axis through the target so
that the intensity of scattered electrons can be observed as a function of angle  .
6
Basic Properties of Nuclei
In the electron-nucleus collision, electron loses kinetic energy and nucleus recoils.
The energy of the recoil nucleus En can be calculated gy treating the electron-nucleus
to be similar to a photon-electron Compton collision.
If
En = Energy of recoil nucleus. E = energy of incident electron,
M = Mass of the nucleus,
 = Scattering angle.
Then from the theory of Compton effect
En 
E2
(1  cos  )
.
2
Mc 1  E (1  cos  )
Mc 2
E’ = E – En.
The energy of scattered electron E is thus given by
Hofstadter and others assumed that the density of charge in the nuclcus is best given
by
P(r ) 
Po
,
1  e k ( r e )
Where p0 and k are constants and c is that value of r where p(r) falls to half its central
value (i.e. p(r)=p0/2). The experimental results with gold were analysed in terms of the
above equation and the results for a wide range of nuclei indicated c to be
proportional to A1/3 and it was found that the results fit well within the formula
R-RoA1/3.
With value of Ro as Ro = 1.32 fermi for A < 50
And
Ro = 1.32 ± 0.01 fermi for A > 50.
1.2.4 Mirror nuclei method
Two nuclides having the same number of nucleons but the number of protons in one
of them being equal to the number of neutrons in the other are called Mirror Nuclei.
For example .
3
1H
3
2He
7
7
4Be
13
6C
13
7N
39
19K
39
20Ca
9
4Be
9
5B
11
4B
6C
3Li
11
Let us consider the mirror pair of nuclei 6C13 and 7N13 . The first member has 7
neutrons and 6 protons while the second one has 7 protons and 6 neutrons.
Let us begin with Z=N nucleus (e.g. 6C12). If we add one proton to such a nucleus we
get one member of the pair formed (7N13). While the addition of one neutron will give
another member(e.g. 6C13). The only difference in the binding of the nucleus for the
7
Properties of Nuclei and Scattering
mirror pair nuclei C13, N13 is that in the case of C13, there are 6 n-n bonds in place of 6
p-p bonds in N13. If we accept the principle of charge symmetry according to which
the nuclear force between a pair of protons is the same as the force between a pair of
neutrons in the same state, than there should be no difference in the nuclear binding
forces of a pair of mirror nuclei. The total binding energy of the two nuclei will not be
the same, however, because of the difference in the coulomb self-energy.
The coulomb energy of a spherical nucleus with a uniform charge distribution could
be calculated as follows:
In a nucleus of charge no. Z and mass number A, Z nucleons are protons, each of
which carries a charge +e. Thus the total charge is +Ze. If we assume that the protons
are uniformly distributed within the nuclear sphere, we can readily calculate the
Coulomb energy through the Gauss’s theorem in electrostatics. We treat the nucleus
as a sphere of radius R=RoA1/3 with a charge q=Ze.
The charge density p 
q
3 Ze
 .
.
4 3 4 Ro 3 A
R
3
The electrostatic energy is simply the work done against electrostatic forces in
assembling such a sphere. Let us suppose we have already assembled a sphere of
radius r and wish to ass on a shell of thickness dr which contains the charge dq
dq = charge density × volume of the shell
 p  4r 2 dr.
Treating the entire charge of the sphere to be concentrated at the center, we calculate
the work done in bringing dq from infinity to r.
4
1
 p  r 3 . .( p  4r 2 dr )
3
r
dEc= potential × charge ,
Integrating this within the limits r-0 to r=R to calculate the total work done in
R
4
1
assembling the sphere we get, Ec   p. r 3 . .( p  4r 2 dr ) ,
3
r
O
Ec 
3 Z 2e 2
.
5 R
If on the other hand each proton remains aloof and discrete entity inside the nucleus
and interacts electrostatic ally with all other protons but not with itself, then the
coulomb energy would be
Ec 
3 e2
Z ( Z  1).
5 R
For a pair of mirror nuclei of radius R, charges Ze and (Z-1) e, the coulomb energy
difference is
8
Basic Properties of Nuclei
Ec 
3 e2
3 e2
[( Z  1) 2  Z 2 ] 
(2Z  1).
5R
5 R
If the mirror nuclei have 2Z±1=A; equation becomes
3
Ec  5
e2
3 e2
.A 
A
R
5 Ro A1/ 3
Ec 
Or
3 e2 2 / 3
.A
5 Ro
From this relation, it is clear that when a graph is plotted between Ec and A2 / 3 it
should give a straight line and from its slope the value of Ro can be calculated. The
value of Ro so calculated come out to be 1.3 × 10-13 cm.
This value is slightly higher than that obtained by high energy electron scattering
method. The energy discrepancy is because of the use of classical principles instead of
quantum mechanical principles in calculating the coulomb energy Ec..
The following table gives the value of Ro for some important nuclei.
TABLE -2
Nucleus
Ro (in fermis)
B11
1.28
C13
1.34
15
N
1.31
O17
1.26
19
F
1.26
Ne21
1.25
23
Na
1.22
Mg23
1.23
27
Al
1.20
Obvously, the mean value of Ro for the remaining mirror nuclei is given by Ro =
1.23±0.3 fermi.
Check Your Progress 1
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1) Assuming that 1 amu = 1.66 1027 kgm and the radius of a nucleus to be
given by R  R0 A1/ 3 where R0  1.2  10 15 m , calculate the density of nuclear
matter.
2) Describe in detail any method for determining the size of the nucleus.
..........................................................................................................................
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9
Properties of Nuclei and Scattering
1.3 BINDING ENERGY CURVE FOR NUCLEI AND ITS
CONSEQUENCES: NUMERICAL PROBLEMS
We have read that when nucleons (protons and neutrons) combine to form the
nucleus, the actual mass of the nucleus is slightly less than its expected mass i.e.,
atomic mass M is less than its mass number A. In other words, there is some mass
defect  M = A – M, when nucleons combine to form the nucleus. Actually this loss in
mass of the nucleus is released in form of energy during the formation of the nuclei,
according to Einstein’s mass-energy equivalence relation E=(  M) c2. This energy is
called the binding energy of the nucleus. Obviously if we want to release all the
nucleons present inside the nucleus, this much amount of external energy is required
for the nucleus. Hence binding energy of a nucleus is the energy released in formation
of nucleus by combining its nucleons. In other words, it is that external energy which
is required to isolate the nucleons from each other . i.e.,
Binding energy
B =(  M ) c2
But from eqn. Mass defect for the nucleus zXA is
 M = {Z × mp + (A-Z) × mn} –M
Binding energy
B = [{Z × mp + (A-Z) × mn} –M] c2
Here Z is the atomic number. (i.e., number of protons inside the nucleus), A is the
mass number (i.e., the total number of nucleons), mp is the mass of proton, mn is the
mass of a neutron, M is the actual mass of the nucleus (or atomic mass) and c is the
speed of light.
Hence binding energy per nucleon of the nucleus (in a.m.u.)
B
B Z
M
 Z
   m p  mn  1    mn  a.m.u.
A A
A
 A
Z

  m p  mn   mn  1  f  a.m.u.
A

Where f 
M
 1 is the packing fraction.
A
But mp=1.00728 a.m.u. and mn =1.00867 a.m.u.
 Z

Hence B    0.00139  0.00867  f a.m.u.
 A

Remember that 1 a.m.u. = 167× 1027 kg
= 167× 1027 ×(3×108)2 joule ,
= 931 MeV
10
Basic Properties of Nuclei
For example, in formation of a deuteron (1H2) nucleus, there is 0.002397 a.m.u. mass
defect. Hence binding energy of deuteron nuclei
=0.0002397 × 931 MeV = 2.23 MeV.
Binding energy curve:
It is clear that the binding energy per nucleon B is maximum when the packing
fraction f is minimum. Since for the nuclei from A>20 to A<200, the packing fraction
f is negative, hence the binding energy per nucleon for these nuclei is more. Higher
the binding energy per nucleon, more stable is the nuclei. For the nuclei from Ca40 to
Sn110 , the value of Z/A changes from 0.50 to 0.42 i.e., the mean value of Z/A is 0.46
and packing fraction in this range is nearly – 6×10-4 a.m.u. per nucleon.
Hence binding energy per nucleon in this range is
B
= 0.00863 a.m.u. = 0.00863 × 931 = 8.03 MeV
Fig. 2. Binding energy curve
Graph shows between the binding energy per nucleon B for the different nuclei
versus their nucleon number A. This is called the binding energy curve.
From the graph, it is clear that
(1)
Except for the nuclei He4, Be8 , C12 , O16 and Ne20, the binding energy
curve for tall other nuclei is generally continuous and linear. The
binding energy per nucleon of some light nuclei such as H1, H2 and
He3 is very low, but for all the other nuclei, binding energy per nucleon
is nearly 8 MeV.
11
Properties of Nuclei and Scattering
(2)
The binding energy per nucleon B is positive for all the values of
mass number A. It means that all the nuclei are stable i.e., the nuclear
(3)
forces acting between the nucleons found inside the nuclear volume are
attractive, but as the nucleus does not collapse, therefore at very short
distances, the nuclear forces between the nucleons become repulsive.
(4)
For the higher nuclei, the binding energy per nucleon B changes very
irregularly form one nuclei to other nuclei and its value for the nuclei
4
8
12
16
20
2He , 4Be , 6C , 8O , 10Ne
is more as compared to that for the other
nearby nuclei (i.e., for these nuclei there are peaks obtained in the
binding energy curve).
(5)
For the nuclei of mass number A>20, the binding energy per nucleon
B is nearly constant (=8.04 Mev). It concludes that inside the nucleus,
the nucleons exert forces on the other neighboring nucleons only in a
limited number i.e., the nuclear forces are saturated .
Near the mass number A = 56, the binding energy per nucleon B is
(6)
maximum (nearly 8.8 Mev) and then it gradually decreases with
increase in mass number A and ultimately for the heavy nuclei of mass
number A>200 (such as uranium etc.), its value becomes nearly 7.6
Mev. It shows that the nuclei near the mass number A = 56 are
relatively more stable and the nuclei of mass number A>200 are
relatively less stable (i.e., they are radioactive).
(7)
When the number of protons or neutrons inside the nucleus is 2, 8, 14,
20, 28, 50, 82, 126, the binding energy curve becomes discontinuous.
These numbers 2, 8, 14, 20, 28, 50, 82, and 126 are called the magic
numbers.
Ex. 1 Calculate the nuclear energy per nucleon and packing fraction for helium
nucleus (2He4).
(mp = 1.007285 a.m.u., mn = 1.008665 a.m.u., mass of helium nuclei=
4.0083873 a.m.u. ).
Sol.
nuclei.
Helium nuclei contains 2 protons and 2 neutrons. Hence mass defect in helium
 M = (2×mp + 2 × mn ) – M
= ( 2 × 1.007285 + 2 × 1.008665 ) – 4.0083873
12
Basic Properties of Nuclei
= 4.0319 – 4.0083873 = 0.02351127 a.m.u.
Hence binding energy (or nuclear energy ) or helium nuclei
= 0.0235127 × 931 MeV = 21.890324 MeV
Nuclear energy per nucleon =
And Packing fraction f
21.890324
MeV = 5.473 MeV (nearly).
4
M
4.0083873
1 
 1  2.968  10 3
A
4
Ex. 2. Calculate : (i) mass defect, (ii) packingfraction , (iii) binding energy, and (iv)
binding energy per nucleon for nickel Ni (Z = 28, A = 64) nuclei Given : atomic mass
of
28Ni
64
= 63.9280 a.m.u. mp = 1.007285 a.m.u. mn = 1.008665 a.m.u., me =
0.000550 a.m.u., 1 a.m.u. = 931 MeV.
Sol. Number of protons in nickel Ni 9Z = 28, A = 64 ) atom is Z = 28 , number of
neutrons N = A – Z = 64 – 28 = 36 and number of electrons Z = 28.
Mass of nickel nucleus
ZM
A
= mass of nickel atom – mall of 28 electrons
= 63.9280 – (28 × 0.000550 ) = 63.9126 a.m.u.
(i)
Mass defect
 M = [Z × mp + (A – Z) mn]- zMA
= [28 × 1.007285 + 36 × 1.008665 ] – 63.9126
= [28.20398 + 3631194] – 63.9126 = 0.60332 a.m.u.
f
M
63.9126
1 
 1  0.0013656 = -13.656 × 10-4
A
64
(ii)
Packing fraction
(iii)
Binding energy B =  Mc2 = 0.60332 × 931 MeV = 561.7 MeV
(iv)
Binding energy per nucleon B 
B 561.7

 8.78MeV
A
64
_____________________________________________________________________
1.4 NUCLEAR SPIN, MAGNETIC AND QUADRUPOLE
MOMENTS OF NUCLEI
When a charge particle moves in a closed oath it produces both angular momentum
and a magnetic field. The magnetic field at large distances may be described as due to
magnetic dipole located at the centre of current loop. Thus the orbital and spin angular
momenta of protons produce extra nuclear magnetic field which can be assumed as
due to magnetic dipole located at the centre of the nucleus.
13
Properties of Nuclei and Scattering
Consider a charge particle of mass M and charge e, When charge revolves in a
circular orbit it is equilent to a current of the strength
i
e
2
(1)
Where  is angular frequency of revolution. The magnetic field of this current is
equilent to that of a magnetic dipole moment of the value
  r 2i
(2)
Where r is the radius of the circular path. The orbital angular momentum of the
particle is
 l (l  1) and its component in the direction of magnetic field is
quantised. Thus the angular momentum is
Mr 2  ml 
(3)
Where ml is the projection of l in the field direction. The maximum value of
magnetic moments along the field direction due to orbital motion is given by ml = l
 max (orbital ) 
el
2M
(4)
and due to intrinsic spin
 max ( spin ) 
es
M
(5)
The spin angular momentum is twice as large as the expected value because spin
frequency is double that of orbital frequency. As the total angular momentum of the
nucleus is contribution due to orbital and spin motions and that it why is the magnetic
dipole moment is conveniently written as    N g N I
where g is in the nuclear g factor and   e / 2 M p , nuclear magneton with Mp as

proton rest mass,  is vector in a direction of I.
Determination of nuclear magnetic moment
The methods based of (1) hyperfine structure of spectral lines (2) alternating
intensities in molecular spectra (3) microwave spectra (4) magnetic resonance and
deflection of atomic and molecular beams (5) nuclear magnetic resonance, in bulk
and (6) optical detection of nuclear magnetic resonance are employed to study the
mechanical and magnetic moments of nuclei. One of the most common method is
14
Basic Properties of Nuclei
Rabi’s Magnetic Resonance Method:
In the presence of the magnetic field the nuclear spin precesses round the field
direction and this precession results into splitting of an energy levels into (2l+1)
levels. The energy of each of these magnetic energy is given by
E  gl M l  N B0
(6)
Where B0 is the magnetic field present and Ml can assume any of the values from Ml
= I to – I with a difference of one. The selection rule involved for nuclear spin
quantum number Ml is M l  1 and energy difference involved in any such transition
may be represented as
h  g l  N B0
(7)
In Rabi’s method the value of  is estimated by establishing resonance with
oscillating external magnetic field and ther by the value of magnetic moment for the
Lithium nucleus is determined from the following relation.
  gl  N I
(8)
A schematic arrangement of the experimental set up used for the determination of
nuclear magnetic moment is shown in the fig. The LiCl sample is first vaporized in
the electric oven O. The vaporized molecular beam emerging out from the electric
oven O is collimated with the fine slit system S. It then enters a highly evacuated
chamber where flux of the beam and pressure are both low so that there is no
significant scattering of the molecules of the beam.. The molecular beam will be
recorded by detector D.
Fig. 3. Rabi’s magnetic resonance method
A fine , hot tungsten wire is used as a detector for LiCl. The work function of
tungsten is greater than the first ionization potential of alkali atoms and when a
15
Properties of Nuclei and Scattering
neutral lithium atom strikes the hot tungsten wire, its balance electron sticks to the
wire and residual positive ion is left free. Thus molecular beam reaching the wire is
detected by ionic current.
Thus the value of  is known from relation (7) and value of magnetic moment of the
nucleus is calculated from relation (8) with known values of gl =2.167 and I=3/2 for
Li nucleus. his method gives magnetic moment for the nucleus the value 3.250
nuclear magneton.
The magnetic moment is computed from the expectation value of the magnetic
moment operator in the state with maximum z projection of angular momentum.The
expectation value of < Sz > can be quickly computed by recalling that j is the only
vector of interest in this problem – the l and s vectors are meaningful only in their
relationship to j. The instantaneous value of Sz varies, but its component along j
remains constant. Thus for
j = l+ ½,  S z   / 2 ,
while for
j = l-1/2
we have
 S z  j / 2( j  1)
.
Fig. 4.
Schmidt lines
Fig.4. shows a comparison of calculated values of magnetic moments with measured
values for shell-model odd-A nuclei. The computed values are shown as solid lines
and are known as the Schmidt lines after the name of the inventor scientist Schmidt.
16
Basic Properties of Nuclei
Quadrupole moments
Till the discussion we have assumed that nucleons inside the nucleus are so
distributed that spherical symmetry is maintained in them i.e., the shape of the nucleus
is spherical . But actually the shape of the nucleus is not spherical, but it is elliptical.
If the positive charge of the nucleus is distributed perfectly spherical, its quadruplle
moment should be zero. If the quadrupole moment of nucleus is positive, it means that
the charge distribution inside the nucleus is elliptical with major axis along the spin
axis and if the quadrupole moment is negative, it means that the minor axis of
elliptical charge distribution inside the nucleus is along the spin axis. Generally the
quadrupole moment of all the nuclei (except for the nuclei of some mass numbers) is
found to be positive.
Calculations of Quadrupole Moment of Nucleus
Fig. 5. Charge distribution in a nucleus
Let charge distribution inside the nucleus is not completely symmetrical i.e., the
charge of nucleus is not at the centre of the nucleus O, but it is at any other point A,
which has coordinated (x,y,z) from the centre O of the nucleus and distance OA = r
Obviously, the electric potential at the point P on Z-axis , which is outside the nucleus
at a distance R from the centre O of the nucleus , due to charge (+q) from the nucleus
is
Vp 
1
q
4o R0
(9)
Where R0 = distance of point P from the charge centre A of the nucleus .
It is clear that ‘R02 = R2 + r2 – 2rR cos 
17
Properties of Nuclei and Scattering
 2r
r2 
R0  R 1  cos   2 
R 
 R
Or
1/ 2
1
1  2r
r2 
 1  cos   2 
R0 R  R
R 
, Or
1 / 2


 1  3 
    
2
2 2


1
1
r 
r 
2  2    2r
 1  2r

 1      cos   2   
cos   2   ........
R0 R   2  R
R 
2
R 

 R



Or

1 r
r 2  3 cos 2   1  r 3  5 cos 3   3 cos  

  3 
  .......
1

cos



2 
2 R
R 
2
2
 R 


From eqn.no.(9), potential at the point P
Vp 
 r

r 2  3 cos 2   1  r 3  5 cos 3   3 cos  

  3 
  .......
1

cos



2 
40 R  R
R 
2
2
 R 



q

q
40 R
q
40 R

qr cos 
qr 2  3 cos 2   1 



40 R 2 40 R 3 
2


qr cos 
qr 2  3 cos 2   1 
qr 3  5 cos3   3 cos  


  .......


 4 R 4 
40 R 2 40 R3 
2
2



0
In the above equation, the first term represents the potential due to a single charge,
hence the coefficient of
1
40
(i.e., total charge of the nucleus q = Ze, where Z is the
total number of protons inside the nucleus or it is the atomic number of the nucleus) is
called the monopole moment. In the second term, since r cos  = Z component of the
distance OA, therefore the coefficient of
1
40 R 2
is qr cos  = charge × Z
component of distance. Hence qr cos  , the coefficient of
1
40 R 2
is called the dipole
moment and the second term represents the potential due to the dipole. Similarly the
third
term
is
the
potential
due
to
quadrupole
and
the
coefficient
 3 cos 2   1 
1
of
is called the quadrupole moment.
qr 2 
3
2

 40 R
Thus quadrupole moment of the nucleus in Z-direction at the point P(0, 0, R) is
Q


qr 2 3 cos 2   1
,
2
Substituting
cos  
z
,
r
18
Basic Properties of Nuclei
qr 2
we get Q 
2
  z  2  q3z 2  r 2 
3   1 
2
  r 

Special Cases
(i)
(ii)
(iii)
If total charge of the nucleus is on Z-axis at a distance equal to the
radius R of the nucleus (i.e., x=y = 0, z = r = R),the quadrupole
moment of the nucleus will be Q = qR2.
If the charge of the nucleus is on nuclear equator (i.e., z = 0 and r
1
= R ), the quadrupole moment of the nucleus will be Q =  qR 2 .
2
If the charge of the nucleus is spherically symmetric i.e., is at the
centre O of the nucleus (i.e., z = 0, r = 0), then its quadrupole
moment will be zero.
Fig. 6. Shape of nucleus according to the quadrupole moment
If the nuclear charge is elongated in the direction of spin axis , the quadrupole
moment will be positive and if the nuclear charge is flat in the direction of spin axis
quadrupole moment is negative.
Check Your Progress 2
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1)
Find the energy release, if two 1 H 2 nuclei can fuse together to form
2
He 4
nucleus. The binding energy per nucleon of 1 H 2 and 2 He 4 is 11 Mev and 7.0
Mev respectively.
b) What do you mean by quadrupole moment of a nuclei. Explain in detail.
..........................................................................................................................
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19
Properties of Nuclei and Scattering
_____________________________________________________________________
1.5 SEMIEMPRICAL MASS FORMULA AND ITS APPLICATION
TO MASS PARABOLAS
We have seen that the nuclei can be taken to be spherical with radius R =R0A1/3. On
the basis of this concept and some other classical concepts such as surface tension,
electrostatic repulsion etc., a formula for the atomic mass of a nuclide in terms of
binding energy correction terms was set up by Weizsacker in 1935 which was later on
modified by Bethe and others . This formula can be used to predict the stability of
nuclei against particle emission, energy release and stability for fission.
The mass of a nucleus is given by the formula
M(z,A) = zMH + NMn – B,
(1)
Where B is the binding energy expressed in mass units. It were possible to calculate B
from a general formula, all nuclear masses could be evaluated theoretically.
Weizsacker and others made an attempt in this direction and developed an empirical
formula assuming the liquid drop model of the nucleus regarding B as similar to latent
energy of condensation. Some of the properties of nuclear forces (saturation, short
range etc.) which have been deduced from the approximate linear dependence of the
binding energy on the number of particles in the nucleus are analogous to the
properties of the forces which hold a liquid drop together. Hence there is ample
justification in considering the nucleus to be analogous to a drop of incompressible
fluid of very high density (~1014 gm./cm3.). The value of B
was calculated
empirically as made up of a number of correction terms given as
B = B1 + B2 + B3 + B4+ ……………..
(2)
We shall now proceed to find out empirically the values of B1 , B2 etc.
Volume Energy Correction
The major contribution to B i.e., B1 comes from the mutual interactions of the
nucleons under the influence of nuclear forces.The nuclear binding energy is
proportional to the volume of the nucleus or to the total number of nucleons Z and so
we can write
B1vA
 B1  a1. A
where
a1> 1
20
Basic Properties of Nuclei
Surface energy.
Those nucleons which are situated in the surface region of the nucleus are necessarily
more weakly bound than those in the nuclear interior because they have fewer
immediate neighbors, The number of such nucleons is proportional to the surface area
of the nucleus and therefore to R2 and so is proportional to A2/3 (because R  A1/3).
B2  a2 A2 / 3
Thus we have
a2  0.
Where
The sign of surface energy B2 must be opposite to that of B1 since this effect which
corresponds to the surface tension of a liquid drop, represents a weakening in the
binding energy.
Coulomb energy
Assuming that the nuclear charge Ze is uniformly distributed throughout the nuclear
volume,
Ee 
the
coulomb
energy
of
the
nucleus
has
been
calculated
as
3 Z 2e 2
.
5 R
The assumption that protons are uniformly distributed is far from correct. Moreover
Protons obey Pauli’s Exclusion Principle and two of them can not occupy the same
place and this effect must be considered in a more accurate determination of Ec. The
effect of coulomb self energy on binding energy is diminutive i.e.,
B2  a3
Z2
A1/ 3
The negative sign indicates the diminution of energy due to repulsion effect. Some
time and particularly in the case of light nuclei having comparatively small values of
Z, the above formula is slightly modified as
B2  a3
Z ( Z  1)
A1/ 3
Asymetry energy
It has been found that for light nuclei, the condition for stability is N=Z. This is
called symmetry effect. Any deviation from N=Z reduces the stability of the nuclei
and hence reduces the binding energy. The deficit in binding energy depends on the
neutron excess (N-Z) and is proportional to
( N  Z )2
. This symmetry effect is purely
A
21
Properties of Nuclei and Scattering
a quantum mechanical effect in contrast to the surface energy effect and coulomb
energy effect. This correction term is given as
( N  Z )2
( A  2Z ) 2
B4  a4
  a4
A
A
The minus sign again represents the weakening in binding caused by asymmetry in N
and Z since beyond a certain stage the neutron cases to act as binding agent within the
nucleus.
Pairing energy
It has been found that even even nuclei are most stable , even odd and odd even nuclei
are less stable and odd odd nuclei are most unstable. To take account of this pairing
effect, an additional term is incorporated into the mass formula
B5
 

 0
 

For e –e nuclei
For e-0 nuclei and 0-e nuclei
For 0-0 nuclei
Where  is empirically found to be given by
  a5 A3 / 4
Combining all the above correction terms, the semi empirical mass formula is given
by
M(Z, A) = ZMH+(A-Z)MN-a2A1/3 +a3Z2/A1/3  a4
( A  2Z ) 2

A
(3)
Where the binding energy correction terms are taken in mass units .
The empirical formula for binding energy is given as
B  a1 A  a2 A
2/3
Z2
( A  2Z ) 2
 a3 1/ 3  a4

A
A
Dividing this expression by A, we get the binding energy per particle. Thus the
binding energy per particle is given by
B
a
Z2
( A  2Z ) 2 
 a1  12/ 3  a3 . 4 / 3  a4

A
A
A
A2
A
(4)
The empirical values of the coefficient evaluated by comparison of the above equation
with the masses of stable nuclides and energetics of nuclear reactions are listed below
a1= 14.1 Mev, a2 = 13.0 Mev, a3 = 0.505 Mev,
a4 = 19.0 Mev,
a5 = 33.5 Mev
From eq.(3) It is obvious that the mass ZMA is a quadratic function of Z for a given
mass number A. Thus a graph of ZMA versus Z will be a parabola, the minimum of
which will represent the most stable isobar. Experimentally it has been found that for
22
Basic Properties of Nuclei
odd-A nuclides, there is only one stable isobar. For even-A nuclides, there are often
two and sometimes three stable isobars.
Mass Parabolas for Isobaric Nuclei
Isobaric nuclides are characterized by same mass number A but different masses. To
examine their behavior we write down eq. In the abbreviated form as.
Fig. 7. Mass parabola for odd A nuclei.
ZMA=Z1A+K2Z+K3Z2  


Where K1  M n - (a 1 - a 4 - a 2 A - 1/3) 

K 2  [4a4  ( M n  M H )]


2/3

4a4 
A



K3 
1

A  4a4 / a3 
 in the above expression is independent of A and Z and has 0 value for odd-A nuclei.
If we differentiate eq. And equate it to zer4o, the condition for most stable Z is
obtained
 M 

  0  K 2  2 K 3 Z stable
 Z 
K
 Z stable   2
2K3
The mass of the stable isobar is therefore, written as
ZSMA = K1A+K2ZS+KSZS2
= K1A – 2K3ZS2+K3ZS2
= K1A – K3ZS2
ZMA = K1A-2K3ZSZ+K3Z2
Subtracting from we eliminate K1A and thus we get
ZMA-ZSMA=K3(Z2-2ZSZ+ZS2)
=K3(Z-ZS)2.
23
Properties of Nuclei and Scattering
This gives the parabolic mass relation for odd-A isobaric nuclei and contains only the
coefficient of K3. This equation is very much useful in calculating the transition
energies in reactions where Z changes to Z±1.
Isobaric nuclei with even A.
the mass – energy profiles for odd-A and even A isobars have a marked distinction for
odd-A,  =0 and are therefore represented by a single parabola while for even A,  is
positive for even –even nuclei and negative for odd-odd nuclei so that the masses of
e-e isobars fall on a separate lower parabola than that for the O-O isobars.
Fig. 8.
Mass parabolas for even- A nuclei.
Writing equation for even –A, odd-Z nuclei, we have
ZSMA=K1A-K3ZS2+ 
And for even-A, even-Z isobars
ZSMA=K1A-K3ZS2- 
In  -decay, Z increases by unit and in  +decay it decreases by 1 unit. Hence in the
case of odd-A isobars, only  -decay can take place for the nuclides which lie along
lie along the left arm of the parabola and  + decay for those lying along the right
arm, the decay scheme in each case finishes with only one isobar as shown in the
following fig.
Fig. 9. Mass parabolas for odd –A nuclei showing beta decay
steps on the left branch and   steps on the right branch
24
Basic Properties of Nuclei
In the case of even-A isobars,  decay changes an even-even nucleus into an oddodd nuclide or vice-versa such that the decay steps zig-zag between the two parabolas.
___________________________________________________________
1.6 MIRROR NUCLEI AND ISOTOPIC SPIN FORMALISM
Mirror nuclei
Two nuclides having the same number of nucleons but the number of protons in one
of them being equal to the number of neutrons in the other are called Mirror Nuclei.
For example .
3
3
1H
2He
7
7
3Li
4Be
13
13
6C
7N
39
39
19K
20Ca
9
9
4Be
5B
11
11
4B
6C
Let us consider the mirror pair of nuclei 6C13 and 7N13 . The first member has 7
neutrons and 6 protons while the second one has 7 protons and 6 neutrons.
Let us begin with Z=N nucleus (e.g. 6C12). If we add one proton to such a nucleus we
get one member of the pair formed (7N13). While the addition of one neutron will give
another member(e.g. 6C13). The only difference in the binding of the nucleus for the
mirror pair nuclei C13, N13 is that in the case of C13, there are 6 n-n bonds in place of 6
p-p bonds in N13. If we accept the principle of charge symmetry according to which
the nuclear force between a pair of protons is the same as the force between a pair of
neutrons in the same state, than there should be no difference in the nuclear binding
forces of a pair of mirror nuclei. The total binding energy of the two nuclei will not be
the same, however, because of the difference in the coulomb self-energy.
Isotpoic spin
A study of the nature of nuclear forces shows that these are charge independent which
means that the forces between n-n, p-p and n-p are all alike. We my therefore regard
the nucleus to be constituted of a single entity called nucleon, and neutron and proton
simply represent the two states of the nucleon. The charge of the nucleon can,
therefore
be treated as variable. The charge on the nucleon can be easily
distinguished by a new variable known as isotopic spin or to be more accurate
isobaric spin. The latter term is preferred because neutron and proton are isobaric and
not isotopic. The isobaric spin t behaves exactly the same way as the electron spin i.e.,
25
Properties of Nuclei and Scattering
it distinguishes between the neutron and proton. Thus for neutron t may be assigned a
value +1/2 and for proton –1/2 and the charge on the nucleon may be expressed as
(1/2, t). The isobaric spin is given a quantum number T and there are 2T+1
independent nuclear states associated with T, each state belonging to a different value
of the quantity TZ where TZ =T, T-1,.......,-T. The quantity TZ may be called the
component of T in the direction of positive charge. TZ is given by TZ   t ,
where the summation extends over all the nucleons. Thus if the nucleus has N
neutrons and Z protons,
TZ = ½ (N-Z) = ½ Neutron excess of the nucleus.
it has been shown that isobaric spin is conserved in nuclear interactions in the same
manner as the conservation of total nuclear angular momentum.
Check Your Progress 3
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1) Explain each term of Semi-empirical mass formula and state its limitation.
2) Write short notes on:
(a)
Mirror nuclei
(b)
Isotopic spin
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___________________________________________________________
1.7 Let Us Sum Up
After going through this unit, you would have achieved the objectives stated earlier in
the unit. Let us recall what we have discussed so far.

The nuclear matter is incompressible and has a constant density for all nuclei .
Deviations from this conclusion appear to be of the order of 10% or so .The
nuclear radius is given by
R = R0A1/3
The variation in R0 is from 1.2 to 1.5 fermi
(1 fermi= 10-13 cm).
26
Basic Properties of Nuclei

On the basis of classical concepts such as surface tension electrostatic
repulsion etc.a formula for the atomic mass of a nuclide in terms of binding
energy correction terms was set up by Weizsacker in 1935 which was later on
modified by Bethe and others.This formula can be used to predict the stability
of nuclei against particle emission, energy release and stability of fission.

Weizsacker and others developed an empirical formula assuming the liquid
drop model of the nucleus regarding
B
as similar to latent energy of
condensation.

The electric quadrupole moment measures the departure of a nucleus from
spherical symmetry. So far we have assumed that nuclei are spherically
symmetrical. But it is not always necessary to make this supposition. Nuclei
can have +ve or –ve quadrupole moments. Positive moments correspond to an
elongation of the nuclear charge distribution along the angular momentum axis
while –ve moments corresponds to flattened or oblate distribution.

Two nuclides having the same number of nucleons but the number of protons
in one of them being equal to the number of neutrons in the other are called
Mirror Nuclei.
___________________________________________________________
1.8 Check Your Progress : The Key
2.2 1017 kgm.m 3
1. i)
ii)
2. i)
ii)
3. i)
See the section 1.2.
1
H 2 1H 2 2 He 4  Energy( E )
2(1.1)  2(1.1)  4(7.0)  E
E  28  4.4  23.6Mev
See the section 1.3.2.
See the section 1.4.
ii) See the section 1.5.
27
Properties of Nuclei and Scattering
REFERENCES AND SUGGESTED READINGS
1. The “Particles of Modern Physics” by J. D. Stranathan, Philadephia: Blakiston.
2. ”Modern Mass Spectroscopy” Advances in Electronics by M. G. Inghram,
Academic Press, New York.
3. Nuclear Physics by E. Fermi, Willey Publications.
4. Concept of Modern Physics by Baiser, TMH.
5. Nuclear Physics by Irving Kaplan, Narosa Publishing House.
6. Handbook of Physics by Condon and Odishaw, TMH NewYork.
7. Concept of Nuclear Physics by Cohen, McGraw Hill.
8. Theory and Problems of modern Physics (Schaum’s outline Series)
9. Modern and Atomic Physics by Kleppner & Kolenkow McGraw Hill.
***********
28