fys3410 - vår 2015 (kondenserte fasers fysikk)...fys3410 lectures (based on c.kittel’s...

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

http://www.uio.no/studier/emner/matnat/fys/FYS3410/v14/index.html

http://www.uio.no/studier/emner/matnat/fys/FYS3410/v14/index.html

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.


• Relevance of condenced matter physics to modern technologies;


• 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


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



• Relevance of condenced matter physics to modern topics of science and technology;






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

Metallic bonding

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

http://en.wikipedia.org/wiki/File:Braggs_Law.svg

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 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 hkl

2

nsin


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 hkl

2

nsin


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




• Waves as a probe to study crystals and idea of useing of reciprocal space;




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




(001)




(001)

(100)




(001)

(100)

(002)




(001)

(100)

(002)

(101)




(001)

(100)

(002)

(101)

(101)




(001)

(100)

(002)

(101)

(101)

(102)


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





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





(010)

(100)

(110)

000

Reciprocal lattice





222 lkh

adhkl

200

000

Reciprocal lattice





222 lkh

adhkl




• Waves as a probe to study crystals and use of reciprocal space;


• 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



• Laue condition;




(hkl) plain indices

a1

a2

(hkl) plain indices

a1

a2

(110)

(hkl) plain indices

a1

a2

(110) (230)

(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,

http://en.wikipedia.org/wiki/File:Bragg's_law.svg

k′

-k

k K k hkl

hkl 2

2| |sin

sin

Laue condition


experiment



• Laue condition;




k′

-k

k

Laue condition

K k hkl

hkl 2

2| |sin

sin



K hkl

2

nsin


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


experiment



• Laue condition;




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


experiment



• Laue condition;




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


experiment



• Laue condition;




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:


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:


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:


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:


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:


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


research


• Bragg planes



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׀.


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


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


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?


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



research


• Bragg planes



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

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

sin2dn

Evolution of x-ray

diffraction

k′

k

2θ

2θ

Evolution of x-ray

diffraction

k′

k

2θ

sin2dn 2θ

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

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


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


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


– 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


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


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


• 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


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?


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


S p

ara

mete

r

a

30/1.60

Rp(Li)

bulk reference


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


S p

ara

mete

r

a

30/1.60

Rp(Li)

bulk reference


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


• Microscopic diffusion mechanism


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



• Microscopic diffusion mechanisms


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.


• Also called energy barrier for diffusion

Initial state Final state Intermediate state

Energy Activation energy




• Microscopic diffusion mechanism


Strained silicon

• How does it work?

• Basic idea: Change the lattice constant of

material

fys3410 - vår 2015 (kondenserte fasers fysikk)...fys3410 lectures (based on c.kittel’s...

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