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Chapter 19
Solid State
19.1 CSIR NET PREVIOUS YEARE
A. Crystal Structure XAM Q.UESTIONS

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Answer to Question no. 2.

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In a fac:e-,centered cubi c (FCC) structurie,
1

t:he numb er ,of atoms per unit ceU can be

1

determined:b,y co,nsidering the

,contribiuti,0 n of atoms at differe,nt
1

p,ositions in the unit cell:

• Th:,ere are 8 corin:er atoms, each

contributing of an atom to the unit
celL

• There are 6 fa ce--c.enter e,d atorn.s,

1 1

each contributing of an atom to the

unit
,ceU.

Th,e total numb ,er of ato1ms in th ,eunit ceU

1

is.:·

Total atoms - 8 x +6x

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Total atoms 1 + 3
Total .atoms =--- 4

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 • T'here are ,8oc rner atomsJ each
contributing ! of an atom to the unit
,celL

• Th,ere are 6 face...centered atoms,,each

contributing of an atom to the unit
,celL

Th,e total n,u1mber of atoms in the unit c:ell is:

Total atoms = 8 x + 6 x
Total atoms == 1 + 3
Total atoms - 4

So, the n umber of atoms in an element

1

packe,d in an FCC structure is:

(3) 4

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Answer to Question no. 4:

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Answer to,Questii,on no..6: (3) 25

Explanation:

To,so!lve this pro,blem we c,an use the

1
,

fo,Uowing steps:

1. CaJculate tihe volume of the !Un t cell: V ,=

(50i0 pm)"3 ;--125 x 1QiA(-30) mA3
2. .c,alculate the mass of the unit cell::mi
= density x vo,lume - 1.3,3 Qi/cc x.125x
1QiA(-3 Q) ffiA3 = 1.662.5 X 1Qi"(-26) Q
3. Since the fee unit ceU co,ntains 4 ato,ms
(on ,e at each corner a1nd one at the c,enter
of each fa• c,e), the-mass of one ato,m
is:: m_.atom = m I 4 = 1.6625 x 1i 0A(-26)1

Qi/ 4
=-4.,15625 X 1QA(-2,7) g;
4. . FinaUy, we caIn callcuiiate thie molar :mas:s
(M) usiingrth1e ,atomi;c mass unit (amu):
M = m_atom x (6.022 x 1QA23am1u1/g) /
(4.1562.5 x 1 0,\(-27) g) 25,g/:mol
1

r·herefore, tihe moli.ar ma.s.s of th!e meta is

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appro,ximately 25 g/moL 9:31 am

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An,swer to Questiion no 8: (2) 2VBCC : VFCC
1
1
••

E.xplanation:
1

Let's co nsider the foUowing:

1

• iFCC (face-ce·ntered cubic) structure has

4
• atoms
BCC per uni:t
-c ceU
(body ,entered
t ·Cellicub.ic) structu!re h:as
2 atoms
The massperOf unii . . ... .-. ·· - -- -
1

• both1
. • each:atom1 .is t.he sa.me for
structures
S:ince th,eden.sity is de·f1:ne•d as mass per unit
v·olume, we can wriite:

Density= Mass/ Volume

For the F'CC structutre:

Density_F·cc = (4 * ato:mic mass)/ VFCC

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For the B,CC structure:
Density_BCC (2 * atomic mass)/ VBCC

Now,,we c,an f1:nd the ratio,of the densities:

Density_FCC / Density_BCC = ((4 *atomic

mass)/ VFCC) / ((2 * atomic mass)/
VBCC)
SimpUfying the expression, we get::

Density_. FCC/ Density_. !BCC == 2VBCC / VFCC

1

Th erefore, the ratio of the densit:ie,sof the

1

FCC and BC C modiftcat:ions is '2VBCC: VFCC.

1

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Ans ,wier t,oQuestion n10. 10: (2) (0, 0, 0) and 1

(1,/2,,, /2, 1/2)

Ex:planation:

In a body-centered cubic (BCC) unit cell,

there
are two,atomis:

. One m ,at e corner .of the 1be,

ato,
represented cu
tih1by the coordinates (0, 0,
. 0)
Q,ine atom at the body ,nter of the c1ub,e , 1

r ce
epresented by the coordinates (1/2,
1/2)

These coordii, nates reflect the , fact th,at the

body-centered atom is located at the center
of cube, halfway along each axis (x, y, and
the
z)
from i the origin (0, 0, ,. 9:42 am
0)

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Answer to qu,esti1on no. 12: (3)

When Frenkel defects:are created in an

otherwise perfect ionic crystal, the density of
the crystal remains the same@

Here's why:
1

• The mass of the crystal remains the

same, :as no atoms .a,dded or removed..
are of the crystal also
• The volume
remains
the sa!me, as the interstitial ion
a very occupies
small volume and the
vacancy
also does;n't affect the overall u .
sg vol m e
i !nif1cantly.
Since the mass and volume remai.n e
th. same, the d,ity .(mass/volume)
ens
unchanged. remains

T herefore, the correct

1

is (3) remains
answer
the same. 10:24

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Here's the expllanation:

,. A atom1s o,ccup,y au c,orners (8 corners)

and face cejnters (6 fac e cent ers),
1 1

m,aking a total of 4 + 6 - 10 A
atoms,.,However, since each corner A
,atom is sihared by ,g, unit ceUs,and ea,ch 1

face C enter A atom is shared !by 2 unit

1

c,eUs, the total number

of A ,atom,s per unit ceU is (4 x 1/8) + (6
x 1 /2) := 4.
,. B ,atoms occup!y 4 tetrah1edral voids,
which means 4 B atoms per unit celL

S0 the formuta is A484, hence x = 4 and y = 4.

1
,

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Answer to,q, uestion no. 17: (4) PF= 0a74 and
N=4

In a Fac,e-Centered ;Cub,ic (FC C) crystal

1

system::

s 0.74,
which
• Th,e packing factor i nit ceU vo i ume
1

mea{PF)
1ns that 74% of the 1 l is
u
occupied by atoms.
, • There are 4 atomic sites per unit ceU
= 4):: 1 atom at the ,o,f each. (Nt
center :ared between 2 unit o,f the
6 faces
(sh cells) and
1 atom at each cornier (shared between
8 unit ceUs).
Th,erefore, ,correct answer is (4) P'F-:- 0:74 1

andthe
N = 4. 10:32 pm

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Answer to question no. 1;8:

To find the angle between the [2 0 1]

plane and the xy plane in a simple cu bic 1

crystal l'attice, we use the d irectiion

1

cosines of the plane normalls. The angle

lbetw,een the two
planes is given by the dot product :o,f their
normal vectors..

1. Normal V,ectoir of the [2 0 1] Plan e: 1

• The Miller indices [h k l]

represent a plane with a normal

vector (h, k, l).

• For [2 0 1], the normal vector is
(2, 0, 1).

2. Normal Vector o,f the xy Plane:

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• The xy plane has a no rmal vector
1

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In a cubic crystal, the [100] plane is equally
inclined to the planes [010] and [011], since
th ey aH intersect at the origin and form a
1

mutu,aUy perpendicul.ar set of pilanes.

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Answer to question no_ 20:

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th!e correct answer is ind'ee,d (2) three circ!les..
1

In a two-dimensional hexagon!al ciose-

pac,ked (HCP) layer of circles, each void (or
hole)
is surrounded by thir.ee circles..This is
because the HCP str1u·ct urre h,as a he,xagonal
1

,arrange·m:ent of c:ircles, wh!ere each circle

1

is ,,n contact with thre ·e neighboring circles,

form!ing ,a triangular ,arrangement aro•und ea,ch
vo.id.

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Answer to question no. 21: (4) -J3n/8

In a body-centered cubic (BCC) structure, each

unit cell contains one atom at the center and
eight at the corners. Assuming the
atoms
atoms are hard spheres touching each other,
the volume occupied by the atoms in the
unit
cell is:

(1 central atom x (4/3)nrA3) + (8 corner

x (1/8) x (4/3)nr"3) atoms

where r is the radius of the atoms.

Simplifying the expression, we get

V_occupied = (4/3)rrr"3 + (4/3)nr"3

V_occupied = (8/3)nr"3

The total volume of the unit cell

is:
V_unit cell = aA3

where a is the lattice parameter.

The fraction of volume occupied is:

V_occupied/ V_unit cell = (8/3)nrA3 / aA3

Since the atoms are touching, the radius r is

related to lattice parameter a by:
V
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r (v'3/4)a

Substituti:ngrth:is expressio:n for r, ·we get:

V_o,c:cup,ied / V unit cell = (-J3rr/8)

Therefore, the correct answer is (4) v'3n/8.

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Answer to question no. 23.

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'T·he correct answer is (2) face centred cubic 1

In a cubic crystal, the presenc,e,of (111) a1n,d

(222) reflectio ns in,dicates that the crystal
1

has a lattic e pllane with a rep,eating pattern

1

of atoms in the directio1n perpendicu1l,ar to

th e plane. The abs,ence of the (001)
1

reflectio·n sug,g ests th,at there is Ino l.attice

1

plane
perpendicular to the z-axis (or [0 01,d] irect 1

o,n),
w·hich is.a ch.aracteristic of Fa,ce-·,Centeredl
c,uribic (F CC) lattice.
1

In an FCC lattice, the lattice points are located

at th e c:orners and[fa.ce centers,of each
1

u:niit cell, but not at the C ornrers.alon,e,·whii,c'h

1

is the case for Siimpl,e Cubic (SC) tattice.

1

80 dy-·Centered Cubic (BC C) !lattice has

1 1

lattice po ints .at the corners and bo,dy center

1

of each un t ·ceU, b ut n,ot at th1e face

1

centers.
Sid!e--Centered C:ubic {SCC) is not a valid
Brava:is:i,attice.
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Therefore, th e correct
1

iis (2) fac,e

answer
entred c 10:45 pm
,c: ·ubic:

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 Answer to Questro!n no 24:

Th,,e,correct answer is:(4) ,8.

B,e2C3 h:as a structure similar to CaF2,

wh:ich
is a face-centered cubic (FCC) lattice. In this
struct ure, each carbon
1

,atom is surro;un,ded
by eight(C)beryl m {Be) atoms, resu'lting in a
1

liu.
coordination , of
num ber1

8.
In CaF2, each Ca2+ ,n is by i ght
io ions, and ,sim:Lla.rly, surrounded
F- n e2C3, each Ceatom
1

: is
surro
i: ·d by eight Be B, 1s, resulting in a
inatioatom
co ord;unde 1
n num ber of ,.
10:46 pm
8

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LJ.

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Answer to question no. 2.

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Given the,diffracti:o,n patterns prov1 ded: 1

• Crystal A djffracts fro,m (1 1 1) a,nrd (2 O

0) planes but n:ot from (1 1 0).

, • Crystal :8 diffracts fromi (1 1 0) ,an,d (2 0

1
1

0) plan:es but not from (1 1 ·1)

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Now, considering th, ediffraction patterns
1

provide,d:

• Crystal A d'iffracts from (1 1' 1) and (2 O

0) but not from (1 1 0). This is
con,sistent with ,an FCC lattice because
it d1iffracts from (2 0 0) (aU,owed.for
both FCC and BCC) and {1 1 1)
(specific to FCC).

• Crystal B diffracts from {1 1 0) a.nd (2 O

O) :but not fro:m1(1 1 1). Thiis is
• Co,:nsistent with a. 8 CC lattice
1

becau,se it diffracts from (1 1 0)

(specific:to B,C C) and (2 0 0) (a,llowed
1 1

for both FCC and 8,CC).

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Therefore, from the,gi:ven diffraction data,
we can con clude:
1

• Crystal A has, an FCC latti ce..

1

• Crystal B has a BCC Iattice.

1

Hen,ce, the,corre,ct option is:

(1) A has FCC latti,ce while B has BCC

li,attiice.

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Answer to question no. 3.

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Answer to Question no. 4:

The correct answer is (4) shows Bragg

diffraction at sin-1 (sin8/2).

According to Bragg's law, the angle of

diffraction (8) is related to the spacing
(d) and
wavelength (A) by:

nA = 2d sin8

For first-order diffraction (n=1), we have:

A= 2d sin8

Now, if the spacing is doubled to 2d, the

new angle of diffraction (8') can be found
rearranging Bragg's law: by

2(2d) sin8' = A

Substituting A = 2d sin8 from the first

equation, we get:

4d sin8' = 2d sin8

Dividing by 2d, we get

2 sinB' = sin8

8' = sin-1 (sin8/2)

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 Answer to Question no. 5:

TheK+
is (1) correct answer
and Cl- are
isoe;lectronic.

K+ ain1d Cl- have the same number of

electrons (18), making them isoelectronic.
As a result, t:h1ey have s:imiilar scattering
factors for X-rays, which leads to a
sim1ple
:cubic diiffraction p,attern for KCL On the other
hand, Na+ and CII- have different n1umbers o,f
electron,s (1Oiand 1,8, resp,ectively)1resulting
in different scatteriing fa,ctors and an FCC
diffracti•on pattern for NaCL 1o:s9 pn1

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Answer to Question no.. 6,:

The correct ,answer is (4) Amorpho,us.s,olids.

X-ray diffraction (X RD) is a tech.:nique used to

1

d,etermin,e the crystal structure of materials. It

relies on the peri,odic arrangement of atoms in
a crystal·lattic,e, which produc,es a ,diffraction
p,attern. Amorphous solids, however, lack .a
periodic crystal structure, and their atoms are
arranged randomly. As a result, XRD d•oe·s.not
produce a diffraction pattern for amorph,ous
solids, and therefore, does not provide any
structural information.

XRD can provide structural information for

metalHc soUds, ionic solids, and
mol.ecular solids,.which have a crystalUne
structure. 1 • () nm

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Answer to Question no. 7:

The correct answer is (3) 0..82.. 1

To•find the ratiio sine, / si1n82, we can use the

followin!g steps:

1i. Determine the interplanar sp,aci1n,gs (d)

for the [101] and (1 ·11] planes in a 1

simple cubic crystal:

- d,01=: a/../2 (for the [101] plane)

- d1,1 =a/,/3(for the [111] plane)

2. Use Bragg's law to re'late the sin of th,e

1

diffr,action angle (8) to the wa·velength (A)

and interpl .anar spacing (d):
-.sin81 = A/ (2d,:0,)
- siin8 -=A/ (2.d
2 11
, )

3 Divide sine, by sin82 to :get the ratio:

- sin:8 sin8 =(A/ (2d
1
/
2
/(A/ (2d 101
))
111
))

- sin8, / sin8 = (2d 2

(2d, 111
) /
01
)

- sin8, / si n82 = (2a / "3) / (2a / '12)

1

- sin8, / sin:82 = (-J:3 / ..J2)

- sine, / sin 82 ::: 0.8,2
1

Therefore,.the ratiio sin181/ si n82 is

approximately 0.8,2.. 17 ::01 pm
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Answer to Question no. ,8:
0

The correct answer is (4) a = .3..46,4 A.

Using s ,aw ,d the values,

Bragg
1
f anthegiveni
we can calculate uni t cell length (a) as
1

foll,ows:

d nA / (2..sine)

For :e (111) pl!ane in an FCC structure,

th
interpl theis related to ,e unit
,aniar spac:ing {d) 1

cell lengthth(a) by:

d =a/

Substitutling the val:u:es,.we get:

d = (2 x 1 A) / (2 sin
= 1 A/ sjn 30!
30°)1
0

= 1 A; o.s
=2A

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Now, substituted into the !equation for th!e
(111) plane:

·2 A= a; ,/3

Solve for a:

a= 2 Ax
= 2.Ax
-./3 0

1..732
== 3.464 A

There·fore, the unit ceU length of the crystal

0

is approximately 3.464 A, whiich matches

option
(4). 11:02 pm

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Answer to question no. 10.

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Answer to question no. 11.

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To f nd the .atomic radiiu.s !of polonium
given its :density and the mass of a
poloniu1m, atom, we ,can follow these
steps::

1. Calculate the volu,me o,f the un,itcell:

• IPolon;ium is know·n to crystalUze in

a si.mple •Cubic lattice..

• The density (p) is given as 10.00

g/cm3

• Th,e m:olar mas:s.o,f po:lon:ium (

MPolonium) is not directl·y
provided, but we are given
the
mass of a polonium atom
(- -
2
matom == - ·
7-- 1 -
X- - -
0-22
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Answer to question no. 12.

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Introduction
In a simple cubic lattice, the ratio between
the unit cell length and the separation of
two adjacent parallel crystal planes is given
by Miller indices. The Miller indices are a
set of three integers that define a plane in a
crystal lattice. The ratio of the unit cell
length and the separation of two adjacent
parallel crystal planes cannot have a value
of some particular numbers, as discussed
below.

Explanation
The Miller indices for the simple cubic
lattice are (hkl), where h, k, and I are
integers. The separation between two
adjacent parallel crystal planes is given by
the formula d = a/
v'(h"2+k"2+1"2), where a is the unit cell
length. Therefore, the ratio of the unit cell
length to the separation between two
adjacent parallel crystal panes is given by
aid= -J(h/\2+k/\2+1"2).

We need to f11nd the values of h, k, and I

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for which the ratio a/d cannot have a value
of 51\ 1/2, 7" 1/2, 11/\1/2, or 13" 1/2.

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- For a/d to be equal to SA1/2 hA2+kA.2+IA2
must be equal to 5. This is not possible for
integers h k and I.
- For a/d to be equal to 7" 1/2, h".2+k"2+I"2
must be equal to 7.This is not possible for
integers h, k, and I.
- For a/d to be equa l to 11" 1/2,
1

hA2+k"2+IA2 must be equal to '11. This is

1

not possible for integers h, k, and I.

- For a/d to be eq al to 13A 1/2
h"2+k"2+ "2 must be equal to 13. This is
possible for integers h=2, k=3, and 1-2.
1

Therefore, the ratio a/d can have a value of

13" 1/2.

Conclusion
Therefore, the rati:,o of the unit cell length
to the separation between two adjacent
parallel crystal planes in a simple cubic
lattice cannot have a value of SA1/2 or
7A 1/2 or 11" 1/2. However, it can have a
value of 13A1/2 for a spe·cific set of Miller
indices.

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The correct answer is (1) 1.56 nm and 1.04
nm.

In an orthorhombic unit ceU, the separation

between planes (hkl) is given by:

d(hkl) =a/ ( (hA2/aA2 + kA2/bA2 + I"2/cA2))

where a, b, a!nd c are the lattice parameters.

Given the separation of the (123) planes:

d(123) = 3.12 nm

We can calculate the separation of the

(246)
and (369) planes using the same formula:
d(246) = a/ (v'(4hA2/aA2 + 4k"2/bA2 6IA2/
c"2)) +
= a I (v'(4(1)"2/aA2 + 4(2)"2/bA2 + 6(3)"2/c/\
2)) + 54/c"2))
= a I (v(4/aA2 + 16/b"2 ·
Since the lattice parameters a, b, and c
1

are the same for all planes, we can

simpl:ify the expression:

d(246) = d(123) / (f(4 + 16 + 54))

= 3.12 nm / ('17 4)
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= 1.56 nm

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Similarly,

d(369) = a/ (♦.J(9hA2/aA2 + 9kA2/bA2 + 91/\2/

CA2))
=a/ (.J(9(1)A2/a"2 + 9(2)A2/b"2 +
= a I 9(3)A2/c"2))
(.J(9/aA2 + 36/bA2 + 81/cA2))

d(369) = d(123) / (.J(9 + 36 +

81))
= 3.12 nm/ (-
J126)
= 104
1

nm
Therefore, the correct answer is (1) 1.56 nm
and 1.!0, 4nm1.
1
11 :1s pm

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Answer to Question no. 14: (1) 22

Explanation:

Given:

• FCC crystal
• (002) plane diffracts X-rays at 90° (Bragg
angle)
• Wavelength (A) = 0.154 nm
• Density (p) = 4 x 1QA4 kg/m"3

We can use the following steps to solve

the
problem:

1. Calculate the lattice parameter (a) using

the Bragg's law:
a =A/ (2 sin(8)) ='A./ 2 (since 8 = 90°)
a = 0.154 nm/ 2 = 0.077 nm
2. Calculate the atomic volume (V) using the
density:
V = m / p = (molar mass) / (density)
3. Since it's an FCC crystal, there are
4 atoms per unit cell. So, the atomic
.IS:
volume
V = (molar mass)/ (density) = (4 x atomic
mass)/ (4 x 1Q/\4 kg/mA3)
4. Equate the atomic volume to the volume
of the unit cell:
V = a"3 = (0.077 nm)/\3
5. Solve for the atomic mass (m):
m = (a/\3 x density)/ 4 = (0.077 nm)"3 x (4 x
101\4 kg/m"3) I 4
m == 22 u (unified atomic mass units)
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Answer to Question no. 15.

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Answer t,oQuestion no. 2:

The corre,ct is (2) Th,e widt,hof

answer
c,onductio,n band decreases and a band gap
i
generated:M

When the lattice dimens,io,n is doubled to

2a,
th,e interatomic,distan.ce increases, which
leads,to ,a,decrease in the ov,erlap between
atomic orbi1tal1s. As ,a re,sult, the widt1h' of the
conduction band decreases, and a band
,g,ap may open up,, potenti,aUy transforming
the metal into a semiconductor or ins1ulator.
This is known as ,a P'eierls transition.. 1 :10 n

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Answer to Question no. 1:

The correct answer is (2) A<2d.

The condition for diffractio;n from a crystal

surface is given by the Bragg's law: nA =
2d sine, where A is the wavelength of the
electron, d is the spacing between the (100)
planes, and 8 is the angle of incidence. Since
the electron is im!pinging normally on the
surface (8 = 90°), the condition reduces to A<
2d.
Answer to Question no. 2:

The correct answer is (2) 300.

The minimum value of cp (V) for the electron

to diffract from the (100) planes corresponds
to the minimum wavelength (maximum
energy) required for diffraction. Using the
Bragg's law and the given val!ue of d, we
can calculate the minimum wavelength A =
2d
= 703 x 10"-12 m. The kinetic energy of the
electron (E) is related to the wavelength by
the de Broglie relation: E = (hc)/A, where h is
Planck's constant and c is the speed of light.
Substituting the values, we get E 300 eV,
which corresponds to a potential difference of
<.p (V) :::: 300 V. 1133 pm
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correct answer is:inde,e,d (1) a :-t b ;t: c;a.
y 9 0°,·"#. 90°.
1

.In a monoi clinic crystal, lattice parameters

are c!haracterizthe
e·d by:

• Three unequal lattice constants (a, b, and

c)
• Tw·o angles (a and y) that .are equ,al to
and one igle ( ) that is not equal 90°,
an to•90°.
Thiis m eans that the crystal has a
orthogonality in two dimensions (a and y),
but not in the third dimension ( ). This is a
characteristic feature of monoclinic crystals.

So, the correct option is (1) a ;c b ;c c; a - y '-

9100, :'#, goo. '1 ,-. ;-

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Answer to Question no. 4:
0

The correct answer is indeed (2) 2.82 A.

Giiven:

• 8 (Bragg angle) = 15.87°

• A (X-ray wavelength) = 1.54 x 1QA-a.cm=
0

1.54 A

Using Bragg's law:

nA.= 2d sine

where d is the spacing between the planes.

Rearranging to solve for d:

d = nJ\ / (2 sine)

Since it's a first-order diffraction (n=1), we get

d =A/ (2 sinB)
= 1.54 A/ (2 sin 15.87°)
0

= 2.a2 A

Therefore, the spacing between the planes

of the NaCl crystal is 2.82 A, which matches
option (2).
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Answer to Question no. 5:

The correct answer is (3) overlap of 1l.led 3s

fan d empty 3p·orbital.
1

Magnesium (Mg) has an atomic configuration

of 1s2 2s2 ·2p6 3.s2. In!solid·-state physicsJ.the
b.andtheory predicts that Mg should b,e an 1

insulator due to its full valence shell (3s2).

However, in reaility. Mg acts as a co,nductor.

This discrepancy is resolved by considering

the overlap of the filled 3s orbital with
the
empty 3p orbital, which creates a hybridization
effect. This overlap leads to the formation of
a partially filled conduction band, making Mga
c,onductor.

So, the correct option is (3) overlap of filled 3s

and empty 3 _ orbital.

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Answer to Question no. 6:

The densit,yof Molybdenum (Mo) can

be calculated using the following
steps:
1. Calculate the volume of th,e unit cell (V):

V = aA3 = (0.314 nm)"3 = 0,.03736,nm/\3

1. Calculate the number of atoms per unit

cell (N) for a bee structure:

N = 2 (since each lattice point has 1/8 of an

atom, and there are 8 lattice points unit
cell)
ina

1. Calculate the mass of a single atom (m):

m = atomic mass/ Avogadro's

number
= 96 g/mol I 6.022 x 10"23 atoms/mol
= 1.59 x 10"-25 kg

1. Calculate the density (p):

p == (N X m) IV
= (2x 1.59x 1 0"-25 kg) / 0.037 36 nm"3
= 7 0200
kg/m"3
Theref?re, the density of Molybdenum.
approximately 10200 kg/rn"3. is
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 Answer to Question no.,7:

The c orrect answer is (2) Frenkel defect is an

1

anion vacancy and ,a cation interstitiali.

Frenkel defect is a ,e of oin1t defect that

1

involves a cation typ p

(positive io,n) vacancy and a
c,atii o,n interstitial, not an anion (negative ion)
vacancy.

The correct statement ,ab,o ut Frenkel defects

.
1s:

(1) Frenkel defect is a cation vacancy and ,a

cation interstitial.

The other options are

correct:
(3) Frenkel defects do not change the overall
density of the solid, as the number of ,ato,ms
remains the same.

(4) Schottky in ,olve the

defects
o,f vacancies in bot,hcatJo
v formation
•n and anion sites,,
lea,dingrto a decre,ase in density.

So, the inc,orrect statement is option (2).

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Answer to Q:uestion no. 8:

The correct answer is (4) a/../2.

In simple cubic lattice, the lattice parameter

a'a' is the distance between two adjacent
atoms. The distance between two successive
{110) planes is equal to the diagonal of the
square formed by the lattice parameters.

Using the Pythagorean theorem, we can

calculate the diagonal:

diagonal= V(aA2 + aA2) = ..J(2a"2)= av2

Sinc .. . ..
O)e the
I distance
• . between
• two .s cc essive
(11 P anes is half of the diagon al.th .
correct answer is:
u ' ,
. e
a/v2.,

So, the cor,rect , . .

• Option IS (4) a/ V2.
A

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 Answer to Question no. 9:

The correct answer is {3) F-centres.

When sodium chloride (NaCl) crystals are

heated in the presence of sodium vapor,
the sodium atoms can replace some of
the sodium ions in the crystal lattice, creating
vacancies (holes) in the lattice. These
vacancies can trap electrons, forming what
are known as F-centres (or color centres).
F-centres are responsible for the yellow
coloration of the crystals.

F-centres are a type of point defect that

involves the presence of excess electrons in
the crystal lattice, which can lead to changes
in the optical properties of the material.

The other options are not correct:

(1) Schottky defects involve the formation

of vacancies in both cation and anion sites,
but do not result in the formation of F-
centres.
(2) _ Frenkel defects involve the
displacement of ion from their lattice sites,
but do not result in the formation of F-
centres.

(4) H-centres are not a type of point defect

that occurs in NaCl crystals. -.
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