Binding Energy and Nuclear Forces
The force that binds the nucleons together is
called the strong nuclear force. It is a very
strong, but short-range, force. It is essentially
zero if the nucleons are more than about 10-15
m apart. The Coulomb force is long-range; this
is why extra neutrons are needed for stability
in high-Z nuclei.
Calculate that at 1 fm:
VCoulomb 
e2
40 r
 1.44MeV
Structure and Properties of the Nucleus
Nuclei: protons and neutrons.
Proton has positive charge;
mn = 1.67262 x 10-27 kg
Neutron is electrically neutral;
mn = 1.67493 x 10-27 kg
Number of protons: atomic number, Z
Number of nucleons: atomic mass number, A
Neutron number: N = A - Z
Symbol:
Isotopes

mp
me
 1836.152 672 61 (85)
No theory
Extreme density
Because of wave–particle duality, the size of the nucleus is somewhat
fuzzy. Measurements of high-energy electron scattering yield:
Masses scale: carbon-12 atom, 12 u
u is a unified atomic mass unit.
Magnetic moments
For a classical point particle

L

  gLB

g L 1
Define nuclear magnetic moments
(unit = the nuclear magneton)
N 
e
2M p
For a relativistic spin (point)

S

S   g S B

Proton magnetic moment
gS 2
 p  2.7928 N
Neutron magnetic moment
n  1.9135 N
Protons and neutrons are not point particles
Binding Energy and Einsteins E=mc2
Binding energy affects mass of composite particle
mH-atom < mproton + melectron
mH-atom(n=1 state) < mH-atom(n=2 state)
Eb atom   M nucleus   Zme  M atom c 2
(mind the signs !)
These effects are much larger in the realm of nuclei
than in that of electromagnetism


Eb nucleus   ZM p  NM n  M nucleus  c 2
Binding Energy and Nuclear Forces


Eb nucleus   ZM p  NM n  M nucleus  c 2
The total mass of a stable nucleus is always less than the sum of the
masses of its separate protons and neutrons.


Eb nucleus   ZM p  NM n  Zme  M atom  c 2
Here we neglect of the binding energy inside the atom (eV).
Eb
A X  ZM 1H NM n  M A Xc2
Binding of nucleus
Mass of neutral atom
This can be used to calculate Eb/A: binding energy per nucleus
Binding Energy and Nuclear Forces
From observation: binding energy per nucleon: Eb/A
Note:
a is point of
relative stability
Binding Energy and Nuclear Forces
From observation:
The higher the binding
energy per nucleon, the
more stable the nucleus.
More massive nuclei
require extra neutrons to
overcome the Coulomb
repulsion of the protons
in order to be stable.
Addition
The semi-empirical mass formula
Von Weizsäcker
Liquid drop model
Eb
A X  a1 A  a2 A2 / 3  a3 A1/ 3  a4
Z2
 A / 2  Z 2  
A
5
Note:
R ~ A1 / 3
Volume term
3
~R ~A
a1  15.76Mev
a2  17.81Mev
Good fit for:
a3  0.7105Mev
a4  94.80Mev
a5  39Mev
Surface correction:
on the outer surface (4R2)
the binding is less, because
There are no particles to
contribute
~ R 2 ~ A2 / 3
Addition
The semi-empirical mass formula
Von Weizsäcker
Liquid drop model
Vrep
Eb
A X  a1 A  a2 A2 / 3  a3 A1/ 3  a4
Z2

3 Ze 2

5 40 R
a5
A3 / 4
0
Coulomb energy stored
in a uniform solid sphere
of charge Ze and radius R
(negative binding)
 A / 2  Z 2  
~
Z2
A1 / 3
A
ee
oo
eo
oe
“pairing energy”
(not further discussed)
5
Addition
Protons and Neutrons in the Fermi-gas model
 a4
 A / 2  Z 2
A
Both protons and neutrons cannot
fill the lowest “orbitals”
because the have to follow the
Pauli exclusion principle
“ Symmetry
Energy”
preference for
ZN
Based on this concept:
the “nuclear shell model”
"for the discovery concerning
nuclear shell structure"
Maria Goeppert-Mayer
Addition
Isobars and the mass equation
Mass of nucleus:
M

 2

Z2
A / 2  Z 2
2/3
X  ZM H  ( A  Z ) M n  a1 A  a2 A  a3 1 / 3  a4
 5  / c
A
A


 
A
Isobar:
 
1
Z  A / 2
M 2
Z
c  M 1 H  M n c 2  2a3 1/ 3  a4
Z
A
A
  
Minimum charge Z:
  

A a4  M 1 H  M n c 2
ZA 
2
a4  a3 A2 / 3
Valley of stability

(even-odd
separate
due to 5 term)
Unstable elements
t = 2 x 106 y
99Tc t = 2 x 105 y
97Tc
t = 18 y
146Pm t = 5.5 y
145Pm
Stable/Meta-stable elements
126Te
stable, 18.95 %
128Te t > 5 x 1024 y
130Te
t = 2.5 x 1021 y
t = 1.6 x 105 y
235U t = 7 x 108 y
238U t = 4.5 x 109 y
233U
Why do nuclei decay  radioactivity