Upload
supyyyy
View
223
Download
0
Embed Size (px)
Citation preview
8/11/2019 Crystal Structure Sem 5
1/123
Crystal Structure of Solids
Dr. Sukanta De
Books for Solid state physics Course: -
Introduction to Solid State Physics by Charles Kittel
Solid State Physics By A J Dekker
Solid State Physics By Ashcroft and Mermin
8/11/2019 Crystal Structure Sem 5
2/123
How do atoms assemble into solid structures?
Crystalline Solids : Atoms are arranged in regular manner , form 3-D
pattern. (by 3-D repetition of a certain pattern unit.)
PERIODIC ARRANGEMENT OF ATOMS/IONS OVER LARGE ATOMI C
DI STANCES
Leads to structure displaying
LONG-RANGE ORDER that is
Measurable and Quantifiable
All metals, many ceramics, and some polymers
8/11/2019 Crystal Structure Sem 5
3/123
When the periodicity of the pattern extends throughout a
certain piece of material Single Crystal
When the periodicity of the pattern interrupted at grain boundary
and grain size is at least several Angstroms
Polycrystalline materials
If the grain size is comparable to the size of pattern unit
Amorphous materials
Materials Lacking Long range order
Example: Ceramic GLASS and many plastics
8/11/2019 Crystal Structure Sem 5
4/123
POLYCRYSTALLINE MATERIALS Nuclei form during solidification, each of which grows into crystals
8/11/2019 Crystal Structure Sem 5
5/123
Ideal Crystal
An ideal crystal is a periodic arrayof 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.
8/11/2019 Crystal Structure Sem 5
6/123
Bravais Lattice
An infinite array of discrete points with anarrangement and orientation that appearsexactly 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 bewritten as a linear combination ofprimitivevectors. The expansion coefficients must beintegers.
8/11/2019 Crystal Structure Sem 5
7/123
Lattice-infinite,perfectly periodicarray ofpointsin a space
8/11/2019 Crystal Structure Sem 5
8/123
Not a lattice:
8/11/2019 Crystal Structure Sem 5
9/123
We abstracted points from the shape:
Now we abstract further:
8/11/2019 Crystal Structure Sem 5
10/123
Now we abstract further:
This is a UNIT CELL
8/11/2019 Crystal Structure Sem 5
11/123
Now we abstract further:
This is a UNIT CELL
Represented by two lengths and an angle
.or, alternatively, by two vectors
8/11/2019 Crystal Structure Sem 5
12/123
Basis vectors and unit cells
T = ma + nb
a and b are the basis vectors for the lattice
a
b
T
8/11/2019 Crystal Structure Sem 5
13/123
In 3-D:
a
b
a, b, and c are the basis vectors for the lattice
c
8/11/2019 Crystal Structure Sem 5
14/123
In 3-D:
T = m1a + m2b + m3c
a
b
T
a, b, and c are the basis vectors for the lattice
c
8/11/2019 Crystal Structure Sem 5
15/123
Crystal SystemsSome Definitional
information
7 crystal systems of varying
symmetry are known
These systems are built by changing
the lattice parameters:
a, b, and care the edge
lengths, , and are interaxial
angles
Fig. 3.4, Callister 7e.
Unit cell: smallest repetitive volume which contains
the complete lattice patternof a crystal.
8/11/2019 Crystal Structure Sem 5
16/123
8/11/2019 Crystal Structure Sem 5
17/123
Crystal Systems
Crystal structures are divided
into groups according to unit
cell geometry (symmetry).
8/11/2019 Crystal Structure Sem 5
18/123
8/11/2019 Crystal Structure Sem 5
19/123
CRYSTAL SYMMETRY
Symmetry defines the order resulting from how atoms
are arranged and oriented in a crystal
The definite ordered arrangement of the faces and edges
of a crystal known as Crystal Symmetry
The Symmetry Operation is one that leaves the crystal
and its environment invariant. i.e., actions which result inno change to the order of atoms in the crystal structure
8/11/2019 Crystal Structure Sem 5
20/123
Imagine that this object will be rotated (maybe)
8/11/2019 Crystal Structure Sem 5
21/123
Was it?
8/11/2019 Crystal Structure Sem 5
22/123
The object is obviously symmetricit hassymmetry
8/11/2019 Crystal Structure Sem 5
23/123
The object is obviously symmetricit has symmetry
Can be rotated 90 w/o detection
8/11/2019 Crystal Structure Sem 5
24/123
so symmetry is really
doing nothing
8/11/2019 Crystal Structure Sem 5
25/123
Symmetry is doing nothing -or at least doing
something so that it looks like nothing was done!
8/11/2019 Crystal Structure Sem 5
26/123
What kind of symmetry does this object have?
8/11/2019 Crystal Structure Sem 5
27/123
What kind of symmetry does this object have?
4
8/11/2019 Crystal Structure Sem 5
28/123
What kind of symmetry does this object have?
4
m
8/11/2019 Crystal Structure Sem 5
29/123
What kind of symmetry does this object have?
4
m
8/11/2019 Crystal Structure Sem 5
30/123
What kind of symmetry does this object have?
4
m
8/11/2019 Crystal Structure Sem 5
31/123
What kind of symmetry does this object have?
4
m
4mm
8/11/2019 Crystal Structure Sem 5
32/123
Another example:
8/11/2019 Crystal Structure Sem 5
33/123
Another example:
6
m
6mm
8/11/2019 Crystal Structure Sem 5
34/123
Why Is Symmetry Important?
Identification of Materials
Prediction of Atomic Structure
Relation to Physical Properties
Optical
Mechanical
Electrical and Magnetic
8/11/2019 Crystal Structure Sem 5
35/123
Symmetry operations performed about a point or a line are
calledpoint group symmetry operations
Point group symmetry elements exhibited by crystals
are
1. The inversion (centre of symmetry)
2. Reflection symmetry (The mirror reflection)3. Rotation Symmetry
8/11/2019 Crystal Structure Sem 5
36/123
Inversion
11
A crystal will possess an inversioncentre if for every lattice point
given by a position vector rtherewill be a corresponding lattice point
at the position -r
Crystallographic symmetry
element : Centre of Symmetry
8/11/2019 Crystal Structure Sem 5
37/123
In this operation, the reflection of a structure at a mirror plane m
passing through a lattice point leaves the crystal unchanged. The
mirror plane may or may not be composed of the atoms lying on
the concerned imaginary plane.
Reflection Symmetry
Crystallographic symmetry
element : Plane of Symmetry
8/11/2019 Crystal Structure Sem 5
38/123
Rotational Symmetry
If a crystal left invariant after a rotation about an axis ,
is said to possess rotational symmetry.
The axis is called Axis of Symmetry
The axis is called `n-fold, axis if the angle of
rotation is 3600/n .
If equivalent configuration occurs after rotation of 180,
120 and 90, the axes of rotation are known as two-fold, three-
fold and four-fold axes of symmetry respectively.
8/11/2019 Crystal Structure Sem 5
39/123
8/11/2019 Crystal Structure Sem 5
40/123
Symmetrythe rules behind the
shapes
8/11/2019 Crystal Structure Sem 5
41/123
8/11/2019 Crystal Structure Sem 5
42/123
Symmetry elements For a simple Cubic Lattice
One CENTRE OF SYMMETRY
8/11/2019 Crystal Structure Sem 5
43/123
Plane of Symmetry:
There are threeplanes of symmetry parallel to the faces ofthe cube and sixdiagonal planes of symmetry
9
planesofSymmetry
If 4 f 90 t ti i id i hi d d th i
8/11/2019 Crystal Structure Sem 5
44/123
If n=4, for every 90 rotation, coincidence is achieved and the axis
is termed `tetrad axis.It is discussed already that a cube has `three
tetrad axes.
If n=3, the crystal has to be rotated through an angle = 120
about an axis to achieve self coincidence. Such an axis is called is`triad axis. In a cube, the axis passing through a solid diagonal
acts as a triad axis. Since there are 4 solid diagonals in a cube, the
number of triad axis is four.
8/11/2019 Crystal Structure Sem 5
45/123
If n=2, the crystal has to be rotated through an angle = 180
about an axis to achieve self coincidence. Such an axis is called a`diad axis.Since there are 12 such edges in a cube, the number of
diad axes is six.
Total 13 axes of rotational symmetry for a Cube
8/11/2019 Crystal Structure Sem 5
46/123
(a) Centre of symmetry 1
(b) Planes of symmetry 9
(Straight planes -3,Diagonal planes -6)
(c) Diad axes 6(d) Triad axes 4
(e) Tetrad axes 3
----
Total number of symmetry elements = 23----
Thus the total number of symmetry elements of a cubic structure is 23.
SYMMETRICAL ELEMENTS OF CUBE
8/11/2019 Crystal Structure Sem 5
47/123
N-fold axes with n=5 or n>6 does not occur
in crystals
Adjacent spaces must be completely filled (no gaps, no overlaps).
8/11/2019 Crystal Structure Sem 5
48/123
ABSENCE OF 5 FOLD SYMMETRY
We have seen earlier that the crystalline solids show only
1,2,3,4 and 6-fold axesof symmetry and not 5-fold axis of
symmetry or symmetry axis higher than 6.
The reason is that, a crystal is a one in which the atoms or
molecules are internally arranged in a very regular and
periodic fashion in a three dimensional pattern, and
identical repetition of an unit cell can take place only
when we consider 1,2,3,4 and 6-fold axes.
8/11/2019 Crystal Structure Sem 5
49/123
P Q R S
a
MATHEMATICAL VERIFICATION
Let us consider a lattice P Q R Sas shown in figure
Let this lattice has n-fold axis of symmetryand the
8/11/2019 Crystal Structure Sem 5
50/123
lattice parameter be equal to a.
Let us rotate the vectors Q Pand R Sthrough an
angle = 3600
/n , in the clockwise and anti clockwisedirections respectively.
After rotation the ends of the vectors be at x and y.
Since the lattice PQRS has n-fold axis of symmetry,
the points x and y should be the lattice points.
P Q R S
a
x y
Further the line xy should be parallel to the line PQRS.
8/11/2019 Crystal Structure Sem 5
51/123
y p Q
Therefore the distance xy must equal to some integral
multiple of the lattice parameter a say, m a.
i.e., xy = a + 2a cos = ma (1)
Here, m = 0, 1, 2, 3, ..................
From equation (1),
2a cos = m aai.e., 2a cos = a (m - 1)
(or) cos = (2)
Here,
N = 0, 1, 2, 3, .....
22
1 Nm
8/11/2019 Crystal Structure Sem 5
52/123
since (m-1) is also an integer, say N.
We can determine the values of which are allowed in a
lattice by solving the equation (2) for all values of N.
For example, if N = 0, cos = 0 i.e., = 90o
n = 4.
In a similar way, we can get four more rotation axes
in a lattice, i.e., n = 1, n = 2, n = 3, and n = 6.
Since the allowed values of cos have the limits1
to +1, the solutions of the equation (2) are notpossible for N > 2.
Therefore only 1, 2, 3, 4 and 6 fold symmetry axes
can exist in a lattice.
8/11/2019 Crystal Structure Sem 5
53/123
Crystal Structure 53
Lattice Sites in Cubic Unit Cell
8/11/2019 Crystal Structure Sem 5
54/123
Crystal Structure 54
Crystal Directions
Fig. Shows
[111] direction
We choose one lattice point on the line as anorigin, say the point O. Choice of origin iscompletely arbitrary, since every lattice point isidentical.
Then we choose the lattice vector joining O toany point on the line, say point T. This vectorcan be written as;
R = n1a + n2b + n3c
To distinguish a lattice direction from a latticepoint, the triple is enclosed in square brackets [
...] is used.[n1n2n3] [n1n2n3] is the smallest integer of the same
relative ratios.
8/11/2019 Crystal Structure Sem 5
55/123
Crystal Structure 55
210
X = 1 , Y = , Z = 0
[1 0] [2 1 0]
X = , Y = , Z = 1
[ 1] [1 1 2]
Examples
8/11/2019 Crystal Structure Sem 5
56/123
Crystal Structure 56
Negative directions
When we write the
direction [n1n2n3] dependon the origin, negative
directions can be writtenas
Y direction
(origin) O
- Y direction
X direction
- X direction
Z direction
- Z direction
][ 321 nnn
][ 321 nnn
8/11/2019 Crystal Structure Sem 5
57/123
Crystal Structure 57
X = -1 , Y = -1 , Z = 0 [110]
Examples of crystal directions
X = 1 , Y = 0 , Z = 0 [1 0 0]
8/11/2019 Crystal Structure Sem 5
58/123
Crystal Structure 58
Crystal Planes
Within a crystal lattice it is possible to identify sets ofequally spaced parallel planes. These are called
lattice planes.
b
a
b
a
The set of
planes in
2D lattice.
8/11/2019 Crystal Structure Sem 5
59/123
Crystal Structure 59
Miller IndicesMiller Indices are a symbolic vector representation for the
orientation of an atomic plane in a crystal lattice and are definedas the reciprocals of the fractional intercepts which the planemakes with the crystallographic axes.
To determine Miller indices of a plane, take the following steps;
1) Determine the intercepts of the plane along each of the threecrystallographic directions
2) Take the reciprocals of the intercepts
3) If fractions result, multiply each by the denominator of thesmallest fraction
8/11/2019 Crystal Structure Sem 5
60/123
Crystal Structure 60
Axis X Y Z
Intercept
points 1 Reciprocals 1/1 1/ 1/
Smallest
Ratio 1 0 0
Miller ndices (100)
Example-1
(1,0,0)
8/11/2019 Crystal Structure Sem 5
61/123
Crystal Structure 61
Axis X Y Z
Intercept
points 1 1 Reciprocals 1/1 1/ 1 1/
Smallest
Ratio 1 1 0
Miller ndices (110)
Example-2
(1,0,0)
(0,1,0)
8/11/2019 Crystal Structure Sem 5
62/123
Crystal Structure 62
Axis X Y Z
Intercept
points 1 1 1Reciprocals 1/1 1/ 1 1/ 1
Smallest
Ratio 1 1 1
Miller ndices (111)(1,0,0)
(0,1,0)
(0,0,1)
Example-3
8/11/2019 Crystal Structure Sem 5
63/123
Crystal Structure 63
Axis X Y Z
Intercept
points 1/2 1 Reciprocals 1/() 1/ 1 1/
Smallest
Ratio 2 1 0
Miller ndices (210)(1/2, 0, 0)
(0,1,0)
Example-4
8/11/2019 Crystal Structure Sem 5
64/123
Crystal Structure 64
Axis a b c
Intercept
points 1 Reciprocals 1/1 1/ 1/()
Smallest
Ratio 1 0 2
Miller ndices (102)
Example-5
8/11/2019 Crystal Structure Sem 5
65/123
Crystal Structure 65
Axis a b c
Intercept
points -1 Reciprocals 1/-1 1/ 1/()
Smallest
Ratio -1 0 2
Miller ndices (102)
Example-6
8/11/2019 Crystal Structure Sem 5
66/123
Crystal Structure 66
Miller Indices
Reciprocal numbers are:2
1,
2
1,
3
1
Plane intercepts axes at cba 2,2,3
Indices of the plane (Miller): (2,3,3)
(100)
(200)
(110)(111)
(100)
Indices of the direction: [2,3,3]a
3
2
2
bc
[2,3,3]
8/11/2019 Crystal Structure Sem 5
67/123
Crystal Structure 67
Example-7
8/11/2019 Crystal Structure Sem 5
68/123
Crystal Structure 68
Indices of a Family or Form
Sometimes when the unit cell has rotational symmetry, severalnonparallel planes may be equivalent by virtue of this symmetry, inwhich case it is convenient to lump all these planes in the sameMiller Indices, but with curly brackets.
Thus indices {h,k,l} represent all the planes equivalent to the
plane (hkl) through rotational symmetry.
)111(),111(),111(),111(),111(),111(),111(),111(}111{
)001(),100(),010(),001(),010(),100(}100{
Coo din t on N mbe
8/11/2019 Crystal Structure Sem 5
69/123
Crystal Structure 69
Coordinaton Number
Coordinaton Number (CN) : The Bravais lattice pointsclosest to a given point are the nearest neighbours.
Because the Bravais lattice is periodic, all points have
the same number of nearest neighbours or coordinationnumber. It is a property of the lattice.
A simple cubic has coordination number 6; a body-centered cubic lattice, 8; and a face-centered cubiclattice,12.
8/11/2019 Crystal Structure Sem 5
70/123
Atomic Packing Factor
Atomic Packing Factor (APF) is defined as thevolume of atoms within the unit cell divided by
the volume of the unit cell.
8/11/2019 Crystal Structure Sem 5
71/123
Crystal Structure 71
1-CUBIC CRYSTAL SYSTEM
Simple Cubic has one lattice point so its primitive cell.
In the unit cell on the left, the atoms at the corners are cut
because only a portion (in this case 1/8) belongs to that cell.The rest of the atom belongs to neighboring cells.
Coordinatination number of simple cubic is 6.
a- Simple Cubic (SC)
a
bc
http://www.kings.edu/~chemlab/vrml/simcubun.wrlhttp://www.kings.edu/~chemlab/vrml/simcubun.wrl8/11/2019 Crystal Structure Sem 5
72/123
APF for a simple cubic structure = 0.52
APF =
a3
4
3p (0.5a) 31
atoms
unit cellatom
volume
unit cell
volume
Atomic Packing Factor (APF)
APF =
Volume of atoms in unit cell*
Volume of unit cell
*assume hard spheres
Adapted from Fig. 3.23,
Callister 7e.
close-packed directions
a
R=0.5a
contains (8 x 1/8) =
1 atom/unit cell Here: a = Rat*2
Where Rat is the handbook atomic
radius
8/11/2019 Crystal Structure Sem 5
73/123
Crystal Structure 73
b-Body Centered Cubic (BCC)
BCC has two lattice points so BCC
is a non-primitive cell.
BCC has eight nearest neighbors.
Each atom is in contact with its
neighbors only along the body-
diagonal directions.
Many metals (Fe,Li,Na..etc),
including the alkalis and several
transition elements choose the
BCC structure.a
b c
8/11/2019 Crystal Structure Sem 5
74/123
Atomic Packing Factor: BCC
a
APF =
4
3 p( 3a/4 )32
atoms
unit cell atom
volume
a3
unit cell
volume
length = 4R=
Close-packed directions:
3 a
APF for a body-centered cubic structure = 0.68
aR
Adapted from
Fig. 3.2(a), Callister 7e.
a2
a3
8/11/2019 Crystal Structure Sem 5
75/123
Crystal Structure 75
c- Face Centered Cubic (FCC)
There are atoms at the corners of the unit cell and at the center ofeach face.
Face centered cubic has 4 atoms so its non primitive cell. Many of common metals (Cu,Ni,Pb..etc) crystallize in FCC
structure.
3 F C d C b
8/11/2019 Crystal Structure Sem 5
76/123
Crystal Structure 76
3 - Face Centered Cubc
Atoms are all same.
A i P ki F FCC
8/11/2019 Crystal Structure Sem 5
77/123
APF for a face-centered cubic structure = 0.74
Atomic Packing Factor: FCC
The maximum achievable APF!
APF =
4
3p ( 2a/4 )34
atomsunit cell atom
volume
a3
unit cell
volume
Close-packed directions:
length = 4R= 2 a
Unit cell contains:
6 x1/2 + 8 x1/8
= 4 atoms/unit cella
2 a
Adapted from
Fig. 3.1(a),
Callister 7e.
(a = 22*R)
8/11/2019 Crystal Structure Sem 5
78/123
Crystal Structure 78
Atoms Shared Between: Each atom counts:
corner 8 cells 1/8
face centre 2 cells 1/2
body centre 1 cell 1
edge centre 2 cells 1/2
lattice type cell contents
P 1 [=8 x 1/8]I 2 [=(8 x 1/8) + (1 x 1)]
F 4 [=(8 x 1/8) + (6 x 1/2)]C 2 [=(8 x 1/8) + (2 x 1/2)]
Unit cell contents
Counting the number of atoms within the unit cell
h l
8/11/2019 Crystal Structure Sem 5
79/123
Theoretical Density, r
where n= number of atoms/unit cell
A=atomic weight
VC= Volume of unit cell = a3for cubicNA= Avogadros number
= 6.023 x 1023atoms/mol
Density = r =
VCNA
nAr =
CellUnitofVolumeTotal
CellUnitinAtomsofMass
Th ti l D it
8/11/2019 Crystal Structure Sem 5
80/123
Ex: Cr (BCC)A=52.00 g/mol
R= 0.125 nm
n= 2
a= 4R/3 = 0.2887 nma
R
r=a3
52.002
atoms
unit cell mol
g
unit cell
volume atoms
mol
6.023x1023
Theoretical Density, r
rtheoretical
ractual
= 7.18 g/cm3
= 7.19 g/cm3
8/11/2019 Crystal Structure Sem 5
81/123
8/11/2019 Crystal Structure Sem 5
82/123
The Concept of the reciprocal lattice devised to tabulate two
i i f l l Th i l d h i i
8/11/2019 Crystal Structure Sem 5
83/123
important properties of crystal planes: Their slopes and their inter
planer distance.
The reciprocal space lattice is a set of imaginary points
constructed in such a way that the direction of vector from one
point to another coincides with the direction of a normal to the
real space planes and the separation of those points (absolute
value of the vector) is equal to the reciprocal of the real inter-planer distance.
8/11/2019 Crystal Structure Sem 5
84/123
Reciprocal Lattice Vectors
The electronic number density is a periodic function inspace with a period equal to the lattice translationvector T, i.e.
This means that one can use a Fourier seriesexpansion to represent in 1D n(x) as
where:
)()( rTr nn
apxip
pp
pp enapxSapxCnxn /2
00 )/2sin()/2cos()(
p
pp
p
aapxi
ap exdxnn
0
/21 )(
In 3D, we have a
ii 1 GG
8/11/2019 Crystal Structure Sem 5
85/123
The set of reciprocal lattice vectorsthat lead to electrondensity invariant under lattice translations is found from thecondition:
The reciprocal lattice vectors that satisfy the aboverequirement are of the form
where v1, v2and v3are integers and
i
Vi edVnnenn
c0
1 )()( rGGrG
G
G rr
1)()( )( TGTGrG
G
GTrG
G
G rTr iiii ewhenneenenn
332211 bbbG vvv
ijjikjikj
i zyxi p
p 2,,,2 ab
aaa
aab
8/11/2019 Crystal Structure Sem 5
86/123
8/11/2019 Crystal Structure Sem 5
87/123
8/11/2019 Crystal Structure Sem 5
88/123
8/11/2019 Crystal Structure Sem 5
89/123
8/11/2019 Crystal Structure Sem 5
90/123
8/11/2019 Crystal Structure Sem 5
91/123
8/11/2019 Crystal Structure Sem 5
92/123
8/11/2019 Crystal Structure Sem 5
93/123
8/11/2019 Crystal Structure Sem 5
94/123
8/11/2019 Crystal Structure Sem 5
95/123
8/11/2019 Crystal Structure Sem 5
96/123
HW:
Reciprocal of a reciprocal lattice is the direct lattice
Wigner Seitz Method
8/11/2019 Crystal Structure Sem 5
97/123
Crystal Structure 97
Wigner-Seitz Method
A simply way to find the primitive
cellwhich is called Wigner-Seitz
cell can be done as follows;
1. Choose a lattice point.2. Draw lines to connect these
lattice point to its neighbours.
3. At the mid-point and normal tothese lines draw new lines.
The volume enclosed is called as a
Wigner-Seitz cell.
Wigner Seitz Cell 3D
8/11/2019 Crystal Structure Sem 5
98/123
Crystal Structure 98
Wigner-Seitz Cell - 3D
8/11/2019 Crystal Structure Sem 5
99/123
8/11/2019 Crystal Structure Sem 5
100/123
8/11/2019 Crystal Structure Sem 5
101/123
8/11/2019 Crystal Structure Sem 5
102/123
8/11/2019 Crystal Structure Sem 5
103/123
8/11/2019 Crystal Structure Sem 5
104/123
X-ray Diffraction
/hcE
Typical interatomic distances in solid are of the order of an angstrom.
Thus the typical wavelength of an electromagnetic probe of such distances
Must be of the order of an angstrom.
Upon substituting this value for the wavelength into the energy equation,
We find that E is of the order of 12 thousand eV, which is a typical X-ray
Energy. Thus X-ray diffraction of crystals is a standard probe.
X-Rays to Determine Crystal Structure
8/11/2019 Crystal Structure Sem 5
105/123
y y
X-rayintensity(fromdetector)
c
d
n
2sinc
Measurement of
critical angle, c,allows computation of
planar spacing, d.
Incoming X-rays diffractfrom crystal planes.
Adapted from Fig. 3.19,
Callister 7e.
reflections mustbe in phase fora detectable signal!
spacingbetweenplanes
d
extradistancetraveledby wave 2
2 2 2hkl
ad
h k l
For Cubic Crystals:
h, k, l are Miller Indices
Figure 3.34 (a) An x-ray diffractometer. (Courtesy of Scintag,
8/11/2019 Crystal Structure Sem 5
106/123
Inc.) (b) A schematic of the experiment.
X-Ray Diffraction Pattern
8/11/2019 Crystal Structure Sem 5
107/123
y
Adapted from Fig. 3.20, Callister 5e.
(110)
(200)
(211)
z
x
ya b
c
Diffraction angle 2
Diffraction pattern for polycrystalline -iron (BCC)
Intensity
(relative)
z
x
ya b
c
z
x
ya b
c
8/11/2019 Crystal Structure Sem 5
108/123
8/11/2019 Crystal Structure Sem 5
109/123
INTER TOMIC FORCES
Energies of Interactions Between Atoms
Ionic bonding
NaCl
Covalent bonding
Comparison of ionic and covalent bonding
Metallic bonding
Van der waals bonding
Hydrogen bonding
What kind of forces hold the atoms together in
a sol id?
8/11/2019 Crystal Structure Sem 5
110/123
Energies of Interactions Between toms
The energy of the crystal is lower than that of the freeatoms by an amount equal to the energy required to pullthe crystal apart into a set of free atoms. This is called thebinding (cohesive) energy of the crystal.
NaCl is more stable than a collection of free Na and Cl. Ge crystal is more stable than a collection of free Ge.
Cl Na NaCl
Types of Bonding Mechanisms
8/11/2019 Crystal Structure Sem 5
111/123
Types of Bonding Mechanisms
It is conventional to classify the bonds betweenatoms into different types as
Ionic Covalent
Metallic Van der Waals Hydrogen
All bonding is a consequence of the electrostatic interaction betweenthe nuclei and electrons.
8/11/2019 Crystal Structure Sem 5
112/123
IONIC BONDING
Ionic bonding is the electrostatic force of attraction between
positively and negatively charged ions (between non-metals
and metals).
All ionic compounds are crystalline solids at room
temperature.
NaCl is a typical example of ionic bonding.
Metallic elements have only up to the valence electrons in their outer
8/11/2019 Crystal Structure Sem 5
113/123
Metallic elements have only up to the valence electrons in their outershell.
When losing their electrons they become positive ions.
Electronegative elements tend to acquire additional electrons tobecome negative ions or anions.
Na Cl
When the Na+ and Cl- ions approach each other closely
8/11/2019 Crystal Structure Sem 5
114/123
When the Na and Cl ions approach each other closelyenough so that the orbits of the electron in the ions beginto overlap with each other, then the electron begins to
repel each other by virtue of the repulsive electrostaticcoulomb force. Of course the closer together the ions are,the greater the repulsive force.
Pauli exclusion principle has an important role in repulsiveforce. To prevent a violation of the exclusion principle, thepotential energy of the system increases very rapidly.
8/11/2019 Crystal Structure Sem 5
115/123
COV LENT BONDING
Covalent bonding takes place between atoms with small differencesin electronegativity which are close to each other in the periodictable (between non-metals and non-metals).
The covalent bonding is formed when the atoms share the outershell electrons (i.e., s and p electrons) rather than by electrontransfer.
Noble gas electron configuration can be attained.
Each electron in a shared pair is attracted to both nuclei
8/11/2019 Crystal Structure Sem 5
116/123
Each electron in a shared pair is attracted to both nucleiinvolved in the bond. The approach, electron overlap, and
attraction can be visualized as shown in the following figure
representing the nuclei and electrons in a hydrogen molecule.
e
e
Comparison of Ionic and Covalent
d
8/11/2019 Crystal Structure Sem 5
117/123
Bonding
8/11/2019 Crystal Structure Sem 5
118/123
MET LLIC BONDING
Metallic bonding is found in metalelements. This is the electrostaticforce of attraction between positivelycharged ions and delocalized outerelectrons.
The metallic bond is weaker than theionic and the covalent bonds.
A metal may be described as a low-density cloud of free electrons.
Therefore, metals have high electricaland thermal conductivity.
+
+
+
+
+
+
+
+
+
8/11/2019 Crystal Structure Sem 5
119/123
V N DER W LS BONDING
These are weak bonds with a typical strength of 0.2 eV/atom.
Van Der Waals bonds occur between neutral atoms andmolecules.
Weak forces of attraction result from the natural fluctuations inthe electron density of all molecules that cause smalltemporary dipoles to appear within the molecules.
It is these temporary dipoles that attract one molecule toanother. They are called van der Waals' forces.
The shapeof a molecule influences its ability to form temporarydipoles. Long thin molecules can pack closer to each other thanmolecules that are more spherical The bigger the 'surface area' of
8/11/2019 Crystal Structure Sem 5
120/123
molecules that are more spherical. The bigger the 'surface area' ofa molecule, the greater the van der Waal's forces will be and thehigher the melting and boiling points of the compound will be.
Van der Waal's forces are of the order of 1% of the strength of acovalent bond.
Homonuclear molecules,such as iodine, develop
temporary dipoles due to
natural fluctuations of electron
density within the molecule
Heteronuclear molecules,
such as H-Cl have permanent
dipoles that attract the opposite
pole in other molecules.
These forces are due to the electrostatic attraction between
8/11/2019 Crystal Structure Sem 5
121/123
the nucleus of one atom and the electrons of the other.
Van der waals interaction occurs generally betweenatoms which have noble gas configuration.
van der waals
bonding
8/11/2019 Crystal Structure Sem 5
122/123
HYDROGEN BONDING
A hydrogen atom, having one electron, can be covalently bonded toonly one atom. However, the hydrogen atom can involve itself in an
additional electrostatic bond with a second atom of highly
electronegative character such as fluorine or oxygen. This second
bond permits a hydrogen bond between two atoms or strucures.
The strength of hydrogen bonding varies from 0.1 to 0.5 ev/atom.
Hydrogen bonds connect water
molecules in ordinary ice.Hydrogen bonding is also very
important in proteins and
nucleic acids and therefore in
life processes.
8/11/2019 Crystal Structure Sem 5
123/123