Lecture notes on Solid State as per the revised syllabus of S.Y.B.Sc class laid down by the University of Mumbai which is in force from June 2017.

1. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
1
04/02/2021
CRYSTALLOGRAPHY :
Crystallography is the branch of science which deals with geometry, properties
and structure of crystals and crystalline substances.
It is based on three fundamental laws :
1) The law of constancy of interfacial angles
2) The law of rational indices
3) The law of symmetry
1) The law of constancy of interfacial angles:
According to this law, the shape & size of crystals of a certain substance
may vary with the conditions under which crystallization occurs but the
angles between the corresponding faces are always constant.
Example : A crystal may appear in a two dimensional space as
perfect hexagone or irregular hexagon, irrespective
of appearance, the angle of intersection of
corresponding faces remains same.
Similarly the interfacial angle of all NaCl crystals
are 900
irrespective of their size & shape.
Goniometer is used to measure interfacial angle of a crystal.
2) The law of rational indices:
It states that “ Ratio of intercept made by the crystal on three co-ordinate
axis remains same irrespective of the size of the crystal.”
Consider a crystal plane intercepting the X, Y, Z axisat A, B and C
respectively. Let the ratio of intercept be a : b : c.
Consider
2
1
1 times larger crystal, than the same crystal plane will make
intercept on X, Y, & Z-axis at K, L and M respectively.
OK : OL : OM
c
2
3
:
b
2
3
:
a
2
3
 a : b : c
Thus the ratio of intercept of larger crystal is same as intercept of
smaller crystal on three co-ordinate axes.
Hence the law of rational indices can also be stated as :
“It is possible to choose along three co-ordinate axes unit distances
a ,b , c , not necessary of the same length, such that the ratio of any
plane in the crystal is given by ma : nb : pc where m , n, p are either
whole number including infinity or fraction.

2. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
2
3) The law of crystal symmetry :
It states that “ all the crystal of the same substance possess same
element of symmetry.”
Various types of symmetry in a crystal is called as element of symmetry.
There are three possible types of symmetry in a crystal.
1) Plane of symmetry
2) Axis of symmetry
3) Centre of symmetry
i) Plane of symmetry : A crystal is said to have plane of symmetry when
an imaginary plane divides the crystal into two equal parts which are
mirror image of each other.
In a cubic crystal there are two types of plane of symmetry rectangular
and diagonal planes.
Rectangular planes are parallel to faces of the crystal & diagonal planes
are obtained by joining opposite edges of the crystal.
A cubic crystal has 3 rectangular & 6 diagonal planes i.e 9 plane of
symmetry.
ii) Axis of symmetry : A crystal is said to have axes of symmetry when an
imaginary line passing through the crystal, about which crystal can be
rotated presents same appearance more than once during the complete
rotation of 3600
.
Depending upon the number of times crystal appears unchanged during
a rotation of 3600
; following 3 types of axes of symmetry are possible in
cubic crystal.
a) Axis of four fold symmetry (Tetrad axis ) : Imagine a line passing
through centers of two opposite faces of a cube. On rotating the
cube about this line it is found that cube presents exactly the
same appearance 4 times during a rotation of 3600
. Such axis is
known as axis of 4 fold symmetry. There are 6 faces in a cube ,
hence 3 such axes are possible.
b) Axis of three fold symmetry (Triad axis ): Imagine a line passing
through diagonally opposite corners. On rotating cube around this
line, it is found that cube presents same appearance 3 times
during a rotation of 3600
. Such axis is called 3 fold symmetry.
There are 8 corners hence 4 triad axis are possible.

3. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
3
c) Axis of two fold symmetry (Diad axis ) : Imagine a line passing
through diagonally opposite edges. On rotating cube around this
line, it is found that cube presents same appearance twice during
a complete rotation of 3600
. Such axis is called 2- fold symmetry.
There are 6 diad axes possible in a cube.
iii) Centre of symmetry : It is an imaginary
point within the crystal such that every face
has an identical face at equal distance
opposite side of the point. A cubic crystal
has only one centre of symmetry.
Thus a cubic system has 23 elements of
symmetry
(9 planes of symmetry, 13 axes of symmetry & 1 centre of symmetry.)
Explain “Weiss & Miller indices” of a plane.
Consider a crystal lattice which is represented in terms of 3 axes OX, OY, OZ
which are perpendicular to each other.
Let a , b , c , be the distance along these axes which represents the unit cell.
The plane ABC is known as unit plane of space lattice and a , b , c are known as
unit intercept.
Let plane LMN makes intercept 2a : 2b : 3c. The ratio of intercept in terms of
standard is 2 :2:3. This ratio can be used to represent the plane LMN and is
called as Weiss indices.
Thus the ratio of coefficient of a , b , c are known as Weiss indices.
But Weiss indices are not always simple whole numbers. There may be
fractional values or infinite values.
Therefore Weiss indices are replaced by Miller indices.
The Miller indices of a plane are obtained by taking the reciprocal of Weiss
indices & multiplying by the LMC.

4. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
4
Eg. Consider a plane which makes intercepts 2a , 3b , 3c
Weiss indices : 2 : 3 : 3
Reciprocal : ½ : 1/3 : 1/3
Multiplying by LCM 6/2 : 6/3 : 6/3
Miller indices : 3 : 2 : 2
Plane is designated as (322) plane
Terms used in crystallography:
1) Lattice points / sites :
The position of the particles (atom, ion or molecule ) in the crystalline solids ,
relative to one another is represented by points which are called lattice points.
2) Space Lattice :
The regular pattern of points which describes the three diamensional
arrangement of particles ( atoms, ions or molecule ) in a crystal is known as
space lattice or crystal lattice
 The essential property of lattice is that each point in the lattice has exactly
same environment as any other point representing same atom or ions.
 Since a point is an imaginary and infinitesimal spot in space, therefore
three dimensional arrangements of points in space is an imaginary
concept.
3) Unit Cell :
 The smallest portion of space lattice which can generate the complete
crystal by repeating its own dimension in various direction is called a unit
cell.
 The shape of the crystal depends upon the shape of the unit cell. Eg. NaCl
crystal is cubic & its unit cell is also cubic.
 Unit cell contains small number of atoms, ions or molecules.
 Any point placed in unit cell must occupy the same relative position in large
unit cell.
Types of Unit cells :
a) Primitive or simple unit cell : In this particles are present only at the
corner of the unit cell.
b) Non primitive or multiple unit cell : In this particles are present at the
corners as well as at some other sites. They are of following types:
i) BCC : In this patricles are present at every corner & one at the centre of
the cube.
ii) FCC : In this particles are present at the centre of each face in additon to a
particle at each corner.
iii) End Face centred : In this particles are present at the cetntre of the two
opposite faces along one axis in addition to particles at the corners.

5. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
5
Characteristics of cubic system :
I) Simple cubic lattice:
 In this type one structural unit (atom, molecule or ion ) is present at each
corner of the cube.
 This unit is shared by 8 cubes, hence only 1/8th mass of the structural unit
belongs to this cell.
 There are 8 such units at the corner of the cube therefore 8 x 1/8th = 1
structural unit belongs to this cell.
II) Face centred cubic lattice (FCC):
 In this type one structural unit (atom, molecule or ion ) is present at the
centre of each face in addition to units present at the corner of the cube.
 The unit at the centre of each face is shared by 2 cubes, hence only ½ mass
of the structural unit belongs to this cell.
 There are 6 such units at the centre of the face of the cube
therefore 6 x ½ = 3 structural unit belongs to this cell.
 The unit at the corner is shared by 8 cubes, hence only 1/8th mass of the
structural unit belongs to the cell.
 There are 8 such units at the corner of the cube therefore 8 x 1/8th = 1
structural unit belongs to this cell.
 Thus in all 3 + 1 = 4 units belong to the cell.
III) Body centred cubic lattice (BCC):
 In this type one structural unit (atom, molecule or ion ) is present at the
centre of the cube in addition to units present at each of the corner. This
unit belongs to this cell only and is not shared by any other cube.
 The unit at the corner is shared by 8 cubes, hence only 1/8th mass of the
structural unit belongs to the cell.
 There are 8 such units at the corner of the cube therefore 8 x 1/8th = 1
structural unit belongs to this cell.
 Thus in all 1 + 1 = 2 units belong to the cell.

6. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
6
Inter planar distance in cubic crystal :
The distance between two lattice planes is known as inter planar distance or lattice
constant.
I) Simple cubic lattice:
0.577
:
0.707
:
1
3
a
:
2
a
:
a
d
:
d
:
d 111
110
100 

II) Face centered cubic lattice:
 Distance between (100) plane : In this an additional plane containing the
atom in the centre of faces appear midway between the two faces of the
crystal. Therefore interplanar distance is a/2.
 This is possible in case of (100) & (110) plane, but incase of (111) plane it is
exactly similar to simple cubic lattice
1.154
:
0.707
:
1
3
a
:
2
2
a
:
2
a
d
:
d
:
d 111
110
100 


III) Body centered cubic lattice:
 In this type one more plane is present mid way between side planes which
contains body centered point. This is possible in case of (100) & (111) plane.
 But in case of (110) plane, no additional plane is possible.
0.577
:
1.414
:
1
3
2
a
:
2
a
:
2
a
d
:
d
:
d 111
110
100 


Q : What are X- rays how are they produced?
 X-rays are electromagnetic radiation of very short wavelength - order of 10-10m.
 X-rays of higher wavelength are called soft X-rays & X-rays of lower wavelength
are called hard X-rays.
 X-rays are not reflected by electric or magnetic field.
 They affect photographic plate
 They can penetrate through matter & they also ionizes the gases.
 They also produce fluorescence in certain substances.
Production of X-rays :
Principle : X- rays are produced in Coolidge tube by bombarding electrons on a
metal target. As a result of electron bombardment the target electrons in the inner
shell are displaced & then electron moves from higher energy level to vacant inner
level. During this transition energy is given out in the form of X-rays.

7. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
7
 The Coolidge tube contains a cathode which is made up of Aluminium & is
concave in shape
 The target tube (anticathode) is connected to anode & its front faced is
inclined at an angle of 450 to axis of cathode rays.
 The target face is usually made up of metal having high M.P like W, Mo, Pb.
 X-rays are emitted from anticathode.
Diffraction in crystals
Interference :
"Interference" is the property of waves to overlap each other and, under certain
circumstances, to cancel out or amplify each other.
Diffraction :
We know from ocean waves that when a wave series travels through a hole smaller
than the wavelength, the waves exiting the hole spread out to the sides. Light
displays the same wave characteristic. The deviation of light from its travel in a
straight line is called diffraction.
X-ray Diffraction From a Crystal Lattice, Bragg's Equation
Q: Derive Braggs equation .
 A crystal is composed of series of equally placed atomic planes which can be
considered as reflection grating ( monochromator ).
 When X-rays are incident on a crystal face they penetrate into the crystal & strikes
the atom in successive plane.
 From each plane X-rays are reflected. The reflection may take place in such a way
that it will either cause interference or reinforcement of reflected beam from 1st
plane. As a result, whole beam will appear as it has been reflected from the surface
of the crystal. Based on this assumption Bragg’s gave an equation which gives the
relation between wavelength, angle of incident & the distance between the
successive planes.

Diffraction of X-rays

8. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
8
Derivation :
 Let a parallel beam of monochromatic X-rays ( AB & DE ) of wavelength ‘λ’ strikes
first plane at an angle ‘θ’ i.e glancing angle.
 Consider ray ‘AB’ reflected from atom ‘B’ in the direction ‘BC’ from the 1st plane and
another ray ‘DE’ reflects from 2nd plane in the direction ‘EF’.
 From atom ‘B’ draw two perpendicular ‘BP’ & ‘BQ’ on ‘DE’ and ‘EF’ respectively. The
two reflected beam will be inface or out of face depending on the path difference.
 When the path difference (PE + EQ) is an integral multiple of ‘λ’ , the two rays will
reinforce each other and the intensity of reflected beam will be maximum.
 Thus, condition for reinforcement is
( PE + EF ) = n λ where ‘n’ is an integer 1 , 2 , 3 , .
 From the diagram angle PBE = angle QBE = θ
Also in Δ BPE , Sin θ = PE / BE
. . PE = BE Sin θ = d Sin θ ( since BE = d = interplanar distance )
Also, PE = EQ
Therefore EQ = d Sin θ
Thus path difference, PE + EQ = d Sin θ + d Sin θ
. . 2 d Sin θ = n λ
The above equation is known as Bragg’s equation. In this equation ‘n’ is known as
order of reflection, Intensity of reflected beam decreases as order of reflection
increases.
Q: Explain Bragg‟s X-ray spectrometer method to determine interplanar distance.
Bragg’s X-ray spectrometer consists of 3 parts-
i) Source of X-ray
ii) Graduated crystal turn table
iii) A detecting device
 X-rays are produced in Coolidge tube by bombarding cathode rays on target such
as molybdenum , W , Cr.
 These X-rays are allowed to pass through metal screen M to get monochromatic
light. Slit S1 & S2 are used to get narrow beam of X-rays.
 These X-rays are then made to strike on a crystal at an angle „θ‟ . The reflected
beam is passed into an ionization chamber containing SO2 or methyl bromide. The
reflected X-rays ionize the gas in proportion to their intensity. The extent of
ionization & hence the intensity of reflection can be determined with the help of
an electrometer.

9. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
9
Determination of interplanar distance:
 A beam of X-rays is allowed to enter in ionizing chamber & table is adjusted in
such a way that electrometer shows maximum current. This angle is recorded as
initial angle.
 A current is then mounted at the centre of the turn table & X-rays are allowed to
strike on plane of crystal. The table is again adjusted to get maximum current &
the angle is recorded.
 The difference between the two reading is equal to twice the angle „θ‟ for 1st
order
reflection. This is because the turn table & ionization chamber are adjusted in
such a way that the ionization chamber rotates twice the angle through which
turn table rotates.
 When the chamber is again moved slowly current initially decreases, after some
time increases & reaches to maximum value. The angle recorded now is for 2nd
order reflection.
 X-ray spectrum is obtained by plotting the ionization current i.e intensity of the
diffracted X-ray beam against reflection angle „θ‟ . Knowing values of „θ‟ , „n‟ and
„λ‟ it is possible to calculate interplanar distance using Bragg‟s equation.
 If X-rays of same wavelength is used then for 1st
order of reflection , the angle of
maximum reflection are inversely proportional to their interplanar distance.




 sin
1
d
Thus
,
sin
1
:
sin
1
:
sin
1
d
:
d
:
d
3
2
1
111
110
100

From the values of ratio of interplanar distance it is possible to know the
nature of crystal.
Q: Write note on ‘crystal structure of NaCl’ .
1) For NaCl crystal the angles obtained for (100), (110) & (111) planes are
5.90, 8.40 & 5.20 respectively.
1.134
:
0.703
:
1
2
.
5
sin
1
:
.4
8
sin
1
:
.9
5
sin
1
d
:
d
:
d 111
110
100



The ratio belongs to FCC lattice hence structure of NaCl is of FCC type.
2) The relative arrangement of Na and Cl atom in lattice can be explained on the
basis of relative intensity of reflected beam from different planes & order.
a. It is observed that there are no sodium chloride molecules as structural units
in the lattice, but the units are of Na+ & Cl- ions and not of atoms.
b. In (100) & (110) plane intensity decreases as the order of reflection increases,
which shows that these planes contain equal number of structural units.
c. In (111) plane there is an alternate decrease & increase in intensity of
reflected beam. This is possible only if (111) plane containing Na+ & Cl- are
alternately arranged. Also plane containing Na+ ions must be mid way
between the planes with Cl- ions.

10. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
10
d. The intensity of X-rays reflected by (111) plane containing all Cl- ion is very
high as compared to intensities of X-ray reflected by (111) plane containing
all Na+ ion. This is because Cl- has 18 electrons and Na+ has 10 electrons.
e. Also each sodium ion is surrounded by six chloride ions and each chloride
ion is surrounded by six sodium ions.
Figure 1 : Crystal structure of NaCl
Q : Prove that a unit cell of NaCl contains 4 molecules of NaCl.
Ans .
 Draw Crystal structure of NaCl ( Fig. 1)
 In a unit cell of NaCl there are 14 Na+ and 13 Cl- ions
 Out of 14 Na+ , 8 Na+ ions are situated at the corner of the cube. Each corner
ion is shared by 8 cubes
Therefore , 8 x 1/8 = 1 Na+ is associated with the unit cell
 There are 6 Na+ ions at the centre of each face and each of which is shared
by 2 cubes
Therefore , 6 x 1/2 = 3 Na+ is associated with the unit cell
Thus unit cell of NaCl has total 1 + 3 = 4 Na+ ions
 Out of 13 Cl- ions , 12 Cl- ions are present at the centre of each edge which
is shared by 4 cubes
Therefore , 12 x 1/4 = 3 Cl- is associated with the unit cell
 Out of 13 Cl- ions , 1 Cl- ion is present at the centre of unit cell which
belongs to this unit cell only
Thus unit cell of NaCl has total 3 + 1 = 4 Cl- ions
 Therefore 4 Na+ ions and 4 Cl- ions i.e 4 NaCl molecules are associated with
one unit cell of NaCl.

11. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
11
Isomorphism : Substances are said to be isomorphous if
i) They crystallize with the same lattice
ii) They have the same chemical composition and arrangement of atoms or ions
i.e they are represented by the same type of chemical formula.
iii) The atomic or ionic ardii of the analogous atoms or ions are very similar, so
that the unit cell dimensions are very nearly the same.
Structure of Sylvine ( KCl ) :Since NaCl & KCl are isomorphous compounds both
of them should have a FCC structure but KCl shows a simple cubic structure.
1) For KCl crystal the angles obtained for (100), (110) & (111) planes are
5.220, 7.30 & 9.050 respectively.
0.575
:
0.704
:
1
38
.
9
sin
1
:
sin7.61
1
:
.38
5
sin
1
d
:
d
:
d 111
110
100



This ratio is almost identical to that for a simple cube hence KCl crystal has a
simple cubic lattice.
2) The atomic number of potassium & chloride are 19 & 17 respectively. As these
numbers are not very much different, the X-rays are unable to detect any
difference between the two kinds of atoms.
3) The (100) & (110) planes in KCl same number of K+ and Cl- ions are present.
Hence these planes show similar typr of X-ray reflections.
4) In case of (111) plane of KCl , the intensities of X-rays reflected by Cl- planes
are K+ planes are equal. Hence these two reflected beams balance each other
and no net reflection is obtained from (111) planes. As a result the KCl unit
cell behaves like a simple cubic structure.
5)
1.11
1
5.9
sin
sin5.3
)
(
)
(
100
100


KCl
d
NaCl
d
The above expression shows that the fundamental units are more widely
spaced in the crystal of KCl than in the crystal of NaCl.
Unit cell of KCl

12. SOLID STATE - (Sem IV – Paper II) Dr. Aqeela Sattar , Royal College
12
Determination of Avogadro’s Number :
Show how Avogadro’s number is determined using X-rays.
Note : If question is of 5 marks - Draw the diagram of NaCl unit cell & prove that
there are four NaCl molecule in one unit cell.
Let ‘a’ be the edge length of the cube i.e ‘a’ us distance between two Na+ or two Cl-
ions. This edge length is twice the distance between two 100 plane.
Therefore, a = 2 d100
Unit cell is cubic, hence its volume of one unit cell can be written as
V = a3 or V = ( 2d100)3 or V = 8 d3 …….. (1)
If ‘M’ is molecular weight of NaCl abd ‘D’ is density then the molar volume of NaCl
will be,
3
cm
D
M

V
Avogadro’s number of molecules are present in molar volume i.e im M/D cm3
Therefore, 4 molecules will be present in ‘x’ cm3
(2)
-
-
-
4
DN
M
x 

Also, volume of 1 unit cell = volume occupied by 4 molecules of NaCl
Therefore,
DN
4M
8 3

d
Therefore,
D
D
N 3
3
2d
M
d
8
4M


************