fys3410 - vår 2015 (kondenserte fasers fysikk)...fys3410 lectures (based on c.kittel’s...
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FYS3410 - Vår 2015 (Kondenserte fasers fysikk) http://www.uio.no/studier/emner/matnat/fys/FYS3410/v15/index.html
Pensum: Introduction to Solid State Physics
by Charles Kittel (Chapters 1-9 and 17, 18, 20)
Andrej Kuznetsov
delivery address: Department of Physics, PB 1048 Blindern, 0316 OSLO
Tel: +47-22857762,
e-post: [email protected]
visiting address: MiNaLab, Gaustadaleen 23a
FYS3410 Lectures (based on C.Kittel’s Introduction to SSP, chapters 1-9, 17,18,20)
Module I – Periodic Structures and Defects (Chapters 1-3, 20) 26/1 Introduction. Crystal bonding. Periodicity and lattices, reciprocal space 2h
27/1 Laue condition, Ewald construction, interpretation of a diffraction experiment,
Bragg planes and Brillouin zones 4h
28/1 Elastic strain and structural defects in crystals 2h
30/1 Atomic diffusion and summary of Module I 2h
Module II – Phonons (Chapters 4, 5 and 18) 09/2 Vibrations, phonons, density of states, and Planck distribution 2h
10/2 Lattice heat capacity: Dulong-Petit, Einstien and Debye models
Comparison of different models 4h
11/2 Thermal conductivity 2h
13/2 Thermal expansion and summary of Module II 2h
Module III – Electrons (Chapters 6 and 7) 23/2 Free electron gas (FEG) versus Free electron Fermi gas (FEFG) 2h
24/2 Effect of temperature – Fermi- Dirac distribution
FEFG in 2D and 1D, and DOS in nanostructures 4h
25/2 Origin of the band gap and nearly free electron model 2h
27/2 Number of orbitals in a band and general form of the electronic states 2h
Module IV – Semiconductors and Metals (Chapters 8, 9, and 17) 09/3 Energy bands; metals versus isolators 2h
10/3 Semiconductors: effective mass method, intrinsic and extrinsic carrier
generation 4h
11/3 Carrier statistics 2h
13/3 p-n junctions and optoelectronic devices 2h
Lecture 1
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern technologies;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea using reciprocal space;
• Lattice planes and Miller indices;
• Introduction of the reciprocal space;
• Formal description of crystal structures.
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern technologies;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea of using reciprocal space;
• Lattice planes and Miller indices
• Introduction of the reciprocal space
• Formal description of crystal structures
Semiconductor physics at UiO
NEC ion implantor
HRXRD
SIMS
ZnO MOCVD
UiO clean room area
Labs
temperature/time
resolved PL
DLTS
6 Professors
4 Adm/technical staff
~ 10 Post docs
~ 15 PhD students and ~ 10 Msc students
Micro- and Nanotechnology Laboratory (MiNaLab)
Halvlederfysikk ved UiO / MiNa-Lab
SiC ZnO
ITO Si
Cu2O
• Solar cells
• Detectors
• IC’s
• Power electronics
• High temperature sensors • LED’s
• Transparent electronics
• Displays
• Electronic ink
• Optical cavities
• Multi-junction solar cells
• Thermoelectric generators
• Thermoelectric cooling
Thermo-
electric
materials
Semiconductor physics at UiO
Multiple Quantum Wells (MQWs) for advanced LEDs
repetitions of
ZnO/ZnCdO/ZnO
1.5nm
Quantum prorties electrons at the excited state
Bulk prorties electrons at the ground state
hν
”blue shift”
Vishnukanthan, et.al Solar Energy, 106, 82(2014)
”blue shift”
PL: optical excitation and subsequent radiative carrier
recombination
Semiconductor physics at UiO
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern topics of science and technology;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea of using reciprocal space;
• Lattice planes and Miller indices
• Introduction of the reciprocal space
• Formal description of crystal structures
Ionic bonding
It costs 5.1 eV for Na to ionize
and 3.6 eV for Cl to
accomodate an extra electron
so that the ”balance” is:
5.1 - 3.6 = 1.5 eV.
Ionic bonding
What is the driving force for the bonding than?!
Coulomb attraction, of course!
𝑬 = −𝒆𝟐/𝟒𝝅𝜺𝟎𝒂
Ionic bonding
Separation, nm
1,0
Ionic bonding
Ionic bonding
Ionic bonding
Ionic bonding
Metallic bonding
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern topics of science and technology;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea of using reciprocal space;
• Lattice planes and Miller indices
• Introduction of the reciprocal space
• Formal description of crystal structures
Bragg diffraction – constructive interference for the wave
interacting with crystal planes
The conditions leading to diffraction are given by the Bragg's law, relating the
angle of incidence of the radiation (θ) to the wavelength (λ) of the incident
radiation and the spacing between the crystal lattice planes (d):
2 d sin (θ) = n λ
k′
-k
k K k hkl
hkl 2
2| |sin
sin
K is perpendicular to the (hkl)
plane, so can be defined as:
K hkl
2
nsin
Laue condition
k′
-k
k K k hkl
hkl 2
2| |sin
sin
K is perpendicular to the (hkl)
plane, so can be defined as:
K hkl
2
nsin
G is also perpendicular to (hkl) so n G
Ghkl
hkl
Laue condition
k′
-k
k K k hkl
hkl 2
2| |sin
sin
K is perpendicular to the (hkl)
plane, so can be defined as:
K hkl
2
nsin
G is also perpendicular to (hkl) so n G
Ghkl
hkl
KG
Ghkl
hkl hkl
2
sin and G
dfrom previoushkl
hkl
1
Laue condition
k′
-k
k
Laue condition
K k hkl
hkl 2
2| |sin
sin
K is perpendicular to the (hkl)
plane, so can be defined as:
K hkl
2
nsin
G is also perpendicular to (hkl) so n G
Ghkl
hkl
KG
Ghkl
hkl hkl
2
sin and G
dfrom previoushkl
hkl
1
Kd
Ghkl hklhkl
2 sin
But Bragg: 2dsin =
K = Ghkl the Laue condition
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern topics of science and technology;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea of useing of reciprocal space;
• Lattice planes and Miller indices
• Introduction of the reciprocal space
• Formal description of crystal structures
Miller indices of lattice planes
• The indices of a crystal plane (h,k,l) are defined to be a set of integers with no common
factors, inversely proportional to the intercepts of the crystal plane along the crystal
axes:
Indices of Planes: Cubic Crystal
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
(002)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
(002)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
(002)
(101)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
(002)
(101)
(101)
Miller indices of lattice planes
We will use a monoclinic unit cell to avoid orthogonal axes; define a plan and
consider some lattice planes
(001)
(100)
(002)
(101)
(101)
(102)
Miller indices of lattice planes
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern topics of science and technology;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and idea of using reciprocal space;
• Lattice planes and Miller indices
• Introduction of the reciprocal space
• Formal description of crystal structures
Reciprocal lattice
Crystal planes (hkl) in the real-space or direct lattice are characterized
by the normal vector and the interplanar spacing :
Defining a different lattice in reciprocal space whose points lie at positions
given by the vectors
hkln̂ hkld
x
y
z
hkld
hkln̂
hkl
hklhkl
d
nG
ˆ2
These vectors are parallel
to the [hkl] direction but
has magnitude 2/dhkl,
which is a reciprocal
distance
The reciprocal lattice is composed of all points lying at positions
from the origin, so that there is one point in the reciprocal lattice for
each set of planes (hkl) in the real-space lattice.
This seems like an unnecessary abstraction. What is the payoff for defining
such a reciprocal lattice? No, the reciprocal lattice simplifies the
interpretation of x-ray diffraction from crystals because:
hklG
• Diffraction pattern is not a direct
representation of the crystal
lattice
• Diffraction pattern is a
representation of the reciprocal
lattice
Reciprocal lattice
Vc = a1•(a2 x a3) – volume of a unit cell
Definition of reciprocal translation vectors
b3 = (a1 x a2) 2π/Vc
G = v1b1 + v2b2 + v3b3
Reciprocal lattice
Generallizing,we introduce a set of new unit
vectors so that they are normal to the plains
determined by the previously introduced
translation vectors
b1 = (a2 x a3) 2π/Vc
a3
a2
b1
b2 = (a3 x a1) 2π/Vc
Reciprocal lattice is nothing with ”anti-matter” or ”black holes” to do – it is
determined by a set of vectors with specific magnitudes just having a bit unusual
dimentions – 1/length. It is actually relatively straightforward – as long as we
understood the definitions – to schetch the reciprocal lattice.
d100
a2
a1
γ
d010
Reciprocal lattice
The important part is that b1 should be normal to a plain
determined by [a2 x a3] and having a magnitude of 1/a1 –
just by definition - or 1/d100, where d100 is the interplain
distance between (100) family of plains. NB, for any plain
from (100) familly the point in the reciprocal space is
exactly the same meaning that any reciprocal lattice point
represents its own family of plains in the real space.
2π/a1 = 2π/d100
b1 a2
a1
Reciprocal lattice
Reciprocal lattice is nothing with ”anti-matter” or ”black holes” to do – it is
determined by a set of vectors with specific magnitudes just having a bit unusual
dimentions – 1/length. It is actually relatively straightforward – as long as we
understood the definitions – to schetch the reciprocal lattice.
b1 = (a2 x a3) 2π/Vc
Vc = a1•(a2 x a3)
(100)
The important part is that b1 should be normal to a plain
determined by [a2 x a3] and having a magnitude of 1/a1 –
just by definition - or 1/d100, where d100 is the interplain
distance between (100) family of plains. NB, for any plain
from (100) familly the point in the reciprocal space is
exactly the same meaning that any reciprocal lattice point
represents its own family of plains in the real space.
Similar excercise can be done with vector b2 which points
out to a reciprocal lattice point representing (010) family of
plains.
In adition (110) family of plaines in the real space would
naturally result in to (110)-points in the reciprocal space.
The procedure can be repeated any type of plain cuts in the
real space
1/d100
1/d010
b1 a2
a1
Reciprocal lattice
b2
Reciprocal lattice is nothing with ”anti-matter” or ”black holes” to do – it is
determined by a set of vectors with specific magnitudes just having a bit unusual
dimentions – 1/length. It is actually relatively straightforward – as long as we
understood the definitions – to schetch the reciprocal lattice.
(010)
(100)
(110)
000
Reciprocal lattice
Reciprocal lattice is nothing with ”anti-matter” or ”black holes” to do – it is
determined by a set of vectors with specific magnitudes just having a bit unusual
dimentions – 1/length. It is actually relatively straightforward – as long as we
understood the definitions – to schetch the reciprocal lattice.
222 lkh
adhkl
200
000
Reciprocal lattice
Reciprocal lattice is nothing with ”anti-matter” or ”black holes” to do – it is
determined by a set of vectors with specific magnitudes just having a bit unusual
dimentions – 1/length. It is actually relatively straightforward – as long as we
understood the definitions – to schetch the reciprocal lattice.
222 lkh
adhkl
Introduction. Crystal bonding. Periodicity and lattices, reciprocal space.
• Relevance of condenced matter physics to modern topics of science and technology;
• Why elements bond together? Why in crystals?
• Waves as a probe to study crystals and use of reciprocal space;
• Lattice planes and Miller indices
• General description of crystal structures
Ideal Crystal
• An ideal crystal is a periodic array of structural units, such as atoms or molecules.
• It can be constructed by the infinite repetition of these identical structural units in space.
• Structure can be described in terms of a lattice, with a group of atoms attached to each
lattice point. The group of atoms is the basis.
Bravais Lattice
• An infinite array of discrete points with an arrangement and orientation that
appears exactly the same, from any of the points the array is viewed from.
• A three dimensional Bravais lattice consists of all points with position vectors R that
can be written as a linear combination of primitive vectors. The expansion
coefficients must be integers.
Primitive Unit Cell • A primitive cell or primitive unit cell is a volume of space that when translated
through all the vectors in a Bravais lattice just fills all of space without either
overlapping itself or leaving voids.
• A primitive cell must contain precisely one lattice point.
Primitive (a1,a2) and not primitive (a1’’’,a2’’’) translation vectors
Crystal structure II
Wigner-Seitz Primitive Cell: Full symmetry of
Bravais Lattice
2-D lattices
3-D lattices Cubic
a=b=c
abg90°
Hexagonal
a=b≠c
ab 90° ; g120°
Tetragonal
a=b≠c
abg90°
Rhombohedral
a=b=c=
abg≠90°
Orthorhombic
a≠b≠c
a=b=g=90°
Monoclinic
a≠b≠c
ag90°≠b
Triclinic
a≠b≠c
a≠b≠g≠90
Primitive Cell:
FCC Lattice
Lecture 2
Laue condition, Ewald construction
and interpretation of a diffraction experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
Laue condition, Ewald construction and interpretation of a diffraction
experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
(hkl) plain indices
a1
a2
(hkl) plain indices
a1
a2
(110)
(hkl) plain indices
a1
a2
(110)
(hkl) plain indices
a1
a2
(110) (230)
(hkl) plain indices
a1
a2
(110) (230)
(hkl) plain indices
a1
a2
(110) (230)
(110)
(hkl) plain indices
a1
a2
(110) (230)
(110)
(hkl) plain indices
a1
a2
(110) (230)
(110) (010)
Distance between (hkl)-planes in cubic lattices
222 lkh
adhkl
The two separate waves will arrive at a point with the same phase, and hence undergo constructive
interference, if and only if this path difference is equal to any integer value of the wavelength, i.e.
.
Therefore,
Putting everything together,
k′
-k
k K k hkl
hkl 2
2| |sin
sin
Laue condition
Laue condition, Ewald construction and interpretation of a diffraction
experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
k′
-k
k
Laue condition
K k hkl
hkl 2
2| |sin
sin
K is perpendicular to the (hkl)
plane, so can be defined as:
K hkl
2
nsin
G is also perpendicular to (hkl) so n G
Ghkl
hkl
KG
Ghkl
hkl hkl
2
sin and G
dfrom previoushkl
hkl
1
Kd
Ghkl hklhkl
2 sin
But Bragg: 2dsin =
K = Ghkl the Laue condition
Laue condition, Ewald construction and interpretation of a diffraction
experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
Laue assumed that each set of atoms could radiate the incident radiation in all
directions
Constructive interference
only occurs when the
scattering vector, K (Δk in
the Kittel’s notations),
coincides with a reciprocal
lattice vector, G
This naturally leads to the Ewald Sphere construction
Diffraction pattern as representation of the reciprocal lattice
We superimpose the imaginary “sphere” of radiated radiation upon the reciprocal lattice
Draw sphere of radius
1/ centred on end of
ko Reflection is only
observed if sphere
intersects a point
i.e. where K=G
Ewald construction
This means that when a lattice point intersects the Ewald sphere, the reflection
corresponding to that family of planes will be observed and the diffraction
angle will be apparent.
Starting with an indexed reciprocal lattice, an incident x-ray beam must pass
through the origin (000) point, corresponding to the incident beam of x-rays.
Ewald construction
The Ewald sphere for this case is defined by making a sphere of radius 1/
having its diameter on the X-ray beam that intersects the origin point. In the
diagram on the left, no other reciprocal lattice points are on the surface of the
sphere so the Bragg condition is not satisfied for any of the families of planes.
To observe reflections, the reciprocal lattice must be rotated until an
reciprocal lattice point contacts the surface of the sphere. Note: it would be
easier to rotate the sphere on paper, but in practice, we rotate the crystal
lattice and the RL.
Ewald construction
When a reciprocal lattice point intersects the Ewald sphere, a reflection will
occur and can be observed at the 2 angle of the inscribed triangle. To be able
to collect as many different reflections as possible, it is thus necessary to be
able to rotate the reciprocal lattice to a great extent…
Ewald construction
Laue condition, Ewald construction and interpretation of a diffraction
experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
The reciprocal lattice is composed of all points lying at positions
from the origin, so that there is one point in the reciprocal lattice for
each set of planes (hkl) in the real-space lattice.
hklG
• Diffraction pattern is not a direct
representation of the crystal
lattice
• Diffraction pattern is a
representation of the reciprocal
lattice
Laue condition, Ewald construction and interpretation of a diffraction
experiment
• (hkl) indices and distance between plains in cubic, tetragonal and
orthorombic lattices;
• Laue condition;
• Ewald construction
• interpretation of a diffraction experiment
• Some consequences: how many lines = reciprocal lattice point will we see
222 lkh
adhkl
In cubic crystal:
sin2
222 lkh
aNow put it together with Bragg:
sin2222 a
lkh Finally
In the experiment we just correlate the increased intencity with the angle
Some consequences:
how many lines = reciprocal lattice point will we see
h k l h2 + k
2 + l
2h k l h
2 + k
2 + l
2
1 0 0 1 2 2 1, 3 0 0 9
1 1 0 2 3 1 0 10
1 1 1 3 3 1 1 11
2 0 0 4 2 2 2 12
2 1 0 5 3 2 0 13
2 1 1 6 3 2 1 14
2 2 0 8 4 0 0 16
sin2222 a
lkh
Some consequences:
how many lines = reciprocal lattice point will we see
Is there anything limiting
(h2 + k2 + l2) values of the
“last” reflection?
Yes it it’s the wavelength.
Why?
2
2
2222 4
sina
lkh
sin2 has a limiting value of 1, so for this limit:
2
2222 4
alkh
2
2222 4
alkh
1 1
1
0 2
2
1 1
3
2 2
2
0 0
4
1 3
3
2 2
4
1 1
53 3
3
1 0 20 30 40 50 60
SP IN EL : LA MB DA = 1 .54
Lambd a: 1 .541 78 Magn if: 1 .0 FW HM: 0 .2 00
Sp ace g rp : F d -3 m:2 Direct cell: 8 .080 0 8 .080 0 8 .080 0 9 0 .00 9 0 .00 9 0 .00
1 1
1
0 2
2
1 1
3
2 2
2
0 0
4
1 3
3
2 2
4
1 1
53 3
3
0 4
4
1 3
52 4
4
0 2
6 3 3
5
1 0 20 30 40 50 60
SP IN EL : LA MB DA = 1 .22
Lambd a: 1 .220 00 Magn if: 1 .0 FW HM: 0 .2 00
Sp ace g rp : F d -3 m:2 Direct cell: 8 .080 0 8 .080 0 8 .080 0 9 0 .00 9 0 .00 9 0 .00
= 1.54 Å
= 1.22 Å
Some consequences:
how many lines = reciprocal lattice point will we see
h k l h2 + k
2 + l
2h k l h
2 + k
2 + l
2
1 0 0 1 2 2 1, 3 0 0 9
1 1 0 2 3 1 0 10
1 1 1 3 3 1 1 11
2 0 0 4 2 2 2 12
2 1 0 5 3 2 0 13
2 1 1 6 3 2 1 14
2 2 0 8 4 0 0 16
Still if one knows the lattice it
should quite stright to index the
peaks, but…
Some consequences:
how many lines = reciprocal lattice point will we see
Let’s take an example: The unit cell of copper is 3.613 Å. What is the
Bragg angle for the (100) reflection with Cu Ka radiation ( = 1.5418
Å)?
hkld2sin 1
= 12.32o, so 2 = 24.64o
BUT…. 10 20 30 40 50 60 70 80
Copper, [W. L. Bragg (Philosophical Magazine, Serie 6 (1914) 28, 255-360]Lambda: 1.54180 Magnif: 1.0 FWHM: 0.200Space grp: F m -3 m Direct cell: 3.6130 3.6130 3.6130 90.00 90.00 90.00
222 lkh
adhkl
• Due to symmetry, certain reflections cancel each other out.
• These are non-random – hence “systematic absences”
• For each Bravais lattice, there are thus rules for allowed reflections:
Some consequences:
how many lines = reciprocal lattice point will we see
Relation to real diffraction experiment
The presence of translational symmetry elements and centering in the real lattice
causes some series of reflections to be absent – can be accurately derived from the
expressions of the structure factors.
e.g. the (001) reflection in a BCC lattice
is absent.
Consider the additional path lengths
vs. beam “1”:
For “2” it is 2d sin(q);
For “3” it is 2(d/2) sin(q), thus the rays
from “3” will be exactly out-of-phase
with those of “2” and no reflection will
be observed.
Relation to real diffraction experiment
So for each Bravais lattice:
PRIMITIVE
BODY
FACE
h2 + k2 + l2
All possible
h+k+l=2n
h,k,l all
odd/even
1
1 0 0
2
1 1 0
1 1 0
3
1 1 1
1 1 1
4
2 0 0
2 0 0
2 0 0
5
2 1 0
6
2 1 1
2 1 1
8
2 2 0
2 2 0
2 2 0
9
2 2 1, 3 0 0
10
3 1 0
3 1 0
11
3 1 1
3 1 1
12
2 2 2
2 2 2
2 2 2
13
3 2 0
14
3 2 1
3 2 1
16
4 0 0
4 0 0
4 0 0
Some consequences:
how many lines = reciprocal lattice point will we see
Bragg plains and Brillouin zones. Use of diffraction experiment in
research
• Repetition of Laue condition and Ewald construction;
• Bragg planes
• Introduction and interpretation of Brillouin zones;
• Use of diffraction experiment in research
Bragg plains and Brillouin zones. Use of diffraction experiment in
research
• Repetition of Laue condition and Ewald construction;
• Bragg planes
• Introduction and interpretation of Brillouin zones;
• Use of diffraction experiment in research
Laue assumed that each set of atoms could radiate the incident radiation in all
directions
Constructive interference
only occurs when the
scattering vector, K (Δk in
the Kittel’s notations),
coincides with a reciprocal
lattice vector, G
This naturally leads to the Ewald Sphere construction
Ewald construction
We superimpose the imaginary “sphere” of radiated radiation upon the reciprocal lattice
Draw sphere of radius
1/ centred on end of
ko Reflection is only
observed if sphere
intersects a point
i.e. where K=G
Ewald construction
Bragg plains and Brillouin zones. Use of diffraction experiment in
research
• Repetition of Laue condition and Ewald construction;
• Bragg planes
• Introduction and interpretation of Brillouin zones;
• Use of diffraction experiment in research
Bragg planes and Brillouin zone construction
The construction of Bragg Planes in the context of Brillouin zones can be
understood by considering Bragg’s Law λ = 2dsinθ. As we now know, in
reciprocal space this can be expressed in the form
k' – k = g
where k is the wave vector of the incident wave of magnitude 2π/λ,
k' is the wave vector of the diffracted wave, also of magnitude 2π/λ, and
g is a reciprocal lattice vector of magnitude 2π/d:
As we also know, this can be illustrated graphically
using the Ewald sphere construction –
with 000 to be the origin of the reciprocal lattice
and O is the centre of the sphere of radius ׀k׀.
Bragg planes and Brillouin zone construction
If the angle subtended at O
between 000 and G on the
diagram is 2θ, simple geometry
shows that
The equation
k' – k = g
can be rearranged in the form
k' = k + g so that
k'.k' = (k + g).(k + g) = k.k + g.g + 2k.g
But k'.k' = k.k because diffraction is
an elastic scattering event,
g.g + 2k.g = 0
To construct the Bragg Plane, it is
convenient to replace k by -k in this
equation so that both k and g begin at
the origin, 000, of the reciprocal lattice.
Hence, the equation can be written in the
form
Bragg planes and Brillouin zone construction
k.(½g) = (½g). (½g)
g.g + 2k.g = 0
Constructing the plane normal to g at the
midpoint, (½g),
then means that any vector k drawn from
the origin, 000, to a position on this plane
satisfies the Bragg diffraction condition.
Do we undrestand this? Let’s repeat
Bragg planes and Brillouin zone construction
k.(½g) = (½g). (½g) When this holds – diffraction occurs – that’s the law
Let’s considering when this ”dot” products will do
coincide? What the dot product by the way?
Bragg planes and Brillouin zone construction
k.(½g) = (½g). (½g) When this holds – diffraction occurs – that’s the law
Let’s considering when this ”dot” products will do
coincide? What the dot product by the way?
Fundamental conslusion is:
A wave with a wafe vector < k has no chance to get diffracted
1/2G
The vector kin (also kout) lies
along the perpendicular bisecting
plane of a G vector
G
k
• Brillouin Zone formed by perpendicular bisectors of G vectors
• Consequence: No diffraction for any k inside the first Brillouin Zone
• Special role of Brillouin Zone (Wigner-Seitz cell of reciprocal lattice) as
opposed
to any other primitive cell
Bragg planes and Brillouin zone construction
Bragg planes and Brillouin zone construction
Bragg planes and Brillouin zone construction
Bragg planes and Brillouin zone construction
Bragg planes and Brillouin zone construction
Bragg planes and Brillouin zone construction
Bragg plains and Brillouin zones. Use of diffraction experiment in
research
• Repetition of Laue condition and Ewald construction;
• Bragg planes
• Introduction and interpretation of Brillouin zones;
• Use of diffraction experiment in research
Use of diffraction experiment in research
ZnCdO
C-Al2O3
Zn1-xCdxO x: max
C-Al2O3
Zn1-xCdxO
Zn1-xCdxO
Zn1-xCdxO x: min
ZnO
single layer
multilayer
Schematic of the structures studied
FIG.1. (a) Typical photoluminescence (PL) spectra from ZnCdO films as a function of Cd content; (b) PL spectrum of a ML-structure as
recorded at 8K with a schematics of the band gap in the inset; (c) Cd profile through ML-structure as measured by RBS; (c) our results in the
context of literature.
Use of diffraction experiment in research
Use of diffraction experiment in research
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
stress = load W
area A
strain = increase in length x
original length L
σij =Cij•εij
εij =Sij•σij
Is there a good reason to introduce complications with so many different? Yes it is, becase,
for example elastic waves in crystals often propagate in different directions, specifically
can be longitudinal or transverse (share) waves
Hooke’s law
F
bonds
stretch
return to
initial
1. Initial 2. Small load 3. Unload
Elastic means reversible!
Elastic strain (deformation)
F
linear elastic
linear elastic
plastic
1. Initial 2. Small load 3. Unload
Plastic means permanent!
Plastic deformation
Elastic/plastic deformation
• Below the yield stress
Elasticity of Modulus
or Modulus Youngs
E
E
• Strength is affected by alloying,
heat treating, and manufacturing
process but stiffness (Modulus of
Elasticity) is not.
Elastic/plastic deformation
How is strain applied to the electronic chips?
Si1-xGex
p-type MOSFET
Si1-xGex
Strained Si
45 nm
Evolution of x-ray
diffraction
k′
k
Evolution of x-ray
diffraction
k′
k
2θ
Evolution of x-ray
diffraction
k′
k
2θ
Evolution of x-ray
diffraction
k′
k
sin2dn
Evolution of x-ray
diffraction
k′
k
2θ
2θ
Evolution of x-ray
diffraction
k′
k
2θ
sin2dn 2θ
Evolution of x-ray
diffraction
k′
k
2θ
sin2dn 2θ
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
Vacancy: A point defect
Defects Dimensionality Examples
Point 0 Vacancy
Line 1 Dislocation
Surface 2 Free surface,
Grain boundary
There may be vacant sites in a crystal
Surprising Fact
There must be a certain fraction of vacant
sites in a crystal in equilibrium.
Fact
Vacancies
• Crystal in equilibrium
• Minimum Gibbs free energy G at constant
T and P
• A certain concentration of vacancy lowers
the free energy of a crystal
Vacancies
Gibbs free energy G involves two terms:
1. Enthalpy H
2. Entropy S
G = H – T S
=E+PV
=k ln W
T Absolute temperature
E internal energy
P pressure
V volume
k Boltzmann constant
W number of microstates
Vacancies
D H = n D Hf
Vacancy increases H of the crystal due to energy required to break bonds
Vacancies
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
Configurational entropy due to vacancy
Number of atoms: N
Number of vacacies: n
Total number of sites: N+n
How many distinguished configurations,
so called microstates?
We calculate this explicitly
Configurational entropy due to vacancy
Configurational entropy due to vacancy
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
DG = DH TDS
neq
G of a
perfect
crystal
n
DG
DH
fHnH DD
TDS ]lnln)ln()[( NNnnnNnNkS D
Equilibrium concentration of vacancies – temperature dependence
0
D
eqnnn
G
D
kT
H
N
n feqexp
Equilibrium concentration of vacancies – temperature dependence
Measure a material property
which is dependent on neq/N vs
T
Find the activation
energy from the
slope
neq
N
T
exponential dependence
1/ T
N
neq ln
1
- QD/k
slope
D
kT
H
N
n feqexp
Equilibrium concentration of vacancies – temperature dependence
– Copper at 1000 ºC
Hf = 0.9 eV/at ACu = 63.5 g/mol = 8400 kg/m-3
First find N in atoms/m-3
3
3
23
//
Check units
0635.0
840010023.6
m
at
molkg
mkgmolatN
N
A
NN
Cu
A
Equilibrium concentration of vacancies – temperature dependence
328 /1097.7 msitesatN
• Now apply the Arrhenius relation @1000 ºC
325
5
28
/1018.2
1273/1062.8
/9.0exp1097.7
exp
mvacN
KKateV
ateV
kT
HNN
v
f
v
Equilibrium concentration of vacancies – temperature dependence
D
kT
H
N
n feqexp
Al: DHf= 0.70 ev/vacancy
Ni: DHf= 1.74 ev/vacancy
n/N 0 K 300 K 900 K
Al 0 1.45x1012
1.12x104
Ni 0 5.59x1030 1.78x10-10
Equilibrium concentration of vacancies – temperature dependence
• Neighboring atoms tend to move into the
vacancy, which creates a tensile stress field
• The stress/strain field is nearly spherical
and short-range. ao
Equilibrium concentration of vacancy – pressure dependence
ΔGf=Ef+PVf - TSf
kTVkTEkSkTGeq
Vffff eeeeC
//// D
Hf=Ef+PVf
Vf = Ω + relaxation volume
fHD
Equilibrium concentration of vacancy – pressure dependence
How big the pressure should be to make
a measurable effect on vacancy concentration?
fV
Compare
kTVkTEkSkTGeq
Vffff eeeeC
//// D
101.325 kPa is “one standard atmosphere” and 1 Pa = 1 N/m2
1 eV = 1.602176487×10−19 Joule
As we calculate the effect of pressure/stress on vacancy concentration
starts to be significant at quite high values – in the range of 100 MPa.
Are these conditions available in real ”life” or happens only in a
laboratory experiment?
Equilibrium concentration of vacancy – pressure dependence
Lecture 4: Mechanical properties and structural defects
• Mechanical properties of solids
• Structural defects in crystals
• Configurational entropy due to vacancy
• Equilibrium concentration of vacancies – temperature and presure dependences
• Watching empty lattice sites – i.e. vacancies – with positrons
Positron beam
Trapping and annihilation
Potential
diagram
g
g
Eg =511 keV
+ Doppler broadening depending on the amount of electron momentum
10-4
10-3
10-2
10-1
INT
EN
SIT
Y (a
rb.
un
its)
30 20 10 0 ELECTRON MOMENTUM (10-3 m0c)
lattice vacancy
W
S
S-parameter characterizes annihilation with low momentum valence
electrons. Increase in S-parameter is naturaly interpreted as an
increase in vacancy concentration
W-paprameter characterizes annihilation with high momentum core
electrons and increase in vacancy concentration results in decrease
of W-parameter
Positron probing of vacancies in semiconductors
Positron beam
Trapping and annihilation
Potential
diagram
g
g
Eg =511 keV
+ Doppler broadening depending on the amount of electron momentum
1.00 1.02 1.04 1.06 1.08
0.8
0.9
1.0
b Zn vacancy
W p
ara
mete
r
S parameter
Bulk
1.03
1.08
10/0.3 20/0.8
Zn vacancy
Energy (keV)/Mean positron implantation depth (µm)
S p
ara
mete
r
a
30/1.60
bulk reference
Experimental points group around a line in the W-S plane if
there are only two annihilation states vailable in the sample
Clustering of ion implantation induced vacancies in ZnO
1.00 1.02 1.04 1.06 1.08
0.8
0.9
1.0
b Zn vacancy
W p
ara
mete
r
S parameter
Bulk
1.03
1.08
10/0.3 20/0.8
Zn vacancy
Li, as implanted
Energy (keV)/Mean positron implantation depth (µm)
S p
ara
mete
r
a
30/1.60
Rp(Li)
bulk reference
Clustering of ion implantation induced vacancies in ZnO
1.00 1.02 1.04 1.06 1.08
0.8
0.9
1.0
b Zn vacancy
W p
ara
mete
r
S parameter
Bulk
1.03
1.08
10/0.3 20/0.8
Zn vacancy
Li, as implanted 500oC, 1h
Energy (keV)/Mean positron implantation depth (µm)
S p
ara
mete
r
a
30/1.60
Rp(Li)
bulk reference
Clustering of ion implantation induced vacancies in ZnO
Lecture 5: Atomic diffusion
• Phenomenology of diffusion: describing diffusion in terms of diffusion flux
• Microscopic diffusion mechanisms
• Sb diffusion as a function of temperature as stress in Si/SiGe heterostructures
Lecture 9: Diffusion
• Phenomenology of diffusion: describing diffusion in terms of diffusion flux
• Microscopic diffusion mechanism
• Sb diffusion as a function of temperature as stress in Si/SiGe heterostructures
Phenomenon of material transport by atomic or particle
transport from region of high to low concentration
• What forces the particles to go from left to right?
• Does each particle “know” its local concentration?
• Every particle is equally likely to go left or right!
• At the interfaces in the above picture, there are more particles going right than left this causes an average “flux” of particles to the right!
• Largely determined by probability & statistics
Diffusion
• Glass tube filled with water.
• At time t = 0, add some drops of ink to one end
of the tube.
• Measure the diffusion distance, x, over some time.
Diffusion
• Flux: amount of material or atoms moving past a unit area in unit time
Flux, J = DM/(A Dt)
• Directional Quantity
• Flux can be measured for: --vacancies
--host (A) atoms
--impurity (B) atoms
Describing diffusion in terms of diffusion flux
• Concentration Profile, C(x): [kg/m3]
• Fick's First Law:
Concentration
of Cu [kg/m3]
Concentration
of Ni [kg/m3]
Position, x
Cu flux Ni flux
Describing diffusion in terms of diffusion
flux
• Steady State: Steady rate of diffusion from one end to the other.
Implies that the concentration profile doesn't change with time. Why?
• Apply Fick's First Law:
• Result: the slope, dC/dx, must be constant
(i.e., slope doesn't vary with position)!
Jx D
dC
dx
dC
dx
left
dC
dx
right
• If Jx)left = Jx)right , then
Describing diffusion in terms of diffusion
flux
• Concentration profile,
C(x), changes
w/ time.
• To conserve matter: • Fick's First Law:
• Governing Eqn.:
Fick’s second law
Describing diffusion in terms of diffusion flux
Lecture 9: Diffusion
• Phenomenology of diffusion: describing diffusion in terms of diffusion flux
• Microscopic diffusion mechanisms
• Sb diffusion as a function of temperature as stress in Si/SiGe heterostructures
vacancy Interstitial
impurity
Substitutional
impurity
Type of point defects
Conditions for diffusion:
• there must be an adjacent empty site
• atom must have sufficient energy to break bonds with its
neighbors and migrate to adjacent site (“activation” energy)
Diffusion mechanisms
Diffusion at the atomic level is a step-wise migration of atoms from
lattice site to lattice site
Higher the temperature, higher is the probability that an atom will have
sufficient energy
hence, diffusion rates increase with temperature
Types of atomic diffusion mechanisms:
• substitutional (through vacancies)
• interstitial
Substitutional Diffusion:
• applies to substitutional impurities
• atoms exchange with vacancies
• rate depends on:
-- number of vacancies
-- temperature
-- activation energy to exchange.
Diffusion mechanisms
• Also called energy barrier for diffusion
Initial state Final state Intermediate state
Energy Activation energy
Diffusion mechanisms
Lecture 8: Diffusion
• Phenomenology of diffusion: describing diffusion in terms of diffusion flux
• Microscopic diffusion mechanism
• Sb diffusion as a function of temperature as stress in Si/SiGe heterostructures
Strained silicon
• How does it work?
• Basic idea: Change the lattice constant of
material