Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

18 1 Crystal Structures

Fig. 1.18. Scattering of X-rays by a pair of atoms in a crystal

1. Let ˆs

be a unit vector in the direction of the incident wave, and ˆs be a 319

unit vector in the direction of the scattered wave. 320

2. Let R

and R

be the position vectors of a pair of atoms in a Bravais 321

lattice, and let r

= R

− R

. 322

Let us consider the waves scattered by R

and by R

and traveling diﬀerent 323

path lengths as shown in Fig. 1.18. The diﬀerence in path length is | R

A − 324

|. But this is clearly equal to |r

· ˆs − r

· ˆs

|. We deﬁne S as S =ˆs−ˆs

; 325

then the diﬀerence in path length for the two rays is given by 326

Δ=|r

· S|. (1.11)

For constructive interference, this must be equal to an integral number of

327

wave length. Thus, we obtain 328

· S = mλ, (1.12)

where m is an integer and λ is the wave length. To obtain constructive inter-

329

ference from every atom in the Bravais lattice, this must be true for every 330

lattice vector R

. Constructive interference will occur only if 331

·S = integer × λ (1.13)

for every lattice vector R

in the crystal. Of course there will be diﬀerent 332

integers for diﬀerent R

in general. Recall that 333

= n

+ n

. (1.14)

The condition (1.13) is obviously satisﬁed if

334

· S = ph

λ, (1.15)

where h

is the smallest set of integers and p is a common multiplier. We can 335

obviously express S as 336

S =(S · a

) b

+(S · a

) b

+(S · a

) b

. (1.16)

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

1.4 Diﬀraction of X-Rays 19

Fig. 1.19. Relation between the scattering vector S =ˆs −ˆs

and the Bragg angle θ

Therefore, condition (1.13) is satisﬁed and constructive interference from every 337

lattice site occurs if 338

S = p (h

+ h

) λ, (1.17)

339

= pG

, (1.18)

340

where G

is a vector of the reciprocal lattice. Equation (1.18) is called the 341

Laue equation. 342

Connection of Laue Equations and Bragg’s Law 343

From (1.18) S must be perpendicular to the planes with Miller indices 344

). The distance between two planes of this set is 345

d(h

2π

= p

|S|

. (1.19)

We know that S is normal to the reﬂection plane PP



with Miller indices 346

). From Fig. 1.19, it is apparent that |S| =2sinθ. Therefore, (1.19) 347

can be written by 348

2d(h

)sinθ = pλ,

where p is an integer. According to Laue’s equation, associated with any

349

reciprocal lattice vector G

= h

, there is an X-ray reﬂection 350

satisfying the equation λ

−1

S = pG

,wherep is an integer. 351

1.4.3 Ewald Construction 352

This is a geometric construction that illustrates how the Laue equation works. 353

The construction goes as follows: See Fig. 1.20. 354

1. From the origin O of the reciprocal lattice draw the vector AO of length 355

−1

parallel to ˆs

and terminating on O. 356

2. Construct a sphere of radius λ

−1

centered at A. 357

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

20 1 Crystal Structures

Fig. 1.20. Ewald construction for diﬀraction peaks

If this sphere intersects a point B of the reciprocal lattice, then AB =

ˆs

is 358

in a direction in which a diﬀraction maximum occurs. Since A

O =

ˆs

and 359

ˆs

ˆs−ˆs

= OB

is a reciprocal lattice vector and satisﬁes the 360

Laue equation. If a higher frequency X-ray is used, λ

, A

,andB

replace λ

, 361

,andB

. For a continuous spectrum with λ

≥ λ ≥ λ

, all reciprocal lattice 362

points between the two sphere (of radii λ

−1

and λ

−1

) satisfy Laue equation 363

for some frequency in the incident beam. 364

Wave Vector 365

It is often convenient to use the set of vectors K

=2πG

. Then, the Ewald 366

construction gives 367

+ K

= q, (1.20)

where q

2π

ˆs

and q =

2π

ˆs are the wave vectors of the incident and 368

scattered waves. Equation (1.20) says that wave vector is conserved up to 2π 369

times a vector of the reciprocal lattice. 370

1.4.4 Atomic Scattering Factor 371

It is the electrons of an atom that scatter the X-rays since the nucleus is so 372

heavy that it hardly moves in response to the rapidly varying electric ﬁeld of 373

the X-ray. So far, we have treated all of the electrons as if they were localized 374

at the lattice point. In fact, the electrons are distributed about the nucleus of 375

the atom (at position r = 0, the lattice point) with a density ρ(r). If you know 376

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

1.4 Diﬀraction of X-Rays 21

Fig. 1.21. Path diﬀerence between waves scattered at O and those at r

the wave function Ψ (r

, r

,...,r

) describing the z electrons of the atom, ρ(r) 377

is given by 378

ρ(r)=





i=1

δ (r − r

)





Ψ(r

,...,r

)





i=1

δ(r − r

)



Ψ(r

,...,r

)



(1.21)

Now, consider the diﬀerence in path length Δ between waves scattered at O

379

and those scattered at r (Fig. 1.21). 380

Δ=r · (ˆs − ˆs

)=r · S. (1.22)

The phase diﬀerence is simply

2π

times Δ, the diﬀerence in path length. 381

Therefore, the scattering amplitude will be reduced from the value obtained 382

by assuming all the electrons were localized at the origin O by a factor z

−1

f, 383

where f is given by 384

f =



rρ(r)e

2πi

r·S

. (1.23)

385

This factor is called the atomic scattering factor.Ifρ(r) is spherically sym- 386

metric we have 387

f =



∞



−1

2πr

dr d(cos φ)ρ(r)e

2πi

Sr cos φ

. (1.24)

Recall that S =2sinθ,whereθ is the angle between ˆs

and the reﬂecting 388

plane PP



of Fig. 1.19. Deﬁne μ as

4π

sin θ;thenf can be expressed as 389

f =



∞

dr4πr

ρ(r)

sin μr

μr

. (1.25)

If λ is much larger than the atomic radius, μr is much smaller than unity

390

wherever ρ(r) is ﬁnite. In that case

sin μr

μr

 1andf → z,thenumberof 391

electrons. 392

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

22 1 Crystal Structures

1.4.5 Geometric Structure Factor 393

So far we have considered only a Bravais lattice. For a non-Bravais lattice the 394

scattered amplitude depends on the locations and atomic scattering factors 395

of all the atoms in the unit cell. Suppose a crystal structure contains atoms 396

at positions r

with atomic scattering factors f

. It is not diﬃcult to see that 397

this changes the scattered amplitude by a factor 398

F (h



2πi

·S(h

)

(1.26)

for the scattering from a plane with Miller indices (h

). In (1.26) the 399

position vector r

of the jth atom can be expressed in terms of the primitive 400

translation vectors a

401



. (1.27)

Forexample,inahcplatticer

=(0, 0, 0) and r

) when expressed 402

in terms of the primitive translation vectors. Of course, S(h

)equalto 403



,whereb

are primitive translation vectors in the reciprocal lattice. 404

Therefore,

2πi

·S(h

)isequalto2πi (μ

+ μ

), and the 405

structure amplitude F (h

) can be expressed as 406

F (h



2πi



. (1.28)

If all of the atoms in the unit cell are identical (as in diamond, Si, Ge, etc.)

407

all of the atomic scattering factors f

are equal, and we can write 408

F (h

)=fS(h

). (1.29)

409

The S(h

) is called the geometric structure amplitude. It depends only 410

on crystal structure, not on the atomic constituents, so it is the same for all 411

hcp lattices or for all diamond lattices, etc. 412

Example 413

A useful demonstration of the geometric structure factor can be obtained by 414

considering a bcc lattice as a simple cubic lattice with two atoms in the simple 415

cubic unit cell located at (0,0,0) and (

). Then 416

S(h

)=1+e

2πi

(

)

. (1.30)

If h

is odd, e

iπ(h

)

= −1andS(h

)vanishes.Ifh

+ 417

is even, S(h

) = 2. The reason for this eﬀect is that the additional 418

planes (associated with the body centered atoms) exactly cancel the scattering 419

amplitude from the planes made up of corner atoms when h

+ h

is odd, 420

but they add constructively when h

+ h

is even. 421

The scattering amplitude depends on other factors (e.g. thermal motion 422

and zero point vibrations of the atoms), which we have neglected by assuming 423

a perfect and stationary lattice. 424

Uncorrected Proof

BookID 160928 ChapID 01 Proof# 1 - 29/07/09

1.4 Diﬀraction of X-Rays 23

1.4.6 Experimental Techniques 425

We know that constructive interference from a set of lattice planes separated 426

by a distance d will occur when 427

2d sin θ = nλ, (1.31)

where θ is the angle between the incident beam and the planes that are scat-

428

tering, λ is the X-ray wave length, and n is an integer. For a given crystal 429

thepossiblevaluesofd are ﬁxed by the atomic spacing, and to satisfy (1.31), 430

one must vary either θ or λ over a range of values. Diﬀerent experimental 431

methods satisfy (1.31) in diﬀerent ways. The common techniques are (1) the 432

Laue method,(2)therotating crystal method, and (3) the powder method. 433

Laue Method 434

In this method a single crystal is held stationary in a beam of continuous wave 435

length X-ray radiation (Fig. 1.22). Various crystal planes select the appropri- 436

ate wave length for constructive interference, and a geometric arrangement of 437

bright spots is obtained on a ﬁlm. 438

Rotating Crystal Method 439

In this method a monochromatic beam of X-ray is incident on a rotating 440

single crystal sample. Diﬀraction maxima occur when the sample orientation 441

relative to the incident beam satisﬁes Bragg’s law (Fig. 1.23). 442

Powder Method 443

Here, a monochromatic beam is incident on a ﬁnely powdered specimen. The 444

small crystallites are randomly oriented with respect to the incident beam, 445

so that the reciprocal lattice structure used in the Ewald construction must 446

be rotated about the origin of reciprocal space through all possible angles. 447

This gives a series of spheres in reciprocal space of radii K

, K

, ... (we 448

include the factor 2π in these reciprocal lattice vectors) equal to the smallest, 449

COLLIMATOR

SAMPLE

FILM

X - RAY

BEAM

SPOT

PATTERN

Fig. 1.22. Experimental arrangement of the Laue method

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BookID 160928 ChapID 01 Proof# 1 - 29/07/09

24 1 Crystal Structures

ROTATABLE

SAMPLE

MONOCHROMATIC

X-RAY BEAM

FILM

Fig. 1.23. Experimental arrangement of the rotating crystal method

POWDER

SAMPLE

FILM

X - RAY BEAM

DIFFRACTION

RINGS

Fig. 1.24. Experimental arrangement of the powder method

next smallest, etc. reciprocal lattice vectors. The sequence of values

sin(φ

/2)

sin(φ

/2)

450

give the ratios of

for the crystal structure. This sequence is determined 451

by the crystal structure. Knowledge of the X-ray wave length λ =

2π

allows 452

determination of the lattice spacing (Fig. 1.24). 453

1.5 Classiﬁcation of Solids 454

1.5.1 Crystal Binding 455

Before considering even in a qualitative way how atoms bind together to 456

form crystals, it is worthwhile to review brieﬂy the periodic table and the 457

ground state conﬁgurations of atoms. The single particle states of electrons 458

moving in an eﬀective central potential (which includes the attraction of the 459

nucleus and some average repulsion associated with all other electrons) can 460

be characterized by four quantum numbers: n, the principal quantum number 461

takes on the values 1, 2, 3,..., l, the angular momentum quantum number 462

takes on values 0, 1,...,n−1; m, the azimuthal quantum number (projection 463

of l onto a given direction) is an integer satisfying −l ≤ m ≤ l;andσ,the 464

spin quantum number takes on the values ±

. 465

Uncorrected Proof

BookID 160928 ChapID 01 Proof# 1 - 29/07/09

1.5 Classiﬁcation of Solids 25

The energy of the single particle orbital is very insensitive to m and σ (in 466

the absence of an applied magnetic ﬁeld), but it depends strongly on n and l. 467

Of course, due to the Pauli principle only one electron can occupy an orbital 468

with given n, l, m, and σ. The periodic table is constructed by making an 469

array of slots, with l value increasing from l = 0 as one moves to the left, and 470

the value of n + l increasing as one moves down. (Table 1.4) Of course, the 471

correct number of slots must be allowed to account for the spin and azimuthal 472

degeneracy 2(2l +1)of a givenl value. One then begins ﬁlling the slots from 473

the top left, moving to the right, and then down when all slots of a given 474

(n + l) value have been used up. See Table 1.4, which lists the atoms (H, 475

He, ...) and their atomic numbers in the appropriate slots. As the reader can 476

readily observe, H has one electron, and it will occupy the n =1,l =0(1s) 477

state. Boron has ﬁve electrons and they will ﬁll the (1s)and2s states with the 478

ﬁfth electron in the 2p state. Everything is very regular until Cr and Cu. These 479

two elements have ground states in which one 4s electron falls into the 3d shell, 480

giving for Cr the atomic conﬁguration (1s)

(2s)

(2p)

(3s)

(3p)

(4s)

(3d)

, 481

and for Cu the atomic conﬁguration (1s)

(2s)

(2p)

(3s)

(3p)

(4s)

(3d)

. 482

Other exceptions occur in the second transition series (the ﬁlling of the 4d 483

levels) and in the third transition series (ﬁlling the 5d levels), and in the rare 484

earth series (ﬁlling the 4f and 5f levels). Knowing this table allows one to 485

write down the ground state electronic conﬁguration of any atom. Note that 486

the inert gases He, Ne, Kr, Rn, complete the shells n =1,n =2,n =3,and 487

n = 4, respectively. Ar and Xe are inert also; they complete the n =3shell 488

(except for 3d electrons), and n = 4 shell (except for 4f electrons), respec- 489

tively. Na, K, Rb, Cs, and Fr have one weakly bound s electron outside these 490

closed shell conﬁgurations; Fl, Cl, Br, I and At are missing one p electron from 491

the closed shell conﬁgurations. The alkali metals easily give up their loosely 492

bound s electrons, and the halogens readily attract one p electron to give a 493

closed shell conﬁguration. The resulting Na

− Cl

−

ions form an ionic bond 494

which is quite strong. Atoms like C, Si, Ge, and Sn have an (np)

(n +1s)

495

conﬁguration. These four valence electrons can be readily shared with other 496

atoms in covalent bonds, which are also quite strong.

Compounds like GaAs, 497

In Table 1.4, we note exceptions (i)–(vii):

i Dropping a 4s electron into the 3d shell while ﬁlling 3d shell(Cr,Cu)

ii Dropping a 5s electron into the 4d shell while ﬁlling 4d shell(Nb,Mo,Ru,Rh,

Ag)

iii Dropping both 5s electrons into the 4d shell while ﬁlling 4d shell (Pd)

iv Dropping both 6s electrons into the 5d shell while ﬁlling 5d shell (Pt)

v Dropping one 6s electron into the 5d shell while ﬁlling 5d shell (Au)

vi Adding one 5d electron before ﬁlling the entire 4f shell(La,Gd)

vii Adding one 6d electron before ﬁlling the entire 5f shell(Ac,Pa,U,Cm,Cf)

viii (h) Adding two 5d electrons before ﬁlling the entire 5f shell (Th, Bk)

Uncorrected Proof

BookID 160928 ChapID 01 Proof# 1 - 29/07/09

26 1 Crystal Structures

t4.1 Tabl e 1. 4. Ground state electron conﬁgurations in a periodic table: Note that n =1, 2, 3,...; l<n; m = −l, −l +1,...,l; σ = ±

t4.2 ← ln + l ↓

t4.3

l =0

t4.4

(1s)

t4.5 l =1

(2s)

t4.6

(2p, 3s)

t4.7 l =2

(3p, 4s)

t4.8

(i)

(4p, 5s)

t4.9 l =3

(ii)

(iii)

(ii)

(4d, 5p, 6s)

t4.10

(vi)

(iv)

(v)

(4f, 5d, 6p, 7s)

t4.11

(vii)

(viii)

(vii)

(viii)

(vii)

100

101

102

(5f, 6d, 7p, 8s)

We note exceptions (i)–(vii):

i Dropping a 4s electron into the 3d shell while ﬁlling 3d shell(Cr,Cu)

ii Dropping a 5s electron into the 4d shell while ﬁlling 4d shell(Nb,Mo,Ru,Rh,Ag)

iii Dropping both 5s electrons into the 4d shell while ﬁlling 4d shell (Pd)

iv Dropping both 6s electrons into the 5d shell while ﬁlling 5d shell (Pt)

v Dropping one 6s electron into the 5d shell while ﬁlling 5d shell (Au)

vi Adding one 5d electron before ﬁlling the entire 4f shell(La,Gd)

vii Adding one 6d electron before ﬁlling the entire 5f shell(Ac,Pa,U,Cm,Cf)

viii (h) Adding two 5d electrons before ﬁlling the entire 5f shell (Th, Bk)

Uncorrected Proof

BookID 160928 ChapID 01 Proof# 1 - 29/07/09

1.6 Binding Energy of Ionic Crystals 27

GaP, GaSb, or InP, InAs, InSb, etc., are formed from column III and column 498

V constituents. With the partial transfer of an electron from As to Ga, one 499

obtains the covalent bonding structure of Si or Ge as well as some degree of 500

ionicity as in NaCl. Metallic elements like Na and K are relatively weakly 501

bound. Their outermost s electrons become almost free in the solid and act as 502

a glue holding the positively charged ions together. The weakest bonding in 503

solids is associated with weak Van der Waals coupling between the constituent 504

atoms or molecules. To give some idea of the binding energy of solids, we will 505

consider the binding of ionic crystals like NaCl or CsCl. 506

1.6 Binding Energy of Ionic Crystals 507

The binding energy of ionic crystals results primarily from the electrostatic 508

interaction between the constituent ions. A rough order of magnitude estimate 509

of the binding energy per molecule can be obtained by simply evaluating 510

V  =



4.8 × 10

−10

esu



2.8 × 10

−8

 8 × 10

−12

ergs ∼ 5eV.

Here, R

is the observed interatomic spacing (which we take as 2.8

A,the 511

spacing in NaCl). The experimentally measured value of the binding energy 512

of NaCl is almost 8 eV per molecule, so our rough estimate is not too bad. 513

Interatomic Potential 514

For an ionic crystal, the potential energy of a pair of atoms i, j can be taken 515

to be 516

= ±

. (1.32)

Here, r

is the distance between atoms i and j.The± sign depends on 517

whether the atoms are like (+) or unlike (−). The ﬁrst term is simply the 518

Coulomb potential for a pair of point charges separated by r

. The second 519

term accounts for core repulsion. The atoms or ions are not point charges, and 520

when a pair of them gets close enough together their core electrons can repel 521

one another. This core repulsion is expected to decrease rapidly with increas- 522

ing r

. The parameters λ and n are phenomenological; they are determined 523

from experiment. 524

Total Energy 525

The total potential energy is given by 526

U =



i=j

. (1.33)