Crystal structure notes

Engineering Physics Notes: Author: Praveen N Vaidya, SDMCET Dharw ad .
Crystal Structure:
Solid state physics is branch of physics, which
deals with the study of structure, properties and
applications of solids in crystalline and amorphous
state.
The solids are classified into basically two types
those are,
1) Amorphous 2) Crystalline
Amorphous solids are solids with a random and
shortly-oriented atomic or molecular arrangement.
Examples of amorphous solids are, glass and
plastic. They are considered as super cooled
liquids because of their random molecular
arrangement just as in the liquids. Amorphous
solids do not have definite melting points.
In materials science, a crystal is a solid in which
the constituent atoms, molecules, or ions are
packed in a regular fashion, repeating pattern
extending in all three spatial dimensions
1) Crystalline solids are arranged in fixed
geometric patterns or lattices.
2) They have orderly arranged units and are
practically incompressible.
3) Crystalline solids also show a definite melting
point and so they pass rather sharply from
solid to liquid state.
4) They are cubic, tetragonal, hexagonal,
Rhombic, monoclinic, and triclinic.
5) The units that constitute these systems can be
atoms, molecules, or ions.
6) Ionic and atomic crystals are hard and brittle
with high melting points. Examples of
crystalline solids are CuSO4, NaCl, KCl, etc.
7) Molecular crystals are soft and have low
melting points. For ex. Ice, methanol, dry ice
etc.
8) Metallic crystals are composed of positively
charged ions in a field of electron gas or freely
moving electrons. Metals are good conductors
of electricity because of the free movement of
electrons in the lattice.
9) There are two forms of crystal called Single
crystals and Polycrystals.
10) Single crystals are artificially grown by
different methods. These types of crystals
are characterized by regular arrangement
of atoms or molecules over entire volume of
crystal.
11) The Polycrystalline crystals are naturally
grown by nature. These types of crystals are
made of number of domains of single crystals.
They are separated well defined boundaries
but oriented in different directions.
1

12) Crystal defects: Irregularity of atoms in
crystal due to missing atoms or presence
foreign atoms etc. This is known as crystal
defects. The crystal with zero defects is
known as Ideal Crystal. The crystal defect is
minimized in case of single crystals or grown
crystals.
Space Lattice:
The group of points in three dimensions is
arranged in various patterns and in regular fashion,
so that environment around every point remains
identical. Such group of points is called a space
lattice and points are called lattice points.
Therefore, A three dimensional collection of
points, so that the environment about a point is
identical to environment of any other point is
called Space Lattice.
Basis: The crystal lattice has no any significance
unless it is filled by atoms or molecules. .
Therefore, the atoms or molecules those fills
position of lattice point in three dimensions are
called bases. The lattice and bases together
gives crystal structure.
+ =
Therefore, Lattice + Bases =Crystal structure
Bravais Lattice:
If the lattice points in a space lattice represent the
location of bases of same type, then the lattice
termed as Bravias lattice.
If a space lattice contains, the basis of different
crystals then the lattice is called as non bravias
lattice.
2
Square lattice Hexagonal lattice
Bravias Lattice
Lattice having
same type of
atoms
Non Bravias
Lattice
Lattice having
different type of
atoms
Black and hollow
circles together
form non-Bravias
and individually
form Bravias
lattice resp.
Lattice + Basis = crystal Structure

In the above figure, if lattice is imagined
considering only one type of atoms (either solid
circles or empty circles individually) the lattices
are Bravais lattices but together forms a non
bravais lattice as because of having two types of
atoms.
Unit Cell:
Unit Cell is a smallest group of lattice points or
basis arranged three dimension, so that
repetition of which will give the actual crystal
lattice.
In the above diagram dark block is unit cell in
simple cubic structure, repeatation of which three
dimensional, defines the ultimate structure of the
crystal with symmetry. It also defines secondary
properties such as cleavage, electronic band
structure, and optical properties like color and
translucence.
There are four different types of unit cells,
depending on the number of atoms molecules
associated with it.
a) If the atoms are present only at the corners of
the unit cell then unit cell is called as primitive
cell or Simple cubic (SC). It has one atom per
unit cell.
b) If a unit cell contains one atom at its centre in
addition to the atoms at the corners, then unit cell
is called as Body Centered Cubic (BCC). It has
two atoms per unit cell.
c) If the unit cell contains an atom at the centre of
each of the four faces of unit cell along with the
atoms at the corners then unit cell is called as Face
Centered Cubic (FCC). It has four atoms per
unit cell.
d) If a unit cell has two atoms one at the upper
face and another at lower face along with the
atoms at the corners then unit cell is called as Base
Centered Cubic. It has two atoms per unit cell.
Primitive Cell:
The primitive cell is defined as unit-cell which
contains lattice points at corners only.
Primitive cell is a unit cell with an atom at each of
the eight corners. Every corner atom of primitive
cell shares its volume with eight other unit cells or
3
SC BCC FCC BASC

c
b
a
each atom of a primitive cell contribute only one
eighth of its volume to a cell. There fore, total
eight atoms each contributes 1/8th
of its fraction to
unit cell, hence total contribution is given by
atomatoms 18
8
1
=× .
Hence a primitive cell has one atom per unit
cell.
Lattice Parameters:
γ β
The geometrical parameters like the sides
(a, b, c) and the angles (α β γ) between any two
adjacent sides together called as the Lattice
Parameters. Also in above diagram the
directions specified along a, b and c are called as
crystallographic axes.
Lattice constants are parameters of a unit cell
specifying the dimensions (a, b, c) along three
crystallographic axes. These are also called as cell
edges.
Crystal translational vectors or basis vectors are
the vectors along the lattice constants a, b and c.
The dimensions of all lattice constants of all unit
cells along crystallographic axes remain same
throughout the crystal.
One can locate any lattice point in a crystal from a
point chosen as origin is given by,
R = n1a + n2b + n3c
Where R is called as position vector. This
determines the position of any lattice point from
the origin.
Crystal Systems:
The crystal systems are the grouping of crystal
lattices according to their axial system and used to
describe their lattice. Each crystal system consists
of a set of three axes in a particular geometrical
arrangement.
There are seven unique crystal systems. The
simplest and most symmetric one is cubic system,
the other six systems, in order of decreasing
symmetry are, hexagonal, tetragonal,
rhombohedral (trigonal), orthorhombic,
monoclinic and triclinic.
The 14 Bravais Lattices:
When the crystal systems are combined with the
various possible lattice centering, one can arrive at
the Bravais lattices. They describe the geometric
arrangement of the lattice points, and thereby the
translational symmetry of the crystal.
4
Seven crystal systems with 14
bravais lattices

In three dimensions, the seven crystal system
together form 14 unique Bravais lattices which
are distinct from one another in the translational
symmetry they contain.
Cubic:
In this system the crystals are made up of three
equal axes at right angles to each other,
i.e., a = b = c and α = β = γ =90o
. These crystals
have three types of lattice depending upon the
shape of unit cells those are, SC, BCC and FCC.
The three lattices types of Cubic as shown in
below fighre.
Example: Diamond, Zinc Sulphide, Anhydrous
chlorides of alkali metals etc.
Tetragonal:
In this system the three cell edges are mutually
perpendicular to each other (α=β=γ=90°) and cell
edges are a=b≠c.
It has two bravais lattice i.e. SC and BCC.
Examples: Indium (s), Tin (s) SnO2, indium,
Titanium dioxide (TiO2) NiSO4 etc.,
Orthorhombic:
In this type of crystals, the three lattice constants
along three crystallographic axes are unequal
(a≠b≠c), and angles between them are α=β=γ=90°
This system contains four type of bravais lattices,
those are SC, BCC, FCC and BASC
5
Simple cubic BCC FCC
a = b = c and α = β = γ =90o
.
Simple (SC) Base Centered
Body centered (BCC) Face centered (FCC)

Examples: Chlorine (g), Galium (l), Bromine (l),
Iodine (s), Uranium (s),
Monoclinic:
In this type of crystals, the crystallographic axes
are unequal (a ≠ b ≠ c) and angles β = γ = 90, α ≠
90. This system has two bravais lattice are SC and
BASC.
Ex: Plutonium, Borax (Na2B4O7, 5H2O),
Cryolite (Na3AlF6)
Triclinic:
i.e., a = b = c and α = β = γ≠ 90
It has only one bravais lattice i.e. SC
Examples: Potassium dichromate (K2Cr2O7),
Copper Suphate (CuSO4 ,5H2O)
Rhombohedral or Trigonal
It has only one bravais lattice i.e. SC
Ex: Quartz and Calcite crystals, As, Sb, Bi
Hexagonal: There is only one hexagonal Bravais
lattice, which has six atoms per unit cell.
Graphite is an example of the hexagonal crystal
system.
Ex: SiO2, ZnO, Mg and Cadmium, Carbon (s)
Directions and planes in crystal Lattice:
The properties of a crystal may be different along
different direction inside the crystal (crystallo-
graphic directions). For example the electrical
conductivity, thermal conductivity, elastic
modulus depends on the direction in some
crystals. The velocity of sound may also be
different along different directions in a crystal.
Thus there is a need to be able to specify
directions within a crystal.
6
a
b
c
In triclinic system, the
crystal is described by
vectors of unequal length.
In addition, all three
vectors are not mutually
orthogonal but are equal
with each other
SC BASC
In this system, three cell
edges are equal, i.e., a = b
= c, and the three
interfacial angles are at
α = β = γ 90 < 120
In the basic hexagonal
structure, a = b = c and α
= β = 90o
and γ =120o
to
them. It is made of three
simple cubic lattice
attached together.

Properties may also depend on the planes of atoms
inside the crystal. For instance when a crystal is
subjected to a stress, it may slip along a particular
plane. Also X-rays incident on a crystal will be
reflected by different planes in different
directions. Thus we need to be able to specify the
planes of a crystal as well.
The crystal direction in a crystal structure is
defined as the vector direction from the origin
(chosen corner) to a lattice point in a crystal.
For ex.: One can formulate various directions in
crystal taking intercepts along X, Y and Z axes
from a origin.
Let the intercepts along X, Y and Z axes are given
by (x, y, z).
These are put in the square brackets ([ ])
011
−
M
N
Q P
S R
In above diagram, consider the corner ‘O’ is origin
of unit cell, hence intercepts are (0, 0, 0).
The direction along OX, x=1, y = 0 and z = 0,
hence crystal direction along OX is (1, 0, 0), and
along negative X-axis i.e. along OX' crystal
direction is (1 , 0, 0). Similarly directions along Y
and Z axes are (0 1 0) and (0 0 1) respectively and
along –ve Y and Z axes (0 1 0) and (0 0 1) resp.
Now consider point P is in XZ axis, it covers ½
distance along Z-axis and 1 along X axis hence
direction is written as (1, 0, ½ ). Similarly along
OQ it is (1, 0, ½)
In XY plane the direction OR is (1, 1, 0) and in –
ve XY- plane directions are OS is (1, 1, 0)
Crystal directions along OM and ON is (1, 1, 1)
and (1, 1, 1) respectively.
Crystal Planes: The crystal is made up of
number of equidistant parallel planes.
Any plane in the crystal passed through the
atoms of crystal, formed by taking the
intercepts along three crystallographic axes, is
called as Crystal Plane.
The characteristic properties of a crystal in the
direction of a set of parallel planes is same, hence
any set of parallel planes is named by taking three
indices along three crystallographic axes are called
as miller indices.
The perpendicular distance between two
successive planes is called as inter planar
spacing. It is represented by ‘d’.
7
O
Y
XX'

In fig. above the shade region is given by crystal
planes. The fig.1 represents the crystal plane (1 0
0), as this plane is in x-intercept and its
coordinates along y and z intercept are zero.
In fig. 2 the shaded area is crystal plane whose
coordinates along Z axis is zero and along X and
Y axes are 1, 1 respectively. Therefore crystal
plane is (1 1 0).
Similarly in fig. 3 the plane formed due to
coordinates along all three intercepts Therefore
crystal plane is (1 1 1).
In fig. 4 the plane the coordinate of x – intercept is
½ and Y and Z intercepts are zero. Therefore
crystal plane is (1/2 , 0, 0)
Crystal planes are located in round brackets ( )
Miller indices:
Miller indices of a crystal plane are the
reciprocals of its intercepts along three
crystallographic axes, when reduced to smallest
integers.
Features of Miller indices:
*Miller indices are equally spaced parallel planes
having the same index number (h, k, l).
*If a plane parallel to one of the coordinate axis
has an intercept of infinity and index of zero for
that axis.
*They do not define a particular plane but set of
plane.
*The characteristic properties of all the planes
having same index is similar.
*In general the miller indices crystal plane is
explained by following diagram.
Consider a crystal lattice with bases vectors (a b c)
along the X, Y and Z axes respectively.
The Miller indices of the given plane are found
by following procedure.
1. A plane ABC is given by the intercepts, x, y, z
respectively along the bases vectors a, b, c
respectively.
2. Express the x, y and z in terms of fraction or
multiple of basis vectors, then
c
z
b
y
a
x
,, .
3. Take the reciprocals of the three fractions
hence, z
c
y
b
x
a
,,
4. Find the LCM of the denominator, and
multiply with the above three ratio. This will
8
Fig. 2
Fig. 3
Fig. 4
Z
X
Y
c b
a

Normal to plane ABC
reduce them into three integers (h, k, l) are
called as Miller Indices.
For Example, In above diagram,
• a, b and c are the length of the basis
vectors.
• x = a
2
1
, y = 4b, z = 2c
•
c
z
b
y
a
x
,, =
2
1
, 4, 2
• Reciprocal of above is (2,
4
1
2
1
)
• LCM of above fractions is 4, is multiplied
with above three numbers.
• (8, 1, 2) is the miller indices of above
plane.
Expression for Inter planar spacing:
Consider ABC is any plane in the crystal
lattice of origin ‘O’. x, y and z are the intercepts
of plane along the three crystal axes X, Y and Z
respectively. The line OP is drawn perpendicular
to the plane ABC.
Z
C Y
P B
O A X
Let us imagine another plane parallel to ABC
passes through point ‘O’, then OP is taken as inter
planar spacing (d h k l ). Where {h, k, l} are the
miller indices of the given planes.
Let α, β, γ are the angle made by OP, with respect
to intercepts x, y and z, also as OP is normal to the
plane. This arrangement creates three right angles
triangles; each contains one intercept as
hypotenuse then,
d h k l = x cos α = y cos β = z cos γ -------- 1
Consider the a, b and c are the lengths of basis
vectors, then miller indices can be written as
{h, k, l} = [ z
c
y
b
x
a
,, ]
Or (x, y, z) = [
l
c
k
b
h
a
,, ] -----------------2
Substitute the value of x, y, z in equation (1),
d h k l =
h
a
cos α =
k
b
cos β =
l
c
cos γ --
3
For Orthogonal co-ordinates we have
cos2
α + cos2
β + cos2
γ =1
Substitute the cosine values from equation (3)
d 2
h k l { 2
2
2
2
2
2
c
l
b
k
a
h
++ } = 1
or d h k l =
2
2
2
2
2
2
1
c
l
b
k
a
h
++
This is the expression for the inter planar
spacing.
If a = b = c, i.e. for the cubic crystal lattice
d h k l =
2
222
1
a
lkh ++
9

or d h k l = 222
lkh
a
++
F o r t h e t e t r a g o n a l a = b , c
2
2
22
2
l
c
kh
a
d lkh +
+
=
F o r t h e O r t h o r h o m b i c a , b , c
2
2
2
2
2
2
l
c
k
b
h
a
d lkh ++=
Coordination number:
The coordination number is defined as the
number of nearest neighbours that an atom has
in a given crystal.
In fig atoms A, B, C, D, E and F i.e. six atoms
are the nearest neighbours of O. hence
coordination number of simple
cubic lattice is Six (6).
Atomic packing factor (APF):
In an unit cell the space occupied its atoms is
always less than the volume of the unit cell.
Therefore, the atomic packing factor is defined
as the fraction of the total volume of a unit cell
occupied by the atoms.
It is given by ratio of total volume occupied by the
atoms (N x Va) to the volume of unit cell (V)
V
VN
APF a×
= Va Volume of an atom
V – Volume of Unit cell
To Determine the APF of some unit cells
1. Simple Cubic structure
Let lattice constant of simple cubic is ‘a’,
Volume of the unit cell is V= a3
= (2R)3
= 8R3
,
Where R is radium of atom, ‘a’ is lattice constant
a =2R
The eight atoms of simple cubic, each contribute
1/8th of its volume to unit cell, therefore,
No. of atoms per unit cell of simple cubic is
N = atomatom 1
8
1
8 =





×
Volume of one atom is
3
3
4
RVa π=
Therefore,
V
VN
APF a×
= =
3
3
8
3
4
1
R
Rπ×
= 0.52 or
= 52%
2. Body centered Cubic:
To calculation lattice constant (a)
(AC)2
= (AB)2
+(BC)2
10
B
F
E
O
A
D C
A B
C

A
(4R)2
= a2
+a2
+(a)2
= 3a2
2
3
2
3
4
3
16 2
2 a
Ror
R
aor
R
a ===
Volume of the unit cell cube is
V =
33
64 3
3 R
a =
Where R is radius of atom, ‘a’ is lattice constant
For BCC, the eight atoms at corners, each
contribute 1/8th of its volume to unit cell, and one
atom at the centre, therefore,
No. of atoms per unit cell
N =
atomscentretheatatomatom 21
8
1
8 =+





×
Volume of one atom is
3
3
4
RVa π=
Then,
V
VN
APF a×
= =
33
64
3
4
2
3
3
R
Rπ×
= 0.68 or
68%
Face centered Cubic:
To find lattice constant ‘a’
In figure 2, AC2
= AB2
+ BC2
= a2
+ a2
= 2a2
Or aAC 2= also AC = 4 R
Or Ra 22=
Where R radius of the atom
Volume of the unit cell cube is
V = 33
216 Ra =
For FCC, the eight atoms at corners, each
contribute 1/8th of its volume to unit cell and six
atoms at the corners contribute half of their
volume, there fore
No. of atoms per unit cell is
N = atomsatomatom 46
2
1
8
1
8 =×+





×
Volume of one atom,
3
3
4
RVa π=
Therefore,
V
VN
APF a×
= =
3
3
216
3
4
4
R
Rπ×
= 0.74
= 74%
X-Ray Diffraction
The wavelength of x-ray is of the order of
Angstrom (Å). Hence optical grating cannot be
used to diffract X-rays. But the dimension of
atoms is of the order of few angstroms and also
atoms are arranged perfectly and regularly in the
crystal. Hence crystals provide an excellent
facility to diffract x-rays.
Bragg’s X-Ray Diffraction and Bragg’s Law: -
Bragg considered crystal in terms of equidistant
parallel planes in which there is regularity in
arrangement of atoms. These are called as Bragg
planes. There are different families of such planes
exist in the crystal and are inclined to each other
with certain angle.
In Bragg’s Diffraction the crystal is mounted such
that an X-ray beam is inclined on to the crystal at
an angle θ. A detector scans through various
angles for the diffracted X-rays. It shows peaks for
11
C
A
B

(maximum current) for those angles at which
constructive interference takes place. Bragg’s law
gives the condition for constructive interference.
2 d sinθ = nλ.
Where‘d’ is interplanar spacing.
‘λ’ is wavelength of X-rays.
‘θ’ is glancing angle.
‘n’ is integer and can take values 1,2,3. ……
Derivation of Bragg’s Law:
Consider Monochromatic beam of X-Rays. It is
incident on the crystal with glancing angle ‘θ’.
Ray AB, which is a part of the incident beam, is
scattered by an atom at ‘B’ along BC. Similarly
the ray DE is scattered by an atom in the next
plane at ‘E’ along EF.
The two scattered rays undergo constructive
interference if path difference between the rays is
equal to integral multiple of wavelength (nλ).
Construction: -BP and BQ are the perpendiculars
drawn as shown in the fig.
The path difference PE + EQ = nλ---------- (1)
From Right angled triangle PBE
Therefore PE= BE Sinθ = d Sinθ
Similarly From Right angled triangle QBE
EQ =PE = BE Sinθ = d Sinθ
Substituting in (1)
d Sinθ+d Sinθ = nλ
2 d Sinθ = nλ
Therefore the condition for constructive
interference is integral multiple of Wavelength of
X- Rays is
2 d Sinθ = n λ _____________ 1
Hence Bragg’s Law.
Since Bragg diffraction satisfies the laws of
reflection it is also called Bragg reflection.
Bragg’s X-ray Spectrometer
(Determination of wavelength and
Interplanar spacing): -
12

It is an instrument devised by Bragg to study the
diffraction of X-Rays using a
crystal as Grating. It is based on the principle of
Bragg Reflection.
Construction: - Monochromatic X-Ray Beam
from an X-Ray tube is collimated by slits s1 and s2
and is incident on the crystal mounted on the
turntable at a glancing angle q. The crystal can be
rotated using the turntable. The reflected X-Ray
beam is again collimated by slits s3 and s4 and
allowed to pass through ionization chamber fixed
on the Mechanical Arm. Due to ionization in the
medium current flows through the external circuit,
which is recorded by the Quadrant Electro Meter
(E). In order to satisfy the laws of reflection the
coupling between the turntable and the mechanical
arm is so made that, if the turntable is rotated
through an angle q then mechanical arm rotates
through an angle 2θ.
Experiment: Rotating the turntable increases
glancing angle. Ionization current is measured as a
function of glancing angle. The Ionization current
is plotted versus glancing angle. It is as shown
below.
The angles corresponding to intensity maximum
are noted down. The lowest angle θ, w.r.t to
maximum intensity corresponds to the path
difference λ.
If eqn. 1 is satisfied for θ1, θ2 & θ3, then bragg’s
law verified.
By determining θ using Bragg’s Spectrometer and
by knowing the value of Interplanar separation (d),
Wavelength (λ) of X-Ray beam can be calculated.
By determining θ using Bragg’s Spectrometer and
by knowing the value of Wavelength (λ) of X-Ray
beam, interplanar separation (d) can be calculated.
Diamond
13

Fig. Tetrahedral sp3
bonding (very hard!)
The diamond structure has an FCC space lattice
and a basis comprising two identical atoms.
Alternatively, the structure may be viewed as two
interpenetrating FCC lattices, one displaced
relative to the other along a body diagonal by a
quarter of its length.
The atoms on four atoms shown inside the main
FCC sub lattice are from other interpenetrated
FCC sub lattices.
Relatively few elements exist in the diamond
structure. Carbon is the most obvious example,
but the importance of this structure in solid state
physics is mainly due to the fact that this is the
natural state of the semiconductors silicon and
germanium.
Many two-element compounds form in a
diamond-like structure in which the first element
occupies one FCC sublattice and the second
element the other. This structure is named after its
prime example, ZnS.
Position of Corbon atoms are at
First kind: (000), (100), (010), (001), (101), (110),
(011), (111), (½,0, ½) (½,½,0). (1, ½,½), (½,½,1)
Second kind: (¼, ¼, ¼), ( ¾ , ¾,¾) (¾ , ¾,¼),
((¼, ¼, ¾,)
CsCl
It contains atoms at the corners of a cube and at
the body centre of the cube. Note that it is NOT a
body centered cubic structure (as the atom at the
centre is different to the atom at the corners of the
cube).
In the cesium chloride (CsCl) structure, the space
lattice is simple cubic. There are two such simple
cubic lattices which are separated by one half of
the body diagonal of the unit cube. One lattice is
occupied by Cs atoms, the other by Cl atoms.
Each atom has 8 nearest neighbors of the
opposite kind and is bound to these 8 atoms which
are arranged at the corners of the surrounding
hexahedron (cube). The bonding is typically ionic.
The dimensions of the CsCl unit cell are a = b = c
= 4.123 Å
Problems:
Obtain the miller indices of a plane which
intercepts at a, b/2, 3c in a simple cubic unit cell.
Solution x= a, y = b/2 and z = 3c
14







c
z
b
y
a
x
,, = 





c
c
b
b
a
a 3
,
2/
, =
( )3,2/1,1
Take reciprocals ( )3/1,2,1
Miller indices are (3, 6, 1)
1. Find the miller indices of a set of parallel
planes which makes intercepts in the ratio 3a,
5b on X and Y axes and parallel to Z axis.
x= 3a, y = 5band z = ∞ (parallel plane meet z=axis
at infinity)






c
z
b
y
a
x
,, = 




 ∞
cb
b
a
a
,
5
,
3
=( )∞,5,3
Take reciprocals 





∞
1
,
5
1
,
3
1
2. Lead is FCC with an atomic radius of
R=1.746Ǻ Find the spacing of, (i) (200) (ii)
(220) and iii) (111) planes.
The interplanar spacing is given by
222
lkh
a
d
++
= For FCC,
2
4R
a = = m10
10
1093.4
2
10746.14 −
−
×=
××
(i) for (200) plane 222
10
002
1093.4
++
×
=
−
d = 2.465 Ǻ
Similarly calculate of ii and iii.
3. Calculate the interplanar spacing for (212)
plane in a simple cubic lattice where lattice
constant is 4.6Ǻ ---- Do your self
4. In a Crystal whose lattice translation vectors
(primitive vectors) are 1.2 Ǻ, 1.8 Ǻ and 2.0 Ǻ.
A plane (231) cuts an intercepts 1.2 Ǻ on x-
axis. Find the corresponding intercepts on y
and z axis.
(h:k:l) = 





z
c
y
b
x
a
:: therefore (x:y:z) =






l
c
k
b
h
a
::
a = 1.2 Ǻ, b =1.8 Ǻ and c = 2.0 Ǻ, (hkl) = (231)
substititute in above equation
therefore (x:y:z) = 





1
2
:
3
8.1
:
2
2.1
= 0.6 : 0.6 : 2
(x:y:z) = 0.6 : 0.6 : 2
Given x=1.2 Ǻ
Therefore x : y = 0.6 : 0.6 or
6.0
6.0102.1 10
××
=
−
y
Or y= 1.2 Ǻ
Similarly z = 4 Ǻ
5. A sodium chloride crystal is used as a
diffraction grating with X-rays. For the d111
spacing of the chloride ions the angle of
diffraction 2θ is 27.5o
. If the lattice constant
of the crystal is 0.563 nm what is the
wavelength of x-rays?
Lattice constant a= 0.563 x 10-10
m
For d111 plane we have, h=k=l = 1
2θ =27.5 o
or θ = 13.75o
We know that wavelength of X-ray λ = 2d sin θ
222
lkh
a
d
++
= = 222
9
111
10563.0
++
× −
= 3.25 x 10-10
m
Therefore λ = 2 x 3.25 x 10-10
sin 13.75
= 1.545 x 10-10
m
6. The Bragg’s angle corresponding to first
order reflection from 111 planes in a
crystal is 300
, when X-rays of wavelength
15

1.75Å are used. Calculate the interatomic
spacing.
θ = 30, (h k l) = (1, 1, 1), λ = 1.75Ǻ
a =?
n λ = 2d sin θ
( n=1 first order diffraction)
d =
θ
λ
sind
= 1.75 x 10-10
m
dhkl = 222
lkh
a
++
a = 222
lkh ++ x dhkl
= 3.03 x 10-10
m
1. Find the miller indices of a plane which has
intercepts at 3a, 3b/2, 2c in a simple cubic unit
cell.
(h:k:l) = 





z
c
y
b
x
a
::
2. Obtan the miller indices of a plane with
intercepts a/3, b/2, c along x,y and z axis in a
simple cubic structure.
3. Find the miller indices of a set of parallel planes
which make intercepts in the ratio 3a:4b on X
and Y axes and are parallel to z axis, a, b, c
being primitive vectors of the lattice.
4. Draw the planes in cubic unit cell. (1 ½, 2),
(3/4, 1, 0), (1, 1, 1), (2, 2, 2), (3, 2, 1),
5. Copper has fcc structure of atomic raidus
0.1278 nm cal. The interplanar spacing for (3, 2
1) plane.
6. Find the miller indices of a set of parallel
planes which makes intercepts in the ratio 3a,
5b on X and Y axes and parallel to Z axis.
x= 3a, y = 5band z = ∞ (parallel plane meet z=axis
at infinity)






c
z
b
y
a
x
,, = 




 ∞
cb
b
a
a
,
5
,
3
=( )∞,5,3
Take reciprocals 





∞
1
,
5
1
,
3
1
7. Lead is FCC with an atomic radius of
R=1.746Ǻ Find the spacing of, (i) (200) (ii)
(220) and iii) (111) planes.
The interplanar spacing is given by
222
lkh
a
d
++
=
For FCC ,
2
4R
a = =
m10
10
1093.4
2
10746.14 −
−
×=
××
(i) for (200) plane 222
10
002
1093.4
++
×
=
−
d = 2.465 Ǻ
Similarly calculate of ii and iii.
8. Calculate the interplanar spacing for (212)
plane in a simple cubic lattice where lattice
constant is 4.6Ǻ ---- Do your self
9. In a Crystal whose lattice translation vectors
(primitive vectors) are 1.2 Ǻ, 1.8 Ǻ and 2.0 Ǻ.
A plane (231) cuts an intercepts 1.2 Ǻ on x-
axis. Find the corresponding intercepts on y
and z axis.
(h:k:l) = 





z
c
y
b
x
a
:: therefore (x:y:z) =






l
c
k
b
h
a
::
a = 1.2 Ǻ, b =1.8 Ǻ and c = 2.0 Ǻ, (hkl) = (231)
substititute in above equation
therefore (x:y:z) = 





1
2
:
3
8.1
:
2
2.1
= 0.6 : 0.6 : 2
(x:y:z) = 0.6 : 0.6 : 2
Given x=1.2 Ǻ
16

Therefore x : y = 0.6 : 0.6 or
6.0
6.0102.1 10
××
=
−
y
Or y= 1.2 Ǻ
Similarly z = 4 Ǻ
10. A sodium chloride crystal is used as a
diffraction grating with X-rays. For the d111
spacing of the chloride ions the angle of
diffraction 2θ is 27.5o
. If the lattice constant
of the crystal is 0.563 nm what is the
wavelength of x-rays?
Lattice constant a= 0.563 x 10-10
m
For d111 plane we have, h=k=l = 1
2θ =27.5 o
or θ = 13.75o
We know that wavelength of X-ray λ = 2d sin θ
222
lkh
a
d
++
= = 222
9
111
10563.0
++
× −
= 3.25 x 10-10
m
Therefore λ = 2 x 3.25 x 10-10
sin 13.75
= 1.545 x 10-10
m
1. The points arranged in space in regular
patterns are called as___________.
2. The Group of points arranged in space in
regular pattern is called as___________.
3. The space lattice contains same type of atoms
is called as____________.
4. The space lattice contains different type of
atoms is called as____________.
5. The smallest number of points repetition of
which gives crystal is called as______.
6. The Unit cell which having atoms only at
corners is called as__________.
7. The unit cell that contains one atom per unit
cell is called as___________.
8. SC contains ________atoms/unitcell.
9. FCC contains ________atoms/unitcell.
10. BCC contains ________atoms/unitcell.
11. BaCC contains ________atoms/unitcell.
12. Lattice parameters determine the________ of
unit cell.
13. The number of crystal systems are ______.
14. The number of known bravais lattices are
______.
15. The co-ordination number of FCC is____.
16. The atomic packing factor of BCC is____.
17. The expression for the Bragg’s law is____.
18. The Cesium Cloride contains _______unit
cell.
19. The diamond has_______ crystal structure.
Long Questions:
1. Draw the different type of unit cells.
2. What are crystal planes? Draw the crystal
planes whose intercepts are (2, 2/3, 1), (2, 1,
2).
3. Write different crystals systems available.
4. Explain the method of determining the miller
indices of a set of crystal plane.
5. Derive an expression for the interplanar
spacing.
6. What is atomic packing factor? Detemine the
APF of BCC lattice.
7. What is bragg’s law. Derive an expression for
the bragg’s law of X-ray Diffraction.
8. Give the working Bragg’s X-ray spectrometer.
9. Explain the crystal structure of Diamond and
Cesium chloride.
Problems:
1. Determine the Miller indices of the cubic
crystal plane that intersects the position
coordinates (1, ¼, 0), (1, 1, ½), (3/4, 1, ¼).
2. Iron at 20◦C is BCC with atoms of atomic
radius 0.124 nm. Calculate the lattice constant a
for the cube edge of the iron unit cell.
3. A sample of BCC iron was placed in an x-ray
diffractometer using incoming x-rays with a
wavelength λ = 0.1541 nm. Diffraction from the
17

{110} planes was obtained at 2θ = 44.704◦ .
Calculate a value for the lattice constant a of BCC
iron. (Assume firstorder diffraction with n = 1.)
18

Crystal structure notes

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Crystal structure notes

Similar to Crystal structure notes (20)

More from Praveen Vaidya

More from Praveen Vaidya (20)

Recently uploaded

Recently uploaded (20)

Crystal structure notes