Structure of the nucleus
A bit of History…
~400 BC:
Greek Philosopher Democritus believed that each kind of
matter could be subdivided into smaller and smaller bits until
one reached the very limit beyond which no further division
was possible.
“atomos” = “that cannot be cut”
1895:
Discovery of X-rays by Wilhelm Röntgen
1896:
Discovery of the radioactivity in Uranium by Henri Becquerel
1898:
Isolation of Radium by Pierre et Marie Curie
 Around 1896: Birth of Nuclear Physics
Radioactivity ?
Emission of energy stored in the material by the mean of “mysterious” rays
~1900: The structure of the atom is not yet known... But Chemistry is !
(Mendeleiv 1834 – 1907)
The radiation emitted by Radium is identified to be the element Helium
1911: Discovery of the atomic nucleus by Ernest Rutherford
The first experiment with a beam of particles:
The α-particles are most of the time not deflected,
but sometimes they are scattered, even in the
backward hemisphere.
 the foil is almost “transparent” to the α’s
But when interaction occurs, it is on a heavy partner
in the foil (the nucleus)
 full analysis: size of the nucleus, mass, etc…
Elementary Constituents
of the Nucleus
All the nuclei can be made with:
p
proton
n
neutron
Z
N
A
positively charge (+q)
neutral (James Chadwick, 1932)
number of protons in the nucleus
number of neutrons in the nucleus
atomic number = Z + N
The atoms are neutral:
Charge of the nucleus: + Z.q
Electron cloud is made of Z electrons of charge (–q)
The electrons determine the chemical behavior  Z defines the Element
Two nuclei with a same Z but different N (or A) are isotopes (of the same element)
Discovery of the neutron
•
In 1930 the German physicists Bothe and Becker used a radioactive
polonium source that emitted α particles. When these α particles
bombarded beryllium, the radiation penetrated several centimeters of lead.
In 1932 Chadwick proposed that the new
radiation produced by α + Be consisted of
neutrons. His experimental data estimated
the neutron’s mass as somewhere between
1.005 u and 1.008 u, not far from the
modern value of 1.0087 u.
1 atomic mass unit (a.m.u. or u) = 1.6605 x 10-27 kg = 931.49 MeV/c2 = (1/12) m(12C)
Mass of the proton: mp = 1.6726 x 10-27 kg = 938.27 MeV/c2
Mass of the neutron: mn = 1.6749 x 10-27 kg = 939.57 MeV/c2
Notations
The symbol of an atomic nucleus is
.
where Z = atomic number (number of protons)
N = neutron number (number of neutrons)
A = mass number (Z + N)
X = chemical element symbol
A
Z
XN
1
Ex:
1
H0
Hydrogen
2
1
H1
Deuterium
12
6
C6
Carbon
U143
Uranium
Other examples:
235
92
Isotopes of the same element
12
Simplified:
C
235
U
In practice, the Z number is redundant with the element symbol & the N number is obsolete
The Chart of Nuclei
N=Z
Stable isotopes (~270), up to 209Bi (Z=83)
Unstable (radioactive) isotopes (>2000 and counting)
Beyond the nucleon…
•
Both neutrons and protons, collectively called nucleons, are
constructed of other particles called quarks.
•
The (non-zero) magnetic moment of the neutron is evidence of an
underlying structure in the nucleon.
Mass of
the proton
Units and Dimensions
-15
Size: the order of magnitude is 1fm = 10 m (a femtometer or fermi)
1/3
Radius (assuming nucleus = sphere): R = R0.A with R0 = 1.2 fm
[ Nucleus made of A nucleons, each of them has a R0=1.2fm radius
3
 Vnucleon = (4/3)πR0 ; Vnucleus = A . Vnucleon ]
17
Nuclear Density: 10
kg/m3  100 million tons per cm3 !!!
density found in the core of a neutron star
Nuclear matter is uncompressible (properties of the strong force)
Binding energy =mass of the constituents – mass of the product
Atom
Nucleus
Force
Coulomb
Strong
Binding Energy
The hydrogen
atom: 13.6 eV
2H:
2.2 x 106 eV
 2.2 MeV
Enrico Fermi
Exercise
• Calculate the density of 238U (M=238.05 u)
238
3
1/3
Assuming that the U nucleus is a sphere: V = (4/3)πR and R = R0.A ,
3
the mass is: M = ρ . (4/3)AπR0
Density of the
nuclear Matter
Volume V of
the nucleus
-27
-25
M = 238.05 x 1.6605 x 10 = 3.953 x 10 kg
-15 3
-42
3
V = (4/3) x 238 x 3.14159 x (1.2 x 10 ) = 1.723 x 10 m
17
3
 ρ = M/V = 2.3 x 10 kg/m
• Based on this density, what would be the weight
of a 1cm radius sphere of nuclear matter?
3
3
-6
3
1cm radius sphere: V = (4/3)πR = (4/3) x 3.14159 x (0.01) = 4.2 x 10 m
11
Mass of the sphere with the nuclear matter density: M = ρ x V = 9.6 x 10 kg
That’s about the mass equivalent of a 1-km sphere of ordinary matter!
The Nuclear / Strong Force
 Short Range, only a few fm
Nuclear Matter is uncompressible ; repulsive at very short range (< 1 fm)
Attractive over a range of a few fm
a given nucleon only interacts with its next neighbors in the nucleus
 Negligible at long distances
 Same force for protons and neutrons
Nuclear properties
• Protons and Neutrons are Fermions (half-integer spin
particles)  Fermi-Dirac statistics
• Density ~ constant
“Skin” e.g. the nucleus has
a diffuse surface
(Valid for light nuclei)
How to calculate
the binding energy ?
The hydrogen atom:
we win energy by bringing together the proton and the electron (13.6 eV)
Equivalence Energy  Mass: E = m.c2
Binding energy (B) = mass of the constituents – mass of the product
= (mpc2+mec2) – mHc2
= 13.6 eV
with: mpc2 = rest mass of the proton = 938.28 MeV
mec2 = rest mass of the electron = 0.511 MeV = 511 keV
 the Hydrogen atom weighs less than its constituents
 One needs to provide 13.6 eV to take the hydrogen atom apart
Binding energy is also called Mass deficiency
Any atom (defined by Z): B = mNucleusc2 + Z. mec2 – mAtomc2
 How do we calculate mNucleusc2 ?
Atomic Mass Unit (amu)
•
The atomic mass unit (amu or u) is defined as 1/12th of the mass of
12C with all its electrons, that is:
– 1uc2 = (1/12) x M(12C)c2 = (1/12) x [6mp+6me-+6mn]c2
•
All masses given in amu are the ATOMIC mass, i.e. they include Z
electrons as well.
•
In order to take into account the electron mass in the calculations,
we replace (mp+me-) by mH, the mass of the hydrogen atom
including the electron.
– B(AZX) = (Z.mpc2 + Z.mec2 + N.mnc2) – mXc2 becomes:
– B(AZX) = (Z.mHc2 + N.mnc2) – mXc2
•
Instead of using: mp = 1.007276u, use mH = 1.007825u
Binding energy of the nucleus
The nucleus:
1H nucleus = free proton  Binding energy = 0 (but still has a rest mass)
2H (deuterium) = one proton and one neutron
 B = (mHc2 + mnc2) – mDc2 = 2.2 MeV
A
ZX
: B = (Z.mHc2 + N.mnc2) – mXc2
Example:
56Fe
(Z=26)
m56Fe = 55.934969 u (or amu, atomic mass unit = 931.5 MeV/c2)
with: mH = 1.007825 u & mn = 1.008665 u
B(56Fe) = 26 x 1.007825u.c2 + 30 x 1.008665u.c2 – 55.934969u.c2
= 0.528461u.c2 = 492.3 MeV
Very large number compared to B’s in atomic physics
Manufacturing 1g of 56Fe from protons and neutrons per second would generate 8.5.1011 Watt !!!
 850 GWatt
Average Binding Energy
Average binding energy produced by the strong force can be expressed by
dividing the total Binding Energy of the nucleus by its mass number (B/A)
B/A ~ 7-8 MeV is a typical value
The Liquid Drop Model (I)
• Goal: estimate the binding energy of a given nucleus
with a “semi-empirical” model.
– Nucleus = Collection of interacting particles in a liquid drop of
nuclear matter
Symmetry term: in the absence of the Coulomb
force, the nucleus prefers to have N ≈ Z
Volume term:
The binding energy is
approximately the sum
of all the interactions
between the nucleons.
Surface term:
Correction for the
nucleons at the
surface of the nucleus
(interact with less
neighbors)
Pairing term
Coulomb term:
Adding more protons
increases the Coulomb
repulsion within the nucleus
The Liquid Drop Model (II)
One can show that the Coulomb term can
also be written (See example 12.5 p436):
-1/3
Ec = 0.72 Z(Z-1) A MeV
-¾
With Δ = 33 A
MeV
The Liquid Drop Model (III)
Example:
Binding energy
per nucleon
(Ex. 12.7 p439):
Liquid Drop Model
20Ne
(Z=10, N=10)
8.03 MeV / nucleon
8.19 MeV / nucleon
56Fe
(Z=26, N=30)
8.79 MeV / nucleon
8.35 MeV / nucleon
238U
(Z=92, N=146)
7.57 MeV / nucleon
6.84 MeV / nucleon