1_amtech

AMTECH ACADEMY, New Panvel

Notes by Prof. A.R.P

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Amtech Academy 103, Sai Paradise, Plot No 100, Sector 1S, New Panvel

Contact Nos. 9321215282/ 9029023210 1. 2.

SYLLABUS Module 1: Crystallography and X-rays [ 22- 30 marks]

� Lattice, basis, crystal axes, unit cells, lattice parameters & crystal systems SC. BCC.

FCC. Diamond, NaCI, zinc blend and HCP crystal structures, Miller indices, planes &

directions, Liquid crystals & phases, LCD displays & specifications.

� X-rays - Origin of x-rays and x-ray spectra, x-ray & Bragg law and determination of

crystal structure.

� Real crystals - crystal imperfections point defects and dislocations.

Module 2: Physics of Semiconductors [ 15- 20 marks] � Classification of solids, Fermi-Dirac stastics , Concept of Fermi level & its variation

with temperature, impurity and applied voltage

� Intrinsic & extrinsic carrier concentrations carrier concentration, Carrier drift, mobility,

resistivity and Hall effect, carrier diffusion, Einsteins relations, current dens & continuity

equations

� Energy band diagrams of p-n junction, formation of depletion region. Derivation for

depletion layer width

Module 3: Superconductivity [ 10- 15 marks]

� Critical temperature, Critical magnetic field, Type -I and type –II super conductors, high

Tc super conductors.

� Meissner effect, Josephson effect.

� SQUIDS, Plasma confinement Maglev.

Module 4: Acoustics [ 10- 15 marks]

� Acoustics of building, Absorption , Importance of Re verberation Time, Units of

Loudness, Decibel, Phon

� Condition for Good Acoustics methods of designs for good Acoustics Determination of

Absorption coefficient ,Noise Pollution.

Module 5: Ultrasonics [ 10- 15 marks]

� Principles of production. Piezo-electric & magnetostriction effect.

� Piezoelectric & magnetostriction oscillator; ultrasonic materials - quartz & ferroelectric

materials, cavitations effect.

� Application based on cavitations effect and Echo sounding, ultrasonic Imaging & medical

diagnosis.

Module 6: Electron Optics [ 10- 15 marks]

� Electrostatic & Magnetostatic focusing system

� Construction & working of CRT, CRO and its application

Theory Examination ( 75 marks – 2 Hours)

1; Question paper will comprise of total 7 questions, each of marks 15

2 Only 5 questions need to be solved.

3. Q.1 will be compulsory and based on entire syllabus.

4. Remaining questions will be mixed in nature.



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Chapter 1

CRYSTALLOGRAPHY AND X-RAY

1. Define Space lattice Ans. The infinite regular, periodic arrangement of points in two or three dimensions to

represent the crystal structure is called the Lattice. The environment about each point is the

same. Two dimensional lattice is called as NET.

SPACE LATTICE

2. Explain Basis with examples? Ans. An atom or a group of atoms associated with the lattice point is called the Basis. The atoms are considered to be identical in composition, orientation and arrangement. Eg:- In a crystal like Al, Cu,Ag, Au,etc. the basis is Single atom.In NaCl, KCl ZnS the basis is diatomic.In BaF2, CaF2 etc the basis is triatomic.

3. Define Unit Cell and Lattice Parameters? Ans. The unit cell is a smallest geometric figure which by repetition can generate the crystal

structure. The repetition shall take place in 3- dimensions.

Space lattice + Basis = Crystal



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UNIT CELL is characterized by lengths of its three sides (i.e. lengths of the associated unit

vectors) i.e a, b, c and angles between the three sides called the interfacial angles say α, β and γ. These size parameters are called as lattice parameters. Lattice parameters are extremely useful in categorizing crystalline solids. For example specific choice of lattice parameters viz. a = b = c and α = β = γ = 90° gives the cubic crystal

system.

W. Bravai had shown that based on the point symmetry in space, there are only

fourteen different lattice types required to describe all possible arrangements.

Each one of them is called the Bravais lattice.

5. Draw the unit cell for SC, BCC and FCC and calculate the average number of

atoms per unit cell for SC, BCC and FCC?

Ans.

The average number of atoms per unit cell is represented by n

n = 1/8 X NC + 1/2 X NF + 1 X NB

where Nc = Number of atoms at the corners of unit cell NF = Number of atoms at the face centre positions

NB = Number of atoms at the body centre positions or atoms completely

lying inside.

For SC n = ⅛X 8 + ½ X 0 + 0 = 1

For BCC n = ⅛X 8 + ½ X 0 + 1 = 2

For FCC n = ⅛X 8 + ½ X 6 + 0 = 4

6. Define atomic radius and derive the relation between lattice constant (a) and

atomic radius (r) for SC , BCC and FCC?

Ans. Atomic Radius:- It is defined as half the distance between nearest neighbours in a crystal of

pure element. It is normally expressed in terms of lattice constant “a”. Atomic radius is denoted

by r.



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Simple Cubic Cell Face Centered Cubic Body Centered Cubic

Diagram

In this structure atoms

touch each other along

the lattice

In this structure atoms touch

each other along the diagonal

of any face of the cube

In this case atoms touch each other

along the diagonal of the cube.

2r = a

Therefore, r = a/2

By Pythagoras theorem

a2 +a

2= (4r)

2

2 a2= 16 r

2

Therefore, r = a / 2√2

In ∆ ABC

By Pythagoras theorem

AB2+BC

2= AC

2

a2+a

2= AC

2

Therefore face diagonal AC= a√2

In ∆ ACD; AC2+CD

2= AD

2

2a2+ a

2 = AD

2

Therefore Body diagonal AD= a√3

Hence 4r= a√3

r = (a√3) / 4

7. Define Coordination number and calculate for SCC , BCC and FCC with

explanation?

Ans. • Coordination number of an atom in a crystal is the number of nearest neighboring

atoms which are simultaneously in contact with that atom.

• It signifies the tightness of packing of atoms in crystal.




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Around any atom A in a SC cell,

there would be six equally

spaced nearest neighbour atoms

each at a distance ‘a’ from that

atom. Four atoms lie in the plane

of the atom, while one is

vertically above it and one is

vertically below as shown in

figure

CN of Simple Cubic cell= 6

In a FCC cell, each corner atom is

in contact with the face-centre

atom. It would be simultaneously

touching 4 atoms in the xy-plane, 4

atoms in the yz-plane and 4 atoms

in the xz-plane, making up a total

of 12 atoms.

Thus, an FCC cell has the

maximum value for the

coordinatIon

CN of Face Centered Cubic= 12

In the BCC cell the corner

atoms do not touch each

other. However, each corner

atom is in contact with the

body centre atom. As there

are 8 unit cells around each

corner, the atom located at

each corner would be

simultaneous touching the 8

Body centre atoms around it.

The same is the case for a

body centre atom as shown in

figure Therefore, the

coordination number is eight.

CN of Body Centered

Cubic= 8.

8. Define Atomic Packing Factor and calculate the APF for SC, BCC and FCC?

Ans • Atomic Packing Fraction (APF) is related with finding how efficiently the unit cell

volume is filled by atoms (or ions in ionic solids).

• It is concerned with the stability of the structure.

• It is defined as the ratio of the volume of the atoms in the unit cell to the total volume of

unit cell.

• Derivation Let ‘n’ represent the number of atoms per unit cell

Let ‘v’ be the volume of single atom = (4/3)π r3

Let ‘V’ the total volume of the unit cell

= n X.v

V

Amtech Academy

Prof. ARP- 902 902 3210



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ATOMIC PACKING FACTOR


9. Define Packing Efficiency, VOID Space and calculate the same for SCC,BCC

and FCC?

Ans.

PACKING EFFICENCY • Packing efficiency is the APF expressed as a percentage term.

SC = 0.52x 100 = 52%

BCC = 0.68 x 100 = 68 %

FCC = 0.74 x 100 = 74 %

VOID SPACE • It refers to the volume of unit cell remaining vacant after it is filled by atoms in different

types.

• VOID SPACE when expressed in percentage it is referred to as VOID SPACE

PERCENTAGE

VOID SPACE % = (100 – PE)%

Thus, void space for SC = 100-52 = 48%

BCC = 100-68 = 32%

FCC = 100-74 = 26%



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6. Calculate the relation between the density of a solid, molecular

weight and lattice constant? Ans. Let us consider a solid belonging to cubic system with molecular weight M and density ρ.

Since all sides are equal for a cubic system, i.e ‘a’ the lattice constant.

Mass of the cubic unit cell (m) = Density X Volume

= ρ a3 . . . . . . .(1)

Mass of each molecule of the solid is M/N where N is the Avgadro’s Number.

Let the unit cell contains n molecules. Then mass of the unit cell would be

m = n X M/N . . . . . . .(2)

Since we are referring to the same mass m, equating steps (1) and (2), we get

ρ a3 = n X M/N

Note: For monoatomic solids i.e. solids having single atom as the basis, take atomic weight (A) instead of molecular weight.

NOTE FOR PROBLEMS


n=1

Therefore, ρ = M / N a3

n=4

Therefore, ρ = 4 M / N a3

n=2

Therefore, ρ = 2 M / N a3

7. Explain the atomic arrangement in HCP along with its number of atoms, atomic

radius, APF, PE and density?

Ans. EXAMPLES : Zn, Gd, Zr, Ti, Mg, Co, etc.

Layer 1 Layer 2 Layer 3

• In a plane, six spheres completely surround the central sphere.

• HCP structure is obtained by placing such layers on each other but in a twisted fashion

i.e. the spheres in the 2nd

layer are fitted in the valleys formed in 1st layer marked by

A-A-A.



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• The 3rd

layer is identical in fitting to the 1st layer and so on.

• The unit cell of HCP structure is hexagonal pyramid

Figure: - Unit Cell of HCP

• Isolated HCP unit cell is shown in above Figure.

• Each corner atom of the hexagonal face is by six unit cells. Consequently it contributes

1/6th of its volume and mass to one unit cell.

• There are six such corner atoms in the base. Further there is an atom at the centre of the

each hexagonal face, which is shared by two adjacent cells. Three atoms forming a

triangle in the middle within the body of the cell and cannot be shared by adjacent cells.

• The total contribution to the effective number of atoms in the cell is (n)

= 1/6 X Nc + ½ X Nf + 1 X 3 = 6 n = 6 atoms/unit cell.

ATOMIC RADIUS

• The atoms are in contact along the edges of the hexagon.

2 r = a

Therefore, r = a /2

COORDINATION NUMBER

• Each atom in the structure is positioned in valleys formed by three adjacent atoms of the

top layer and by three adjacent atoms in the bottom layer; and is surrounded by six

neighbour atoms in the middle layer.

• All these twelve atoms are in contact with the atom under consideration (CN= 12)




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Packing efficiency= 0.74 X100 = 74%

VOID space percentage = (100-74)%= 26 %

DENSITY OF HCP

ρ = 6 M____

N (3√2 a 3

)

8. Explain the atomic arrangement in diamond unit cell and calculate its average

number of atoms per unit cell, coordination number atomic radius, APF, PE ,VS

and VSP?

Ans. Diamond like structure is also exhibited by Si, Ge, C, ZnS, CdS, etc.



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ATOMIC ARRANGEMENT

• It is an extension to the FCC structure

• The FCC structure is further added with 4 more complete atoms within the unit cell

• Out of these 4 atoms, 2 are placed along the two opposite body diagonals at a distance

of (a√3)/ 4 from respective corners in the lower half of the unit cell.

• The remaining 2 atoms are placed along the remaining pair of opposite body diagonals at

a distance of (a√3)/ 4 from respective corners in the upper half of the unit cell.

TOTAL No. of ATOMS per UNIT CELL (n) = 4 +4= 8

ATOMIC RADIUS

2r = ( a √3) / 4

Therefore, r = ( a √3) / 8


4 π r3

APF = ___n X_ 3 ____

a3

8 X 4 π r3

= ______3_______

a3

8 X 4 π X ( a √3)3

= ___3__X__83__ = ( π√3) / 16.

a3

= 0.34

Packing Efficiency = 34 %

VOID space = (100 - 34) % = 66 %

CORDINATION NUMBER= 4

9. Explain the atomic arrangement in NaCl unit cell and calculate its average

number of atoms per unit cell, coordination number atomic radius, APF, PE ,VS

and VSP?

Ans.

Examples: Many chlorides and oxides such as LIF, MgO, FeO, BaO exhibit this structure



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AVERAGE NUMBER of ATOMS per UNIT CELL

In the unit cell, of NaCl

FOR CHLORINE There are 8 chlorine ions in the corners, shared by 8 unit cells

and 6 ions at the faces each of which is shared by two unit cells

Therefore, total number of chlorine ions per unit cell is 1/8 X 8 + ½ X 6=4

FOR SODIUM There are 12 Sodium ions on the edges of the cube; each sodium ion is shared by 4 unit cells

and 1 Na located at the centre of the unit cell

Therefore, the number of sodium ions per unit cell = ¼ X 12 +1 = 4

Therefore there are 4 molecules of NaCl per unit cell.

CO-ORDINATION NUMBER

Every Na ion is surrounded by 6 Cl ions and vice-versa.

Therefore co-ordination number of NaCl is 6

ATOMIC RADIUS

(rNa+ + rCl-) = a/2

rNa+ = 0.98 AU and rCl- = 1.81 AU

Therefore a = 2 (rNa+ + rCl-)

= 2 X (0.98 +1.81)

= 5.58 AU for NaCl

4 π (rNa+3 + rCl-

3)



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APF = ___n X_ 3 ____

a3

4 X 4 π ((0.983

+1.813)

= ______3_______ = 0.66

5.583

Packing Efficiency = 66 %

VOID space = (100 - 66) % = 34 %

10. Explain the procedure to determine Miller indices of crystal plane which

intercepts all the three axis?

Ans. • The crystal lattice may be regarded as made up of an aggregate of a set of parallel

equidistant planes passing through the lattice points which are known as lattice planes.

• Miller evolved a method to designate a plane in a crystal by three numbers (h, k, 1)

known as Miller indices.

• This method is now universally employed.

• The Miller indices are the three smallest possible integers which have the same ratios as

the reciprocals of the intercepts of the plane concerned on the three axes.

• The rules for fmding the Miller indices are as follows:

1. First of all determine the intercepts of plane on the three cooordinate axes.

2. Secondly take the reciprocals of these inter-cepts.

3. Find the L.C.M. of denominators

4. Multiplying the LCM with reciprocal shall give the Miller indices of this plane.

EXAMPLE 1:- Find the Miller indices where the plane cuts the intercepts of 2, 3 and 4 units

along the three axes.

• Intercepts are 2, 3 and 4.

• Reciprocals are (1/2, 1/3, ¼)

• LCM= 12

• Therefore Miller Indices are (6,4,3)

EXAMPLE 2:- Find the Miller indices for the plane shown below with intercepts (2a,2b,c)?

• Intercepts are 2,2,1.

• Reciprocals are (1/2, 1/2, 1)

• LCM= 2

• Therefore Miller Indices are (1,1,2)

Amtech Academy Prof. ARP- 902 902 3210



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11. Derive the formula for interplanar spacing for a family of planes having a

Miller Indices h, k, and l ?

Ans. FIGURE 1 FIGURE 2

FIGURE 3

• Consider the plane ABC of a a cubic crystal as shown in Fig. 1.

• This plane belongs to a family of planes, whose Miller indices are (hkl) because Miller

indices represent a set of planes.

• Here ON is the perpendicular drawn from the origin to this plane.

• The distance ON represents the interplaner spacing d of the family of planes.

• Let α’,β’ and γ’ be the angles between coordinate axes X, Y, Z and ON respectively

• The intercepts of the plane on the three axes are

OA= a/h

OB= a/k

and OC=a/l



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• This is the relation between interplanar spacing d and the edge of the cube a.

PART –II

X- Rays

12. Explain construction and working of Coolidge tube for production of X-rays?

Ans.

• When fast moving electrons (e

-) are suddenly stopped then X-rays are produced.

• The modern X-ray tube used for the production of X-rays is known as Coolidge tube, It

consists of highly evacuated hard glass bulb with a cathode (filament) and anti-cathode or

anode

• The cathode i.e. metal filament (F) is surrounded by molybdenum metal cylinder kept at

negative potential to the filament.

• Hence, the electrons emitted from the filament are concentrated into fine beam of

electrons. The target T consists of copper block in which a piece of tungsten or

molybdenum is fitted.



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• The anode should possess following properties. 1. High melting point: So that it is not melted due to bombardment of fast moving

electrons which produces large amount of heat.

2. High atomic weight: To reduce the hard X-rays.

3. High thermal conductivity: To carry away the generated heat.

• The target is placed at an angle of 45º with the path of electron beam

• When the cathode (filament) is heated by an electric current, it produces electrons due to

process known as thermionic emissions.

• This beam of electrons is then focused on the anode (target). The electrons from the

cathode are accelerated by application of high voltage between the cathode and anode by

using step-up transformer.

• When these fast moving electrons are suddenly stopped by the tungsten anode they loose

their kinetic energy and X-rays are produced from the target.

• However 98 % of K.E. is converted into large amount of heat. By controlling the filament

current, the thermionic emission of electrons and hence the intensity of X-rays can be

controlled.

• The penetrating power of X-rays determines the quality of X-rays which can be

controlled by changing voltage between cathode and anode.

• The X-rays of high penetrating power and higher frequency are called hard X-rays and those with low penetrating power and low frequency are called soft X-rays.

• Intensity of X-ray depends on filament current.

• Penetrating power of X-ray depends upon potential difference (P.D.) between

cathode and anode.

13. Explain Characteristic X-ray (LINE spectra) spectra? Ans.

• The electrons in the outer orbit are loosely held by nucleus.

• Inner orbit electrons being closer to nucleus are more tightly held by nucleus.

• When the incident electron penetrates deep into an atom. They knock out one of the inner

shell electron of target atom, which creates a vacancy i.e. hole (lack of electron) in that

orbit. For eg. in K shell.

• Later on electron from higher orbit , i.e from L shell jumps into the hole of K shell.

• Then the energy difference EK — EL is emitted in the form of characteristic X rays.

• The Kα spectral line originates when an electron from L shell jumps into the vaccancy of

K shell.

• The Kβ spectral line originates when an electron from M shell jumps into the vacancy of

K shell

• When electron from higher orbit i.e. L, M, N jumps into K shell which gives rise to K

series. X-rays i.e. Kα,K β K, γ and K δ



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THE WAVELENGTHS OF X-RAYS IN LINE SPECTRA ARE THE

CHARACTERISTIC OF TARGET MATERIAL AND HENCE THEY ARE

CALLED AS CHARACTERISTIC X-RAYS

14. Explain Continuous X-ray spectra?

• Some of the high-velocity electrons penetrate the interior of the atoms of the target

material and are attracted by the positive charge of their nuclei.

• As an electron passes close to the positive nucleus, it is deflected from its path as shown

in Figure above.

• The electron experiences de-acceleration during its deflection in the strong field of the

nucleus.

• The energy lost during this de-acceleration is given off in the form of X.rays of

continuously varying wavelength (and hence frequency).

• These X.rays produce continuous spectrum.

15. Derive the formula for minimum wavelength of X-rays produced in continuous

X-ray spectra (Duane Hunt Law)?

Ans. • The striking electron has its velocity reduced from v to v’ during its passage through the

atom of the target material,

• Then its loss of energy is.( ½ mv2

- ½ mv’

2) and this must be equal to the energy of the

X-ray photons emitted.

½ mv2

- ½ mv’

2 = h.υ

• The highest or maximum frequency of the emitted X.rays corresponds to the case

when the electron is completely stopped i.e. when v’ = 0.

• In that case

½ mv2

= h.υmax……………………………………. (1)

• If the electron is accelerated through a potential of V volts,

½ mv2

= e. V………………………………. (2)

From (1) and (2)

h.υmax = e.V

• Since υmax= c/ λ min

h. c = e.V

λ min

• Therefore λ min = h. c / e.V



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• Since h,c,e are constants , therefore λmin is inversely propotional to V

• λ min = (12400 / V) A◦

16. Derive Bragg’s law ?

Ans.

• An 2-dimensional view of the crystal is represented.

• It shows a beam of monochromatic X rays incident at a glancing angle θ on a set of

parallel planes of crystal.

• The successive layers are rich in atoms hence selected. Ray No. 1 is reflected from atom

B in plane 1 whereas Ray No. 2 is reflected from atom E lying in plane 2 immediately

below atom B.

• Whether two reflected rays will be in phase or antiphase with each other will depend on

their path difference.

• This path difference can be found by drawing perpendiculars BP and BQ on ray No 2 .

Since the 2 rays travel the same distance from A and N onwards, we say that No.2 travels

an extra distance = PE + EQ

• Hence, the path difference between the two reflected beams is

= PE + EQ

= d sin θ + d sin θ = 2d sin θ where d is the interplanar spacing, i.e., vertical distance

between two adjacent planes belonging to the same set.

• The two reflected beams will be in phase with each other· if this path difference equals

an integral multiple of λ and will be antiphase if it equals an odd multiple of λ / 2



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Hence, the condition for producing maxima becomes 2 d sin θ = n λ where n = 1, 2, 3, etc. for the first-order, second-order and third-order maxima respectively.This relation is known as Bragg's Law. 15. Explain Braggs Spectrometer and its use to analyze crystal structure?

Ans. BRAGG’s SPECTROMETER

• X-rays from a tube are collimated into a narrow beam by two fine slits SI and S2 normally

made of lead.

• The beam is incident at a glancing angle of θ on the face of a crystal plate (of calcite, rock

salt, NaCl or KCI etc.) which is mounted on the turntable T of the spectrometer.

• This turntable is capable of rotation about a vertical axis passing through its centre and its

rotation can be read from the circular scale.

• Most of the incident beam passes straight through the crystal plate but some X-rays au

scattered by the regularly-arranged atoms lying in different crystal planes.

• These scattered X-rays can be looked upon as having been reflected from the crystal

planes particularly those which au rich in atoms.

• The reflected X-ray beam enters an ionization chamber carried by an arm R which is

capable of rotation about the same axis as the turntable.

• The turntable and arm R are so linked together that when the turntable (and hence the

crystal) rotates through an angle θ, the arm R (and hence the ionization chamber) turns

through double the angle, i.e., 2θ. • In this, way the beam is always reflected into the ionization chamber whatever its incident

or glancing angle at the crystal surface. The ionization current produced by the reflected

X-ray beam can be measured by a sensitive electrometer E (or the reflected beam can be

recorded on a photographic film in which case the spectrometer is known as

spectrograph)



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Determination of crystal structure using Bragg’s Law Bragg used a KCI crystal (which is cubic) and found first maxima of reflected X-rays to occur at values of θ equal to 5'22°, 7·30° and 9·05° respectively when the three planes (100), (110) and (111) are used in turn as reflecting planes. Now, for first-order spectrum, 2d sin θ =λ 1 = 2 Sin θ d λ

Therfore



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CRYSTAL DEFECTS

16. What do you mean by crystal defects?

Ans. Any deviation from the regular periodic arrangement of atoms in a crystal is called as

crystal defect. Ideal crystals neither occur in nature nor can be produced artificially.

CRYSTAL DEFECTS are of three types

POINT DEFECTS LINE DEFECTS PLANAR DEFECTS

(a) Vacancy i. Edge dislocation a. Grain Boundries

(b) Interstitial ii. Screw dislocation

(c) Substitution impurities

(d) Interstitial impurities

(e) Schottky defect

(f) Frankel defect

17. Explain the various point defects in crystals? Ans. The various types of point defect are Vacancy defect, interstitial defect, Substitution impurities, Interstitial impurities, Schottky defect and Frankel defect

Vacancy is produced due to the removal of an atom from its regular position in the lattice The removed atom

does not vanish. It travels to the surface of the material. For low concentration of vacancies, a relation is

n = N exp (-Ev/ KT)

Interstitial:- An extra atom of the same type is fitted into the void between the regularly

occupied sites is referred to as interstitial defect. Since in general the size of atom is larger than

the void into which it is fitted, so the energy required for the interstitial formation is higher than

that of vacancy formation.

Substitution impurities:- In this a foreign atom is found occupying a regular site in a crystal lattice



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Interstitial Impurities: A foreign atom is found at a interstitial site.

Schottky defect: - Missing of an ion pair (i.e. an anion and cation each) from the ionic crystal

lattice is called as Schottky defect. .

Frenkel defect:- For ionic crystal when a negative ion vacancy is associated with an interstitial

negative ion or a positive ion vacancy is associated with an interstitial positive ion then it is

called Frankel defect.

18. Explain the various line defects in crystals?

Ans.

A dislocation is a linear disturbance of atomic arrangement caused by the displacement of one

atom or group of atoms in a crystal from an adjacent group. They occur during the solidification

of a crystal. There are two types of dislocations

i. EDGE DISLOCATION ii SCREW DISLOCATION



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i. Edge dislocation: -

• An edge dislocation arises when one of the atomic planes exist only partially and does

not extend through the entire crystal.

• The atomic plane AB abruptly terminates at B and it may be viewed as an extra plane

inserted in between a set of parallel planes.

• The edge of such a plane constitutes a linear defect and is called an edge dislocation.

• The edge dislocation is commonly designated by the ‘inverted T’ symbol (┴).

ii. Screw dislocation

• It is also known as Burgers screw dislocation.

• Dislocations arise in a crystal as a result of growth accidents, thermal stresses

• The points where edge or screw dislocation lies emerge from the crystal surface

represents points of strain and enhanced chemical reactivity.

• It is equivalent to making a cut in the crystal and pulling one side down compared to

other.

19. What do you mean by planar defects?

Ans.

A majority of the materials are polycrystalline and have random orientations.. The boundary

between two adjacent grains forms a disturbance in the periodicity of the lattice. Hence, the grain

boundaries are planar imperfections.

Liquid Crystals 20. What do you mean by liquid crystals?

Ans.

• Liquid Crystal is a new phase of matter

• In solid state, the molecules are highly ordered. Each molecule occupies a fixed rigid

position and is immobile. Therefore, there is a positional & orientation order in solids.

• In liquid state, the molecules neither occupy specific positions nor remain oriented in a

specific manner.

• In liquid crystalline state, there exists liquid crystal phase wherein the molecules are free

to move but are oriented in a particular manner.



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21. Explain “ORDER PARAMETER” and its significance for solids, liquid and liquid

crystals?

Ans.

• Liquid crystals possess some degree of orientation order but not as good as solid material.

• To quantify how much order is present in a material an order parameter (S) is defined.

• The order parameter is given as follows,

S= ½ [3cos2 θ– 1] where θ is the angle made by each molecule with the director.

• Order parameters for solids, S = 1

Liquids, S = 0

Liquid crystals, S = 0.3 to 0.9

• Order parameter depends on temperature.

• As the temperature increases order parameter decreases.

22.Give the classification of Liquid Crystals?

Ans.

Liquid Crystals

Thermotropic Lyotropic

Calamitic (Rod Like) Discotic (Disc Like)

Nematic

Chiral Nematic

Smectic

Nematic discotic

Columnar



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Nematic Liquid Crystal

• The term ‘nematic’ is derived from the Greek word which means ‘thread like’.

• This phase is characterized by the molecules that have no positional order but tend to

orient in the same direction i.e., along the director.

Chiral Nematic Liquid Crystal (Cholesteric)

• This phase is formed by compounds having Chiral centers. • In this structure the directors actually form in a continuous helical pattern.

Smectic A and Smectic C



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• As temperature is further reduced in a nematic type, molecules begin to segregate into

planes giving rise to a smectic phase.

• This phase is more solid-like.

• The molecules are aligned along the director and they have one-dimensional translational

order also.

• Smectic phases are typically classified as Smectic A type and Smectic C type.

• The word ‘smectic‘ is derived from the Greek word which means ‘soapy’.

• In smectic state, the molecules maintain the general orientational order of nematic but

also tend to align themselves in layers or planes.

Nematic Discotic and Columunar

• Liquid crystals are also formed by disc or plate like structured molecules. These are

called discotic or columnar liquid crystals.

• The simplest discotic phase is called discotic nematic phase due to the fact that there is no

positional order but there is orientational order.

• There is a random motion of molecules but on an average , the axis perpendicular to the

plane of each molecule tend to orient along the director.

Lyotropic liquid crystals

• Some compounds transform to a liquid crystal phase when with a solvent.

• The anisotropic solution mesophase for high solute concentration which are formed by

the solution rod-like entities in an isotropic solvent are called lyotropic liquid crystals.

23. Give the applications of liquid crystals?

Ans.

The most common applications of liquid crystals are

1. In liquid crystal displays (LCD). LCDs are found in wrist-watches, calculators,

computers and various other devices where a low-power display is needed.

2. The optical properties can be used to make electronic shutters and other optical switches.

3. Polymer liquid crystals have been extensively studied; liquid crystal phases are important

in the formation of super-strength polymers.

4. Liquid crystals also seem to be intimately involved in the functions of biological

structures, such as living tissue.



26

5. The "mood ring", a popular novelty a few years ago, took advantage of the unique

ability of

6. Liquid crystal temperature sensors can also be used to find bad connections on a circuit

board by detecting the characteristic higher temperature.

7. Special liquid crystal devices can be attached to the skin to show a "map" of

temperatures. This is useful because often physical problems, such as tumors, have a

different temperature than the surrounding tissue.

8. Liquid Crystals are highly sensitive materials that respond to temperature. Liquid

crystals change from black to a rainbow of colors and then back to black again upon

heating. Upon cooling the reverse color change occurs.

9. Liquid crystals can be formulated into thermometers with a wide or narrow temperature

sensitivity, and can be made into any size or shape. Examples of liquid crystal

thermometers are aquarium thermometers, forehead thermometers,etc.

24. Give the disadvantages of Liquid crystal displays?

Ans.

1. Unfortunately, liquid crystals are very difficult to work with and require highly

specialized printing and handling techniques.

2. Liquid crystals are also more expensive.

3. They are adversely effected by high temperatures, ultra violet light, and strong solvents

or chemicals.

IMPORTANT FORMULAE

2. Interplanar Spacing dhkl = a_____

√ h2 + k

2 + l

2

3. Braggs Law 2 d sin θ = n λ 4. λ min = h. c / e.V

5. λ min = (12400 / V) A◦

6. Number of vacancies at T0K n = N exp (-Ev/ KT)



27

Chapter 1 (27 marks)

Crystallography and X- rays Part I (Crystallography and defects)

1.Describe the structure of diamond. Explain its hardness by obtaining the packing efficiency

and name two semi conducting elements having the same structure ( Dec-03 5 M)

2. Miller indices in practice do not define a particular plane but a set of parallel planes. Explain

this statement with the help of example. (Dec 03 -3 M)

3. Explain the procedure to determine Miller indices of crystal plane which intercepts all the

three axis. (May04, Dec04, May 05, Dec.06,May07-5 M)

4. Find the packing density of

(a) BCC

(b) Diamond crystal structure. (May04-5 M)

5. Define lattice parameter and explain how it is related with molecular weight and density in a

single crystal. (Dec. 04-5 M)

6. Find the packing density of BCC crystal and show with diagram that co-ordination Number of

FCC is 12. (May. 05, Dec.05- 6 M)

7. Show that in a crystal of cubic structure, the distance between the planes with miller indices

(hkl) is equal to (Dec. 05- 5 M; May06- 7 M)

8. Explain in short (May06- 2 M)

( a) Single crystal ( b) Lattice parameter

9. What are atomic planes? Sketch (321), (123) and [111] (May-06- 5 M)

10. Discuss diamond crystal structure. (May06- 5 M)

11. What are the Miller indices? How planes having different intercepts can have similar miller

indices? Explain with suitable example. (Dec.06- 5 M)

12. Calculate atomic packing factor for SC structure. (May07- 5 M)

13. Write a short note on Miller indices. (May07-5 M)

14. Describe the HCP structure? What is its coordination number , Atomic radius and number of

atoms owned by the unit cell.Also find the packing factor (Dec.07-8 M)

15. Determine the co-ordination number and packing density for a Hexagonally Closed Packed'

structure. Show that a HCP structure demands an axial ratio of (√8/3) (May08, 09-8 M)

16. Explain following terms:- Phases of liquid crystal (May08, 09-3M,)

17. What is the lattice constant? Find the relation between lattice constant and atomic weight

and density in a single crystal. Draw unit cell of NaCl crystal structure and discuss its structure.

(Dec.08-10M)

18. Write short note on point defects in crystals (Dec.08– 5M)

19. Describe working of liquid Crystal display. (Dec09- 3M)

20. What are liquid crystals ? State its different phases. (May10 -3M)

21. Explain atomic arrangement in diamond structure and calculate. (May10 -10M)

Total number of atoms per unit cell (n), Atomic radius (r), Co-ordination number (CN)

Atomic packing factor (APF), Packing efficiency (PE), Void space and density.

Also write the materials exhibiting diamond structure. 22. What are Crystal imperfections? What is their Significance? (May 10- 5M)



28

Part II- X-rays 1. What is continuous and characteristic X-ray spectrum? Explain their origin.

(Dec03, 6 M, Dec.08- 3M)

2. Short note on: Bragg’s spectrometer. (Dec 03-8 M)

3. Explain construction and working of Coolidge tube for production of X-

rays. (May04-5 M)

4. Explain construction and working of a Bragg’s spectrometer with derivation of Bragg’s

equations. (May04-5 M)

5. Examine the formation of Lβ X-ray line spectrum (Dec.04-5 M)

6. Describe Bragg’s spectrometer in brief, explain the experimental

verification of Bragg’s law. (Dec.04-5 M)

7. Differentiate between continuous X-rays and characteristic X-ray spectra. (May05-5 M)

8. Short note on Coolidge tube (May05-5 M)

9. How Bragg’s law can be used for analysis of crystal structure?

10.Explain the fundamental differences between characteristic X-ray spectrum and continuous

X-ray spectrum. How do they origin? (Dec06-7 M) 11.

11. What are characteristic X-rays of element? How can they be used to determine the atomic

number of an element (May07-8 M, May09-3M)

12. Explain Industrial applications of x-rays. (Dec09-3M)

13. Derive Bragg’s law. Explain Bragg’s Spectrometer and its use to analyze crystal structure.

(May 10 – 10 M)

14. Write short notes (a) Miller indices (b) X-rays in Crystallography (Dec09-10M)

PROBLEMS

1. A crystal lattice plane (3 2 6) makes an intercept of 1 .5 A on X-axis in a crystal having lattice

constant 1 .5 A, 2 A and 4 A on X, Y and Z axis respectively. Find Y and Z axes intercepts.

(Dec03- 5 M)

2. Calculate density of GaAs using following each

Atomic weight of Ga: 69

Atomic weight of As : 74

Lattice constant for GaAs = 5.6 x 10-8

Number of atoms/unit cell in GaAs = 4. (Dec04- 5M)

3. Silver has FCC structure and its atomic radius is 1.441Ao . Find the spacing of (220) and

(111) planes. (Dec. 04- 5M)

4. Silver has FCC structure and its atomic radius is 1.441Ao . Find the spacing of (220) and

(111) planes. (May 05- 5 M)

5. A crystal lattice plane (3 2 6) makes an intercept of 1.5 A on X-axis in a crystal having lattice

constant 1 .5 A, 2 A and 4 A on X, Y and Z axis respectively. Find Y and Z axes intercepts.

(Dec. 05- 5 M)

6. Sketch the planes (123). (12 3) and [213] [110] (Dec.06-5 M)

7. What is lattice constant ? Find the relation of lattice constant wt’ a weight

and density in a single crystal. Gold belongs to cubic monatomic

structure. Its density is 19320 kg/m3 and the lattice constant a = 4.08 A.

Atomic weight . is 197.Find the type of unit cell. (Dec.06-8 M)



29

8. A crystal lattice plane (3 2 6) makes an intercept of 1 .5 A on X-axis a crystal having lattice

constant 1 .5 A, 2 A and 4 A on X, Y and Z axis respectively. Find Y and Z axes Intercepts.

(Dec.06-7 M) 9. Draw the following (100) (110) (123) [121] [110] (May07- 5M)

10. Calculate the number of atoms per unit cell of a metal having the lattice parameter 2.9 A and

density is 7.87 gm/cm3 Atomic weight of a metal is 55.85. Avogadro’s number is 6.0238 x

1023

/gm-mole (May07-5 M)

11. Find the Miller indices for the plane with the following set of intercepts (a/2, b,∞).Draw the

plane (Dec.07- 3 M)

12.Calculate the distance between 2 atoms of a basis of the diamond structure if the lattice

constant of structure is 5 A U (Dec.07-3 M)

13. Ni has FCC structure .Its lattice constant is 3.52 AU.Atomic weight of 58.71,Given that the

Avagordos number is 6.023 X 1026

/kg-mole. Calculate its radius APF and density.

(Dec.07-7M, May08-7M) 14. Describe the NaCl structure and calculate the number of molecules per unit cell and packing

factor assuming radius of Na+is 0.98 AU.and radius of Cl- is 1.81 AU ( May-08- 8 M)

15. Derive the packing factor for FCC. Copper has FCC structure and atomic radius is 1.28 AU.

Calculate its density if its atomic weight is 63.5 ( May-08- 7 M)

16. Draw the following for a Simple Cubic Structure ( May-09- 3 M)

17. What is Frankel defect? How it is developed in a crystal? Estimate the ratio of vacancies

at (i) -119°C (ii) 80°C, where average energy to create vacancy is 1.8 e. V. (May-09-7 M)

16. Draw the following for a Simple Cubic Structure ( Dec.08- 3 M)

17.Silver has FCC structure & its atomic radius is 1.441 A°. Find the spacing of (111) plane

(Dec.08- 3 M)

18. Calculate the number of atoms per unit cell of a metal having the lattice parameter 2.9 A°

and density is 7.87 gm/cm3. Atomic weight of a metal is 55.85. Avagadro’s number is 6.023 x

1023

/gm.mole. (Dec.08- 5 M)

19. Find the Miller Indices for a plane with the following set of intercepts (a/2, b, ∞).Draw the

plane for the same. (May08- 3M)

20. Calculate the distance between two atoms of a basis of the diamond structure, if the lattice

constant of structure is 5AU (May08- 3M)

21. Molybdenum has a BCC structure. Its density is 1 and its atomic weight is 95.94. Determine

the radius of molybdenum atom. (May10- 5M)

22. Draw following planes in Cubic Unit Cell (-1 1 -1), (1 0 -1) ,(-1 0 1) (Dec09- 3M)

23. Draw the following planes and directions in cubic cell. (101), (100), [ (May10-3M)

24. Sodium is a BCC Crystal. It’s density is 9 x 102 kg/rn and atomic weight is 23. Calculate

the lattice Constant for Sodium Crystal . (Dec09-5M)

25. Estimate the number of Frankel defects per mm3 in Silver chloride if energy of formation

of Frankel defects is 1 ev at 700°k. The molecular weight of AgCI is 0.143 kg/mol and specific

density is 556. (Dec09-5M)



30

Problems (Xrays)

1. Calculate the smallest glancing angle at which k-copper line of 1.549 A will be reflected from

crystal having atomic spacing of 4.255 A. What is the highest order of reflection that can be

observed with this radiation? (Dec03 -5 M)

2. Bragg’s reflection of the first order was observed at 21.7o for parallel planes for a crystal

under test. If the wavelength of X-rays used is 1 .54 A find the lattice constant of the crystal.

(May04-5 M) 3. Calculate the velocity and kinetic energy of electrons striking the target in an X-ray tube

operating at 3500 volts. (May04-5 M)

4. Calculate the glancing angle on the cube (100) of a rock salt (a = 2.814 A) corresponding 2

order diffraction maximum for X-rays of wavelength 0.714 AU . (Dec.04-5 M)

5. If the p.d. applied across an X-ray tube is 25 kV and current through is 10

mA. Calculate number of electrons striking the target per second and minimum wavelength limit

for the X-ray generated. (Dec05-5 M)

6. The spacing between the principle planes in a crystal of NaCl is 2.82 A. It is found that first

order Bragg’s reflection occurs at 100

(a) What is the wavelength of X-ray?

(b) At what angle the second order reflection occurs?

(c) What is the highest order of reflection seen? (May06 5 M)

7. Derive Bragg’s law. Calculate the smallest glancing angle at which k copper line of 1.549 A

will be reflected from crystal having atomicspacing of 4.255 A. (May06-8M)

8. In comparing the wavelengths of two monochromatic X-ray lines, it is found that line A gives

a 1st order Bragg reflection maximum at a glancing angle of 30° to the smooth face of a crystal.

Line B of known wavelength of 0.97 A gives a 3 order reflection maximum at a glancing angle

of 60 with the same face of the same crystal. Find the wavelength of the line A.(May07-5M)

9. Find the shortest wavelenght of X-ray produced by an X-ray tube operated at 16 KV potential

(May09-3M)

10. 1 0 KeV electrons are passed through a thin film of a metal for which atomic spacing is 5·5 x

10-11

m. What is the angle of deviation of the first order diffraction minimum? (May09-7M)

11. Electron bombarding the anode of Coolige tube produces X-rays of wavelength 1.5 A°. Find

the energy of the electron at the moment of impact.

(Given h = 6.63 x 1 0-34 J-sec; e = 1.6 x 1 0-19

C) (Dec.08-3M)

12. In comparing the wavelengths of two monochromatic x-ray lines, it is found that line A gives

1st order Bragg’s reflection maximum at a glancing angle of 30° to the smooth face of a crystal.

Line B of known wavelength of 0.97 A° gives 3rd

order reflection maximum at a glancing angle

of 60° with the same face of the same crystal. Find the wavelength of line A. (Dec.08- 5 M)

13. An X-Ray beam of wavelength 0.71A is diffracted by a FCC crystal of density 1•99 x103

kg/rn3.

Calculate the interplanar spacing for (2 00) planes and the glancing angle for the second

order reflection from these planes. Given mol. wt of the crystal is 74.6 and Avagadros No. is

6.023 x 1026

/kg mole. (May08-7M)

14. If P.D. across an X-ray tube is 25kV and filament current is 10mA. Calculate number of

electrons striking the target per second and velocity of electrons striking the target.

(May 10– 5M)

1_amtech

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