Chapter 6
Nuclides  composite particles of
nucleons
A nuclide is a type of atoms whose nuclei
Discovery consists in seeing what everyone else has
have specific numbers of protons and
seen and thinking what no one else has thought.
neutrons (nucleons). The standard model
considers up and down quarks as basic
Albert Szent-Gyorgi
components of nuclei, but nucleons (protons
and neutrons) are convenient units. On the
other hand, nuclides are energy states in the form of masses. Stable states remain unchanged
for an indefinite period whereas unstable ones undergo radioactive decay. For example, the
energy equivalents of protons, deuterons and 238U are 938 MeV, 1.88 GeV and 212 GeV
respectively. For convenience, we discuss nuclides in terms of nucleons.
Discoveries of protons and neutrons infer the existence of isotopes. Since isotopes refer to
atoms of the same element with different number of neutron, the term nuclide is more
appropriate when referring to a type of atom. Nuclear properties are specific for nuclides, but
not necessarily for chemical elements.
The number of protons is the same as the atomic number, Z, and the
mass number, A, is the number of nucleons in the nucleus. Thus the
number of neutrons, N, is A - Z. Any nuclide is an isotope of an
element, and the symbol of the element is used to represent the
nuclide, but the mass number is given as a pre-superscript, such as
16
O, where 16 is the mass number. For clarity and convenience, the
atomic number is given as a post-superscript, 16O8.
A nuclide as an
isotope of
element E.
A
EZ
Stable nuclides exist for an indefinite period of time, and they are the constituents of
ordinary material. Unstable nuclides emit subatomic particles, with 4, n, p being the
most common. Few undergo nuclear fission. However, radioactive nuclides with long halflives are also present in nature.
177
Stable Nuclides
Stable nuclides remain unchanged for an indefinite period, and they are not radioactive. Of
the natural elements on Earth, only elements with atomic number less than 83 have stable
isotopes, except technetium (Tc, Z = 43) and promethium (Pm, Z = 61). Only 81 elements
have at least one stable isotope. However, there are 280 stable nuclides, and many elements
have more than one stable isotope. The composition of isotopes in an element is an important
piece of information in many technologies. Elements with more than 92 protons are all manmade, as are technetium (Tc) and promethium (Pm), because they have no stable isotopes.
Elements with atomic number greater than 83 have no stable isotopes, but they are decay
products from uranium and thorium, and they are constantly produce, but at the same time
they decay and convert eventually to Bi or Pb.
There are many factors affecting the stability of nuclides. In this section, we describe some
common factors of stable nuclides. We can only describe one at a time, but all factors
concertedly affect the stability of nuclides.
Numbers of Protons and Neutrons
Protons are charged but neutrons are not. Although they may not exist as individual nucleons
in the nucleus, their numbers are obvious properties. From accounting point of view, we
consider them as individuals.

How do numbers of protons and neutrons affect the stability?
What is the role of neutrons in the atomic nuclei?
What are the ratios of neutron numbers to proton numbers for stable nuclides?
Do the ratios vary systematically?
We have examined some light nuclides in an Chapter 5 when we introduced the Nuclide Chart
as a means to organize nuclear information. For convenience, we divide the large Nuclide
Chart into three sections: light-weight nuclides with atomic number less than 20 (Z < 20),
medium-weight nuclides with 20 < Z < 50, and heavy- weight nuclides with Z > 50.
Light-weight nuclides are given in the next page. Some general observations regarding their
stability are given below:
Only 1H and 3He have more protons than neutrons in their nuclei. Natural hydrogen contains
0.0015 % of the isotope D, and only a trace of 3He is present in natural helium (mostly 4He).
All other nuclides have either equal numbers of protons and neutrons or more neutrons than
protons. The heaviest nuclide with equal numbers of protons and neutrons is 40Ca20, (20 being
a magic number).
178
Stable Nuclides
 Z
| (Magic numbers
21
20 . . . . .
19
18
17
16
15 . . . . .
14
13
12
11
10 . . . . .
9
8
7
6
5 . . . .
4
3
Li Li
2
He He
1 P D
0 1 2 3 4
and double magic-number nuclides are in bold)
.
.
.
.
.
.
.
.
.
.
.
.
.
S S
P
Si Si Si
Al
Mg Mg Mg .
Na
. . . . . Ne Ne Ne
F
.
O O O
N N
C C
.
.
B B
Be
.
.
.
.
.
.
.
.
5
.
.
.
7
8
9
Ar
Cl
S
Ca
K K
Ar
Cl
S
Sc
Ca Ca Ca
K
Ar
.
Ca
Ca
.
.
6
.
(to be continued)
.
.
.
.
.
.
.
.
.
.
.
.
.
N
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
Most elements with odd atomic numbers, Z, have only one stable isotope. The number of
stable isotopes generally increases as Z increases. Nuclides with odd numbers of neutrons have
only one stable isotone (nuclides with the same number of neutrons), and the number of
stable isotones also increases as the number of neutrons, N, increases.
Four nuclides 2D1, 6Li3, 10B5, and 14N7 have equal and odd numbers of protons and neutrons.
For these light stable nuclides, the ratio N/Z ranges from 1 (D) to 1.22 (40Ar), ignoring H for
which N/Z = 0. All but a few stable nuclides have more neutrons than protons. Uncharged
neutrons probably moderate the proton-proton repulsion, making heavy nuclide stable.
A chart of stable nuclides with 20 < Z < 70 is given on the next page. For convenience in our
discussion, nuclides with atomic numbers range between 20 (Ca) and 50 (Sn), two magic
numbers called medium-weight nuclides.
A line with equal numbers of protons and neutrons is marked by “+” signs. For stable
nuclides, the ratio N/I increases as Z (or N) increases: N/I = 1.14 for Sc (Z = 21), 1.27 for Nb
(Z = 41), 1.41 for Sb (Z = 51), and 1.43 for Tb (Z = 65). The effect of this increase can be
seen from the deviations of the locations of stable nuclides from the line marked by “+”.
Many stable isotopes appear on lines for Z = 8, 20, 28, 50, and 82 (see list below). Similarly,
unusual numbers of stable isotones also appear on vertical lines with N = 8, 20, 28, 50, and 82.
The existence of many stable nuclides with these numbers of protons or neutrons supports
calling them magic numbers.
For the medium-weight nuclides, 1 (40Ca) < N/Z < 1.48 (124Sn50). However, 1 < N/Z < 1.40 for
the six stable isotopes of calcium alone: 40Ca20, 42Ca20, 43Ca20, 44Ca20, 46Ca20, and 48Ca20. There are
ten (10) stable isotopes of tin (112, 114, 115, 116, 117, 118, 119, 120, 122, & 124Sn50), for which 1.24 < N/Z < 1.48.
179
Stable Nuclides (Continue)
 Z
The Line with equal numbers of p and n is marked by “+”.
(Double magic-number nuclides and magic numbers are in bold)
74 W
73 Ta
To be continued in Table form
72 Hf
X XXXX
71 Lu
XX
70 Yb . . . . . . . . . . . . . . . . . . . . . . . . + . . . . . . . . . .
X XXXXX X
69 Tm
X
68 Er
X X XXX X
67 Ho
X
66 Dy
X X XXXXX
65 Tb . . . . . . . . . . . . . . . . . . . . . + . . . . . . . . . .
X
64 Gd
X XXXXX X
.
63 Eu
X X
62 Sm
X XXXX XX
.
61 Pm - - 60 Nd . . . . . . . . . . . . . . . . . . + . . . . . . .
XXXXX X X
.
59 Pr
X
58 Ce
X X X X
.
.
57 La
XX
56 Ba
X X XXXXX
.
.
55 Cs . . . . . . . . . . . . . . . + . . . . .
X
54 Xe
X X XXXXX X X
.
.
53 I
X
52 Te
X XXXXX X X .
.
.
51 Sb
X X
50 Sn . . . . . . . . . . . . . + . . .
X XXXXXXX X X
.
.
.
49 In
X X
48 Cd
X X XXXXX X .
.
.
.
47 Ag
X X
46 Pd
X XXX X X
.
.
.
.
45 Rh . . . . . . . . . . +. .
X
44 Ru
X XXXXX X
.
.
.
.
43 Tc - - 42 Mo
X XXXXX X .
.
.
.
.
41 Nb
X
40 Zr . . . . . . . . + . . . XXX X X
.
.
.
.
.
39 Y
X
38 Sr
X XXX
.
.
.
.
.
37 Rb
X X
36 Kr
X X XX X
.
.
.
.
.
35 Br . . . . . + . .
X X
34 Se
XXXX X X
.
.
.
.
.
33 As
X
32 Ge
X XXX X
.
.
.
.
.
.
31 Ga
X X
30 Zn . . . + . X XXX X
.
.
.
.
.
.
29 Cu
X X
28 Ni
X XXX X
.
.
.
.
.
.
.
27 Co
X
26 Fe
X XXX
.
.
.
.
.
.
.
25 Mn +
X
24 Cr
X XXX
.
.
.
.
.
.
.
23 v
XX
22 Ti XXXXX.
.
.
.
.
.
.
.
21 Sc X
20 Ca
X X
2
2 3
4
5
6
7
8 8
9
10
0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567
180
Because we run out of room in the graph above, the stable isotopes of elements with Z > 70
are listed below with their mass numbers. Number of neutrons (N) can be evaluated by
subtracting the atomic number (Z) from the mass number (A), N = A - Z.
In the table below, the abundance in percentage, %, of the isotopes are given in the
parentheses (% omitted). Gold (Au) and bismuth (Bi) are two elements that have only one
stable isotope, the percentages of 197Au and 209Bi are 100%.
Mass Numbers of Heavy Stable Elements (70 < Z)
Z Symbol Mass (abundance or half life)
71
72
73
74
75
76
77
78
79
80
81
82
83
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
175 (97.4), 176 (2.6)
174 (0.163), 176 (5.21), 177 (18.56), 178 (27.1), 179 (13.75), 180 (35.22)
180 (0.0123), 181 (99.9877)
180 (0.135), 182 (26.4), 183 (14.4), 184 (30.6), 186 (28.4)
185 (37.07), 187 (62.93)
184 (0.018), 186 (1.59), 187 (1.64), 188 (13.3), 189 (16.1), 190 (26.4), 192 (41.0)
191 (38.5), 193 (61.5)
190 (0.0127), 192 (0.78), 194 (32.9), 195 (33.8), 196 (25.2), 198 (7.19)
197 (100)
196 (0.146), 198 (10.02), 199 (16.84), 200(23.13), 201(13.22), 202(29.8), 204(6.85)
203 (29.5), 205 (70.5)
204 (1.4), 206 (25.1), 207 (21.7), 208 (52.3)
209 (100)
90
Th
232 (100% half life 1.4x1010 y)
92
U
235 (0.720, half life 7.04x108 y), 238 (99.276, half life 4.5x109 y)
For more information about nuclides, consult the Handbook of Chemistry and Physics by CRC
Press, and the Handbook of Isotopes by CRC Press. More information about the Nuclide Chart is
also available from the following web sites:
http://www.fysik.lu.se/~nsr/isoexpl/isoexpl.htm
http://www.dne.bnl.gov/~burrows/usndn/usndn.html
http://csa5.lbl.gov/~fchu/isoexpl/man73.htm
The contents in these web sites include the following.
 Description of the US Nuclear Data Network
 Center for Nuclear Information Technology, San Jose State University
 Idaho National Engineering Laboratory
 Isotopes Project, Ernest Orlando Lawrence Berkeley National Laboratory
 Lund Nuclear Data Service, University of Lund, Sweden
 Tandem Accelerator Laboratory, McMaster University, Canada
 National Nuclear Data Center, Brookhaven National Laboratory
 Nuclear Data Evaluation Project, Triangle Universities Nuclear Laboratory
 Nuclear Data Project, Oak Ridge National Laboratory
181



Permanent Mass-chain Evaluation Responsibilities
International Nuclear Structure and Decay Data Network (NSDD)
Glossary of nuclear data evaluation and WWW jargon
Skill Building Questions
1. How does the ratio N/Z vary as Z or N increases?
2. What are the stable nuclides that have equal numbers of protons and neutrons. Which of these have odd
atomic numbers?
3. Which medium-weight elements do not have stable isotopes? What are their atomic numbers? How many
stable isotones are there with N = 19, 31, 35, 39, 61, and 89?
4. Can you calculate the atomic weight of say lead (Pb) from the abundance given to each stable isotope in the
table of heavy stable nuclide, why or why not? If not, what more information is required?
Pairing of Nucleons
Since free nucleons have ½ spin, they obey Pauli's exclusion principle by allowing two protons
or neutrons each with opposite spin to occupy a quantum state (if they are nucleons in a
nucleus). There is a preference for having pairs of protons or neutrons, and it is known as
pairing of nucleons.

How does pairing of protons and neutrons affect the stability of nuclides?
What evidence suggests pairing of nucleons contributes to the stability of nuclide?
The numbers of stable nuclides due to even
or odd numbers of protons and neutrons
seem to suggest support the preference
theory.
Effect of Paring Nucleons
Z
N
No. of stable nuclides
even even
166
even odd
57
There are 166 stable nuclides with both Z and
odd
even
53
N even, 57 with even Z odd N, and 53 with
odd odd
4
odd Z even N. The only 4 light stable
2 1 6 3 10 5
14 7
nuclides with both Z and N being odd are D , Li , B , and N . Note that H, Li, and B have
two stable isotopes, which belong to the odd-even type. The distribution of stable nuclides due
to odd and even number of Z and N can be interpreted as due to pairing of nucleons. We have
mentioned that technetium (Z = 43) and promethium (Z = 61) have no stable isotopes. There
are no stable isotones for N = 19, 31, 35, 39, 61, and 89.
The distinction between stable and long-lived nuclides is vague, but we choose to list the stable
ones only. The effect of pairing also affects the abundance of the isotopes in elements, as well
as the abundance of a nuclide on a planet, galactic or universal scale.
182
Pairing of nucleons also affects the decay or transmutation of unstable nuclides, particularly
transmutation of isobars with even mass numbers A. For this type of nuclide, a beta decay
converts an even-even nuclide into an odd-odd nuclide.
Skill Building Questions
1. Which medium elements do not have stable isotopes?
What are their atomic numbers? How many stable isotones are there with N = 19, 31, 35, 39, 61, and
89?
2. How many stable isotones are there with N = 2, 8, 20, 50, and 82?
Magic Numbers of Nucleons
Magic numbers 2, 8, 20, 28, 50, 82, and 126 have been mentioned in previous chapters in
connection with the shell model of nuclear structures.

How does the distribution of stable nuclides support a number to be a magic number?
How do magic numbers affect the stability of a nuclide?
One of the supports for the shell model comes from the fact that there are more stable
isotopes with magic numbers of proton and more stable isotones with magic number of
neutrons when compared to nuclides of similar mass numbers.
The emission of 4He2 (or  particle) nuclei in the form of radioactive decay supports 2 as a
magic number. Oxygen is a very abundant element, with 16O8 being the most abundant isotope
(99.759%), plus 0.037% of 17O and 0.204% of 18O. There are 2 isotones (15N, 16O) with N = 8,
plus the long-lived 14C. These data suggest 8 as a magic number.
There are 5 stable isotones with N = 20, but 0 with N = 19, 1 with N
= 21, and 3 each for N = 18 and 22. There are 6 stable Ca isotopes (Z
= 20); the abundances of Ca isotopes are: 40Ca 96.97%, 42Ca 0.647%,
43
Ca 0.145%, 44Ca 2.06%, 46Ca 0.0033%, 48Ca 0.185%. The light and
heavy isotopes of Ca have 20 and 28 neutrons. There are also 5 Ni (Z
= 28) isotopes and 4 isotones with 28 neutrons. These data reinforce
the effect of pairing of protons, and pairing of neutrons and the
magic numbers 20 and 28. Note that 4He2, 16O8, 40Ca20, 48Ca20, and
208
Pb82 are double-magic-number nuclides, because they have magic
number of protons and neutrons.
Abundance (%) of
calcium isotopes
40
Ca
Ca
43
Ca
44
Ca
46
Ca
48
Ca
42
96.97
0.647
0.145
2.06
0.0033
0.185
Numbers of stable isotopes and isotones for magic numbers 50 and 82 are very convincing.
Furthermore, three families of radioactive series decay to a stable lead isotope (Z = 82) after
many steps of transmutations.
The heaviest stable nuclide 209Bi83 has 126 neutrons; so has its isotone 208Pb82. You may think
that two stable isotones are not impressive, but no heavier nuclides are stable.
183
So far, elements with Z = 112 have been artificially made. One of the objectives for people
who synthesize heavy nuclides is to test the large magic numbers. Glenn T. Seaborg, a Nobel
laureate in Chemistry (1951) believed that element 126 may have a half-life long enough to be
observed. However, this element has not been observed yet.
Skill Building Questions
1. What are magic numbers?
What are the reasons for them to be magic numbers?
2. Give five double-magic number nuclides.
Abundances of Elements
The abundance of an element or nuclide is its amount in a system.

What is the most abundant element in the universe?
How about the solar system and on the planet Earth?
How are they estimated and based on what evidences?

Is the abundance of a nuclide related to its stability?
Aside from stability, what else affects the abundance?
The sun has 99.9% of the mass of the solar system. Hydrogen atoms contribute 72%, and
helium 4He 26% to all atoms in the Sun according to reliable estimates from spectroscopic,
density, and meteorite studies. Light nuclides H and He are the major components of the Sun,
where nuclear reactions convert H into He. Since stars are by far more massive than planets,
the most abundant element in the universe is also hydrogen, followed by helium.
Taking as a whole, the most abundant element of the planet Earth is iron,
which is the major component of the earth (molten) core. Additional
evidence comes from the many iron meteorites, which are considered debris
from outer space. However, the most abundant element of the Earth crust
is oxygen in terms of number of atoms, but silicon is the most abundant
element by mass. Their physical and chemical properties made them
accumulate in different systems.
The atomic abundance (abundance based on the number of atoms) of elements are made
from mineral and meteorites analyses. For example, the Earth crust is mostly silicate, SiO2, and
the most abundant elements in the crust are O8 and Si14. Taking other data into consideration,
oxygen is the most abundant element of the inner solar system, of which the Earth is a major
part. The abundance of oxygen is given as 1 (log 1 = 0) for reference. Magnesium, silicon, and
iron are the next most abundant elements, but their abundances are less than 0.5. Abundances
of Ni, Ca, S, Al and Na are less than 0.1 whereas those of N and F are 0.0001 as shown in the
diagram below.
184
The abundance of elements is one of the most important parameter in order to understand a
planet, a satellite, or any body in the universe. Thus, space explorations include the probe to
find the abundances of elements in the moon and mars.
The abundances of elements with even atomic numbers are usually higher than those with odd
atomic numbers of comparable masses. The relative high abundances of O, Ca, Fe, Sn and Pb
are due to the magic numbers 8, 20, 28, 50, and 82. Many factors influence the stability of
nuclides and their abundances.
Atomic Abundance (AA) of Elements of the Inner Solar System Excluding the Sun
Log (AA)
0
.
-1
.
-2
.
-3
.
-4
.
-5
.
-6
.
-8
.
-9
.
-10
O
.
. MgSi
.
.
.
Ca
Al S
Fe .
Ni
.
Na
.
C
.
P
.
. Ti CrMCo
V
Cu SeAs
.
.
.
.
.
.
.Sc
.
.
Ga
.
Cl
.
.
.
.
.
.
.
.
Sn
.
.
.
.
.
.
N F
.
.
.
Be
LiB
Sc Zr
Pb
Mn
.
Y.
.
RhAgIn
Br
.
.
.
.
.
.
.
.
. Mo
0
.
Ba
.
Ru
Sb
PdCd.
.
W
I
.
.
CeNdSm
. Dy
YbHf
Pr
La . EuTbHoTu Lu
.
.
.
Pt.
Os
IrAu
.Tl
Re
.
1
2
3
4
5
6
7
8 Mass No.
12345678901234567890123456789012345678901234567890123456789012345678901234567890
A general notion is that stable nuclides are abundant on a planetary or galactic scale. However,
other factors such as the process of their synthesis and sources of nuclides for their
productions also affect their abundances. The history and formation of the Sun, the Earth, and
its satellite are different, and these differences give rise to their composition differences. The
abundance, however, is partially related to their stability.
The group of elements C6, N7, O8, Ne10, Mg12, Si14, S16, Ar18, and Ca20 with even atomic
numbers are relatively more abundant than the group with odd atomic numbers: F9, Na11, Al13,
P15, and Cl17. The abundances of Li3, Be4, and B5 are very low. The abundance decreases from
Ca20 on as the atomic numbers increase, but there is a relatively high peak at iron, Fe26, and
nickel Ni28. Thus, the abundance strongly supports the theory that pairing of nucleons is an
important factor for their stability.
Skill Building Questions
1. What are the most abundant elements in the Earth crust and why?
2. How do the magic numbers affect the abundances of elements or nuclides?
185
Mass and Stability of Nuclides
Unstable nuclides under go radioactive decay or fission, and they are radioactive nuclides.
Some long-life unstable nuclides occur naturally, but many of them have been made artificially,
and their properties well studied. Their making (synthesizing) involves nuclear reactions, which
will be discussed in the next Chapter. We are only concerned with mass or energy regarding
their stability in this section.
Aside from the stable isotopes of 12C and 13C, radioactive carbon
isotopes with mass number 9, 10, 11, 14, 15 and 16 have been made
and identified. Their half-lives are 127 ms, 19.3 s, 20.3 m, 5730 y,
2.45 s, and 0.75 s respectively. Light carbon isotopes 9C, 10C, and 11C
decay by emitting positrons or electron captures whereas heavy
carbon isotopes 14C, 15C and 16C are beta  emitters. Their masses
hold the key for their stability. Radioactivity is a process by which
unstable nuclides convert to stable ones.
9
C
10
C
11
C
12
C
13
C
14
C
15
C
16
C
Half life
127 ms,
19.3 s
20.3 m
stable
stable
5730 y
2.45 s
0.75 s
The Binding Energy
Radioactivity or nuclear decays are spontaneous and exothermic reactions. The decay energy is
the difference in total energy content before and after the decay. Energy and mass are
equivalent, and the relative mass is the key to stability.

Is the mass of a nuclide the sum of masses of its components?
Why or why not?
How is the mass of a nuclide related to its stability?
Masses of nuclides have been carefully measured, and the mass of a nuclide is usually less than
the sum of masses of components. This suggests that the formation of a nuclide from its
component is an exothermic reaction. For convenience in our discussion, a special name is
given to the energy released when a nuclide is made from its components.
The binding energy (BE) of a nuclide is the energy released when the atom is synthesized
from the appropriate numbers of hydrogen atoms and neutrons. Hydrogen atoms and
neutrons are convenient components, because they provide appropriate number of electrons
and nucleons. The binding energies of hydrogen atoms and neutrons are thus zero in this
definition. The concept of BE applies to both stable and radioactive nuclides, and the
definition given above can be represented by a hypothetical equation:
Z H + N n = AEZ + BE
where H, n, and AEZ represent the energy equivalent of hydrogen, neutron, and the nuclide
A Z
E . In terms of masses, we have
Z mH + N mn = mE + BE
186
where mH, mn, and mE are masses of H, n, and the nuclide AEZ respectively. Thus, BE can be
evaluated using:
BE = Z mH + N mn - mE.
The binding energy is thus the minimum energy required in order to decompose nuclide AEZ
into Z H and N n, i.e., AEZ + BE = Z H + N n. The more the binding energy, the more stable
is the nuclide.
To evaluate BE, we need to know the mass of a hydrogen atom (1.007825 amu, more than the
mass of a proton), the mass of a neutron (1.008665 amu) and the mass of the nuclide AEZ. For
example, the binding energies of 3He (mass 3.016031) and 4He (mass 4.0260) are:
BE (3He) = (2 x 1.007825 + 1.008665 - 3.01603) 931.481 MeV
= 7.72 MeV
BE (4He) = (2 x 1.007825 + 2 x 1.008665 - 4.00260) 931.481 MeV
= 28.30 MeV
The estimates here show that formation of 4He releases much more energy than the formation
of 3He, 20 MeV per atom more. Although 4He contains only one neutron more than 3He does.
For comparison purpose, we can look at the amount of energy released when one nucleon is
added to a nuclide. The binding energy of a nucleon (BEn, or BEp) is the energy released
when a nucleon is added to a nuclide. From the above estimates, the binding energy of a
neutron for 3He is 20.58 MeV. As another example, the binding energies of proton and
neutron for 234U are calculated below. For these calculations, the masses of 234U92, 235U, and
235
Np93 are 234.040946, 235.043924 and 235.044056 amu respectively. These values are looked
up from data Table of nuclides. By definition, the binding energy of a neutron, BEn in 234U is
the amount of energy released in the hypothetical reaction:
234
U + n  235U + BEn. Thus,
BEn = (234.040946 + 1.008665 - 235.043924) amu
= 0.00566 amu x 931.5 MeV / amu
= 5.30 MeV
The binding energy of proton (BEp) when it is added to 234U to make a 235Np atom (234U + p
 235Np + BEp) is,
BEn = (234.040946 + 1.008665 - 235.044056) amu
= 0.00555 amu x 931.5 MeV / amu
= 5.17 MeV
The binding energy for a group of nucleons such as an  particle (BE) is the negative energy
of decay Edecay. For example, we can write the equation for the decay of 212Po84 as
212
Po84 = 208Pb82 + 4He + Edecay
But for the reverse reaction, we have
187
208
Pb82 + 4He =
212
Po84 + BE.
Thus,
BE = – Edecay
Note, however, BE is the energy of binding an  particle in the nuclide 212Po84, not the
binding energy of 4He from 2 H and 2 n, neither is it the binding energy of 212Po84. By the way,
the  particle energy from 212Po is 6.09 is MeV (looked up from a table).
Skill Building Questions
1. Evaluate the binding energy of 12C (mass = 12.00000), 14N (mass = 14.003074) and 16O
(mass = 15.994915). Discuss your results.
2. Discuss the binding energies of 16O, 17O (mass = 16.999131), and 18O (mass = 17.999160).
3. Calculate the binding energies of a proton and a neutron to 4He. You need to look up the masses of 5He
and 5Li.
4. Is it possible for 235Np to be a proton emitter and 235U to be a neutron emitter?
Is the binding energy of an alpha particle in an alpha emitter positive or negative?
The Average Binding Energy
The average binding energy, Eab, of a nuclide is its binding energy (BE) divided by the
number of nucleon in it (or mass number A), Eab, = BE/A. The average binding energy is also
called the packing fraction. This term is given to reflect the tighter the nucleons pack in a
nuclide, the more energy is released.

How does binding energy, BE, vary as the mass number A increases?
How does the average binding energy vary as A increase?
Which nuclide has the largest average binding energy, 16O, 56Fe or 235U?
Comparison of BE provides an indication about relative stability for similar nuclides, but BEs
of different nuclides are not based on the same numbers of H atoms and neutrons.
As we evaluate the average binding energy BEav of some of the nuclides, pay attention to the
variation of binding energy as A increases:
BEav (3He2) = 7.72 MeV / 3 = 2.57 MeV / nucleon
BEav (4He2) = 28.3 MeV / 4 = 7.08 MeV / nucleon
BEav (16O8) = 127.6 MeV / 16 = 7.98 MeV / nucleon
BEav (56Fe26) = 492.3 MeV / 56 = 8.79 MeV / nucleon
BEav (54Fe26) = 471.76 MeV / 54 = 8.74 MeV / nucleon
BEav (208Pb82) = 1636.44 MeV / 208 = 7.87 MeV / nucleon
BEav (238U92) = 1801.7. MeV / 238 = 7.57 MeV / nucleon
188
In general, the binding energy BE increases as A increases. For example, the binding energy
for 16O, 56Fe (mass = 55.934939 amu), 208Pb82 (mass = 207.976627) and 238U (mass =
238.050784) are 127.6, 492.3, 1636 and 1801.7 MeV respectively. There is a steady increase as
A increases.
There is a dramatic increase of BEav from 3He to 4He. Note that 4He, 16O and 208Pb are doublemagic-number nuclides, but their BEavs are lower than those of the two iron isotopes. In fact
BEavs for the two iron isotopes are the highest among the group, indicating a very high stability
of iron in terms of BEav. There is a very slight difference between BEavs of the two iron
isotopes, but only 54Fe contains a magic number (28) of neutrons.
A general trend of the variation of average
binding energy as a function of the mass
number is sketched here. As the atomic
number increases, the average binding energy
increases, reaching a peak at Fe, and then it
decreases gradually. From the average binding
energy point of view, nuclides with mass
number around 58 are the most stable.
Variation of the Average Binding Energy
as a Function of Mass Number A
BEa
v
Fe
Since the average binding energy for Fe is the
highest, synthesis of Fe from hydrogen atoms
and neutrons will release the most energy per
3
nucleon compared to all other nuclides. Thus,
He
energy will be released when light nuclides such
as H and D combines to form He, or even
when He combine to form C, O etc.
Combining light nuclides to form heavy ones is called nuclear fusion and it usually is
accompanied by a release of (fusion) energy.
U
A
When a heavy nuclide split up, it is called nuclear fission. Fission also releases energy. The
release of energy in fusion and fission is evident from the sketch.
The BEav is an indicator of energy frozen per nucleon in a nuclide. The more average energy is
released when a nuclide is synthesized, the less energy is frozen per nucleon in a nuclide. Thus,
BEav is a parameter for the stability of a nuclide, stable and unstable.
Skill Building Questions
1. Calculate the average binding energies of 11B (Mass, 11.00931), 30Si (29.97376), 52Cr (51.9405),
100
Ru (99.9030), 150Sm (149.9170), and 208Pb (207.9766).
2. Use a spreadsheet to evaluate the average binding energy of the above nuclides and those given in this section.
Use the spreadsheet to plot or sketch the negative values of BEav as a function of A.
189
Mass Excesses
The mass of 12C is defined as 12 atomic mass units (amu) exactly, and thus the average mass
of a nucleon in 12C is exactly 1 amu. The difference between the mass of a nuclide and its mass
number, A, is called the mass excess (ME),
ME = mass - A.
For example, the mass excesses of H (1.007825 amu) and neutron (1.008665 amu) are
0.007825 and 0.008665 amu respectively. Ironically, mass excess is also called mass deficiency
because many nuclides have negative mass excesses. Like binding energy and average binding
energy, mass excess is a convenient and useful criterion regarding a nuclide's stability.

How is the mass excess (ME) evaluated?
What is the ME for 12C?
What is the relationship between ME and BEav?
Sketch ME as a function of A, and use it to describe fusion and fission energy.
The mass excess of 3He (mass = 3.01603) is 0.01603 amu. This quantity is easily calculated for
nuclides whose average nucleon masses are more than 1 amu. For these nuclides, the mass
excesses are positive. The mass of 54Fe is 53.939612 amu for 54 nucleons, and the mass excess
is (53.939612 - 54.000000 =) -0.060388 amu, a negative value. In fact, MEs are negative for
most nuclides except the light ones (H, He, … C) and heavy ones (Ra, Th, U, … etc).
In general, the variation pattern of negative average binding energy (- BEav), and mass excess
(ME) are the same, but there is no direct way of converting ME into BEav and vice versa. A
comparison is given in a table form below:
Comparison Among ME. Eab.
Nuclide
H
n
3
He
4
He
12
C
16
O
40
Ca
54
Fe
56
Fe
208
Pb82
238 92
U
Mass/amu ME/amu
1.007825
1.008665
3.01603
4.00260
12.000000
15.994915
39.96259
53.939612
55.934939
207.976627
238.050784
0.007825
0.008665
0.01603
0.00260
0
-0.005085
-0.03741
-0.060388
-0.065061
-0.023373
0.050784
-Eab /amu BE/amu
0
0
-0.00276
-0.0076
-0.00825
-0.00857
-0.00917
-0.00938
-0.00944
-0.00845
-0.00813
0
0
0.00828
0.0304
0.09894
0.1369
0.3669
0.5065
0.52851
1.757
1.934
The binding energies of H atoms and neutrons are zero, 0. The more H atoms and/or
neutrons are combined, the more energy is released. Thus, the BE increases as mass number A
190
increases. The average binding energies (BEav) are calculated by treating H atoms and neutrons
in a similar manner.
The mass of 12C is a zero point of reference for ME. Since the mass of H atoms and neutrons
are different, ME values cannot be converted to BE values directly, but the two parameters
vary in a similar manner. Both ME and BEav are good indicators for the stability of a nuclide.
Because ME values are easily evaluated, they are widely used. For example, a Table of Nuclides
may choose to list ME for unstable nuclides instead of masses. In such a list, however, the unit
MeV is often used.
The variation of ME as a
Variation of ME with A
function of mass number A
for Some Stable Nuclides
is sketched here. The MEs
ME amu
for light nuclides oscillate
3
0.01 He
rapidly due to the effect of
nucleon pairing and magic
n
numbers of nucleons.
0.005
However, the variation for
H
medium and heavy nuclides
U
are not as dramatic. In
4He
general, the ME decreases as 0.0
12C
Pb
A increases from 1 to 56, but
Fe
–0.005
it increases as A increases
from 56 to 208. The ME
A
values indicate that nuclides
with A = ~56 are more
stable than light or heavy nuclides. Thus, combining light nuclides such as H, D, 3He, into
heavier nuclides 12C, 20Ca etc. in a nuclear fusion reaction will release energy, and splitting
heavy nuclides such as 235U in a nuclear fission reaction into lighter nuclides such as Sr, La,
Xe etc. will also release energy. The variation of ME as a function of A is an important piece of
information concerning nuclear energy.
Another application of ME is to use it to calculate the energy of radioactive decay. For
example, the ME’s of 40Sc21 and 40Ca20 from a handbook are -20.527 and -34.847 MeV
respectively. Thus the energy of the decay process
Sc21 
40
Ca20 + +
40
or
Sc21 + e– 
40
40
Ca20
is
Edecay = -20.527 - (-34.847) = 14.32 MeV
Of course, 1.02 MeV (2 times the rest mass of an electron) has to be spent in producing the
positron-electron pair in positron decay, but not in electron capture. This example illustrates
the fact that the ME not only vary as mass number A changes, it also vary as the atomic
number Z changes for isobars. Thus, a plot of ME versus Z and A shall produce a 3dimension plot. The three parameters are BE, A, and Z. We shall discuss the variation of ME
as Z changes for the isobars in the next section.
191
Skill Developing Questions
1. The masses are given for the following nuclides. Calculate the mass excesses in MeV:
2
6
D
2.0140 amu
Li
6.015121 amu
9
10
Be 9.012182
B 10.012937
11
13
B 11.009305
C 13.003355
14
14
C 14.003241
N 14.003074
2. Use the ME to calculate the decay energy for the decay of 14C  14N7 + –?
3. The ME’s of 40Sc21 and 40Ca20 from a handbook are -20.527 and -34.847 MeV respectively. What are
the masses of these two nuclides?
4. What is the difference between ME and BEav of a nuclide? Discuss the advantages and disadvantages of
their usage as indicators.
Mass Excesses of Isobars
Isobars have the same number of nucleons, but different numbers of protons and neutrons.
Their mass excesses are usually different. Thus, the mass excess among isobars varies as
atomic number, Z, changes.

How does mass excess vary as a function of atomic number Z?
Why is the variation of mass excess interesting for isobars?
What decay will decrease or increase the atomic number, Z?
What is the relationship between the mass excess the decay energy?
Energy is the driving force for change, and a change usually results in a system containing less
energy. Isobars convert to each other by the beta decay process, which change the atomic
number giving a more stable nuclide. Thus, the variation of mass excess as a function of Z is
interesting, because it shed some light about the beta decay process.
If the mass number is odd, all isobars are even-odd or odd-even type in numbers of protons
and neutrons, whatever Z is. These nuclides do not have a special contribution to ME due to
pairing of protons or neutrons, because in either type, there is one unpaired nucleon.
As an example, the mass excesses of isobars with an arbitrary mass number of 123 are given
below:
In49
-0.0896
Sn50
-0.0943
Sb51
-0.0958
Te52
I-53
-0.0967 -0.0944
Xe54
-0.0915
Cs55
Ba56
-0.0870 -0.0808 amu
Note that all the mass excesses are negative. These values are usually available from a
handbook or from a table of nuclides. However, the units may be in MeV or keV.
192
For these isobars, the two with the lowest
Mass Excesses of Isobars
mass excesses, Sb and Te, are stable. For a
with Mass number 123
comparison, the mass excesses are plotted
as a function of Z as shown here. The
-0.08 amu
further away a nuclide is from its stable
isobars, the higher (less negative) are their
+, EC
mass excesses. If we connect all these points
–

-0.09
to form a curve, its shape looks like a
hyperbola. This meaning that ME varies as a
function of Z2.
-0.10 In
Since the relative mass excesses of the
49
isobars are directly related to their masses,
their locations on the diagram represent
their thermodynamic stability. The lower the
position, the more stable is the isobar.
Sn Sb Te I Xe Cs Ba
50 51 52 53 54 55 56
Although mass excess cannot be converted to binding energy
directly, the two quantities are similar, especially if –BE is
used. For the isobars with mass number 57, the BEs and BEs are listed and plotted here:
Isobars of Mass number 57
Mass
/amu
BE
/amu
-BE
/MeV
Cr24
Mn25
Fe26
Co27
Ni28
56.9434 56.9383 56.9354 56.9363 56.3980
Isobars with Mass
Number 57
-494 –
Cr
-495
-496
-497
0.53031 0.53462 0.53667 0.53493 0.53240
-498
-494
-499
-498
-500
-498
-496
For a small mass number of 57, the number of isobars listed
in handbooks is limited. Isobars with atomic number outside
the 24 - 29 range are too unstable to be observed.
_
Ni
+
EC
–
-500
–
Mn
24
25
–
Co
–
Fe
27
28
For mass numbers 57 and 123, the effect of nucleon pairing is not present, because all these
isobars have either the protons or neutrons paired, but not both. The effect of nucleon pairing
is important for isobars with even mass numbers, because the beta decay process convert a
even-even (paired) nuclides into odd-odd ones, which have two unpaired nucleons. Isobars
with even mass numbers are discussed next.
Isobars with 120 nucleons have two stable nuclides 120Sn50 (mass = 119.902200 amu, mass
excess = -0.097800 amu = –91.1 MeV) and 120Te52 (mass = 119.904048 amu, mass excess =
0.095952 amu). The former has a magic number of protons.
193
The atomic numbers Z, element symbols (Isobar), mass differences (Mass) [compared to
120
Sn50, which is taken as zero] decay energies (Edecay), decay modes, and half lives, are given
below:
Z
47
Isobar
Ag
Mass/MeV 15.32
Edecay/MeV 8.2
Decay mode –
Half life
1.2 s
48
Cd
7.13
1.80
–
50 s
49
In
5.3
5.3
–
44 s
50
Sn
0
stable
–
51
Sb
1.51
2.68
+,EC
15.9 m
The decay energy and mass differences are
consistent for isobars which undergo – decay,
but for those undergo + and EC decays, there
is a little discrepancy, perhaps due to
production of positrons.
The mass differences (Mass) can be
converted to mass excesses by subtracting 91.1
MeV, which is the mass excess of 120Sn (ME =
(Mass - 91.1) MeV). For simplicity, the mass
differences are plotted against the atomic
number Z here. From the graph, the - decay
series Ag (- -)  Cd (- -) In (- -)  Sn,
reduces the energy of a nuclide via several
isobars in steps. So does the decay series by +
emission or electron capture that convert Ba
via Cs, Xe, I, to Te in five steps.
The stable and even-even type isobar 120Te52
contains slightly higher energy than the
radioactive and odd-odd type nuclide 120Sb51,
which is converted to the most stable isobar
120
Sn50, which has the lowest mass (or energy).
The two decay modes + and EC of 120Sb51
occur at 41 % and 59 % respectively.
52
Te
1.72
stable
–
–
53
I
7.12
5.4
+,EC
1.35 h
54
Xe
9.07
2.0
+,EC
40 m
55
Cs
17.3
8.22
+,EC
64 s
56
Ba
22.6
5.3
+, EC
32 s
Relative Mass Differences (MeV) for
Isobars with Mass Number 120.
Ba
22
20
18
Cs
16
Ag
14
12
10
8
6
4
2
0
Xe
Cd
I
In
Sb Te
Sn
47 48 49 50 51 52 53 54 55 56
To see the effect of nucleon pairing, we connect the even-even type and the odd-odd isobars
by lines. These lines form two hyperbolas.
There is a difference in Mass for odd-odd type nuclides and the even-even type nuclides due
to the effect of nucleon pairing. The even-even and odd-odd nuclides appear to belong to two
separate classes. The gap between these two types of nuclides depends on the mass number A.
An average over many nuclides has given a difference of 250/A MeV. For A = 120, the gap is
1 MeV, and this value seems reasonable from the two curves in the plot.
194
The variation of binding energy for mass 57 represents typical isobars with odd number
nuclons, and the variation of binding energy for mass 120 reflects the importance of nucleon
pairing. In both examples, there are two stable nuclides, but more often, there is only one
stable nuclide for a series of isobars. From these graphs, we gain a little more insight to the
energy for the beta decay process.
Skill Building Questions
1. List atomic numbers Z, element symbols (Isobar), mass excess (ME) decay energies (Edecay), decay
modes, and half lives for isobars with mass number A = 140 and 141 respectively. (Please find
the data from a handbook)
2.
Plot the - BE of the nuclides given above versus atomic number.
3. Use a spreadsheet to plot the variation of ME as functions of both Z and A for a selected group of nuclides,
for example Z = 1 to 20 or for Z=21 to 40.
A Semi-empirical Equation for Binding Energy
Binding energy (BE) varies in a systematic manner as a function of mass number A and atomic
number Z.

Can the systematic variation be represented by a formula?
What a theoretical model can be proposed for variations of BE?
The variation of BE is complicated but somewhat systematic and a model for calculating BE
has been proposed by Seeger in 1961. He used a semi-empirical approach. The equation is
based on some theoretical consideration, but mostly based on empirical data (the masses of
nuclides). His theoretical consideration is based on the liquid-drop model, in which the nucleus
volume is directly proportional to A. Its radius is proportional to the cubic root of A (A1/3)
and its surface area is proportional to the square of its radius A2/3. The function BE as a
function of A and Z is,
2/3
BE(A,Z) = 14.1A - 13A
2
0.6Z 2 20( A  2 Z )
- 1/ 3 + Eo.
A
A
For nuclides with A  80, the equation gives reasonable results (in MeV). The rational for
various terms are described below:
1. The 1st term (14.1 A) shows BE being proportional to the number of nucleons A. The
more nucleons that pack into a nucleus, the more energy is released. Thus, BE is
proportional to A.
2. The 2nd (negative) term (- 13A2/3) indicates an increase in surface area causes a decrease in
stability. The factor A2/3 reflects the square of the radius (A1/3).
195
0.6Z 2
represents the instability caused by protons due to
A1 / 3
Coulomb interaction, which is proportional to Z2, but inversely proportional to the radius.
3. The 3rd (negative) term
20( A  2 Z ) 2
4. The 4th (negative) term
attributes to a favorable number-of-neutron to
A
number-of-proton (N/Z) ratio for a nucleus of mass A, and the instability is proportional to
(A - 2 Z)2 but inversely proportional to A.
5. The last term (Eo) is a constant reflecting stability due to pairing of nucleons, Eo = 0 for
odd-even and even-odd nuclides, = -125/A for even-even nuclides, and = +125/A for
odd-odd nuclides.
All isobars have the same value A, and their binding energy is thus a function of Z and Z2. For
isobars with an odd mass number, the nuclides are either odd-even or even-odd type, and the
last term is zero. This equation suggests that a plot of BE for a set of isobars as a function of
Z is a parabola with its vertex at the stable isotope. For isobars with an odd number of
nucleons, there are usually one or two nuclides that contain the least amount of energy. These
are stable isobars.
For isobars with even mass numbers, the equation suggests two parabolas, one corresponding
to the odd-odd type, and one to the even-even type due to the different values of the last term
E0.
Skill Building Questions
1. Use the semi-empirical equation to calculate the binding energies for the isobars with mass number 120
giving earlier. Compare the values with those listed earlier and comment on your results.
2. Apply the semi-empirical equation to calculate the binding energy for Cr24, Mn25, Fe26, Co27, and Ni28 for
mass number 57, and discuss your results.
196
Exercises
1. What are the lightest and heaviest stable nuclides? Which element has the largest number
of isotopes? What elements with atomic number less than 83 are absent and why? What
are some of the common features of these absent nuclides? What roughly is the number
of stable nuclides?
2. What are the ratios of number-of-neutrons to number-of-protons ratio (N/Z) for the stable
nuclides: 1H, 14N, 35Cl, 56Fe, 80Br, 100Rh, 150Sm, 200Hg, 209Bi? Describe the variation of N/Z
ratios as the mass number increases.
3. The heaviest stable nuclide is 209Bi (Z = 83). How many neutrons are there in the nucleus?
What is its N/Z ratio? (Ans. 126 n and 1.52)
4. Give the four stable nuclides that have equal and odd numbers of protons and neutrons?
5. What evidences support the conclusion that nuclides with even numbers of protons and
neutrons are more stable? What theory can you give to explain this fact?
6. List a summary of the considered factors that affect the stability of nuclides.
7. List some features about the stable nuclides. For example, explain why there are six stable
nuclides for the element hafnium but only two each for lutetium and tantalum?
8. Is there a relationship between abundance of nuclides and their stability? If so, what
evidence can you give? If not, provide some arguments.
9. Enumerate the magic numbers. Why are they called magic numbers? Give some double
magic-number nuclides? Comment on the fact that 209Bi (Z = 83) is stable, whereas 209Pb
(Z = 82) is a positron emitter.
10. What are the most abundant nuclides in the solar system and in the planet earth? How are
the estimates made? Comment on your answer.
11. Calculate the binding energies (BE) and average binding energies of 56Fe (Z = 26, mass =
55.9349 amu), and 238U (Z = 92, mass = 238.0508). (Ans. BE = -492.3 MeV and -1801.7 MeV
respectively.)
12. Use a spreadsheet to evaluate the average binding energy of the above nuclides and those
given in this section. Use the spreadsheet to plot or sketch the negative values of Eab as a
function of A (one of the Skill Building Questions).
13. The disintegration energies for isobars with mass number 133 are given below. Make a
plot of energy content as a function of Z for these isobars from the disintegration energy
(in MeV). Te (52), 2.4; I (53), 1.8; Xe (54), 0.43; Cs (55), stable; Ba (56), 0.489; La (57), 1.2l
Ce(58), 1.8. Do the points fall on a parabola?
197
14. Calculate disintegration energies of 133Ba and 133La using the function of binding energy?
15. Plot the binding energy of isobars with mass number 114 as a function of Z. Which
nuclide is stable? (Check the properties of nuclides from a handbook please.)
16. Discuss the stability of isobars in terms of binding energy (BE). In particular, how does
BE vary as a function of the atomic number Z and mass number A? What decay process
is responsible for the transmutation of isobars? How does a function for the calculation of
BE takes the effect of even or odd number of protons and neutrons into account?
Further Reading and Work Cited
Harvey, B.G. (1969), Introduction to nuclear physics and chemistry, 2nd Ed. Prentice Hall.
Kuroda, P.K. (1982), The origin of the chemical elements and the oklo phenomenon, Springer-Verlag.
Lederer, C.M., Hollander, J.M. and Perlman, I. (1986) Table of isotopes 8th Ed.
Noddack, I. and Noddack, W. (1930), Naturwissenschaften 18, 757.
Pagel, B. (1965), New Scientist, 26, 103.
Seeger, P.A. (1961) Nuclear Physics, 25, 1
Suess, H.E. and Urey, H.C. (1956), Rev. Mod. Phys. 28, 53.
Useful Web Sites
The following web sites contain very handy data about nuclides.
http://www.nndc.bnl.gov/
from that, choose the Nuclear Wallet Cards to go to
http://www.nndc.bnl.gov/wallet/walletcontent.shtml
On it is a list of atomic numbers, such as
0-n
10-Ne
20-Ca
1-H
11-Na
21-Sc
2-He
12-Mg
22-Ti
3-Li
13-Al
23-V
4-Be
14-Si
24-Cr
5-B
15-P
25-Mn
6-C
16-S
26-Fe
7-N
17-Cl
27-Co
8-O
18-Ar
28-Ni
9-F
19-K
29-Cu
Each of these items is hyper linked to another file which contains data on all isotopes of the
elements. Access these data is easier than looking up from a handbook.
If you click the NuDat, you will go to the web site http://www.nndc.bnl.gov/nndc/nudat/
and there are various choices of nuclear data.
198