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UNIT-II

CRYSTALLOGRAPHY AND ULTRASONICS

Basic terms and definitions

Crystallography Crystallography is the branch of science that deals with discerning the arrangement and bonding of atoms in crystalline solids and with the geometric structure of crystal lattices.

Crystalline solids Crystal is a solid composed of periodic (regular) arrangement of

atoms and molecules.

Amorphous solid Amorphous solid composed of irregular arrangement of atoms and

molecules.

Space lattice A space lattice or crystal lattice is defined as a regular three dimensional arrangement of identical points in space or it can be defined as an array of points showing how molecules, atoms or ions are arranged at different sites in three dimensional space.

Basis The group of atoms is called the basis; when repeated in space it

forms the crystal structure.

Unit cell The smallest part of a crystal is called as unit cell. It is formed by combination of atoms and molecules. The whole crystal structure can be formed by the repetition of these unit cells.

Primitive cells Primitive cells contain only one lattice point, which is made up

from the lattice points at each of the corners

Multiple cell Multiple cell contains more than one lattice point.

Lattice Parameters The primitives (a, b, c) and interfacial angles (α, β, γ) are the

lattice parameters. Effective number of atoms The total number of atoms in a unit cell by considering the

contribution of corner atoms, centre atoms and face centred atoms

Atomic radius It is half the distance between any two successive atoms in a crystal lattice

Coordination number The number of equidistant neighbours that an atom has in crystal lattice

Nearest neighbour distance The distance between two nearest neighbouring atoms in a crystal lattice

Atomic packing factor no. of atoms present in a unit cell x volume of the one atom

Volume of the unit cell

Interstitial space The empty space in the crystal lattice is called interstitial space or void space.

Miller indices Three smallest integers which have the same ratio as the reciprocals of the intercepts of the crystal plane with the coordinate axes

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Bragg’s law

Bragg’s law states that the x-rays reflected from different parallel planes of a crystal interfere constructively when the path difference is integral multiple of the wavelength of x-ray

Ultrasonic It is defined as "sound at frequencies greater than 20 kHz."

Properties of ultrasonic 1.They are highly energetic and has high frequency 2. It requires a material medium for its propagation 3.They show negligible diffraction due to their smaller wavelength 4.Exhibits disruptive effect in liquids by causing bubbles

Non-destructive testing Non-destructive testing is a method of testing materials for structural integrity without destroying or damaging the material.

Concepts:

CRYSTAL SYSTEMS: Based on geometrical consideration crystals are 32 classes. Taking different symmetries

and structures all these structures are classified into 7 crystal systems. Each system is characterized by the

values of a, b, c and α, β, γ.

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BRAVAIS LATTICES:-

There are 14 bravais lattice in 3 dimensions. Each of these 14 Bravais lattice belongs to 7 crystal system.

Lattice symbol:

1) ‘P’:- Primitive or Simple cell: It has lattice points at corners only.

2) ‘C’:- Base centered cell: It has an extra atom at the center of base only.

3) ‘I’:- Body centered cell: It has an extra atom at the center of body.

4) ‘F’:- Face centered cell : It has extra atoms at the centers of all six faces.

base centered

S.No Crystal System Bravais Lattice Lattice Symbol No. of lattices in

the system

1 Cubic Simple Body centered Face centered

P I F

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2 Tetragonal Simple Body centered

P I

2

3 Orthorhombic

Simple Base centered Body centered Face centered

P C I F

4

4 Monoclinic Simple Base centered

P C

2

5 Triclinic Simple P 1

6 Trigonal

Simple

P

1

7 Hexagonal Simple P 1

4

Note:

Lattice

Parameter

condition

a =b=c

α=β=γ

=900

a=b≠c

α=β=γ=9

00

a≠b≠c

α=β=γ

=900

a≠b≠c

α=β=900

γ≠900

a≠b≠c

α≠β≠γ≠9

00

a =b=c

α=β=γ≠

900<1200

a=b≠c

α=β=900

γ=1200

Crystal

system

(7)

Cubic Tetragon

al

Orthorhom

bic

Monoclinic Triclinic Trigonal Hexagonal

Bravais

Lattice

(14)

(P I F)

3

(P I)

2

(P C I F)

4

(P C)

2

(P)

1

(P)

1

(P)

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Cubic Crystal structures:-

1) Simple Cubic (SC) structure:-

Each corner atom surrounded by 6 equidistant atoms. So co-ordination number is 6.

Here atoms are present at corners of the unit cell. Hence the unit cell of simple cubic structure is a

primitive cell.

Each corner atom is shared by eight neighbouring unit cells so that the share of each unit cell is only

(1/8) th of an atom. There are 8 atoms at 8 corners of unit cell.

Number of atoms = (1/8) *8 =1 atom/cell

I.e. it is a primitive cell.

Volume of atoms per cell = 1*(4/3 ∏r3)

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v = 4/3 ∏r3

➢ Volume of unit cell (v) = a3

From fig(b), nearest neighbor distance 2r = a

Substitute ‘a’ value in the above equation,

Therefore, Volume of unit cell (V) = a3 = (2r) 3 = 8r3

Therefore,

Packing factor =Volume occupied by atoms in unit cell/volume of unit cell

= 1*(4/3 ∏r3)/ 8r3 = ∏/6 =0.52 (or) 52%

i.e. 52% of volume of unit cell is occupied by atoms and remaining 48% is empty.

So, Simple cubic structure is a loosely packed structure.

Eg.:- Polonium at <750c

2) Body Centered Cubic (B.C.C) structure:-

Fig shows unit cell of B.C.C structure. It has 8 corners atoms and one extra atom at the center of the

body. Hence; it is named as “Body centered cubic” structure.

All the corner atoms at the same distance from body centered atom, i.e. there are 8 atoms at

the same distance from the centre of the unit cell.

Coordination number is 8.

Each corner atom is shared by 8 neighbouring unit cells. So, that the share of each unit cell is only 1/8th

of corner atom. There are 8 atoms at 8 corners of unit cell. In addition, there is an extra atom at the center of

body of unit cell.

No. of atoms =(1/8 *8) +1 = 1+1 =2 atoms/cell

B.C.C is a non primitive cell

Volume of atoms ν =2*(4/3 ∏r3)

ν= 8/3 ∏r3

➢ Volume of the unit cell(V) = a3

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To find “a”:-

From ∆ACD

AD2=AC2+CD2

(4r) 2= AC2 +a2

(4r) 2 = 2 a2 + a2

16r2= 3 a2

Therefore Volume of the Unit cell = a3

V = (4r /√3)3 = (64 r3 /3√3)

Packing factor v/V = 8/3 ∏r3/( 64 r3 /3√3)

= ∏/ (8/√3)

= ∏√3/8 ≈0.68 (or) 68%

i.e., 68% volume of unit cell is occupied by atoms and remaining 32% is empty. So, compare

to simple cubic structure B.C.C is closely packed structure.

Eg.:-Lithium(Li) , Sodium(Na) , Potassium(K) , Iron(Fe) , Tungsten(W) , Chromium (Cr) etc.

a =4r/√3

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3) Face Centered Cubic (FCC) Structure:-

The figure shows unit cell of F.C.C structure. It has 8 atoms at 8 corners and 6 atoms at 6

faces.

The atom at the center of face, they are 12 atoms.

[4 corner atoms in its plane and 4 face centered atoms in each of two planes on either side of its plane].

Therefore Coordination number is 12.

Each atom at the corner shared by 8 unit cells and each atom at the center of face shared by 2 unit cell.

Number of atoms = (1/8 * 8) + (1/2 *6)

=1+3 =4 atoms

So, F.C.C is a non primitive cell

Therefore Volume of the atoms (ν) =4*(4/3) ∏r3 = (16/3) ∏r3

The volume of unit cell =a3

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To find “a”:-

AC2=AB2 +BC2

(4r) 2 =a2+ a2

16r2=2 a2

8r2=a2 (or) a=2Ѵ2r

Volume of unit cell (ν) = a3 = (2Ѵ2r) 3 =16Ѵ2r3

˙. Packing Factor =ν/V = (16/3∏r3)/(16 Ѵ2r3)

= ∏ / (3Ѵ2)

= 0.74 (or) 74%

i.e 74% of volume of the unit cell is occupied by atoms and remaining 26% is empty. So, compare to simple

cubic and B.C.C, F.C. C has highest packing factor. So it is more closely packed structure.

Eg:- Aluminum(Al) , Copper(Cu) , Lead (Pb) , Silver(Ag), Nickel (Ni) etc.

LATTICE CONSTAT (a) OF CUBIC CRYSTAL:-

For cubic crystal a=b=c and α=β=γ=900

Let ‘ρ’ be density of material

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Volume of the unit cell = a3

Mass of unit cell =Density X volume

= ρ a3 →(1)

Mass of each atom /molecule in unit cell =M/NA

Where M → Atomic weight /Molecular weight

NA → Avogadro number, NA=6.023X1023 atom/molecule

If n → are number of atoms,

Mass of unit cell = n X (M/NA) → (2)

From eq.(1) and eq. (2)

eq.(1) =eq. (2)

ρ a3 = n X (M /NA)

a3 = (n M) / (ρ NA)

a = {(n M) / (ρ NA)} 1/3

Therefore Lattice constant for cubic structure

a = {(n M) / (ρ NA)} 1/3

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CRYSTAL DIRECTIONS AND PLANES –MILLER INDICES:-

CRYSTAL DIRECTION:-Inside crystal there exists certain directions (lines) along large concentration of atoms are

called Crystal directions. Generally [ ] is used to indicate a direction.

2-D Representation:-In 2 –D crystal direction is represented by coordinates of an integral point [x, y] through

which line passes.

3 –D Representation:-In 3- D direction is represented by a whole number point [x, y, z] through which line

passes.

Crystal Plane:- An imaginary plane which contains large concentration of atoms called as crystal plane.

Generally crystal plane is indicated by using ( ).

2D Representation:

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3D Representation:

MILLER INDICES:-

Miller derives a method to represent crystal directions and planes by three small integers h, k, l known

as miller indices.

[h, k, l ]- indicate crystal direction

(h, k , l ) - indicate crystal plane

Definition:- Miller indices are defined as “reciprocals of intercepts made by a plane along 3 crystallographic

axes when reduced into smallest possible integers”.

Procedure for finding miller indices:

Let us consider a 3-D plane along 3 crystallographic axes as shown in figure.

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Step 1):- Find the coordinates of intercepts made by plane along 3 crystallographic axes

i.e., X Y Z

pa qb rc

Where a, b, c → unit cell dimensions (or) Lattice parameters.

p, q , r → Simple fractions (or) small integers .

Step 2):- Express the intercepts as the multiples of unit cell dimensions

i.e., pa/a, qb/b, rc/c

p q r

Step 3):- Take reciprocals

i.e., 1/p, 1/q, 1/r

Step 4):- Reduce these reciprocals into smallest possible integers.

i.e. 1/p :1/q :1/r =h : k : l

FEATURES OF MILLER INDICES:-

1) All parallel and equidistance planes having the same miller indices values.

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2) If any plane is parallel to one of the axes then its intercept is taken as infinity. Hence miller

indices for that axes is ‘0’.

3) If any plane passes through origin it is defined in terms of parallel plane having non-zero

intercept.

4) If a normal is drawn to a plane (h,k,l) , the direction of normal is [h,k,l].

5) If miller indices of two planes have the same ratio i.e.,(8 4 4) ,(2 1 1) , (4 2 2) then the

planes are parallel to each other.

6) If the intercept of a plane on any axes is negative, a bar (─) is put on corresponding miller

indices.

SEPARATION BETWEEN SUCCESSIVE (h k l) PLANES:-

A crystal is made up of large number of parallel and equidistance planes. The perpendicular distance

between successive planes is called “Inner Planar distance (d)”.

Let (h, k, l) are miller indices of ABC plane let intercepts along x, y, z axes are a/h, b/k, c/l.

That is OA= a/h

OB=b/k

OC=c/l

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Let ON =Normal (d) makes angles α, β, γ with x, y, z respectively

From ∆ NOA,

Cosα =ON/OA=d/ (a/h)

From ∆ NOB

Cosβ = ON/OB =d/ (b/k)

From ∆ NOC

Cosγ =ON/OC =d/(c/l)

According to cosine law

Cos2α + Cos2β + Cos2γ =1

{d/(a/h)}2+{d/(b/k)}2+{d/(c/l)}2

d2 {h2/a2+k2/b2+l2/c2} =1

d=1/√ ( h2/a2+k2/b2+l2/c2)

For cubic system

a=b=c

d=a/√ ( h2+k2+l2)

Diffraction of X-rays by crystal planes:-

X-rays are Electromagnetic rays. Hence X-rays exhibit the phenomenon light interference and

diffraction .The most important condition for observable diffraction effect is that the wavelength λ of X-rays

and inner planar distance (d) should be same in order.

Generally the wavelength of X-rays is of order of 1A0 units and the inner planar distance for different sets

of planes is arranged 2 A0 units.

So, crystals are more suitable to observe diffraction effect with X-rays. Hence X-ray diffraction is utilized to

study crystal structure.

λ ≤ 2d

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BRAGG’S LAW:-

When X-rays incident on atoms of different parallel planes of the crystal, each atom act as

source of scattering of radiation of the same wavelength. Each plane in the crystal act as reflecting plane and

crystal act as series of parallel reflecting planes.

Statement:-

“Intensity of reflected rays at certain angles will be maximum when path difference between two reflected

rays from two different planes is equal to integral multiples of wavelength”.

Let us consider a set of parallel planes as shown in figure. Let d is inner planar distance.

A part of incident X-ray AB is incident on the atom at B point in plane 1, with an angle θ is

called “Glancing angle”, and is reflected along the direction BC. Similarly, a part of incident X-ray DE incident on

atom at E point in the plane II and is reflected along the direction EF. If path difference between reflected rays

is equal to integral multiples of wavelength

i.e nλ (where n=1, 2, 3, 4-------).

The X-rays reflected from adjacent planes are inphase so constructive interference takes place

hence intensity of reflected rays is maximum. If path difference between reflected rays is equal to (2n+1) λ/2

.Then the reflected rays are out of phase so, destructive interference takes place and hence intensity of

reflected rays is minimum.

To find the path difference drawn normal from B two DE and EF lines .the intersecting points

are P, Q respectively

From ∆PBE,

Sinθ=PE/BE

PE=BEsinθ

PE=dsinθ

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Path difference=PE+EQ

=dsinθ + dsinθ

=2dsinθ

For constructive interference 2dsinθ=nλ

POWDER METHOD (Debye-Scherrer method):-

The powder method is used to study structure of crystals which are not obtaining easily in the form of single

crystal . Therefore sample used in the form of fine powder containing large number of tiny randomly oriented

crystals. This method was developed by Debaye and Scherrer so it is called Debye-Scherrer method.

Fig(a) shows experimental arrangement used to produce diffraction pattern. It consist a

cylindrical camera called Debye-Schreer camera. A strip of photographic film arranged along inner periphery of

the camera. The powder specimen is filled in a thin capillary glass tube made up of non-diffracting material and

placed at centre of camera.

By passing a beam of monochromatic X-rays through collimated (parallel) mono chromatic X-

rays (sharp). The X-rays entered into the camera through collimator and strike the powdered sample. Since

specimen contains large no. of small crystals with random orientations all possible diffraction planes will be

available for Bragg reflection to take place.

The reflected rays will form a cone whose axis lies along the direction of incident X-ray beam

and whose semi vertical angle is equal to twice the glancing angle (θ).

If all crystal planes of inter planar distance’d’ reflect at the same Bragg angle(θ) all reflections from a family lie

on the same cone. Hence many cones of reflections are emitted by different sets of planes.

The recorded lines from any cone are a pair of arc’s that form path of circles as shown in figure

(b) and figure(c).

The distance between any two corresponding arcs on the film indicated by‘s’ In this case of

cylindrical camera cone angle proportional to ‘S’.

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Cone angle α s

4θ=S/R

θ =S/4R

Where R is the radius of camera.

If S1, S2, S3 -------------- Sn are the distance between symmetrical lines on the film then θ1=S1/4R,

θ2 = S2/4R --------------θn= Sn /4R

Using these values of θn in Braggs equation that is

2dhkl sinθn =nλ

The inner planar distance dhkl can be calculated.

ULTRASONICS:

Production of Ultrasonics by piezoelectric method:

Piezoelectric effect :The application of mechanical stress to a pair of opposite faces of a dielectric crystal cut

with its faces perpendicular to the optic axis, results in the accumulation of equal and opposite charges on the

other pair of opposite faces. This phenomenon of generation of voltage under mechanical stress is called as

piezoelectric effect.

Piezoelectric oscillator:

When an alternating voltage is applied to one pair of opposite faces of a quartz crystal, alternative

mechanical contractions and expansions are produced on the other pair of opposite faces of the

crystal. The crystal is thus, set into mechanical vibrations.

When the frequency of the applied AC voltage is set equal to the natural frequency of the crystal,

then resonance will occur. Vibrations of larger amplitude of sound are produced. If the frequency of

the AC voltage is in the ultrasonic range, the crystal would produce ultrasonic waves.

Construction

The circuit diagram of tuned base oscillator circuit is shown below. A slice of quartz crystal is

placed between the two metal plates A and B. The coil L1 of oscillator circuit is taken from

primary of transformer and coil L2 from secondary of the transformer.

The plates A and B are coupled to the electronic oscillator through secondary (L3) of the

transformer. The coil L2 and variable capacitor C from tank circuit of the oscillator.

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Working:

When the battery is switched on, the electronic oscillator produces alternating voltages with

frequency f = 1 / 2π√L2C

An oscillatory e.m.f is induced in the coil L3 due to transformer action.The crystal is now under high-

frequency alternating voltage. The crystal expands and contracts alternately due to inverse piezo-

electric effect. The crystal is thus set into mechnical vibrations.

The capacitance of the variable condenser C is adjusted such that the frequency of the applied

AC voltage is equal to the natural frequency of the crystal. Now the crystal vibrates with large

amplitude due to resonance. Thus, high-power ultrasonic waves are produced from the crystal.

Suppose, there is an X-cut crystal plate of length L, then the frequency of its length-wise

vibration is

The frequency of vibration of the crystal is

E

l

pn

2= (Length mode)

The frequency of vibration of the crystal is

E

t

pn

2= (Thickness mode)

Where, p is the pitch of the note ie., p =1,2,3…, E is Young’s modulus of the crystal and ρ is

the density of the crystal plate

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Merits

1. More efficient than magnetostriction oscillator.

2. Frequencies as high as 5 x 108 Hz

3. Not affected by temperature and humidity.

Demerits:

1. Cost is very high.

2. Cutting and shaping is very difficult

PROPERTIES OF ULTRASONIC WAVES:

(1) Ultrasonic waves are having frequencies higher than 20 KHz and hence they are highly energetic

and their wavelengths are small.

(2) Due to their small wavelengths, the diffraction is negligible. Hence, they can be transmitted over a

long distances without any appreciable loss of energy.

(3) When they are passing through a medium, at discontinuities, they are partially reflected and this

property is used in Non-Destructive Technique (NDT).

(4) When the ultrasonic wave is absorbed by a medium, it generates heat.

(5) They are able to drill and cut thin metals.

(6) At room temperature, ultrasonic welding is possible.

(7) They mix molten metals of widely different densities to produce alloys of uniform composition.

(8) Using ultrasonic wave, acoustic grating can be formed in a liquid.

DETECTION OF ULTRASONIC WAVES: Ultrasonic waves propagated through a medium can be detected in a number of ways. Some of the methods

employed are as follows:

(1) Kundt’s tube method:

Ultrasonic waves can be detected with the help of Kundt’s tube. At the nodes, lycopodium powder collects in

the form of heaps. The average distance between two adjacent heaps is equal to half the wavelength. This

method cannot be used if the wavelength of ultrasonic waves is very small i.e., less than few mm. In the case of

a liquid medium, instead of lycopodium powder, powdered coke is used to detect the position of nodes.

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(2) Sensitive flame method:

A narrow sensitive flame is moved along the medium. At the positions of antinodes, the flame is steady. At the

positions of nodes, the flame flickers because there is a change in pressure. In this way, positions of nodes and

antinodes can be found out in the medium. The average distance between the two adjacent nodes is equal to

half the wavelength. If the value of the frequency of ultrasonic wave is known, the velocity of ultrasonic wave

propagated through the medium can be calculated.

(3) Thermal detectors:

This is the most commonly used method of detection of ultrasonic waves. In this method, a fine platinum wire

is used. This wire is moved through the medium. At the position of nodes, due to alternate compressions ad

rarefactions, adiabatic changes in temperature takes place. The resistance of the platinum wire changes with

respect to time. This can be detected with the help of Callendar and Garrifith’s bridge arrangement. At the

position of the antinodes, the temperature remains constant. This will be indicated by the undisturbed

balanced position of the bridge.

(4) Quartz crystal method:

This method is based on the principle of Piezo-electric effect. When one pair of the opposite faces of a quartz

crystal is exposed to the ultrasonic waves, the other pairs of opposite faces developed opposite charges. These

charges are amplified and detected using an electronic circuit.

(5) Piezoelectric Detector: Piezoelectric effect can also be used to detect ultrasonics. If ultrasonics comprising of compressions and

rarefactions are allowed to fall upon a quartz crystal a certain potential difference is developed across the faces

which after amplification by a value amplifier can be used to detect ultrasonics.

Applications in non-destructive testing:

NON-DESTRUCTIVE TESTING (NDT)

It is a method of testing the material without causing any destruction to the material.

Whenever there is change in medium, then the ultrasonic waves will be reflected. This is

the principle used in ultrasonic flaw detector. Thus, from the intensity of the reflected echoes,

the flaws are detected without destroying the material.

A metal slab is the specimen which has to be detected for any flaws. If there is a minute

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crack inside the specimen, the ultrasonic waves falling on it gets reflected.

And on further studying the reflected signals, the crack can be easily detected.

In this manner, the ultrasonics are useful in Non-destructive testing.

APPLICATIONS OF ULTRASONICS IN MEDICAL FIELD:

1. Ultrasonics is used to release the contained enzymes.

2. Meniere’s disease produces hearing loss in the ear and it can be cured by ultrasonic

exposure in the ear through the destruction of diseased tissue in the middle ear.

3. Proliferation: It is used to increase the fibroblasts by stimulation, thereby producing

myofibroblasts.

4. Remodeling: It is used to decrease the strength of the scar by affecting the direction of

elastic fibers which are responsible for the scar.

5. Reversible blood cell stasis: While treating the patient, it is used to gather the blood in

columns separated by plasma.

6. Ultrasonics is also used in cardiology ultrasonic imaging techniques.

ADVANTAGES OF ULTRASONICS OVER OTHER METHODS

1. There is no mutation(or) residual effects

2. There is no ionization

3. The side effects are almost zero

4. By means of controlled focusing the normal tissues situated around diseased

tissues can be saved.

5. Here the physiological effects depend on frequency and amplitude of ultrasonics.

6. It doesn’t affect the fetus in the mother’s womb during diagnosis.

7. It is non-invasive compared to X-ray and laser radiations.

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Important Questions:

1. Describe the seven crystal systems in detail with diagrams.

2. What are Bravais lattice? Describe the 3-dimensional Bravais lattices in combination with

Crystal system?

3. Show that FCC is the most closely packed out of the three cubic structures.

4. Derive an expression for lattice constant for a cubic crystal.

5. What are Miller Indices? How they are obtained?

6. Derive an expression for the inter planar spacing in the case of cubic structure.

7. State and explain Bragg’s law. Describe with suitable diagram , the powder method for determination

of crystal structure.

8. What are ultrasonics? Explain the properties of ultrasonics.

9. Explain the production (or)generation of ultrasonics by piezoelectric method

10. Explain, how the ultrasonics can be detected.

11. What is non-destructive testing? Explain ,how the ultrasonics are used in non- destructive testing.

12. Mention the applications of ultrasonics in non-destructive testing.