Download The Physics of Energy sources Basic Nuclear Physics – The Atom

The Physics of Energy sources
Basic Nuclear Physics – The Atom
This part of the course is based on material from different sources and mainly:
• Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006) – ref 2
• Special thanks to previous lecturer (Pr. King) for providing lecture notes and examples
The atom
1
The major players (but not all of them)
Photograph of participants of the first Solvay Conference (Brussels) - 1911:
First world Physics conference. Photograph taken by Benjamin Couprie
Seated (L-R): Walther Nernst, Marcel Brillouin, Ernest Solvay, Hendrik Lorentz, Emil Warburg, Jean Baptiste Perrin, Wilhelm Wien,
Marie Curie, and Henri Poincaré.
Standing (L-R): Robert Goldschmidt, Max Planck, Heinrich Rubens, Arnold Sommerfeld, Frederick Lindemann, Maurice de Broglie,
Martin Knudsen, Friedrich Hasenöhrl, Georges Hostelet, Edouard Herzen, James Hopwood Jeans, Ernest Rutherford, Heike Kamerlingh
Onnes, Albert Einstein, and Paul Langevin
The atom
2
History of the atom – Early discoveries
Most of these discoveries occurred through experimentations
"   1886 – 1898 Discovery of radioactivity
"   Henri Becquerel discover accidentally radioactivity (1886)
"   Further work by Pierre and Marie Curie over several years
"   They all share the Physics Nobel Prize in 1903. Then in 1911 Marie Curie get
the Chemistry Nobel prize for the discovery of Radium and Polonium
"   1900-1935 The nuclear nature of atoms
"   J.J. Thomson – Discovery of electron – Physics Nobel prize 1906
"   Then most of this work achieved by Ernest Rutherford and his students
"   Early 1900s Rutherford experimental work lead to confirmation of the
planetary model of the atom (positive nucleus surrounded by electrons) 
Chemistry Nobel prize in 1908
"   Under his supervision, his students will
•  Understand the atomic structure (quantum physics) – Bohr: Phys. Nobel prize 1922
•  Discover the existence of Neutron – Chadwick: Phys. Nobel prize 1935
The atom
3
The second epoch - Theory
"   From 1930s Quantum Mechanics
"   The basis of the new theories are developed to explain the atom
"   Schrodinger and Dirac (Phys NP 1933)
"   Pauli – Exclusion principle – Phys. NP 1945
"   Fermi, Einstein + many others
"   Statistical Physics of the atom
"   β decay explanations leading to fission
"   Fermi will lead the Manhattan project  First artificial nuclear reactor (1942)
•  Leading (un)-fortunately to the first fission bomb in 1945
"   After the 2nd World War
"
"
"
"
The atom
 
 
 
 
A lot of effort put in the technological development of fission (cold war)
Particles Physics theories + discoveries (quarks….)
Colliders such as CERN
But no controlled fusion still….
4
The atomic structure
"   An atom consist of a nucleus around which electrons are orbiting
"   Early 20th century: atom contains + and – charges but 2 models
•  Rutherford – planetary system with electrons orbiting a positively charged nucleus
•  Thompson – electrons embedded in region filled with positive charges
Breakthrough happened with back scattering experiment
Rutherford, Geiger and Marsden (Manchester 1909)
Study of interaction of α particles through matter
α particles (positively charged) fired at gold foil
Results: some particles are sometimes deflected
through large angles and even in the backward
direction
α
That can be explained only by the existence of a small
α
but heavy, charged entity less than 1/1000th of the
atom in size.
The atom
Figures adapted from wikipedia
Thompson model:
α part. should deviate
slightly due to small
charges but as an
average no deviation
should occur
Rutherford model:
α part. could deviate
with large angles
(even go back) due to
small but heavy
charged nucleus
5
The electronic structure
"   Theoretical work on quantum Physics (Bohr, Schrodinger,
Pauli…) explain the electronic structure of the atoms
"   They surround the nucleus on stable orbits (with associated energy)
"   These orbits represent the most probable positions of the electrons around
the nucleus
"   We cannot have 2 electrons in exactly the same state of energy
"   Electrons can transit from one orbit to another through energy exchange
•  Photo-electric effect
"   Electronic structure is responsible for the most of the properties
of the atoms and molecules
"   Explain the EM spectrum of emission/absorption
"   The “top layer” of the electronic structure explains many properties
•  Valence electrons
•  Chemical properties: such as binding between atoms, electro-negativity
•  Physical properties such as thermal and electrical conductivity of materials
"   Atomic energies are fairly small in comparison to nuclear ones
"   In eV typically. Ex: ionisation of Hydrogen atom=13.6eV
The atom
6
The nucleus
"   The nucleus is much smaller than the atom:
"   Typical atomic size: 10-10m
"   Nucleus size ~ 10-15m= 1femtometer (fm) or fermi
"   However the nucleus contains most of the mass
"   What is making up the nucleus?
Figure adapted from wikipedia
"   It was know that the electrons have a negative charge
"   The nucleus had to be made of basic positive charges to balance the e- charges
"   That did not explain the masses of the atoms
•  Helium which should have 2 protons has four times the mass of Hydrogen….
•  Some elements have a relative atomic mass which is not a integer…. (see after)
"   There should be another nuclear particle, with about the proton mass but
without charge  Neutron
"   So nuclei are made of 2 types of nucleons: protons and neutrons
The atom
7
Notation
Notation:
Number of protons: Atomic number Z = number of electrons
Number of neutrons: Neutron number N
Atomic mass number A = Z + N = total number of nucleons
We write this as
A
Z
XN
Electron: eProton: p or e+
Neutron: n
where, X: chemical symbol.
Usually abbreviate to AX e.g. 238U. This is enough as the name of the
chemical element will uniquely specify Z
Each nuclear specie is called a nuclide
We define the unified atomic mass unit (u) or dalton (Da) as one twelfth of the
mass of an atom of 12C at rest and in its ground state.
Because by definition 1mole of 12C weights exactly 12g and contains NA atoms
NA being Avogadro s number = 6.022x1023 atoms
1
1u =
g = 1.6605 ×10 − 27 kg
NA
The atom
Mass of proton = 1.007276u
Mass of neutron = 1.008665u
Mass of electron = 5.485799x10-4u
8
Isotopes
"   It was found out that some elements had an anomalous atomic
weight such as chlorine with an atomic mass of 35.5
"   In 1911, Soddy put forward the idea of elements consisting of
different isotopes: nuclei having different mass but the same charge
Same value of Z but different N
Example 1: Chlorine (Z=17)
76% is composed of 35Cl and 24% of 37Cl making an average atomic mass of 35.5
Example 2: Hydrogen has 3 isotopes
1
1
H 0 protium, 99.985% of all Hydrogen
2
1
H1 deuterium, 0.015% of all Hydrogen
3
1
H 2 tritium
"   Isotopes have slightly different physical properties that can be used
to separate them.
The atom
9
Isotones and Isobars
"   Isotones:
Nuclei with the same N (neutrons) and different Z (protons)
"   Isobars
Nuclei with the same A (total number of nucleons)
There are 92 elements but 270 stable nuclei.
The atom
10
Nuclei plot
Line of stable nuclei
Ref 2
Stable and unstable nuclei plotted as a function of proton number Z and neutron number N
The atom
11
Atomic mass determination
Mass of electron << mass of nucleon  Mass of nucleus ~ Mass of the atom
A mass spectrometer will:
• Vaporise the sample
• Ionise the atoms  single charges atoms (q=+1)
• Accelerate and filter the speed of the ions
• Modify the charged particle beam (Magn. Field)
• Spatially separate the various ions (as a function
of mass)
Inside the magnet each charged ion will:
Have a a constant speed v
Be subjected to 2 forces


motion
Fm = ma



field
F f = qv × B


B is perpendicular to v
Forces at equilibrium 
ions will follow a circular path
mv 2
Fm =
r
mv 2
= qvB
r
mv
r: curvature radius
r=
qB
The atom
Figure adapted from wikipedia
Each mass appears at a different position in the detector plan
Typical accuracy: 10-6 of atomic mass
12
What is holding the nucleus together ?
"   Protons are in a confined within the nucleus
"   They all have the same charge  they should be repulsed
"   So it exists a counter-reacting force that holds them together
"   Indeed there is a nuclear force between nucleons (p and n)
•  Nuclear force or nucleon-nucleon interaction or residual strong force
"   This force is a short range one, quite different from the EM forces.
"   It exists only inside the nuclear radius
Ref 2
Repulsive Coulomb Potential
energy (1/r variation)
Nucleus radius
Proton
Neutron
Attractive nuclear
interaction
The atom
B: Coulomb barrier
13
How to determine the nuclear size?
Let s go back to Rutherford s back scattering experiment!
α particles are actually He nuclei
(positively charged)
4
2
He2
When α particles are fired at the gold foil, the ones back
scattered at 180deg reach the closest to the nucleus
Back scattered
at 180deg
α
α part. moves in Coulomb potential of the gold nucleus
Z Au Zα e 2
V (r ) =
4πε 0 r
Figure adapted from wikipedia
Note: Coulomb force varies as r-2
When the particle moves towards the nucleus its kinetic energy will be counter-balanced by
the repulsive Coulomb potential until KE=PE (when the particle is the reach the closest to
nucleus), then goes backwards.
2
2
Z
Z
e
Z
Z
e
Au
α
Au
α
When the particle stops we have: KE =
Z Au = 79
→r =
4πε 0 rc
c
4πε 0 KE
If we have KE=4.7844MeV for instance
−19 2
79
×
2
×
(
1
.
6
×
10
)
−15
rc = 9 ×109
=
47
.
6
×
10
m
6
−19
4.7844 ×10 ×1.6 ×10
The atom
rnucleus < rc
14
Liquid drop model
"   In Rutherford s experiment, as we increase the a particle
energy, we go closer towards the edge of the nucleus.
"   Results of scattering experiments show that the nuclear radius
can be written as:
r = r0 A1/ 3 , where r0 = 1.2fm, A being the atomic mass number
for Gold, 197Au, r = 1.2(197)1/ 3 = 7fm
This A1/3 dependence suggests the idea of a liquid drop model of the nucleus
ro represents the radius taken by one nucleon. So when the nucleons get in contact
with each other (nuclear force dominates), the nucleus radius grows as A1/3
The nucleus behaves like an accumulation of neutrons and protons forming a droplet of
incompressible fluid (constant density). This liquid drop model developed in the 30 s has
shown to account very well the behaviour of the nucleus
The atom
15
Mass - Energy
"   In 1905 Einstein presented the equivalence relationship
between mass and energy
2
E = mc , c being the speed of light
"   That means that the complete conversion of 1g of matter
releases as much energy as 20000 tons of TNT !
We have seen that energy is normally expressed in Joules and mass and kilograms.
E ( J ) = m(kg ) × (2.998 ×108 ) 2
However, these units are large at atomic or nucleus level. Smaller units are preferred
E ( MeV )
− 27
8 2
=
m
(
u
)
×
1
.
6605
×
10
×
(
2
.
998
×
10
)
12
6.2415 ×10
E ( MeV ) = m(u ) × 931.5
And so, particle/nuclear physicists
even express mass in MeV
The atom
1u =
1
g = 1.6605 ×10 − 27 kg
NA
1J = 6.2415 . 1018 eV
Mass of proton = 938.27MeV
Mass of neutron = 939.57MeV
Mass of electron = 0.51MeV
16
Nuclear mass and binding energy
"   The binding nuclear energy keeping the nucleons together has a
direct effect on the mass of an atom
"   The mass of the nucleus is less than the masses of the separated
neutrons and protons
"   The “missing mass” is actually equivalent to the energy needed for the
nucleons to stay together
•  Energy must be added to the nucleus to separate it into neutrons and protons
"   This is the binding energy of the nucleus, noted B or BE
Example
Mass of deuteron = 2.013553u
Mass of proton = 1.007276u
Mass of neutron = 1.008665u
m p + mn = 2.015941u
Δm = 0.002388u
MeV
ΔE = Δm × c = Δm(u ) × 931.5
= 2.224MeV
u
2
We need 2.2MeV to pull apart a nucleus of deuteron (one proton and one neutron)
The atom
17
Nuclear mass and binding energy (2)
If we have an element
A
Z
Using the relation massenergy we can write
XN
B
m( A, Z ) = Z × m p + N × mn + Z × me − 2
c
Note: Here we are neglecting the binding energy of the electrons to the nucleus
which is much smaller that the binding energy of the nucleons.
e- binding energy 1to10s of eV, nucleons several MeV
If we use the smallest atom:
We can write the total
nuclear binding energy
of a nucleus as
1
1
H 0 = m p + me
Neglecting the binding energy
of the electron (13.6eV)
(
)
B = Z ×11H 0 + N × mn − m( A, Z ) × c 2
Mass defect
1
1
The atom
H 0 and m( A, Z ) represent atomic masses
18
Example
Let s calculate the binding energy of Uranium
1
1
H0
U 146
1.007825u
238
92
238.050785u
mn
1.008665u
U 146
238
92
(
)
B = Z ×11H 0 + N × mn − m( A, Z ) × c 2
Mev
= (146 ×1.008665u + 92 ×1.007825u − 238.050785u ) × 931.5
= 1802MeV
u
If we normalise this binding energy to the potential energy residing in a nucleus of Uranium
1802
≈ 1%
238 × 931.5
Energy per nucleon:
The atom
This binding energy represents only 1% of
what a nucleus could release in theory
B 1802
=
= 7.57 MeV
A 238
19
Parameters of the liquid drop model (1)
"   In order to have a representative equation of the B/A curve, we
need to have several terms that will define the liquid drop model
"   We can describe this model with 5 terms
I – Volume energy term
The strong force is a short range one. So each nucleon within the volume of the
nucleus will interact on average with a fixed number its nearest neighbours only.
So for this term B ∝ A
If each nucleon was interacting with all other nucleons then B would be
proportional to A(A-1)
Volume energy term = av.A
The atom
20
Parameters of the liquid drop model (2)
II – Surface correction term
A nucleon near the surface of the nucleus will interact with fewer nucleons  It
is less tightly bound than if it were in the interior. So the volume term is
overestimating B.
Thus a term proportional to the surface area must be subtracted from the volume
term.
Radius ∝ A1/ 3
So surface correction term should be ∝ A2 / 3
Surface correction term = -as.A2/3
The atom
21
Parameters of the liquid drop model (3)
III – Coulomb energy term
Coulomb repulsion of the protons in the nucleus will act against the binding
energy.
Coulomb force is long range  each proton will interact with all others
 Z(Z-1) dependence
Electrical potential energy of a uniform charged sphere of radius R and charge
Q can be written as:
2
3× Q
5 × 4πε o R
Coulomb energy term = -ac.Z(Z-1)/A1/3
The atom
22
Parameters of the liquid drop model (4)
IV – Symmetry term
Stable nuclei tend to have Z~N, i.e. symmetric in N and Z (ref to plot of nuclei
as a function of N and Z – slide # 10)
We represent this fact by the symmetry term:
Symmetry term = -asym. (A-2Z)2/A
The atom
23
Parameters of the liquid drop model (5)
V – Pairing term
There is a tendency for nuclei with nucleons that are coupled pair-wise (p-p or n-n)
to be more tightly bound. e.g. Only 4 nuclei with odd Z and odd N, but 161 with
even N and even Z. This reflects tendency for like-nucleons to form spin-zero pairs
in the same spatial state.
Several forms of A dependence for this term exist we will use the most common one:
+ δ 0 for Z, N even (A even)
pairing term = δ(Ζ,Α) =
0 for odd mass A
- δ 0 for Z, N odd (A even)
Note: other dependence you might come across is
δ0 =
ap
3/ 4
A
δ0 =
ap
1/ 2
A
with a p = 12
with a p = 34
However: this will lead to a slightly different set of values for the a coefficients.
The atom
24
Semi-empirical mass formula
Note: For a more detailed explanation of the last 2 terms, which is out of scope of this
course (needs deeper understanding of quantum and particle physics) refer to possible
readings:
-  Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006)
-  2nd year lectures on Quantum Physics
-  Particle Physics, Martin and Shaw (Manchester Physics series - Wiley)
The overall semi-empirical mass formula is then:
2
ac .Z ( Z − 1) asym ( A − 2Z )
2/3
B( Z , A) = av . A − as . A −
−
+ δ ( A, Z )
1/ 3
A
A
av = 15.5MeV
as = 17.2MeV
ac = 0.72 MeV
asym = 23.2MeV
a p = 12 MeV
The atom
25
Experimental vs Model comparison
Ref 2
The atom
26
B/A curve of stable nuclei
If we perform the same calculation for each stable nuclei we get the following curve
Average binding energy per nucleon as a function of atomic mass number A
12
C
16
O
20
Ne
"  Some stable nuclei at low A
4
He
8
Be
"  B/A reach a maximum of
8.6Mev for A~60 – the most
tightly bound nuclei
"  Then goes down to 7.6MeV
for high A.
"  No stable nuclei above A=238
Ref 2
The atom
27
B/A curve explanations
"   On the B/A curve we have 3 regions
"   Below A=60
•  B/A grows steadily: all nucleons interact with each other. The more
nucleons, the more tightly bounded they are
"   Around A=60
•  The strong interaction has a limited range. At some point the nucleus
becomes so large that nucleons cannot interact with all the others. We
reach a constant binding energy for each nucleon
"   Above A=60
•  The 2 forces with opposite effects (coulomb and strong interaction) do
not have the same range. Coulomb force is more effective when the
nucleus grows and B/A decreases
The atom
28
Fusion – Fission Q factor
"   From the curve we can see:
"   For nucleus A=120 and beyond, there would be a clear energy gain
(energy released) if we were able to split the nucleus into 2
roughly equal nuclei (A~60 each)
•  This is fission: this will occur spontaneously if A is sufficiently large
or if a heavy nucleus (ie 238U) is excited to a higher energy state.
"   On the other hand, for light nuclei, there could be a large gain if
we could fuse them into one single nucleus
•  Fusion: the smallest are the nuclei, the largest gain we could get.
"   The difference in mass energy between the parent nuclei
and the final product is the Q factor
The atom
29
Example of Q value for fusion
Fusion of 2 light nuclei
Fusion of deuterium and tritium
∑ mi → ∑ m j +
i
j
before after
Q
c2
2
1
H1 +13H 2 →24He2 + n
energy released
Mass before: (D+T) = 2.014102u + 3.016049u=5.030151u
Mass after: (He+n) = 4.002603u + 1.008665u=5.01128u
Δm=Mass before-Mass after=0.018883u
Q= Δmx931.5MeV/u Q=17.6MeV
Q is positive, we then have an exothermic process
The atom
30
Mass excess – Mass defect
"   Atomic masses are usually given in u unit
"   However they are sometimes quoted as a difference
between atomic mass and atomic mass number A
Mass excess=Δe= Atomic mass – A
usually quoted in µu (10-6u)
Note: While the mass defect as defined
before (difference between the added masses
of all the nucleons and the atomic mass ) is
always positive due to the mass converted
into binding energy, the mass excess can be
positive or negative.
Ref 2
The atom
31
Examples
(
)
1
1
B = Z × H 0 + N × mn − ( A + Δe) × c
Binding energy
of a nucleus
9
4
2
Mass defect
Be mass excess = 12182 µu
Mass defect = 4 ×1.008665u + 5 ×1.007825u − (9 + 0.012182)u = 0.061603
84
36
Kr mass excess = -88493µu
Mass defect = 36 ×1.008665u + 48 ×1.007825u − (84 − 88493 ×10 −6 )u = 0.776033
The atom
32
Example of Q value for fission
Fission of a heavy nucleus: the other side of B/A curve
Fission of Uranium 236
236
141
92
U → Ba + Kr + 3n
Mass excess:
236U  45562µu
141Ba  -85594µu
92Kr  -73847µu
Mass before = 236u + 0.045562u = 236.045562u
Mass after = 141-0.085594 +92-0.073847+ 3x1.008665
= 235.866554u
Δm=Mass before-Mass after=0.179008u  Q=166.75MeV
Q is positive, we then have an exothermic process
The atom
33
Periodic table of elements
The atom
34
Notation summary
Notation in equations
A
Z
140
58
Ce82
N
12C
gives the definition of:
 A mole = NA atoms of an element
Avogadro s number NA = number of atoms in 12g of 12C
 u  Mass of an atom of 12C is exactly 12u
The mass of 1 atom of 140Ce is (140-0.094566)u
There is then NA atoms in (140-0.094566)g of 140Ce
And NA atoms in 140.116g of natural Ce
Average mass of element
Ex: natural Ce: 88.48% 140Ce, 11.08% 142Ce, 0.25% 138Ce, 0.19%136Ce
Takes into account several isotopes
Average mass = ∑ ( fraci × (Ai + Δei )) = 140.116u
The atom
i
i representing each isotope
35