Applied Physics - I

CHAPTER Crystal Structures
NO.

1

Q. 1. Explain the term lattice parameters of cubic crystal. [Dec’12(3m)]
Ans.:

i. The smallest cell, which when repeated throughout space is called primitive cell or unit cell.

ii. A primitive cell is specified by axes a, b and c called primitive axes.

iii. The magnitude a, b and c are called lattice parameters
iv. In the adjacent figure, the parallelogram ABCD formed by primitive vectors a̅ and b̅, forms a primitive

cell which is called unit cell.

v. The choice of the unit cell is not unique. It may be ABCD or EFGH as

shown in the figure.

vi. In 3-dimensional case, the unit cell is a parallelepiped formed by the
basis vectors a̅, ̅b and c̅ and angles α, β and γ. Together they are called

lattice parameters.

Q. 2. Draw the unit cell of HCP. What are its co-ordination number, atomic radius, and effective

number of atoms per unit cell? Also calculate its packing factor. [Dec’12(7m)]

Ans.:

i. Co-ordination number

1.Each atom in the structure is positioned in the empty space 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.

2.All these twelve atoms are in contact with the atom under consideration.

3.Hence Co-ordination number = 12

ii. Atomic radius a
2
1. The atoms are in contact along the edges of hexagon. i.e. 2r = a or r =

iii. Effective number of atoms per unit cell

1. Each corner atom of hexagonal phase is shared by six unit cells,

contributing 1/6 of its mass and volume to one unit cell. There are two

such hexagonal faces.

Hence corner atoms contribute = 2 × 6 × 1 = 2
6

2. There is an atom at the centre of the each hexagonal face which is shared by two adjacent cells.

There are two such hexagonal faces.

Hence face centered atoms contribute = 2× 1 × 1 =1
2

3. Three atoms are forming a triangle in the middle layer are there in the body of cell and they are not

shared by adjacent cells.

Hence contribution of such atoms to the unit cell = 3 × 1 =3

Total number of atoms per unit cell in HCP = 2 + 1 + 3 = 6

iv. Atomic packing factor

Atomic packing factor = total volume of atoms = n × 4 π r3 …eq (1)
3

total volume of unit cell volume of unit cell

Where, n = number of atoms in the unit cell = 6

r = a and c = √8/3
2 a

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2 Applied Physcics - I

Base of HCP is hexagon i.e. area of the hexagon is 6 times the area of

equilateral triangle

∴ Area of hexagonal base = 6 × Area of equilateral triangle

= 6 × 1 × base × height = 6 × 1 ×a ×asin600
2 2

= 3a2sin600 = 3a2 (√3 ) =3√3 a2

22

∴ volume of HCP unit cell = Area of hexagonal base × height

= 3√3 a2 × c
2

∴ V = 3√3 a3 ( c) but we know that c/a = √8/3

2a

∴ V = 3√3 a3 (√8/3 ) = 3√2 a3
2
4 4
APF = 6 × 3 π r3 (from eq(1)) = 6 × 3 π r3 (substituting a = 2r)

3√2 a3 3√2 (2r)(2r)(2r)
APF = π = 0.74

3√2

Q. 3. With neat diagram of a unit cell, explain the structure of

BaTiO3. [Dec’12(5m)]

Ans.:

i. BaTiO3 is used in the application of Piezo-electricity which

exhibits induced electric polarization on application of

mechanical stress.

ii. The BaTiO3 structure ia a tetragonal type in which a = b ≠ c

and α =β = γ = 900.

iii. In tetragonal structure, Ba+2 ions are at each corner of unit cell,

O2- ions are near the centre of face and Ti+4 ion.

∴ Number of atoms per unit cell = 8 + 6 + 1 = 5

821

iv. BaTiO3 has its centre of gravity for position and negative ions non coincidental hence a permanent

ionic dipole moment is seen here. This makes BaTiO3 a popular ferroelectric crystal.

Q. 4. Draw unit cells showing position of the atoms for: [May 13(5m)]

i. A monoatomic BCC crystal.

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Crystal Structures 3

ii. A monoatomic SC crystal.

iii. CsCl crystal.

Q. 5. Define the terms: space lattice, unit cell and lattice parameter. [Dec’13(3m)]

Ans.:

i. Space lattice is a regular, periodic, repeated three dimensional array of points. Space lattice can also

be defined as the set set of those points selected in space such that irrespective of the reference,

neighbourhood or environment remains the same. It was introduce by Bravais hence also known as

Bravais lattice.

ii. The smallest cell, which when repeated throughout space is called primitive cell or unit cell.
iii. In 3-dimensional case, the unit cell is a parallelepiped formed by the basis vectors a̅,̅b and c̅ and

angles α, β and γ. Together they are called lattice parameters.

Q. 6. Explain Diamond crystal structure with proper diagram and determine its APF. ( 7M, May 2014 )
Ans.:
i. The diamond lattice can be described as being built up from two identical interpenetrating FCC sub
lattice one of which is displaced from the other ¼ of the length along the body diagonal.
ii. The atomic basis is two atoms per lattice point, i.e. each lattice point corresponds to two identical
atoms, one located at (0,0,0) and other at ( ¼ , ¼ , ¼ ).

Fig. (b)

Fig. (a)
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4 Applied Physcics - I

iii. As shown in the above figure, the 4 corner atoms are at diagonally opposite points in the respective

planes with the arrangement, two carbon atoms one at the centre of the tetrahedron and the other at

the corner, form the basis.
iv. The lattice constants a = 3.5 Ȧ . So, in a diamond structure eight atoms are at eight corners, six face

centered atoms at the centers of each six faces and four atoms are positioned on the way along the

body diagonal inside the unit cell, in such a way that each of them is linked tetrahedrally with one

corner and three face centered atoms.

 Number of atoms / unit cell = 8
11

(onside 4 , 2 x 6 on faces and 8 x 8 at corners)
1

The length of body diagonal = √3 a and 4 of body diagonal
= 2r

√3 a
4 = 2r

APF ( Atomic packing fraction ) √3 a
r= 8

(Number of atoms⁄unit cell ) x ( volume of one atom )
= Volume of unit cell
a)3
8x 4 π r3 8 x 4 π (√38
3 3
= = a3 = 0.34
a3

 APF for diamond structure = 0.34

Q. 7. Derive Bragg’s law. Explain why X-rays and not Y-rays are used for crystal structures analysis.

What data about the crystal structure can be obtained from the X-ray diffraction pattern of a

crystal. [( 7M, May 2013, May 14 )]

Ans.:

Fig. : Bragg’s Law
Bragg planes with atoms shown with dots

i. An ordered or regular arrangement of atoms has been depicted. Let the inter-planar spacing be d. A
monochromatic and parallel beam of x-rays at glancing angle  is made incident on planes. Ray AB
will scattered at point B on the first plane. Rays DE and GH which are parallel to AB will also
experience scattering at points E and H respectively at second and third plan. The scattering due to
atoms on crystal plane is in all directions.

ii. Among the scattered rays select rays BC and EF which are parallel to each other. It is assumed that
they have path difference  = n and produce constructive interference. Bragg's law provides the
condition at which  = n. Lets obtain value of path difference  .

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Crystal Structures 5

iii. Draw perpendiculars BP and BQ to the rays DE and EF. Now one can say that up to BP, path covered
by both the incident rays is the same. So as BQ onwards parallel rays BC and EF covers the same
distance
 Path difference between rays 1 and 2 is
 = PE + EQ
From  BPE and  BQE, PE = BE sin  and EQ = BE sin 
 = BE sin  + BE sin  = 2 BE sin 
= 2 d sin  (BE = d)
As we have already assumed that constructive interference is taking place  = n
 n = 2d sin 
The logic can be extended for rays 2 and 3 on the Fig. above. Hence if rays 1 and 2
gives constructive interference and rays 2 and 3 also give constructive interference, then rays 1 and 3
will also provide the same. In this case path difference between rays 1 and 3 will be
' = 4 d sin  = 2 (2 d sin ) = 2 ()
i.e. integer multiple of  .

Q. 8. Reason why X-ray and not Y-ray for crystal structure analysis.
X-ray Diffraction :
Diffraction is defined as bending of ray when waves passing through an object whose dimensions are
of the order of their own wavelength. The explanation needs support of interference also. Secondary
wavelets originate at the location of the object and spread in all directions. The constructive and
destructive interference of such waves result in increase and decrease of intensity at the corresponding
regions which is known as diffraction pattern. If the diffracting objects are located in random fashion
the superposition of individual diffraction effects due to various objects leads to a pattern which will
not have any particular distribution of intensity where as if the diffracting objects are distributed on a
regular pattern, then the diffraction pattern will also have regularity.
Essential aspect of the diffraction is availability of objects whose dimensions are extremely small, i.e.
of the order of the wavelength of incident on it. In crystals we have seen that atoms are arranged in a
perfectly ordered manner. Also the dimensions of atoms are 108 cm which is nearly of the same order
of X-ray wavelength. Hence, when X-rays are made incident upon crystals, we get an ordered, regular
diffraction pattern or one can say that “ Crystal act as three dimensional reflection grating with X-
rays”.

Q. 9. Explain with a neat diagram construction of Bragg’s X-ray Spectrometer. Write the procedure
to determine the crystal structure.

Ans.:

Construction:- Fig. : Bragg’s X-Ray Spectrometer

i. Bragg's Spectrometer consists of a collimator containing two slits S1 and S2 made up of lead, through

which X-ray is passed.

ii. A turn table is situated in-front of the collimator on which crystal is placed.

iii. Ionization chamber collects the reflected X-ray.

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6 Applied Physcics - I

Procedure:-
i. A fine beam of a monochromatic X-ray is made to fall on the crystal.
ii. The crystal reflects the X-rays which are collected by the ionization chamber.
iii. Turn table is rotated till a sharp increase in the intensity is detected.
iv. The sudden increase in intensity suggests that Bragg's Law is satisfied at the given angle .
v. Then the inter-planar spacing can be determined by using Bragg's Law.

n = 2d. sin 
vi. The peak in ionization current occurs more than once for different values of “ n ”.

Q. 10. What is X-ray? Why X-rays are preferred to study crystalline solid?
Ans.: X-ray is a form of electromagnetic radiation with very high frequency and energy. X-rays lie
between ultraviolet radiation and gamma radiation on the electromagnetic spectrum.
( For next part refer Q. 8 )

Q. 11. Describe powder metallurgy to determine crystal structure of powdered specimen.[( 8M , May 14)]
Ans.:
i. In the previous two methods i.e. Laue's method and Rotating crystal method, the limitations are
requirement of single crystal of reasonably larger and moderate size respectively. It is not
necessary that always we get a specimen in a single crystal form.
For such specimens powder crystal method is very useful. Refer Fig. (a)

Narrow Specimen 2 = 0
x – ray beam x – ray film Fig. (b) Typical Diffraction pattern

Fig. (a) Powder Method

ii. In this case, a specimen in finely powdered form is taken in a thin walled capillary tube. Since we are

using powder, specimen is made up of crystallites (tiny crystals) which are oriented randomly.

iii. When narrow beam of monochromatic x-rays enter the set up, it will come across crystallites which

are randomly oriented but few of them are bound to lie with their planes at glancing angle  so that

Bragg's condition is satisfied.

iv. Since all orientations are equally likely, the diffracted rays will form a cone such that its axis is along

the direction of incident beam and semi vertical angle 2 for a particular set of planes. The cone will

intersect cylindrically arranged photographic film of radius R.

v. The film is cut from the point at which X-ray enters and made straight which appears as shown in Fig.

(b). We find arcs on it where base intersect the film. (For higher orders, we get arc at 4, 8 .... etc

which has not been shown on diagram).

vi. If the cone leaves a diameter L on photographic film Fig. (b) and radius of cylindrical drum is R then

L 4θ ( Where D = 2R , diameter )
πD = 360°

90° L
∴ θ = πD

For various orders we get 2 , 3 , ..........

correspondingly L2 , L3 , .......... etc.

Using  one can find inter-planar spacing d and other unit cell parameters.

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Crystal Structures 7

Q. 12. On what sense real crystal differ from Ideal crystals? Explain the point defects in crystals.
[Dec. 12(5m)]

Ans.:
i. We have studied several crystal structures viz. SC, BCC, FCC, Diamond, Nacl, BaTiO3 and HCP
assuming that the crystal were ideal or perfect for the sake of simplicity and understanding.
ii. However, in practice, there are many defects or imperfections present in the practical or actual
crystals as such they are called as real or imperfect crystals.

iii. There are different types of point defects in the real crystals.

Point defects:

1. Vacancy:
Vacancy is produced due to the removal of an atom from its regular position in the lattice.

Fig. Vacancy
The removed atom does not vanish. It travels to the surface of the material. For low concentration
of vacancies, a relation is

n = N−/

Where n = Number of vacancies
N = Total number of atoms
T = Temperature in (K)
EV = Average energy required to create a vacancy

2. Interstitial:
An extra atom of the same type is fitted into the void between the
regularly occupied sites.
Since in general the size of atom is larger than the void into which it
is fitted, so the energy required for interstitial formation is higher than
that of vacancy formation.

3. Substitutional impurities:
In this, a foreign atom is found occupying a regular site in a crystal
lattice.

Q. 13. Explain the difference between three different liquid crystal phases w.r.t. the order in the
arrangement of molecules, with the help of diagram.

Ans.:
Types of liquid crystals:
a) Smectic or soap like liquid crystals
b) Nematic or thread like liquid crystals
c) Cholesteric liquid crystals

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8 Applied Physcics - I
(a) Smectic or soap like liquid crystals:
 Crystalline state, the orientation and periodicity - all these characteristics are
retained.
 On heating the crystal lose periodicity within the plane, but retain the orientation and arrangement in
equispaced planes.

Z

Fig. Smectic
Phase
(b) Nematic or thread like liquid crystals:
 These crystals on heating lose their planer or layer structure but retain a parallel alignment. Thus they
retain orientation but lose periodicity.
 Hence molecules lie parallel to each other but can move up and down or sideways or rotate along the
axes.

(c) Cholesteric liquid crystal :
 These liquid crystals have the same arrangement of molecules as in Nematic type but their optical

activity is many times higher than that of its solid crystalline variety.
 The cholesteric phase of liquid crystals has molecules parallel to each but the direction of alignment

twists gradually and results in a helical structure. Therefore the substance consists of parallel layers.
Molecules are aligned parallel in each layer. The helical structure of cholesteric substance is
responsible for optical activity(optical rotation).

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Crystal Structures 9

Q. 14. What is mesomorphic state of matter? Explain with neat diagram chloresteric phase?
Ans.:
i. In some of the substances, the tendency towards an ordered arrangement of molecules is so
great that crystalline solid does not melt directly in to liquid state but passes first through an
intermediate change that is called liquid crystal ( or mesomorphic phase ) before changing to
liquid state on further application of heat.

Solid Liquid Liquid
State heat Crystal heat State

ii. Liquid crystal is one of the homogeneous phase with properties distinct from either traditional
state. The liquid crystals show some of the properties of solid state and some of the properties
of liquid state.

iii. The temperature at the transition point, provides enough energy to disrupt the binding between
some of the molecules but the energy is not sufficient to break the strong lateral force of
attraction between the long molecules.

iv. The important property in liquid crystals for practical application is that the arrangement of
molecules can be upset by very slight changes in their surrounding i.e. molecules in liquid
crystal rearrange themselves when a small electric field is applied and the change form an
isotropic to more isotropic arrangement changes the way the crystal absorbs light.

v. A small electric field can disturb the alignment of molecules while the large electric field
induces turbulence in the liquid crystal with light scattering.
( Chlosteric phase – refer Q. 13 )

Q. 15. What is liquid crystal state of matter? Draw the diagram to describe molecular arrangement in

their different phase. [(5M , May 14)]

( Refer question 13 and 14 )

Q. 16. Define ligancy and critical radius ratio in case of ionic solid. Write the condition stability of

ionic crystals in 3–D? Determine critical radius ratio for ligancy 6. [(7M, Dec.13)]

Ans.:
Ligancy: In ionic crystals coordination number is defined as “The number of nearest anions surrounding a

central cation.” This is also called the Ligancy. The factor which controls the Ligancy is ionic packing or

packing geometry.

Critical radius ratio:
The ratio of cation radius to anion radius ( rc ) measured is a useful quantity for the limit of stability. It is

ra

called critical radius ratio.

Conditions for stability of ionic crystal in 3–D:
For this particular configuration (Ligancy 3), it is 0.155, which can be calculated from the simple

geometry.

Coordination number Critical radius ratio ( ) Configuration
or Linear

Ligancy 0.0 – 0.155

2

3 0.155 – 0.225 Triangular

4 0.225 – 0.414 Tetrahedral

6 0.414 – 0.732 Octahedral

8 0.732 – 1.0 Cubic

12 1 and above FCC – HCP

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10 Applied Physcics - I

Critical radius for ligancy – 6:
In the said arrangement a cation is squeezed into 4 anions in a plane as shown in Fig. and 5th anion is in

upper layer and 6th in bottom layer (not included in Fig.)

Joining cation anion centres E and B and we complete the  EBF.

In EBF , m F = 90 and EF = BF

 m  B = m  E = 45

and EB = ra + rc and BF = ra
 cos B = cos 45 = ra = 0.7071

rc+ra

 rc = 1−0.7071 = 0.414

ra 0.7071

 the critical radius ratio for ligancy 6 is 0.414
For ligancy 6 , the range of rc is 0.414 to 0.732 . At rc = 0.732 we get

ra ra

ligancy 8 . Fig. Ionic arrangement for Ligancy 6

 Maximum radius ratio for ligancy 6 is given by ,
rc
ra < 0.732

∴ rc < 0.732 2.02

∴ rmax < 1.4786 Å

Numerical :

Q. 1. Find the interplanar spacing between the family of planes ( 1,1,1) in a crystal of lattice
constant 3Ǻ. ( 3M, Dec. 13 )

Ans.:

Given : a = 3 Å

( h, k, l ) = ( 1, 1, 1)

By formula a

dhkl = √h2 + k2 + l2
33
d111 = √12 + 12 + 12 = = √3 Å
√3

Q. 2. Represent [ ̅ ] , ( 0 0 2 ) , [ 1 2 1 ] in a cubic unit cell.
(1) [ ̅ ]
(2) ( 0 0 2 )

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Crystal Structures 11

(3) [ 1 2 1 ] : [ 1 2 1 ] divide it by 2 ( , , ) draw a point ( , , )


⃗⃗⃗⃗⃗⃗ = [ ]

Q. 3. Find the miller indices of a set of parallel planes which makes intercept in the ratio 3a : 4b
on the x and y axes and parallel to z – axis. ( 3M, May 14 )

Ans.:

Intercept of the plane are in the proportion 3a : 4b : ( plane is parallel to z-axis)

as a, b and c are basic vectors, the proportion of intercepts 3 : 4 :
11 1 1 1

 reciprocal 3 4 ∞ = 3 , 4 , 0
Taking LCM and Converting to the integers 4, 3, 0

 Miller Indices ( 4 3 0 )

Q. 4. Calculate the glancing angle on the plane ( 1 0 0 ) for a crystal of rock salt ( = . Å ).

Consider the case of 2nd order maximum and = . Å . [(Dec. 12)]

Ans.:

(i) d for rock salt is
a

d = 2 = 1.0625 Å
(ii) Using Bragg’s law for n = 2

n = 2d sin 

 2 x 0.592 x 1010 = 2 x 1.0625 x 1010 x sin 

sin  = 0.557

  = 33.86

Q. 5. Calculate the maximum order of diffraction if x-ray of wavelength . Å in incident on a

crystal of lattice spacing 0.282 nm . [(Dec. 12)]

Ans.:

Given λ = 0.819 Å , d = 0.282 nm

Formulae: n = 2d sin 

For the given values of  , d and n

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12 Applied Physcics - I

For n to be maximum , sin  = 1
2d 2 x 0.282 x 10−9

∴ n = λ = 0.819 x 10−10 = 6.88
 as order cannot be a fraction and n is maximum.

n=6

Q. 6. If the x-ray of wavelength . Å will be reflected from crystal having spacing of . Å .

Calculate the smallest glancing angle and highest order of reflection that can be observed.( 5M,

May 14 )

Ans.:

Bragg’s law is n = 2d sin 

(i) For smaller glancing angle, n = 1

∴ θ = sin−1 ( λ ) = 10.488°
2d

(ii) For highest order, we know that always sin   1

 Find n which satisfies this condition using relation

nλ sin θ = λ = 0.18
sin θ = 2d
For n = 1 2d

For n = 2 sin θ = 0.36

For n = 3 sin θ = 0.54

For n = 4 sin θ = 0.72

For n = 5 sin θ = 0.9

For n = 6 sin θ = 1.08 ≫ 1

As sin  cannot be greater than 1 , the highest order possible is n = 5

 other way possible to solve this is as take maximum sin  = 1

 n  = 2d

2d 2 x 4.255
 n = λ = 1.549 = 5.49
Discard fraction part to get the answer n = 5

Q. 7. Calculate the number of atoms per unit cell of a metal having the lattice parameter 2.9 A0 and
density is 7.87gm/cm3. Atomic weight of a metal is 55.85. (NA = 6.2038 × 1023 /gm.mole )
[Dec’12(5m)]

Ans.:

Given : a = 2.9 A0 = 2.9 × 10-8 cm, ƍ =7.87 gm/cm3, A = 55.85, NA=6.023×1023/gm-mole

Solution : a3ƍ = nA ∴ n = Na3ƍ
N A
= 6.023 × 1023 ×(2.9 ×10-8)3 × 7.87 = 2, hence it is BCC.

55.85

Q. 8. An elemental crystal has a density of 8570 kg/m3, packing fraction is 0.68. determine the mass
of one atom if the nearest neighbor distance is 2.86 A0. May 13(7m)

Packing fraction = 0.68 ………. ∴ It is a BCC structure. i.e. n = 2

Also nearest atoms distance in BCC = √3 × a
2

∴ 2.86 = √3 × a

2
∴ a = 3.302 A0

Now using the formula ….. a3ƍ = nA a3 ƍ = Mass of unit cell = mass of one atom

N

n Number of atoms/unit cell

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Crystal Structures 13

∴ mass of one atom = (3.303 × 10-10)3 × 8570 = 1.542 × 10-25 kg
2

Q. 9. Find out the critical radius ratio of an ionic crystal in ligancy 6 configuration. What is the
maximum size of cation in ligancy 6 configuration when the radius of anion is 2.02 A0. May

13(5m)

In the said arrangement a cation is squeezed into 4 anions in a plane as shown in figure and 5th anion is

in upper layer and 6th in bottom layer.

Join cation anion centres E and B and complete the ∆EBF, ∠F = 900 and EF = BF
In ∆EBF, ∠B = ∠E = 450

And EB = ra + rc and BF = ra
ra
∴ cos B = cos 450 = rc+ra = 0.7071

∴ rc = 1−0.7071 = 0.414
ra 1.7071

∴ the critical radius ratio for ligancy 6 is 0.414.

Now ra = 2.02 A0 rc = ? rc
ra
We know that…………… rc/ra is 0.414 to 0.732 i.e. at = 0.732 we get ligancy 8

∴ rc < 0.732
ra

∴ rc < 0.732
2.02

∴ rc < 0.732 × 2.02

∴ rc = 1.4786 A0

Q. 10. Identify the crystal structure if its intensity is 9.6 × 102 kg/m3 lattice constant is 4.3 A0 and

atomic weight 23. Dec’13(5)

 Given : ƍ = 9.6 × 102 kg/m3 , a = 4.3 A0 , A = 23

 Solution : a3ƍ = nA
N

∴ n = a3 ƍN = (4.3 × 10-10)3 × 9.6 × 102 × 6.023 × 1026

A 23

Since the number of atoms per unit cell is 2, it is BCC structure.
Q. 11. Find out the intercepts made by the planes (1 0 1) and (4 1 4) in a cubic unit cell. Draw [1 ̅ 1]

and [1 2 4] in a cubic unit cell. May 13(5m)

Let basic vectors be a,b,c and intercepts m,n,p respectively.

On expressing m,n and p in terms of fractional multiples of a,b,c
mnp
a ,b,c

As miller indices are reciprocal of these fractions

(h k l) :: ( a , b , c) ∴ for (1 0 1) : a =1, b=0, c=1
mnp mnp

∴ the intercepts are the reciprocals of 1,0 and 1

i.e. m= 1 ; n = ∞ ; p = 1

similarly for (4 1 4)

the intercepts will be reciprocals

m = ¼ ; n = 1; p = ¼

To draw [1 2 1] ; Draw points [ ½ 1 ½ ] ∴ O⃗⃗⃗⃗P⃗ = [1 2 1]

[1 2 4] ; Draw points [ ¼ ½ 1] ∴ ⃗O⃗⃗⃗Q⃗ =[1 2 4]

 All the best !!!

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