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English Pages 620 [303] Year 2005
Solid State Physics M A Wahab
Contents Preface to the Second Edition Preface to the First Edition Units of Measurement Physical Constants Conversion Factors
1. Atoms in Crystals
v vii ix xiv xv '
,/
1
^
1.1 The Solid State 1.2 Periodicity in Crystals 1.3 Choice of a Unit Cell 1.4 Wigner Seitz Unit Cell 1.5 Number of Lattice Points per Unit Cell 1.6 Symmetry Elements 1.7 Combination of Symmetry Elements ( Point Group ) 1.8 Bravais Lattice in Two Dimensions ( Plane Lattice ) 1.9 Bravais Lattice in Three Dimensions (Space Lattice ) 1.10 Rational Features of a Crystal and Miller Indices 1.11 Interplanar Spacing 1.12 Density of Atoms in a Crystal Plane 1.13 Some Simple and Common Crystal Structures 1.14 Summary 1.15 Definitions
1 3 6 6 7 10 15 16 17 21 31 36 42 49 50
^
2. Atomic Bonding 2.1
^
2 2.3 2.4 2.5 2.6 2.7 u 2.8 i/2.9

•
56
Forces Between Atoms Ionic Bonding Bond Dissociation Energy of NaCl Molecule Cohesive Energy of Ionic Crystals Evaluation of Madelung Constant for NaCl Structure Madelung Potential The Bom  Haber Cycle Covalent Bonding Metallic Bonding
56 58 60 61 71 72 73 74 79
e
"Hi
Contents
,
2 0 Hydrogen Bonding 2 11 Van der Waais Bonding 2 12 Summary

2.13
Definitions
3. Atomic Packing
3.1 3.2
3.3 3.4 3.5 3.6 3.7 3.8 3.9
. 3 r
Packing of Equal Spheres in 2 Dimensions Packing ol Equal Spheres in 3 Dimensions Close Packing of Equal Spheres in 3Dimensions Classification of Close Packings Axial Ratio and Lattice Constants Voids in Close Packing Size and Coordination of Voids Significance of Voids Packing of Unequal Spheres in 3 Dimensions and Effect of Radius Ratio Representation of Close Packings Pauling ’s Rule Application of Pauling’s Rule to Actual Structures Examples of Some Close Packed Structures Notations: Representing Close Packed Structures
3.10 3.11 3.12 3.13 3.14 3.15 Summary 3.16 Definitions
Contents xix
81 82 83
86 86 88 92 93 95 96 97 101
102 104 106 107 109 113 117 118
.
4 Atomic Shape and Size
4.1 4.2 4.3 4.4
Introduction
Bohr Model of the Hydrogen Atom Atomic Shape Wave Mechanical Concept of Atom .4.5 Atomic Size 6 Ionic Radii 4.7 Empirical ionic Radii 4.8 Variation of Ionic Radii 9 Covalent Radii LLA ' v 4 1 0 Metallic Radii jt. Van der Waais Radii 4.12 Summary
> 
n
4.13
Definitions
5. Crystal Imperfections
5.1 5.2 5.3
Introduction Point Imperfections Concentration of Point Imperfections
120 121 123 124 126 127 131 135 138 143 147 148 148
150 150 150 151
5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14
Line Imperfections Burgers Vector and Burgers Circuit Presence of Dislocation Dislocation Motion Energy of a Dislocation Slip Planes and Slip Directions
Perfect and Imperfect Dislocations Dislocation Reaction Surface Imperfections Summary Definition
161 164
N
—
6. Atomic Diffusion
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13
Introduction Fick’s First Law Fick’s Second Law Solution to Fick’s Second Law ( Constant D ) Some Applications of Diffusion Diffusion Mechanisms Random Walk Treatment of Diffusion The Kirkendall Effect Diffusion in Alkali Halides Ionic Conductivity in Alkali Halide Crystal Diffusion and Ionic Conductivity Summary
*
.
165 171 176 178 181 182 185 190 191
194 194 194 197 198 203 213 217 218 219
Definitions
224 225 225
7/iattiee ( Atomic) Vibrations
228
7.1 7.2
7.3 7.4
7.5 7.6
7.7 7.8
Introduction Dynamics of the Cham of Identical Atoms Dynamics of a Diatomic Linear Chain Dynamics of Identical Atoms in Three Dimensions Experimental Measurements of Dispersion Relation Anharmonicity and Thermal Expansion Summary
Definitions
8. Diffraction of Waves and Particles by Crystals
8.1 8.2 8.3 8.4 8.5
Introduction X  rays and Their Generation Moseley’s Law' Absorption of X  rays ( Classical Theory ) Absorption Edge
228 229 233 240 240
242 245 246 249 249 249 254 257 260
xxii
Contents xxiii
Contents
13.7 13.8 13.9 13.10
Hall Effect in Semiconductors Junction Properties Summary Definitions
14. Dielectric Properties of Materials
v^MTl
Introduction
Dipole Moment % 34 /Polari za ti on ~ T h e Electric Field of a Dipole M:5 , Local Electric Field at an Atom M O/'Dielectric Constant and its Measurement
^
4
^O
14 7
Polarizability ^ ^ The Classical Theory of Electronic Polarizability ,
^
^
l f 'Dip lar Polarizability ° , Pyro and Ferroelectric Properties of Crystals 1.4 At ) Piezo , 4 t T J  Ferroelectricity 14.12 Ferroelectric Domain 14.13 Antiferroelectricity and Ferrielectricity 14.14 Summary 14.15 Definitions
15. Optical Properties of Materials 15.1 Absorption Process 15.2 Photoconductivity 15.3 Photoelectric Effect 15.4 Photovoltaic Effect 15.5 Photoluminescence 15.6 Colour Centres 15.7 Types of Colour Centres 15.8 Generation of Colour Centres 15.9 Maser and Laser 15.10 Summary
15.11 Definitions 16. Magnetic Properties of Materials 16.1 16.2 16.3 16.4 16.5 16.6 16.7
Introduction Response of Substance to Magnetic Field Classification of Magnetic Materials Atomic Theory of Magnetism The Quantum Numbers The Origin of Permanent Magnetic Moments Langevin ’s Classical Theory of Diamagnetism
441 442 451 452 455
455 455 456 457 458 461
463 466 468 472 473 475 476 476 479
480 480 481 483 487 489 489 489 491 493 499 500
502 502 502 503 504 505 507 509
16.8
Sources of Paramagnetism
16.9 Langevin’ s Classical Theory of Paramagnetism 16.10 Fundamentals of Quantum Theory of Paramagnetism 16.11 Paramagnetism of Free Electrons 16.12 Ferromagnetism 16.13 The Weiss Molecular (Exchange) Field 16.14 Temperature Dependence of Spontaneous Magnetization 16.15 The Physical Origin of Weiss Molecular Field 16.16 Ferromagnetic Domains 16.17 Domain Theory 16.18 Antiferromagnetism 16.19 Ferrimagnetism and Ferrites 16.20 Summary 16.21 Definitions 17. Superconductivity Introduction Sources of Superconductivity Response of Magnetic Field The Meissner Effect Thermodynamics of Superconducting Transitions Origin of Energy Gap Isotope Effect London Equations London Penetration Depth Coherence length 17.11 Elements of BCS Theory 17.12 Flux Quantization 17.13 Normal Tunneling and Josephson Effect 17.14 High Tc Superconductivity 17.15 Summary
17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9 17.10
17.16 Definitions 18. Anisotropic Properties of Materials 18.1 18.2 18.3 18.4
18.5 18.6 18.7 18.8 18.9
Introduction Classification of Physical Properties Description of Physical Properties Intrinsic Symmetry of Physical Properties Transformation Law for Second Rank Tensor Effect of Crystal Symmetry on Physical Properties Physical Properties of Zero and First Rank Tensors Physical Properties of Second Rank Tensor Physical Properties of Third Rank Tensor
512 513 514 516 518 518 519 520 522 523 525 529 530 532
534 534 535 535 538 539 542 544 544 546 548 549 552 553
555 556
558 560
560 560 562 563 564 567 570 573 576
5 )
xxiv
Contents
18.10 Physical Properties of Fourth Rank Tensor 18.11 The Strain Energy 18.12 Effect of Crystal Symmetry on Elastic Constants 18.13 Summary 18.14 Definitions
580 581 582 585 586
Bibliography
588
Index
5Q1
Chapter 1
Atoms in
Crystals
#
1.1
THE SOLID STATE
Matter is usually regarded to exist in the solid state or the fluid state , the latter being subdi \ ided into the liquid and gaseous states. However, on the basis of modem concept, the matter is more conveniently divided into the condensed state and the gaseous stale the former being subdivided into the solid and jhe liquid states. It is not easy to give a completely satisfactory definition d a solid ; a “ borderline” material can always be found . We shall use “ solid” to mean any material whose constituent particles ( atoms, ions or molecules ) are relatively fixed in position, except for thermal vibration. Fluid on the other hand will mean a material whose particles are in a state of constant translational motion. In order to understand various states of solid, it is useful to consider the process of growth . The growth could be either from solution, melt or vapour, or from a suitable combination of these . If the growth process is slow, then the constituent particles have a tendency to settle down in positions for which the free energy of the configuration is minimum. This leads to an arrangement of longrange order in the solids. However, in certain extreme cases when the grow th process or the phase change is rather rapid , the constituent panicles do not have sufficient time to achieve minimum energy configuration . Consequently a long range order is not obtained in such cases Nevertheless, there is evidence for the preferential ordering of neighbouring atoms aiound a particular atom or ion and hence a short range order is still preserved In the following section , an attempt will be made to classify solids in terms of then degree and type of order. It is convenient to think of three broad classes: crystalline, semi crv stalling and non crystalline. It is to be pointed out here that the extremes such as perfect order oi perfect disorder are rare and are of theoretical interest only.
_
,
Crystalline Solids A solid in general is said to be a crystal if the constituent particles t atoms, ions or molecules) are arranged in a three dimensional periodic manner, oi simply if it has a reticular structure. The regularity in the appearance of crystals found in nature ( Fig. 1 1 ) or grow n in laboratory has led us to believe that crystals are formed by a regular repetition of identical building blocks nvthree dimensional space ( Fig. 1.2). When a crystal grows under constant environment , the external geometrical shape of the crystal often remains unchanged. The shape is a consequence of internal arrangement of constituent particles of which it is build up.
——
Atoms 2
Seniicrvstalline
Solid State Phvsics
iSS boundaries
( Polycrystalline )
Solids
said to be a polycrystalline solid , the grains in such a solid ( Fig  1.4 ) ln general are not related in shape to the crystal structure. tie surface being random in shape rather than well defined crystal planes. A great majority of solids occurring in nature such as rock , sand , metals, salts, etc. are of polycrystalline structure. However, they can be grown as a single crystal under suitable conditions. Due to random distribution of crystallites, a polycrystalline solid , is isotropic , i .e . its properties are same on an average in all directions. is

\
/
2
N
Quartz
.
Fig 1.1 External forms of the nature
crystal
found in
.
repetition Fig. 1.2 Formation of crystals by regular of identical building blocks
sharp straight edges Observations show that crystals are bounded by optically plane and interfacial angles. A relationship among these elements can be expressed by the formula (1) /+ c= e + 2
laces ,
where / is the number of faces, c is the number of angles and e is the number ol edges. Out ot the three elements in eq . 1, if two are known then the third can be easily obtained. Example: Determine the number of edges in a quartz crystal , if there are 18 faces and 14 angles in it
Solution: Given: / = 18 , c = 14 ( also refer Fig. 1.1). Using eq . i , number of edges e = 18 + 1 4  2 = 30. This can be verified from Fig . 1.1 A crystal if more or less regular in shape is called a monocrystal or a single crystal . Quite common crystals such as rock salt , calcite, quartz, etc. ( Fig. 1.3) are some examples of this class. A typical feature of a single crystal is its anisotropy, i .e. the difference in its physical properties in different directions . In addition , a crystalline solid has a sharp melting point .
in Crystals
* smallestj [ f j k l ] t 0 get the required e.
.
.
loTonoTOMOOl
TJ
l2
( b2 +
I
F
< 110>
V2 a
V2 a
a a V2
VI a
V3 a
Tetragonal
J(a
•
Example: Classify the members of < I 00>, and < 111 > families of direction in the tetragonal system.
«
Solution: For a tetragonal system , we know that a = b c . Thus, the unit translation for the * directions [ 100 ] , [ TOO ] , [010] and fOTO] is the same as a, whereas for the directions [001 ] and [00 1 ] 1sc. Consequently the first four directions are the members of 100 family and the last < > two are the members of family. Similarly. family has four members, i.e. [ 110 ], [ 110], [ 1 TO ] and [ I TO], while the direct ons [101], [ T 01 J. [10 T ], [ TOT ], [Oil ], [OTl ], [01 T ] and [OlT ] are the members of < J 0 I > lamily or equivalently the family.
.
otherjngh
j
2 ( /1 2
\ 2 / (
+
M 2 ) + c 2 ( M2 )
)
^
2 + k f ) + c 2 l{ ) { a { h +
\>
fcf ) + c 2 / f ]
J
h\ h2 +
Hexagonal
^
\
2
a 2 h x h 2 + b 2 k x k 2 + c 2 l\ l 2 2 2 2  ( a 2 /i 2 + b 2 k f + c 2 f 2 ) ( a h f + b k $ + c l % )
+
kxk 2 
( h ) k 2 + kxh 2 ) +
*?  Mi +
+ k 2  M2 +
*1 )
&cample: Calculate the angle between [111] and [001] directions in a cubic crystal. JSolution From Table 1.9, we have
_
:
cos and < 111 >.
Solution: The family < 110> comprises the directions [ 110 ], [ 110 ], [ 1 TO ], [ T TO ], [ 101 ], [ T 01 ], HOT ], fTOTJ, [OilJ, [Oil ], [Oil ], and [Off ]. Similarly, the family < 111 > comprises the directions fill ], fill ], [1 T 1], [11 T ], [ T 11], [ Tl I ], [ 1 T J ] and [ T T T ].
2
q 2 ( /» l /l 2
Unit translation
a
HI
/ll / + k k + l ll ,, , *, + /, )* a  k + * J + lf z
Cubic
Orthorhombic
a
IM /zl
Cos 0
Crystal system
directions of a cubic system
p
S
Table 1.9 Angle between two crystal directions ( MiU and
r
. . .translates
Family
c
.
^
Table 1.8 The unit translation for low index
4
Since all < 111 > directions have the same unit translation in the tetragonal system and therefore the family contains the same eight members as it does in the cubic system . In certain calculations (e.g. in resolved shear stresses, Ch . 5), it is necessary to determine the angle between two different crystal directions. The formula for a general case like tnclinic is very complicated . Therefore in Table 1.9 we enlist the formulae for some crystal systems which may be frequently used in calculations. The formula for cubic system is the simplest of all which can be determined by dot product.
.
J^
>
23
Atoms in Crystals
indices
. of that and all parallel directions actual geometry . c rtf direction are quite independent of theany indices the that of point on the noted be It should coordinates the of solely y of the unit cell and are determined given direction to the appropriate unit of directions. Out of these, there are In a crystal, there exists an mfin the atoms (i .e. same unit translations) are tween p identic sets of directions having related by a symmetry operation ^ ajways js a set of angled brackets i.e. calledI equivalent directions Such pe j The shortened notetion us family < 1 (K )> comprises the directions [ 100 ], all the principal directions ( crystallorgraphic shows 1.27 . [ TOOf J and ( 00 . Fig for the low tndex . The um ons t nota orientations ) in a cub c crystal alongwi h thetr arc given in I able . directions in the three Bravais lattices of cubic system
.
mmm
, , .
to the origin in a gtven d reel on , of the lattice site nearest trinsotions .
( i) Note down the cooni na es P ( ii ) Divide the coordinates by PP multiply each o ( iii ) If fractions result, in a set of ( iv ) Put the resulting integers
'
so that
0 = 54.75°
^,
+ kjfc 2
hxh 2
,
( /i 2 + k 2 +
K7
^
= 0 + 0 + 1j V 3 I ) + k\ + mu
Example: Calculate the angle between [ 111 ] and [ I I I ] directions in a cubic crystal .
Solution: From Fig. 1.28, we observe that [001 ] direction bisects the angle between [111] and [ T i l ]. Therefore, the angle 0 between them is 2 x 54.75° = 109.5°. This can be verified from the dot product as mentioned above. COS
,
so that
0
 1 1 + 1 = V3 x V3
0 = 109.5°
o
1 3
X
y
.
Fig 1.28
Angles made by till ] and [ 1 111 directions with respect to [ 001] in a cubic crystal
— r
' 3
b
24
WJMOffi
[ Acc. No ..\ P.5g£3Cl
•
j
'
.
Solid State Physics
Indices of Liittic Plane crystal lattice may be
A
considered as an assem
number of equidistant parallel planes por gvcn au ce. these sets of
>‘
lattice points For simplicity, let us stall with a
up of a number of rows ^ made a number of ways as shown ^^ partjcuar row of alolm )
’
to
jn
,
^
(
respectively ( Fig. 1.30a ) . From
will intercept the coordinate axes x and y at a ana simple geometry the equation of this line is given as
ia + Zb + ic
,
^
ihrce dimensional case. A dimensional lattice is of atoms In this case, the set of rows of atoms sa n F g 1 29 Now, f we fix the coordinate axes
, ,
=
CrjaaLt
25
.
=
(10 )
I
Equation 10 describes the first lattice plane, nearest to the origin , in a set of parallel, identical and equally spaced planes. The resulting set of three integers h. k and l are conventionally enclosed in parentheses ( hkl ), called the Miller indices and describe the lattice plane in question liquation 10 can also be written as
33
JL + 1lh
( d ) ( 110 )
(0 ( 010 )
in
1
hx + ky + lz
( b ) (230 )
a ( 110)
,
In describing crystallographic planes , the axes are taken along three non parallel edges of the unit cell , and the intercepts are measured in term of respective unit length, which is assigned to each of the cell regardless of its actual dimension . The use of the intercepts a. b and c to represent a crystallographic plane has the disadvantage thai the intercepts are often fractions ( less than l ) and may be infinite ( for a plane parallel to an axis ). For this reason, it is ncm common practice to use the reciprocal of the intercepts to designate a cry stallographjc plane Therefore, suppose h l /a , k llb , and / = Me so that eq . 9 becomes
,
( )
s.
_L n_ _ Mk
+ Z ~ \
yi
ac
( 11)
i
where Mh , \lk and Ml are intercepts along the x, y and z axes, respectively. This implies that the plane divides “ a ” into h parts, “ b" into k parts and “ c" into l parts Therefore, to obtain the Miller indices of a plane, we adopt the following procedure:
( hkl )
the intercepts of the plane along x, y , and z axes in terms of lattice parameters Divide these intercepts by the appropriate unit translations. ( iii ) Note their reciprocals. ( iv ) If fractions result, multiply each of them by the smallest common divisor. ( v ) Put the resulting integers in parenthesis ( hkl ) to get the required Miller indices of that and parallel planes . ( i ) Determine
( ti )

.
Fig 1J9 Miller indices of some planes in a two dimensional lattice yi
1/A
.
irnple : In a crystal, a plane cuts intercepts of 2a 3b and 6c along the three cry stallographic axes. Determine the Miller indices of the plane.
i/ k
1/ A
Solution: Following the above proedure. we have
X
a
(i )
( ii ) Division by unit translations i
Fig. 130 Intercepts along ( a) x and y axes ( b) x, y and z axes
( iii ) Reciprocals ( iv ) After clearing fraction
t
,,°n
fa + £b = i
^
,
eqUa W hree d me Si 0nS havin ® I b ) along the axes x, y and z , respectively will be 1.30
‘
"
(8)
intercepts a , ft and c (Fig5.
=>
2a
Intercepts
—
a
=2 1 2 3
6c
3b
3b
,
—
c
b
1 3 2
=6 I 6 l
The required Miller indices of the plane are ( 321 )
Example: Determine the Miller indices of a plane which is parallel to xaxis and cuts intercepts of 2 and 1 /2, respectively along y and zaxes.
Solution: Following the same procedure, we have
V
r
a )
26 5