Chapter 4
The Masses of Nuclei
◎ The naturally occurring nuclei
● The nuclear binding energy
◎ The liquid drop model
● Mass parabolas and the stability line
◎ The implications of the semi-empirical mass formula
4-1 The naturally occurring nuclei
Line of Stability
1. Apart from three or four exceptions, the naturally occurring
elements up to lead are stable and lie on or near a line (the line
of stability) in the Z, N plane.
2. These elements have a neutron number N which is equal to Z,
Z+1,or Z+2 in nuclei up to A = 35 (except for 1H and 3 He,
for which N =Z –1) but which thereafter increases faster
than Z until in lead N ≈ 1.5 Z.
What we would
like to know.
1. We would like to understand why the stable nuclei have this
property and what happens if nuclei are produced in which
N is greater or less than the stable optimum.
2. We would also like to understand why it is that for A > 209,
there are no stable nuclei.
4-2 The nuclear binding energy
For the study of nuclei the nuclear mass (M) is a very
important quantity. It is related to measurements of both
binding energy (B) and separation energy (S).
The convenient unit for measuring the nuclear mass
u
: is called the atomic mass unit or for short amu.
The mass of a 12C atom (including all six electrons) is defined
as 12 amu (or 12 u) exact.
1 u  1 amu  1.6605402 (10) 1027 kg
 931.49432 (28) MeV/c
2
(1)
The mass of a proton M p  1.007276470 (12) u  938.27231 MeV/c 2
The mass of a neutron M n  1.008664898 (12) u  939.56563 MeV/c 2
Total binding energy B(A,Z)
Definition:
B( A, Z )  [ZM p  NM n  M ( A, Z )]c2
(2)
The total binding energy B(A,Z) is defined as the total minimum
work that an external agent must do to disintegrate the whole
nucleus completely. By doing so the nucleus would no longer be
existent but disintegrated into separated nucleons.
This can also be considered as the total
amount of energy released when nucleons,
with zero kinetic energy initially, come close
enough together to form a stable nucleus.
An interesting measured quantity is the
B ( A, Z )
averaged binding energy per nucleon Bave. ( A, Z ) 
A
(3)
Separation energy (S)
(1). The separation energy of a neutron Sn
A
Z
X N  A Z1X N 1  n
Sn  [ M ( A  1, Z )  M n  M ( A, Z )]c2
(4)
M ( A, Z )c2  (ZM p  NM n )c2  B( A, Z )
M ( A 1, Z )c2  [ZM p  ( N 1)M n ]c2  B( A 1, Z )
Sn  {[ ZM p  ( N  1) M n ]c2  B( A  1, Z )  M nc2
- (ZM p  NM n )c2  B( A, Z )}
 B( A, Z )  B( A  1, Z )
Sn  B( A, Z )  B( A  1, Z )
(5)
Separation energy (S)
(2). The separation energy of a proton Sp
A
Z
X N  ZA11YN  p
S p  B( A, Z )  B( A  1, Z  1)
(6)
(3). The separation energy of a α-particle Sα
A
Z
X N  ZA42YN  2  42 He 2
S  B( A, Z )  B( A  4, Z  2)  B(4,2)
(7)
The average binding energy per nucleon versus mass number A
56
26
Fe has 8.8 MeV per nucleon
binding energy and is the most
tightly bound nucleus
Bave = B/A
1. The saturation property observed in the figure is the manifestation
of short range characteristics of nuclear force.
2. The short range nuclear interaction can be studied by examining data
collected from the (p,p) and (n,p) scatterings as well as from the
binding energy of deuteron.
12C
4He
24Mg
16O
Bave = B/A
8Be
3. Overall description of the figure
24Mg
12C
(a).
4He
16O
A ≈ 56
8Be
(b). 30  A  240
The value of Bave varies slowly and
is around 7.5 ~ 8.5 MeV.
(c).
A  56 (56Fe)
The value of Bave reaches its maximum.
0  A  30
The binding energy per nucleon Bave
increases as the mass number A
increases. At some particular values
of A the value of Bave is apparently
larger than neighboring A’s.
(d).
A  60
The value of Bave decreases as the
mass number A increases. This is
due to the Coulomb repulsion.
24Mg
12C
4He
16O
8Be
Nucleons in a
nucleus like to
form α-particle
clusters since a
α-particle is a
relatively stable
4-nucleon system.
For the case when A = 4n
(n = 1,2,3,4,…) the value
of Bave is relatively larger
when compared to
neighboring nuclei.
The 9Be nucleus is
stable and its natural
enrichment is 100%.
If we model A = 4n (for 2 < A < 25) nuclei as comprised of α-particle clusters and
consider the energy difference ΔE defined as
E  B( A, Z )  nB( )
(8)
We will soon discover that ΔE(8Be) = - 0.09 MeV. This shows that the nucleus
8Be is not naturally stable. Indeed the nucleus 8Be is not stable and has a
measured half-life τ1/2 = 7.0 × 10-17 s.
ΔE(MeV)
n
n(n-1)/2
ΔE/[ n(n-1)/2] (MeV/pair)
12C
7.27
3
3
2.42
16O
14.44
4
6
2.41
20Ne
19.17
5
10
1.92
24Mg
28.48
6
15
1.90
1. In fact, if we divide ΔE by the number of α-particle pairs, given by n(n-1)/2, the result is
roughly constant, with a value around 2 MeV.
2. This gives us a picture that, at least for light nuclei, a large part of the binding energy lies in
forming α-particle clusters, around 7 MeV per nucleon, as can be seen from the binding
energy of 4He.
3. The much smaller remainder, around 1 MeV per nucleon or 2 MeV between a pair of αclusters, goes to the binding between clusters.
4. This phenomenon is usually referred to as the “saturation of nuclear force”. That is, nuclear
force is strongest between the members of a group of two protons and two neutrons, and as a
result, nucleons prefer to form α-particle clusters in nuclei.
4-3 The liquid drop model
1. A detailed theory of nuclear binding, based on
highly sophisticated mathematical techniques
and physical concepts, has been developed by
Brueckner and coworkers (1954-1961).
2. A much cruder model exists in which the
finer features of nuclear forces are ignored,
but the strong inter-nucleon attraction is
stressed. It was derived by von Weizsäcker
(1935) on the basis of the liquid-drop
analogy for nuclear matter, suggested by Bohr.
Bave = B/A
Over a large part of the periodic
table the binding energy per
nucleon is roughly constant.
The mass density of nuclear
matter is approximately constant
throughout most of the periodic
table.
These two properties of nuclear matter are very similar to the
properties of a drop of liquid, namely constant binding energy
per molecule, apart from surface tension effect, and constant
density for incompressible liquids.
The essential assumptions of the liquid drop model:
1. A spherical nucleus consists of incompressible matter so that R ~ A1/3.
2. The nuclear force is identical for every nucleon and in particular does not
depend on whether it is a neutron or a proton.
Vpn = Vpp= Vnn (V denotes the nuclear potential)
3. The nuclear force saturates.
The binding energy of a nucleus
Definition:
B( A, Z )  [ZM p  NM n  M ( A, Z )]c2
(9)
From the liquid drop model ̶ Weizsäcker’s formula
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
(10)
Carl Friedrich von Weizsäcker, 1993
A German physicist (1912-2007)
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
aV A
 aS A
2/3
(10)
is the “volume term” which accounts for the binding
energy of all the nucleons as if every one were entirely
surrounded by other nucleons.
is the “surface term” which corrects the volume energy
term for the fact that not all the nucleons are surrounded
by other nucleons but lie in or near the surface.
Nucleons in the surface region are not attracted as much as
those in the interior of a nucleus. A term proportional to the
number of nucleons in the surface region must be
subtracted from the volume term.
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
Z2
 aC 1 / 3
A
(10)
is the “Coulomb term” which gives the contribution to
the energy of the nucleus due to the mechanical
potential energy of the nucleus charge.
Assume a charged sphere of radius r has been built up, as shown in the figure (a). The additional
work required to add a layer of thickness dr to the sphere can be calculated by assuming the
charge (4/3)πr3ρ of the original sphere is concentrated at the center of the shell [see figure (b)].
The electrical potential energy of the nucleus is therefore
where
1
4

VCoulomb    r 3    (4r 2dr ) 
0
r
3

Ze


4 3
16 2 2 5 3 Z 2e2
R
   R 
3
15
5 R
R
and
R ~ A1 / 3
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
(10)
Three terms that were discussed previously are in a sense
classical. The following terms that are to be discussed are
quantum mechanical.
(1) the asymmetry term
These include
(2) the paring term
(3) the shell effect correction term
(N  Z )
 aA
A
2
is the “asymmetry term” which accounts for the fact
that if all other factors were equal, the most strongly
bound nucleus of a given A is that closest to having
Z = N.
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
(10)
The Pauli exclusion principle
states that no two fermions
can occupy exactly the same
quantum state. At a given
energy level, there are only
finitely many quantum states
available for particles.
Different system energies due to asymmetric configurations
1. If Z = N, then both wells
are filled to the same
level (the Fermi level).
2. If we move one step up away from that
situation, say in the direction of N > Z
(or Z > N), then one proton must be
changed into a neutron. All other
things being equal (including equal
proton and neutron mass), this state
has energy ΔE greater than the initial
state, where ΔE is the level spacing at
the Fermi level.
3. A second step in the same direction causes the energy excess to become 2ΔE.
4. A next step means moving a proton up three rungs as it changes from proton to
neutron and the excess becomes 5ΔE.
Cumulative effect
N  Z  2, 4, 6, 8, 10, 12, 14, 16,...
1, 2 , 5, 8, 13, 18, 25, 32,....
unit of E
4. Therefore to change from N – Z = 0 to N > Z, with A = N + Z held constant,
requires an energy of ~ (N – Z)2ΔE/8.
5. This is independent of whether it is N or Z that becomes larger and it means that, if
all other things are equal, nuclei with Z = N have less energy and are therefore more
strongly bound than a nucleus with Z ≠ N.
6. The energy levels of a particle in a potential well have a spacing inversely
proportional to the well volume, thus we put ΔE ~ A-1.
This is the asymmetry term.
( N  Z )2
 aA
A
(11)
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A

(10)
is the “pairing term” which accounts for the fact that a pair
of like nucleons is more strongly bound than is a pair of
unlike nucleons.
1. For odd A nuclei (Z even, N odd or Z odd, N even) →δ = 0.
2. For A even there are two cases;
(a). Z odd, N odd (oo)
→–δ
(b). Z even, N even (ee) → + δ
aP
 ( Z , A)  1 / 2 ,
A
aP  12 MeV
(12)
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A

(10)
is the term accounts for the nuclear shell effect when Z or N is
some magic number. This term is much less important than
other terms. Therefore this term is not included in most of the
applications.
A favorable set of values for the coefficients:
aV = 15.560 MeV
aS = 17.230MeV
aC = 0.6970 MeV
aA = 23.385 MeV
aP = 12.000 MeV
(13)
2
2
Z
(
N

Z
)
B( A, Z )  aV A  aS A2 / 3  aC 1/ 3  a A
  
A
A
(10)
4-4 Mass parabolas and the stability line
If we make some rearrangement the Weizsäcker’s formula can be written as
aS
  M n c  aV  a A  1/ 3
A
2
M ( A, Z )c 2  A  Z  Z 2   
(14)
   4aA  (M n  M p )c2 ≈  4aA
γ=
a
4a A
+ 1C/ 3
A
A
For a fixed mass number A, this is the equation of parabola with respect to the variable Z.
We may differentiate the equation (14) and find the root (Z0, usually not an integer) of the
following equation (15). Z0 is the optimum nuclear proton number for a fixed mass number A.
The nuclear system with a specified mass number A is the most stable with proton number Z0.
(15)

( Mc 2 )  0
Z
β + 2γZ 0 = 0
Z0  

A/ 2

2 1  1 (a / a ) A2 / 3
C
A
4
(16)
Z0  

A/ 2

2 1  1 (a / a ) A2 / 3
C
A
(16)
4
From the equation (15) the most stable nuclear
systems of various mass numbers A are determined
by the value of Z0. By using the relation A = Z0 + N
we are able to plot stability lines on the N-Z plot.
This follows exactly the shape of the empirical
stability line in the figure.
From expression (16), we can recognize that the
deviation of the stability line from N = Z or Z = A/2
is caused by the competition between the Coulomb
energy, which favors Z0 < A/2, and the asymmetry
energy which favors Z0 = A/2.
For odd-A isobars, δ = 0, and equation
(14) gives a single parabola, which is
shown in the figure (a) for a typical
case. We will see later that if
M(A,Z) > M(A, Z+1)
beta (electron) decay takes place
from Z to Z+1
(15)
M(A,Z) > M(A, Z-1)
electron capture and perhaps
positron decay takes place from Z to
Z-1
M ( A, Z )c  A  Z  Z    (14)
2
2
It is clear from the figure (a) that for
odd-A nuclides there can be only one
stable isobar.
For even-A isobars, two
parabolas are generated by the
equation (14), differing in mass
by 2δ. A typical case is given in
the figure (b). Depending on
the curvature of the parabolas
and the separation 2δ, there can
be several stable even-even
isobars. Figure (b) shows that
for certain odd-odd nuclides
both conditions (15) are met so
that electron and positron
decay from the identical
nuclide are possible and do
indeed occur.
M ( A, Z )c 2  A  Z  Z 2   
(14)
   decay
(Z , A)  (Z  1, A)  e  e
  decay
( Z , A)  ( Z  1, A)  e  e
EC (electron capture)
e  ( Z , A)  ( Z  1, A)   e


Three types of
β-decay
4-5 The implications of the semi-empirical mass
formula
1. The volume binding term aVA in the mass formula means that
each nucleon interacts only its nearest neighbors and that the
constant density is equivalent to a separation from nearest
neighbors does not change with A. All this means that nuclear
force saturates and is short range.
2. The nuclear density of nucleons is approximately 1 in every 7
cubic fermis so that the average separation is about 1.9 fm. Thus
the range must be 1-2 fm.
0.5
Hard
core
3. From a study of nucleon-nucleon scattering at energies in the
range 100-400 MeV. It was found that there is a strong repulsive
force between nucleons at separations of less than about 0.5 fm.
4. The liquid-drop model does accommodate the existence of excited
states corresponding to the mode of vibration (vibrational states)
which are one kind of collective motion that nuclei may have, the
collective implying the motion of the constituent nucleons is
correlated in some way to give the shape vibration. But this
phenomenon can not be described by the model.
Water drop vibrations
5. There is another type of collective mode which is in the form of
rotation for non-spherical nuclei. The quantum states associated
with nuclear rotational motion is called rotational states. These
states can be identified from energy spectra of non-spherical nuclei.
A prolate deformed nucleus
~ The End ~