Crystal Structure
Introduction
Atomic Structure
Nomor atom = jumlah proton= jumlah netron
dalam suatu atom
Massa atom(nomor massa) = massa proton + masa
neutron
Massa proton= massa netron= 1,67x10 -27 kg
Massa elekron = 9.11 X 10-31 kg
The atomic mass unit (amu) sebagai satuan berat
atom, sebagai standart adalah
1/12 berat atom 12C =1.66 X 10 -24 gr = 1 amu
Berat atom sering dinyatakan per mol (gr/mol)
Satu mol menyatakan jumlah partikel yang yang
mempunyai massa dalam gram
Jumlah atom dalam satu mol dinyatakan dengan
Avogadro number, N = 6.023 x 1023
Atomic Structure
ATOM = (PROTON+NEUTRON) + ELECTRON
Bohr atom model
Wave-mechanical model
Periodic Table
Atomic Bonding
Primary bonding
Secondary bonding
Influences the physical
properties of material
Atomic Bonding - Primary
Ionic bonding
Composed of both metallic and
nonmetallic elements
Created between two unlike atoms
with different electronegativities, or
Strong Coulomb interaction among
negative atoms (have an extra
electron each) and positive atoms
(lost an electron).
Atomic Bonding - Primary
Ionic bonding
NaCl. When sodium donates its
valence electron to chlorine, each
becomes an ion; attraction occurs,
and the ionic bond is formed
Atomic Bonding - Primary
Covalent bonding
Stable electron configurations are assumed by the
sharing of electrons between adjacent atoms
Two atoms that are covalently bonded will each
contribute at least one electron to the bond, and
the shared electrons may be considered to belong to
both atoms
Directional; that is, it is between specific atoms and
may exist only in the direction between one atom
and another that participates in the electron
sharing.
Atomic Bonding - Primary
Example: H2, Cl2, CH4,
SiC diamond, H2O
Number of covalent
bonds (NCB) that is
possible for a particular
atom is determined by
the number of valence
electrons.
NCB = 8 – N’; N’ is valence
electrons
Atomic Bonding - Primary
Metallic bonding
Metals and their alloys.
Valence electrons are not bound to any
particular atom in the solid and are more or less
free to drift throughout the entire
metalforming sea of electrons
The remaining nonvalence electrons and atomic
nuclei form what are called ion cores, which
possess a net positive charge equal in magnitude
to the total valence electron charge per atom
Atomic Bonding - Primary
Metallic bonding
Nondirectional
free electrons act as
a “glue” to hold the
ion cores together
Atomic Bonding - Secondary
Van der waals bonding
Physical bond
Weak
Secondary bonding exists between virtually all
atoms or molecules, but its presence may be
obscured if any of the three primary bonding
types is present
Secondary bonding forces arise from atomic or
molecular dipoles. In essence, an electric
dipole exists whenever there is some
separation of positive and negative portions of
an atom or molecule.
Atomic Bonding - Secondary
Van der waals bonding
The bonding results from the coulombic
attraction between the positive end of
one dipole and the negative region of an
adjacent
Dipole interactions occur between
induced dipoles, between induced dipoles
and polar molecules (which have
permanent dipoles), and between polar
molecules
Atomic Bonding - Secondary
Van der waals bonding
Hydrogen bond: a special type of
secondary bonding, is found to exist
between some molecules that have
hydrogen as one of the constituents
Example: HF, HCl
Bonding Forces & Energies
Force: attractive and repulsive
Magnitude is a function of the
separation or interatomic
distance
The bonding energy for two
atoms, E0, corresponds to the
energy at minimum point; it
represents the energy that
would be required to separate
these two atoms to an infinite
separation
Furthermore, a number of
material properties depend on
E0, the curve shape, and
bonding type
Bonding Forces & Energies
Crystal Structure
Properties of material directly
related to crystal structure
Classification of solid materials: the
regularity with which atoms or ions
are arranged with respect to one
another
Crystal Structure
Crystalline:
The atoms are situated in a repeating or periodic
array over large atomic distances;
Long-range order exists, such that upon
solidification,
The atoms will position themselves in a repetitive
three-dimensional pattern, in which each atom is
bonded to its nearest-neighbor atoms
Metals, many ceramic materials, certain polymer
Crystal Structure
Non crystalline
Amorphous
No crystallization
Absence of long range
order
Crystal Structure
Illustration of some structure of materials
Crystal Structure
Kristal
Susunan atom-atom dimensi,
mengikuti aturan tertentu dan
berulangulang dalam tiga dimensi
System Kristal
Unit Cell
3 sudut antara 2 sumbu:
α,β,γ
3 rusuk yangmasing2
terletak/searah dengan sumbu x,
y, dan z:
U, v, w
7 System Kristal
-------------------------------14 struktur kristal
Crystal Structure
Atomic hard sphere model
Lattice: three-dimensional
array of points coinciding
with atom positions (or
sphere centers).
Lattice: A collection of points
that divide space into smaller
equally sized segments or 3D
network of imaginary lines
connecting the atoms
Crystal Structure: Unit Cells
Subdivision of crystal structure consist of small
repeat entities
A subdivision of the lattice that still retains the
overall characteristics of the entire lattice
parallelepipeds or prisms having three sets of
parallel faces; one is drawn within the
aggregate of spheres which in this case
happens to be a cube
Basic structural unit or building block of the
crystal structure and defines the crystal
structure by virtue of its geometry and the
atom positions within
Crystal Structure: Unit Cells
Unit cell
Crystal Systems
Division of crystal structure based on
the combination of lattice
parameters
Lattice parameters: unit length (a, b,
c) and interaxial angle (, , )
7 possible crystal system
7 crystal system
Metallic Crystal Structures
Face centered cubic (FCC)
Body centered cubic (BCC)
Hexagonal close-packed (HCP)
Metallic Crystal Structures: HCP
Unit cell : hexagonal form
Total atoms 6 per unit cell
c/a ratio = 1.633
Coordination number 12
APF = 0.74
Mg, Zn, Ti, Cd
Metallic Crystal Structures
Thoretical density:
Polymorphism
The characteristic of some metals or
nonmetals for being able to exist in more
than one crystal structure
Depends on temperature and external
pressure
Allotropy: polymorphism in elemental
solids, reversible polymorphism
(Fe, C,…)
Crystalographic
Points
Directions
Planes
The basis for determining index values
is the unit cell, with a right-handed
coordinate system consisting of three
(x, y, and z) axes situated at one of
the corners and coinciding with the
unit cell edges
Crystalographic – Poin Coordinates
The position of
any point located
within a unit cell
may be specified
in terms of its
coordinates as
fractional
multiples of the
unit cell lengths
i.e. a, b, c.
Crystalographic – Poin Coordinates
Locate the coordinate of ¼ 1 ½
Unit cell length: a=0.48 nm, b=0.46 nm, c=0.40 nm
Fractional length: q=1/4, r=1, s=1/2
Then qa=1/4x0.48=0.12 nm along x, rb=1x0.46=o.46 nm
along y, sc=1/2x0.40=0.20 nm along z.
Sketch the position according to the last calculation
fig. (b)
Crystalographic – Poin Coordinates
Point coordinates of all atom positions for a BCC
Crystalographic – Directions
Line between two points or a vector
A vector of convenient length is positioned
such that it passes through the origin of the
coordinate system. Any vector may be
translated throughout the crystal lattice
without alteration, if parallelism is
maintained.
The length of the vector projection on each
of the three axes is determined; these are
measured in terms of the unit cell dimensions
a, b, and c.
These three numbers are multiplied or
divided by a common factor to reduce them
to the smallest integer values.
The three indices, not separated by
commas, are enclosed in square brackets,
thus: [uvw]. The u, v, and w integers
correspond to the reduced projections
along the x, y, and z axes, respectively
Crystalographic – Directions
Determine the
indices for the
direction shown in
the accompanying
figure
Crystalographic – Directions
Draw a [110] direction
This problem is solved by reversing the
procedure of the preceding example.
The projections along the x, y, and z
axes are a, and 0a, respectively.
This direction is defined by a vector
passing from the origin to point P,
which is located by first moving along
the x axis a units, and from this
position, parallel to the y axis units,
as indicated in the figure.
There is no z component to the vector,
since the z projection is zero
Crystalographic – Directions
Direction A
1. Two points are 1, 0, 0, and 0, 0, 0
2. 1, 0, 0, -0, 0, 0 = 1, 0, 0
3. No fractions to clear or integers to reduce
4. [100]
Direction B
1. Two points are 1, 1, 1 and 0, 0, 0
2. 1, 1, 1, -0, 0, 0 = 1, 1, 1
3. No fractions to clear or integers to reduce
4. [111]
Direction C
1. Two points are 0, 0, 1 and 1/2, 1, 0
2. 0, 0, 1 -1/2, 1, 0 = -1/2, -1, 1
3. 2(-1/2, -1, 1) = -1, -2, 2
4.[ 1 2 2]
Crystalographic – Directions
Several nonparallel directions
with different indices are
actually equivalent the
spacing of atoms along each
direction is the same.
Equivalent
directionsfamilybracket
<>
<100>: [100], [100 ], [010],
[010], [001], and [001]
Crystalographic – HCP Directions
Four axis / Miller-Bravais
Conversion from 3 index
system to 4 index system
[ u ' v ' w ' ] [ uvtw ]
Formulas
1
u ( 2 u ' v ' )
3
1
v ( 2 v ' u ' )
3
t (u v )
w w'
Crystalographic – HCP Directions
Determine direction in fig (a)
Projection: a(a1 axis), a(a2
axis) and c(z axis)
u’=1 v’=1 w’=1
Converting 3 index to 4 index
using the formulas
1,1,-2,3 [1123]
Crystalographic – Planes
Crystallographic planes are specified by three Miller indices as (hkl)
except in HCP
If the plane passes through the selected origin, either another
parallel plane must be constructed within the unit cell by an
appropriate translation, or a new origin must be established at the
corner of another unit cell.
At this point the crystallographic plane either intersects or parallels
each of the three axes; the length of the planar intercept for each
axis is determined in terms of the lattice parameters a, b, and c.
The reciprocals of these numbers are taken. A plane that parallels an
axis may be considered to have an infinite intercept, and, therefore,
a zero index.
If necessary, these three numbers are changed to the set of smallest
integers by multiplication or division by a common factor
Finally, the integer indices, not separated by commas, are enclosed
within parentheses, thus: (hkl).
Crystalographic – Planes
Determine the miller indices of fig (a)
Crystalographic – Planes
Construct a (011) plane within a cubic
Carry out the procedure used in the preceding
example in reverse order
the indices are removed from the parentheses, and
reciprocals are taken, which yields , -1, 1.
x=paralel, y=-1, z=1
Draw it as per fig (b)
Crystalographic – Planes
Plane A
1. x = 1, y = 1, z = 1
2.1/x = 1, 1/y = 1,1 /z = 1
3. No fractions to clear
4. (111)
Plane B
1. The plane never intercepts the z axis, so x =
1, y = 2, and z =
2. 1/x = 1, 1/y =1/2, 1/z = 0
3. Clear fractions: 1/x = 2, 1/y = 1, 1/z = 0
4. (210)
Plane C
1. We must move the origin, since the plane
passes through 0, 0, 0. Let’s move the
origin one lattice parameter in the ydirection. Then, x = , y = -1, and z =
2. 1/x = 0, 1/y = -1, 1/z = 0
3. No fractions to clear
4. (0 1 0)
Crystalographic – Planes
One interesting and unique characteristic of cubic
crystals is that planes and directions having the
same indices are perpendicular to one another
Reversing the directions of all indices specifies
another plane parallel to, on the opposite side of
and equidistant from, the origin
Equivalent planes (above definition) family of
planes { }
s
Crystalographic – Planes
Crystalographic – HCP Planes
Miller-Bravais
index
Four-index
(hkil)
i = -(h+k)
Crystalographic – HCP Planes
Determine the Miller–Bravais indices for the
plane shown in the figure
Consider the plane in the figure
referenced to the parallelepiped labeled
with the letters A through H at its
corners
Intersections: a1=a, a2=-a, z=c 1, -1, 1
h= 1, k= -1, l= 1
Calculate i, i= 0
Thus, [1110] parallel to a3 axis
Crystalographic – HCP Planes
Linear & Planar Density
Directional equivalency is related to
linear density in the sense that, for
a particular material, equivalent
directions have identical linear
densities
The corresponding parameter for
crystallographic planes is planar
density, and planes having the same
planar density values are also
equivalent
Linear Density
Linear density (LD) is defined as the number of atoms per unit
length whose centers lie on the direction vector for a specific
crystallographic direction
Units: reciprocal length (nm-1, m-1)
Linear Density
Example [110] in FCC
LD110=2 atoms/4R = 1/2R
Planar Density
planar density (PD) is taken as the number of
atoms per unit area that are centered on a
particular crystallographic plane
Units: reciprocal area (nm-2, m-2)
Planar Density
Example (110) in FCC
PD= 2 atoms/8R2e2 = 1/4R2e2
Linear & Planar Density
Linear and planar densities are
important considerations relative to
the process of slip—that is, the
mechanism by which metals
plastically deform
Slip occurs on the most densely
packed crystallographic planes and,
in those planes, along directions
having the greatest atomic packing
Closed-Packed Crystal Structures
FCC and HCP have atomic packing
factor 0.74most efficient packing
of equal-sized spheres or atoms
Apart of unit cell representation,
FCC & HCP may be described in
terms of close-packed planes of
atoms (i.e., planes having a
maximum atom or sphere-packing
density)
Closed-Packed Crystal Structures
The real distinction between FCC and HCP lies in
where the third close-packed layer is positioned
FCC: ABCABCABC
HCP: ABABAB
Closed-Packed Crystal Structures
Close-packed planes for HCP are (0001)
Closed-Packed Crystal Structures
Close-packed planes for FCC are (111)
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