. The term scattering is used when the radiation is getting dispersed in all directions by a point-like atom. The word reflection broadly refers to the waves deviating from a surface like crystal planes (collectively made up of periodically arranged atoms). Both the phenomena result in diffraction maxima when constructive interference condition is

UNIT 4 STRUCTURE UNFOLDING TECHNIQUES

Structure 4.1 Introduction

Objectives 4.2 X-ray Diffraction by a Crystal

Bragg Formulation Laue Formulation: Structure Factors

4.3 X-ray Diffraction and Reciprocal Lattice 4.4 Determination of Crystal Structures

X-ray Diffraction Neutron and Electron Diffraction

4.5 Summary 4.6 Terminal Questions 4.7 Solutions and Answers

Appendix A

4.1 INTRODUCTION

In the previous units of this block, you have seen how crystals are classified on the basis of point and space group symmetries. You have also learnt how information about the structure and nature of a crystal is obtained from their appearance1 characteristics, external faces, density and volume. But such data is not sufficiently reliable to define a crystal, at the atomic scale. The discovery of X-rays ied to the development of a fertile area of research to unfold internal arrangement of atoms in a crystal. When X-rays having wavelengths comparable with inter-atomic spacing interact with atoms in a crystal, they produce diffraction patterns which can be recorded with the help of an appropriate device.

In 1912 van Laue suggested that when a beam ofx-rays falls on a crystal, each atom becomes a source of scattered waves, which interfere constructively to produce maximum intensity in particular directions. For a crystalline material such as quartz or rock salt, we obtain a set of discrete sharp diffraction spots on a photographic film, whereas for an amorphous material such as glass, we obtain a few broad diffused spots. On analysing such patterns, we obtain information about the type of atoms present, their numbers per unit volume, relative sizes, positions and arrangement. In Sec. 4.2, we first consider theoretical aspects ofX-ray scattering from electrons in an atom. The scattering from all atoms in the crystal is determined by summing over all the electrons present in the crystal. The conditions for a crystal to diffract X-rays are determined by using Bragg and Laue formulations.

The concept of a reciprocal lattice is a convenient tool to understand diffraction of X-rays by crystal planes. In Sec. 4.3, we re-examine the diffraction condition in view of the reciprocal lattice concept. You will learn that a photograph of the diffraction pattern is simply a record of the reciprocal lattice vectors. In Sec. 4.4, you will learn about various X-ray diffraction techniques used to determine the structure of solids. It may be noted that advent of these techniques opened fertile channels in material characterization and gave birth to a new field of study - crystallography. However, X-rays are highly energetic and may damage the specimen in some cases. In such situations, it is more convenient to use neutrons or electrons as incident probe in the

Objectives After studying this unit, you should be able to:

state why X-rays are appropriate to unfold crystal structure; obtain Bragg condition for X-ray diffraction; show how Bragg law follows from Laue treatment; explain why reflection from certain planes are missing in the diffraction pattern obtained for bcc and fcc lattices; apply the concept of reciprocal lattice to X-ray diffraction; describe different experimental methods of X-ray diffraction to unfold internal structure of crystalline solids; and compare the relative merits and demerits of X-ray, neutron and electron diffraction techniques.

4.2 X-RAY DIFFRACTION BY A CRYSTAL To understand how X-ray diffraction by crystals provides information about their From PHE-09 course on internal structure, it is important to understand how they interact with crystals. The Optics* you may that first aspect we must appreciate is that for achieving diffraction, any probe when visible light is incident

on an arrangement of two or (electromagnetic or otherwise) must have wavelength of the order of inter-atomic more slits and their separation distances (- 10-'~m). For X-rays to satisfy this condition, their energy must be is of the order of wavelength of

light, the diffraction pattern hc ( ~ . ~ ~ X ~ O - ~ ~ J S ) X ( ~ X I O ~ ~ S - ~ ) may be observed. The E = - = diffraction of electromagnetic

(10-'Om) x (1.6 x lo-'' J~V- ' ) waves by atonis in a solid is an analogous phenomenon; the wavelength of electromagnetic

= 1 2 . 4 ~ lo3ev. radiations used should be of the order of inter-atomic distances

That is, X-rays of a few keV will be required for probing microscopic structure of in solids (- ]A). A crystal solids using diffraction phenomenon. You have learnt the praperties of X-rays in differs from a diffraction PHE-11 course on Modem Physics. However, for brevity, we state them here: grating in that the diffracting

centers here are not in one plane! So, we can imagine a X-rays are produced by sudden deceleration of electrons in metals as well as by as a 3-dimens,onal

excitation of core electrons in the atoms of a metallic target. space-grating.

Characteristic X-ray energies are a few keV; i.e. X-rays are ,e.m. waves having about 1 A wavelength and are suitable tools for studying crystal structure.

4.2.1 Bragg Formulation

It had been observed experimentally that crystalline materials exhibit sharp spots in X-ray diffraction patterns. To explain these results, W.L. Bragg put forward a model which resulted into the conditions for diffraction in a very simple way. He proposed that crystals may be imagined to consist of parallel atomic planes; similar to the stack of reflecting glass plates. According to him, when a crystal is irradiated with X-rays, electrons in individual atoms scatter them elastically so that the regular structure of these atoms acts as a three-dimensional grating. Waves (X-rays) scattered from different atoms interfere with each other constructively in some directions and destructively in others. When detected on a X-ray film, the pattern is characteristic of the particular crystal structure.

Let us consider X-rays striking a set of parallel planes of atoms in a lattice. Every crystal has a large set of such planes. Fig. 4.1 shows a two-dimensional representation of a crystal to illustrate some of many possible planes that could reflect X-rays. Note that reflection from some planes are more dominant than others because the density of atoms on them is greater than that on others. (Plane AA' has more number of atoms per unit length as compared to BB'.) The facets seen on the surface of the crystals are such crystal planes. They are the macroscopic manifestation of the microscopic lattice

structure of the crystal. (A cubic lattice structure of NaCl is evident in its big cubic crystals and the faces represent the crystal pltmes in it.)

* * a . . . . . a . . a , ,

. . . . . . . . . . . . O . . , I

To obtain sharp diffraction spots, Bragg considered specular reflection by atoms in any one plane and reflected waves from successive parallel planes separated through

In specular reflection, the distance d (Fig.4.2a) to interfere constructively. The path-difference between two angle of incidence n ~ s t be specylarly reflected waves (X-rays) is just AB + BC. In terms of 8 and d, we can writ equal to the angle of the total path-difference as reflection. Hence, here the diffracted beams are just like reflected ones in optics A = AB+BC=2dsin8. (4.1) term~nology. This leads to the use of terms X-ray As you know, for constructive interference, this path difference should be an integral d@ractron and reflection intnchangeably in the texr multiple of 1. This leads us to Bragg ~ondition for constructive interference:

2ds in8=nh; n = 1,2 ... (4.2)

The integer n defines the order of the corresponding diffraction maxima.

800 70 60 500 40' 300 200 - 20

Fig.4.2: a) X-rays reflected by parallel crystal planes; b) Bragg angle is half the total angle by which incident beam is deflected; and c) diffraction maxima obtained from a KBr crystal. The numbers in brackets indicate reflecting planes for various peaks

Note that the angle of incidence in X-ray crystallography is measured from the plane of reflection rather than from the normal to that plane, as in classical optics. As a result, 8 is just half the angle of deflection of the incident beam (Fig. 4.2b). Since sin 8 5 1, you will note that for diffraction to be observed, we must have h 5 26. It is for this reason that Eq. (4.2) does not hold when waves of larger wavelengths are made to fall on a crystal. This puts a limitation on the wavelength ofx-rays that can be used in diffraction studies with reference to interplanar distances in a crystal. You will appreciate this point further on answering the following SAQ.

SAQ 1 Structure Unfolding Spend Techniques

Radiation of wavelength 4 x 10-~m is incident on a crystal with interplanar separation 2 min. distance of 2 x 1 ~ - ~ c m in a solid. Shall we observe Bragg diffraction pattern? Justify your answer.

When a narrow beam of X-rays is incident on a crystal, diffracted beams emerge from it for each family of parallel planes satisfying Eq. (4.2). Each emerging beam has an intensity depending on the density of atoms in the corresponding set of planes. Fig.4.2~ shows the diffraction maxima obtained at various angles from a KBr crystal. The height of the peaks represents the intensity of diffracted beah. When photographic film is exposed to this X-ray radiation, it displays spots. These are analysed to obtain information about diffraction angle 8 and thereby interplanar distance d to ascertain the crystal structure. The X-ray diffraction pattern from NaCl is shown in Fig.4.3.

You must have noted that Bragg formulation is over-simplified, Since X-rays are scattered by individual atoms, you may not agree to represent crystal planes by a set of continuous pile of reflecting planes. That is to say, sectioning a crystal into lattice planes and the assumption of specular reflection may be done away with, as suggested by Laue. You will learn about it now.

Flg.4.3: X-ray dlflractlon pattern from NnCl crystal

4.2.2 Laue Formulation: Structure Factors According to Laue, the diffracted beam arises due to interference of X-rays scattered by all the atoms in a lattice. While adding the contributions of all these scattered The main contribution to X-ray scattering is made by core waves, we will have to consider their phases. You will appreciate here that the electrons. The valence difference in phases arises because waves travel different distances due to spacing electrons are a) much less in between individual scatterers. We shall discuss diffraction of X-ravs in a more general m~mber and b) can be involved

d u in bonding and may not be way by first considering the scattering of wave from a single atom and then collective associated with the atom whose

effect of all lattice sites in a crystal to obtain the condition for diffraction maxima. lattice position is under consideration.

a) Scattering from an atom: Atomic scattering factor We know that in an atom electrons surround the nucleus. When these electrons are subjected to a monochromatic beam of X-rays, they gain energy but subsequently lose it by emitting radiation in all directions giving rise to a spherical wavefront. You are familiar with the concept of spherical wavefiont from your PHE-02 course on Oscillations and Waves.

From Unit 14 of PHE-07 course you may recall that a plane electromagnetic wave incident on a single electron may be mathematically represented as

where A is amplitude of the wave, ko is wave vector (ko = 2x1 h) and o is angular frequency. The scattered wave is spherical and its amplitude decreases with distance. Mathematically, we may write

wheref, is scattering length of the electron representing the fraction of radiation scattered by the electron, D is radial distance between the electron and the point of observation i.e. where the detector or photographic film is placed, and k is wave number of the scattered wave. For elastic scattering, k has the same magnitude as ko,

In case the incident wave acts on two electrons (Fig.4.4), both will give rise to spherical waves and the scattered wave observed at a distant point in space (at

"1 Crystal Structure distance D) will be obtained by superposition of the waves scattered from each electron. The resultant wave may therefore be represented as where 6 is the phase difference between the scattered waves. Note that the time factor has not been mentioned in the above equation but its presence is implied in 6. Also, in this equation it has been assumed that distance between two electrons is negligible in comparison with D.

Incident waves with plane wavefront

result in spherical wavefronts

I

Fig.4.4: Schematic representation of scattering of X-rays by two electrons

From ~ i g 4 . 4 we note that the path difference between the waves incident on and scattered from two electrons separated by displacement vector r would respectively be

A, = BC = r cos 8, = r.io

and

where io and; are unit vectors along the incident and scattered directions, respectively. For specular scattering, as shown in Fig.4.2bY 8, = O2 = 8 = half of the scattering angle. Hence, net path difference between two scattered waves is given by

If ko and k are the wave vectors in the incident and scatter directions respectively, wAk we can write 2n

28 A k = k - k o =-(i-io). h

Ak is referred to as the scattering vector (Fig.4.5). Far elastic scattering, ]kt = lkol and the magnitude of Ak is given by

Fig.4.5: Scattering vector

A k=2ks in 8, (4.7a)

where 0 is half the scattering angle.

The phase difference between the reflected waves can now be written as

The expression for resultant scattered wave, represented by Eq. ( 4 3 , may now be rewritten as

If the origin is taken as an arbitrary point rather than at one of the electrons, the expression for resultant scattered wave can be written as

where r, and 1-2 are the position vectors of the two electrons with respect to the chosen origin. Note that if the origin were chosen at 'electron l ' , as shown in Fig.4.4, rl = 0, and r2 = r is displacement vector between the two electrons and Eq. (4.10) reduces to Eq. (4.9). When X-rays are scattered by N such electrons, Eq. (4.10) takes the form

Comparing Eq. (4.1 1) with Eq. (4.4), we can define the total scattering length as

This equation represents the fraction of incident radiation scattered by N electrons. Then Eq. (4.11) giving the field observed at a distance D due to scattered X-rays from all electrons in an atom becomes

You know that intensity I of a wave is proportional to the square of its amplitude. Hence the intensity of scattered X-rays is proportional to the square of the magnitude of the scattering length:

We may point out here that this expression conforms to the case of coherent scattering where scatterers maintain a definite phase relationship with each other, and leads to interference between scattered waves. On the other hand, if the scatterers were to oscillate randomly, the waves scattered from them would not interfere because of incoherence. In that case the intensity at the detector would simply be the sum of partial intensities, i.e.

Structure Unfolding Techniques

I where N is the number of scatterers. 65

Crystal Structure Note that electrons do not have discrete positions in an atom; they a re spread as a continuous charge cloud around the nucleus of an atom. Therefore, it will be more :tppropriate to rewrite the discrete sum in Eq.(4.12) as an integral

where p(r) is the density of the charge cloud (in electrons per unit volume), and the integral is over the atomic volume. i i

I

The atomic scattering We then define an atomic scattering factor, fa , which is the integral in Eq. (4.16): factorf, is defined as the ratio of radiation amplitude scattered by the actual J ir.Ak d v f, = v(r>e electron distribution in an (4.17) atom Cf) to that scattered by a single electron we). This is a dimensionless quantity. For spherically symmetric charge distribution, as you will prove in TQ 3, the atomic

scattering factor has the form

R sin rAk

f a = P n r 2 p ( r ) dr , 0

where R is the radius of the atom. This result shows that f, depends on the scattering angle 0 (as Ak = 2k sine) through

sin rAk the factor - in the integrand; in fact atomic scattering factor decreases as

r Ak scattering angle increases due to interference between scattered waves resulting from

-

sin rAk different regions of the charge cloud. The oscillatory behaviour of - is depicted rAk

in Fig.4.6.

fig4.6: Plot of oscillating factor versus rAk

sin rAk For the forward direction, i.e. 0 = 0, Ak = 0 and - 4 1. Then Eq. (4.18) reduces

r Ak

Note that the integrand represents the charge inside a spherical shell of radius r and thickness dr. So we can conclude that the integral gives the total electronic charge inside the atom, i.e. it equals the atomic number 2:

This means that in case of carbon,& (8 = 0) = 6. To check your progress, we now wish that you should answer an SAQ.

SAQ 2 Spend 2 min.

Interpret Eq. (4.20) physically. What will happen if we look in the forward direction?

Structure Unfolding Techniques

You have now learnt how the electrons in an atom lead to X-ray scattering from every lattice site. Let us consider the effect of scattering from all atoms arranged periodically in a crystal lattice.

b) Scattering from a crystal: Lattice and geometric structure factors You now know from Eqs. (4.14), (4.16) and (4.17) that the intensity of an X-ray beam diffracted by an atom depends on its atomic scattering factor. The total scattering from a crystal depends on various atoms constituting the crystal and their arrangement in the crystal, i.e. the number, type and distribution of atoms within the unit cell. Moreover, the scattered waves originating from different atoms of a unit cell may or may not be in phase with each other. It is, therefore, important to know the effect of various atoms (present in a unit cell) on the total scattering amplitude in a particular direction. To this end, by analogy with the atomic case, we define crystal scattering factor as

P where the sum extends over all electrons in the crystal. As discussed above, this may be considered to be made up of two parts: We first sum over all electrons in an atom and define atomic scattering factor& for the yth atom and then sum over all the atoms in the lattice. Then, Eq. (4.21) modifies to

where R, is the position vector ofyth atom whose corresponding atomic factor is&

Fig.4.7: Contribution of m th unit cell to scattering

Crystal Structure To simplify Eq. (4.22) refer to Fig.4.7. It shows a 2-D non-Bravais lattice with basis consisting of two types of atoms: A and B depicted as cross and circle, respectively. The unit cell in this lattice consists of one atom each of type A and B. The origin of the lattice is at 0 while Of is the reference point in the mth unit cell. Vector R', defines the position of this unit cell with respect to 0. Let a, and a2 denote are the position vectors of atoms A and B with respect to Of. Note that we have considered here a lattice consisting of two types of atoms and there are only two (a, and a2) position vectors. But in the general case, where unit cell consists of many atoms, there are those many aj vectors. From the figure it is clear that we can express the position vector Ry of any atom as a combination of R', and aj, i.e.

RY = R: + ai. (4.23)

To account for all atoms in the lattice, we have to sum over all the atoms in the unit mth cell and then sum over all the unit cells. Hence Eq. (4.22) can be rewritten as

(4.24) m i

Rearranging terms, we get

(4.25)

You will observe that the first summation term corresponds to the contribution from the atoms in a unit cell. Let us denote it as

(4.26)

The second summation term in Eq. (4.25) corresponds to the total contribution of all the unit cells in the lattice. Let us denote it by Sc and write

m

Then Eq. (4.22) simplifies to

f c = F x S c .

F is called geometrical structure factor and Sc is the lattice structure factor. Note that F depends on the geometrical arrangement of atoms within the unit cell and SC depends on the arrangement of cells in the lattice, and is determined only by the structural properties of the crystal system. These two factors can be treated independent of each other. Let us now understand what information can be obtained from lattice structure factor.

For X-ray scattering, the lattice structure factor Sc is very important and you will soon discover that for certain values of Ak, Sc will be non-zero and these values form a discrete set which is connected to Bragg law!

Fig.4.8: Scattering from a one-dimensional monoatomic lattice

Let us consider a one-dimensional monoatomic lattice (Fig.4.8). Here each atom constitutes a cell and the interatomic separation a is the lattice constant. Hence the lattice structure factor for the basis vector a takes the form

where N denotes the total number of atoms in the lattice. By changing the sllmmation variables from 1 to N to 0 to N-1, we can rewrite the lattice structure factor as

This is a geometric series with common ratio You can easily calculate the sum to obtain

( 1 - eiNa.Ak sc = 1 - e"," )

You can readily show that N

sin -a.Ak i (N-l)p,Ak 2 < S c = 1 xe 2

Structure Unfolding Techniques

The sum of a finite geometric I-r"

series is given by -, I-r

where r is common ratio.

sin L a . ~ k 2

SAQ 3 Spend 5 min.

Derive Eq. (4.30).

Since (e'J2 =I , we can write 2 N , sin -a.Ak

2 lscl = 2- . 2 1 sin -a:Ak

2

Since, the numerator involves N, which is a large number, the terms in the numerator will oscillate more rapidly than in the denominator. Do you recognise this expression?

2 The plot of 1 versus a.Ak is shown in Fig.4.9. You can see a number of maxima

2 between 0 and 2n. At a d k = 0 and a.Ak = 2n:; Isc I = N'. This defines the principal maximum.

Crystal Structure

I

Fig.4.9: Plot of lattice structure factor versus a.Ak

In between two principal maxima, there are a number of subsidiary maxima which arise due to rapid oscillations of the terms in the numerator. When N is sufficiently large, i.e., there are a large number of atoms, these subsidiary maxima can be ignored. In other words, (sc l2 can be taken to be non-zero only in the immediate vicinity of the

2 principal maximum, Mathematically, we can say that IscI will be finite at a.Ak = 27th where h is an integer. Thus, the condition for constructive interference is

Note that this condition determines all the directions in which diffraction takes place. To understand the physical significance of this equation, we recall the definition of Ak (Eq. (4.7)) and again refer to Fig.4.8:

This means that a.Ak denotes the phase-difference between two consecutive scattered X-rays.

For a given h, the condition for constructive interference defines an, infinite number of I directions forming a cone whose axis lies along the lattice line. To see this, we rewrite

Eq. (4.32) as

where Po is the angle of incidence and P is the angle of diffraction. Thus, for a given h and Po, the beam diffracts along all directions for which P satisfies Eq. (4.32). These diffracted beams form a cone whose axis lies along the lattice, and whose half-angle is p. For h = 0, the diffraction cone contains the forward scattering as well. Diffraction cones corresponding to different values of h are shown in Fig.4.10.

Structure Unfolding Incident

rn

or ward scatter Fig.4.10: Diffraction cones for first (h = 0) and second order (h = 1) maxima

The steps outlined above can be extended to determine the lattice structure factor for the 3-D lattice. Using the fact that in real crystals, 3 basis vectors define the lattice, you wiM find that Eq. (4.32) expressing diffraction condition modifies to

a2.Ak = 2x k, and

a3.Ak = hr I. (4.34)

This set of equations constitutes Laue diffraction condition.

The value of lattice structure factor Sc under these condition is equal to the total number of scattering atoms in the lattice, N , M ~ and the crystal scattering factor fc can be-written as

IC(~, = Fhkl N~~~~~ 9 (4.35) where h, k and 1 are integers. It may be noted that h, k, and 1 are the standard notations for Miller indices. But we have used these for the set of integers because these give the diffraction condition for the set of planes represented by (hkl). Fhkl denotes the geometric structure factor associated with (hkl) family of planes. The geometry of planes, in turn is decided by the structure of unit cell of the crystal. The intensity of the diffracted waves is then

As mentioned earlier, Fhu depends on the shape and contents of the unit cell. Thus, if FM is zero for certain indices, the intensity will be zero even though the corresponding planes satisfy Bragg condition.

Let us now evaluate F M . For identical atoms; let a, be the position vector for the jth atom in the unit cell. Then,

a, =ujal + via2 + wja3

where u,, v, and w, are the position coordinates of the jth atom.

Techniques

Using the condition of constructive interference, we can write

aj .Ak = 2x(u,h+vjk + w,l), so that the generalised form of Eq. (4.26) for geometric structure factor becomes

1 Crystal Structure Let us now obtain expressions for the geometric structure factor for some simple crystals.

Simple cubic crystals

From Unit 2, you will recall that the effective number of atoms in a unit cell of a cubic structure is one. Assuming that this atom lies at the origin, Eq. (4.37) reduces to

That is, for a simple cubic crystal, Fhkl is equal to atomic scattering factor.

Body-centred cubic crystals

The effective number of atoms in a bcc unit cell is two: one occupies the comer position and the other occupies the body-centre of the cube. If comer atom is arbitrarily chosen origin (000) of the coordinate system, the coordinates of the of centre atom will be (: --- : :) . And if both atoms in the unit cell are identical

2

( fa , = faZ = f a ) Eq. (4.37) becomes

From this result, we note that

Fhu= 0 fo rh+k+lodd

= 2f, for h + k + I even,

since e'" = -1 for n odd and +1 for n even.

From this result you must note that for the bcc structure, reflection from certain planes with Miller indices h, k, 1 will be absent. (These reflections would be present for a simple cubic structure having the same edge dimensions.) For example, in bcc crystal (100) reflection will be absent but (200) reflection will be present. Similarly, there will be no (1 11) reflection but (222) reflection will be observed. Physically, this can be understood as follows: Consider a simple cubic crystal and the (100) reflection. For this, the top and bottom faces of the unit cell will give rise to reflected waves, which differ in phase by 2n. But in the case of a bcc structure, in addition to the top and bottom faces, there will be another plane formed by the body-centred atoms of the unit cells. This additional plane of atoms is parallel to and equidistant from the top and bottom faces of cubic unit cell, as shown in Fig.4.11.

Fig.4.11: Phase relations for the (100) reflection from a bcc lattice 72

The concentration of atoms in this intermediate plane is equal to that of the top and Structure Unfolding bottom planes defined by the faces the cube and so this plane will give reflected Techniques waves of equal intensity. However, the phase of the reflected X-rays from this plane will lag by 15 compared to that from the top plane. The reflected waves from the top and the middle planes of atoms, therefore, interfere destructively resulting in zero intensity. However, the reflection corresponding to (200) is observed. It is because in this case reflections from the top and the bottom planes differ in phase by 475 whereas the reflections from the top and the middle planes differ in phase by 2n, giving rise to constructive interference.

If the atoms located at (000) and ( --- k : i) in a bcc lattice are different, as in the case of CsCI, Eq. (4.37) takes the form

This implies that

for h + k + 1 even F = f i +h I

=fi -h f o r h + k + l odd, (4.42) wherefi andh correspond to Cs and Cl, respectively. Then, the relative magnitudes of f i andh would decide whether a reflection would be possible or not! Now you may like to answer an SAQ.

SAQ 4 Determine the condition governing geometric structure factor for fcc lattice and enlist at least 3 missing planes.

For cubic structure, the various possible planes are (loo), (1 lo), (1 1 l), (200), etc. The interplanar distances in these cases can be determined by using the equation (2.13).

where N = h2 + + 1'.

The possible values of N for the planes showing X-ray diffraction for sc structures are given in Table 4.1.

Note here that N value of 7, 15 . . . etc. are missing in case of simple cubic structure.

From Eq. (4.40), you can easily deduce that the only allowed values of N for bcc structureare: 2,4,6,8,10,12,14,16 ... etc.

Similarly fcc structure will show the X-ray diffraction for the planes with N value of 3,4,8,11,12,16 . . . etc. These conditions on N values help us in determining the crystal structure from diffraction pattern analysis, as will be discussed in the Example 1 of Section 4.4.

Spend 5 min.

In the cubic system the interplanar distance dm will be same for a given combination of h, k, I in any order:

dim = dolo = dm1 or

4 1 0 = 4 2 0 = 4 0 2 = dOl2 = dO2, = dZo1 etc.

Table 4.1

Plane N = h2 + k2 + i2

Crystal Structure 4.3 X-RAY DIFFRACTION AND RECIPROCAL LATTICE In the previous unit, you learnt about reciprocal lattice. Can you guess its importance for studying X-ray diffraction?'We now know that X-rays are diffracted by various sets of parallel planes having different slopes and interplanar spacings. It is difficult to visualize all such planes because of their two-dimensional nature. To get over this : difficulty, Ewald put forth the idea of reciprocal lattice such that each point in a reciprocal lattice is representative of a particular set of planes. (You learnt the construction and properties of reciprocal lattice points in Unit 3.) In Appendix A we have shown that i n terms of reciprocal lattice vector G, Bragg condition for constructive interference can be expressed as

where Ak is scattering vector. For elastic scattering, this result simplifies to

Ewald's Construction

The condition Ak = k - ko = G implies that scattering does not change the magnitude of the wave vector, only its direction changes. This also means that the scattered wave differs from the incident wave by a reciprocal lattice vector G.

A. helpful geometrical construction to locate the direction of the scattered wave, known as Ewald's construction, is shown in Fig. 4.12. Note that the points in the

Fig.4.12: Ewald's construction

figure-represent the reciprocal lattice points. Vector ko is drawn from 0 in the direction of the incident beam and terminates at A in the reciprocal lattice. A sphere of

2x radius ko = - is drawn with A as center. If this sphere does not pass through any

h other reciprocal lattice point (except O), the beam will not be diffracted by the crystal. However, if the sphere passes through point B of reciprocal lattice, the diffraction will take place satisfying the condition k = ko + G .

This construction of Ewald's sphere is quite useful in the analysis of X-ray and neutron diffraction patterns to determine the crystal structure.

4.4 DETERMINATION OF CRYSTAL STRUCTURES

I As mentioned earlier, the diffraction techniques which employ X-rays, neutrons or

of crystal structures. Of these, X-ray diffraction techniques were the first to be used. Let in learn about X-ray diffraction methods.

Structure Unfolding Techniques

4.4.1 X-ray Diffraction X-ray diffraction is widely used to determine crystal structure of solids using Bragg law (Eq. (4.2)). You may recall that diffraction can take place for those values of d, 8 and 3L which satisfy this equation. For structural analysis, X-rays of known wavelength are used and the angles for which reflections take place are determined experimentally. This enables us to obtain the value of d using Eq. (4.2). On the basis of this information, we can determine the size of the unit cell. On the other hand, from the measurement of intensity of the diffracted beam we get the structure factor Fhk, (using Eq. (4.37)) which facilitates knowledge of the arrangement of atoms in the unit cell. Laue analysis of structure factor also enables us to trace the missing planes in diffraction pattern and thereby determine the crystal structure. In this section, you will study this analysis by various examples.

In X-ray diffraction studies, one of the following experimental methods may be used:

Laue method; Rotating crystal method; Powder diffraction method.

Let us learn about these now.

The Laue method is of great practical interest and is used very widely by solid state physicists for rapid determination of the symmetry/orientation of a single crystal. The experimental arrangement is shown in Fig.4.13. A single crystal under study is held stationary and an X-ray beam with a spread in the wavelength is made to fall on it in a fixed direction i.e. angle 8 is fixed while 3L varies. Different wavelengths present in the incident beam are reflected by appropriate planes out of the several present in the crystal satisfying Bragg condition.

- -

Diffracted Film

Primary beam

Fig.4.13: The Laue method: a) experimental arrangement; b) typical Laue pattern for a hexagonal crystal, with the X-ray beam parallel to the 6-fold symmetry axis; and c) Ewald construction

Flat photographic films are placed around the crystal. If the spread of wavelengths in the incident beam is sufficiently large, we will obtain ome spots on the film . corresponding to reciprocal lattice points which satis 4 Bragg condition at the given angle. Now refer to Fig.4.13~. It shows Ewald construction for Laue method. Since orientation of the crystal and direction of incident x-rhys are fixed, the spread in wavelengths corresponds to spread in the magnitude of wave vectors between ko and kb. The Ewald spheres for all incident wave vectors will lie in the region between the spheres centred on the tips of vectors ko and kb. However, Bragg peaks can be observed corresponding to all reciprocal lattice points in the shaded region only.

Crystal Structure Note that the wavelength corresponding to a spot cannot be measured. So it is not possible to determine the actual values of the interplanar spacings by this method. This means that we can determine only the shape of unit cell, not its dimensions. However, if the direction of the incident X-ray beam is along the axis of symmetry of the crystal, the diffraction pattern will also exhibit this symmetry. As a result, it facilitates information about the orientation of a single crystal of known structure.

The rotating crystal method is used to analyse the structure of a single crystal. The schematic representation of the experimental arrangement is shown in Fig.4.14. A single crystal sample of a few mm dimensions is mounted on a spindle which can be rotated.

Film

Primary beam

Axis o i rotation

Fig.4.14: Experimental arrangement for the rotating crystal method

The pattern obtained in the rotating crystal method can be easily understood by the Ewald construdtiol~ for this experimental geometry. In this case, the wavelength h and the angle of incidence of X-rays are kept constant and the crystal is rotated. It effectively means that the reciprocal lattice is also rotated. The basic condition for diffraction in Ewald construction is that the reciprocal lattice point lying on the Ewald sphere gives rise to diffraction maxima in that direction. In this experimental configuration, as the crystal rotates, various reciprocal lattice points fall on the Ewald sphere giving rise to diffraction spots.

P. p,

(a) (b) Fig.4.15: Ewald construction for rotating crystal method. a) Only one reciprocal lattice point PI

lies on the Ewald sphere; and b) When crystal is rotated by 15" two reciprocal lattice points Pz and Pj lie on the Ewald sphere

In Fig.4.15a, the Ewald sphere is depicted by a circle with the centre at A. The concentric rings depict the rotating reciprocal lattice points with centre along one of the principal axes of the crystal. All the reciprocal lattice points falling on Ewald sphere give rise to diffraction at a given instant. As the crystal is rotated (Fig.4.15b) different sets of reciprocal lattice points satisfy the diffraction condition (i.e. lie on the Ewald sphere) and give rise to maxima. In this way, all the reciprocal lattice points

(i.e. crystal planes) are recorded one by one as the crystal rotates through 360. The Structure Unfolding resultant photograph can throw light on the symmetry, shape and size of the unit cell. Techniques

The powder diffraction method of X-ray diffraction was introduced in 1916 by Debye and Scherrer. It is the most widely used method in crystal structure analysis. It is modification of the rotating crystal method where the axis of rotation is varied over all possible orientations effectively by moving the sample. In practice, it is achieved by using the sample in the form of a fine powder containing a large number of tiny crystallites randomly oriented with respect to each other. The schematics of the method is illustrated in Fig.4.16a. It consists of a cylindrical camera. The powdered sample is filled in a (non-diffracting) capillary tube and mounted at the centre of the camera. A collimated monochromatic beam of X-rays is made to fall on the sample. Since the specimen may be regarded as comprising a large number of small crystallites (- 10" in 1mm3) whose crystal axes are randomly oriented, almost all

X-rays source Filter

Transmitting beam

-------- pi--.! i

Photographic film

Oriented differently (b)

Entrance hole Exit hole

d3 d2 d, d,

Fig.4.16: a) Schematics of Debye-Scherrer camera; b) a cone produced by reflection of X-rays from identical planes having different orientations; and c) flattened photographic film after developing and indexing diffraction lines

Crystal Structure possible 8 and d values are available. In this method the sample is not moved but planes with various (hkl) value are simultaneously exposed to X-rays due to randon! packing of crystallites. The diffraction peaks are obtained for those values of d and 8 which satisfy Bragg condition. Also, since for a particular value of 0, several orientations of a particular set of planes are possible, the diffracted X-rays corresponding to fixed values of 8 and d lie on the surface of a cone whose apex is at the sample and the semi-vertical angle is equal to 28 (Fig. 4.16b). Different cones are observed for different sets of d and 8 for a particular value of n, and also for different combinations of 8 and n for a particular value of d. The undeviated X-rays move out of the camera through an exit hcle located diametrically opposite to the entrance hole. A photographic film is attached to the inner sides of the curved surface of the camera. Each cone of the reflected X-ray beam leaves two impressions on the film which are in the form of arcs on either side of the exit hole with their centres coinciding with the hole (Fig. 4 .16~) . Similarly, cones produced by back reflected X-rays produce arcs on either side of the entrance hole. The film is exposed for a few hours in order to obtain lines of sufficiently high intensity. It is then removed from the camera and developed.

In the Ewald construction for powder method, each reciprocal lattice point lying on the Ewald sphere results in a circle defined by the cone of angle 28 cutting the sphere. This is because of random orientation of reciprocal lattices due to random packing of crystallities. Further, most of the reciprocal lattice points fall on the Ewald sphere when combined constructions for all the crystals exposed to X-rays is considered.

In present day X-ray diffraction apparatus, the photographic films are replaced by solid state detectors that provide electrical signals whose strength is proportional .to the intensity of X-rays falling on them. Instead of continuous photographic film stripe, the X-ray detector is made to move around the sample in a circular path in order to detect the diffraction maxima around the entire cylinder. The data collected is in the form of diffraction angle 20 versus intensity of the diffracted X-ray. The data acquisition time in this case is reduced drastically due to amplification capacity of electronic circuitry.

In Fig.4.16b, the angle 8 corresponding to a particular pair of arcs is related to the distance D between the arcs as

where R is the radius of the camera. If 0 is measured in degrees, the above expression is modified as

Note that calculations can be made simpler by taking the radius of the camera in multiples of 57.296. For example, by taking R = 57.296 mm, we get

Thus, one-fourth of the distance between the corresponding arcs of a particular pair in millimeter is a measure of the angle 8 in degrees. Knowing all possible values of angle 8 and considering only the first order reflections from all the possible planes, Eq. (4.2) is used to calculate the interplanar spacing for various sets of parallel planes which contribute to these reflections. Thus, we have

The values of dobtained in this way are used to determine the lattice spacing of the crystal structure. A typical X-ray diffraction pattern obtained by the powder method is shown in Fig.4.17.

Structure Unfolding Techniques

Fig.4.17: Typical X-ray diffraction pattern obtained by powder method

Following example will illustrate the analysis done for determining the crystal structure and lattice constant from the diffraction pattern.

The X-ray diffraction maxima obtained from an alkali halide crystal using Cu K, radiation (h = 1.5404A) are 0 = 10.83", 15.39', 18.99', 22.07' and 24-84". Determine the lattice constant and the structure of the crystal under study.

Solution:

To determine the crystal structure, we first determine the interplanar distances represented by the given 8 values. From Eq. (2.13) we know that

But by Bragg condition,

2 Substituting for d from Eq. (4.48)

so that

sin 0 h2 =-

h 2 + k 2 + 1 2 4a2

For the given experimental set up, h = 1.5404 A and a, the lattice parameter; both are constants. Hence RHSof this equation is a constant i.e. for every 0, there must be a

set of h, k, 1 values that result into the fraction sinZ 0 = constant. hZ + k2 + l 2

To determine this constant factor we first determine sin20 corresponding to every 0 value and then divide by different values of h2 + k2 + P = N to find the constant fraction. This procedure is followed to prepare Table 4.2,

/ Crystal Structure Table 4.2

The common factor in above table = 0.0353. It corresponds to N = 1,2,3 ,4 and 5. I

Kinetic energy of a neutron is given by

E = p2/2m,, where m, is = Mass of neutron

(= 1.67 x 1 0 - ~ ' k ~ )

Hence from the discussion in Section 4.2, the crystal structure is simple cubic. Now,

This concludes our discussion of crystal structure determination using X-rays. However, you must have realised that X-ray diffraction studies can be used only to know the structure of crystalline substances; these cannot be used for amorphous solids, liquids or gases.

You also know that X-rays are diffracted by electrons in an atom. Therefore, it fails completely to distinguish isotopes. Also, for a lighter substance such as hydrogen, the number of electrons are few, and hence X-ray diffraction intensities are very poor. Moreover, since we require highly energetic X-rays for diffraction studies, it is possible that the specimen may be damaged. In view of these limitations, we have to look for alternative techniquedProbes/tbols. Neutrons and electrons are two such probes which have been used extensively in crystallography. Let us now learn about these.

4.4.2 Neutron and Electron Diffraction

According to de Broglie principle, every moving particle has a wave associated with it:

If the energy of the particle is such that its wavelength h - 1A it can be diffracted by crystals in the same way as X-rays. In the case of neutrons,

Neutrons of such low energy, called slow neutrons, can be obtained from nuclear research reactors. Use of neutron diffraction and spectroscopy has provided vital information about atomic structure. Since neutron-matter interaction is weak, the information is undistorted. And in some cases, the information cannot be obtained in any other way.

It is noteworthy that details of neutron diffraction are precisely the same as those for X-rays and hence, Bragg Law and Laue equations hold as such. These are the direct conseqqences of the structure factor which, being a lattice sum, depends only on the lattice structure and not on the atomic scattering factor. The only difference lies in the fact that neutron scattering length replaces electron scattering length in the relation of scattered wave given in Eq. (4.4). Some of the salient features of neutron diffraction are:

It can distinguish between different atomic isotopes since neutrons interact with nuclei.

It can be used to study magnetic properties of materials. The magnetic crystals possess magnetic moment due to the net spin ofelectrons in atomic orbitals. A neutron also behaves like a tiny magnet and "feels" the moment generated by the electrons.

One of the major drawbacks of the neutron diffraction technique is that it requires high intensity beams. But the intensity of even the most powerful neutron sources is less than the intensity obtained from common X-ray sources by about lo5. Therefore, we require nuclear research reactors for obtaining neutron beams of appropriate intensitylenergy. Also, we have to use large crystals.

You may now like to answer an SAQ.

SAQ 5 Spend 2 min.

Which technique will you use to study the structure of ZnHz crystal and why?

From your twelfth classes, you will recall that electrons also exhibit wave-particle duality, and can undergo diffraction. The wavelength of an electron is given by

where h is -in angstroms, and V is in eV. For h = 1 A, the potential V= 150V, or E = 150 eV. There is one important difference between X-rays and electrons - the ekctrixw re scattered much more efficiently by atoms than X-rays. In fact, atoms scatkr more strongly by several powers of ten for the energies involved.

You will understand other difference between these techniques by following SAQ.

SAQ 6 a-+ 2-

The experiments with electrons need to be carried out under ultra high vscura~ conditions, while X-ray studies can be performed in air. Can you telf, why?

Structure Unfolding Techniques

Crystal Structure 4.5 SUMMARY

The wavelength associated with a probe used to unfold crystal structure should be of the order of interatomic spacing in the crystals (- 1 A). X-rays of a few keV and neutrons of a few meV are useful probes.

Bragg condition for diffraction maxima is 2d sine = nh where d is the interplanar spacing, h is the wavelength, 8 is the diffraction angle and n is the order of diffraction.

Laue proposed the generalised method of determining crystal structure based on the atomic scattering factor and geometrical structure factor of the crystal and the elements co~istituting it.

Laue method is particularly useful in determining the orientation of a single crystal.

Powder method is most widely used for crystal structure determination.

Ewald construction determines the diffraction condition based on the reciprocal lattice vectors.

In Laue method, the radius of Ewald sphere varies as white X-rays are used. In rotating crystal method, different reciprocal lattice points satisfy diffraction condition as crystal rotates. In powder diffraction, many of the reciprocal lattice points satisfy diffraction condition simultaneously due to randomly oriented crystallites.

4.6 TERMINAL QUESTIONS Spend 30 min. 1. Determine the geometric structure factor of monoatomic diamond structure built

around fcc lattice with the bases located at (0,0,0) and (I) - , - I) , - I)) . State the conditions on h, k, I.

2. Calculate the energy of the X-ray beam of wavelength 0.2 nm. For what separation between the lattice planes, when the angle of incidence is 42", we will obtain first order diffraction maxima.

3. Prove that for spherical charge distribution the atomic scattering factor is given by sin rAk dr

fa = J4nr2p(r)-

4. In Debye-Scherrer experiment, the following values of sin28 were obtained for Fe K, radiation (h = 1.932 A) : 0.1843,0.2450,0.4887, 0.6707, 0.73 14, 0.9739. Assuming that all'these values are for first order diffraction determine the crystal structure and lattice constant of the substance under study.

4.7 SOLUTIONS AND ANSWERS Self-Assessment Questions 1. Incident radiation wavelength is 4 x 10-~m while inter-planar distance is

2 x 10-'Om. The diffraction condition of equation (4.2) holds only if h I 2 d . Here h is much larger than d, hence, we will not get any diffraction pattern.

i 2. In the forward scattering direction (0 = 0) all the scattering waves will be in phase and hence interfere constructively.

N -i-a .Ak i-a .Ak

-e 2 -e 2 + e 2 " 1

N sin -a.Ak ifia,Ak

- - 2 1 e 2 ,

sin -a.Ak 2

4. An fcc unit cell has four atoms. One of these atoms is contributed by corners and may arbitrarily be assigned the coordinates (OOO), whereas the other three are contributed by face centers and may have the coordinates as

I 1 (-0-), ( 1 1 0 ) and ( 0 1 1 ) . if all the atoms are identical, we can write 2 2 2 2 2 2

From this it readily follows that

F = 4$, if h, k and 1 are all even or all odd = 0 for any other even-odd combinations of h, k and I .

This means that reflections like (1 1 1), (200), (220) etc. will be present, whereas those of the type (loo), (1 10) and (2 1 1) etc. will be absent for a fcc crystal.

F

5. Neutron diffraction is more suitable to study the structure of hydrogen containing crystals such as ZnH2.

b 6. The electrons while travelling through air get collided with the particles in air and

get scattered. Hence they cannot reach the detector. Under ultra high vacuum, the path of electrons towards detector is unobstructed. On the other hand the X-rays can travel in air very effectively without getting scattered significantly.

Terminal Questions 1. Diamond structure is built around fcc lattice. Further, the bases are at (0, 0,O) and

Structure Unfolding Techniques

(-!-,l,l] . The fcc bases are (0,0, O), ( -,O,- ' ) , ( O , ~ , ~ ) , ( l , ~ , O ) . Hence 4 4 4 2 2 2 2 2 2

Crystal Structure

The structure factor will be non-zero, only if both the parenthesis are non-zero.

(a) From SAQ 4, you may recall that for fcc, the value of second parenthesis is 4 for all h, k and I odd or even; and 0, otherwise.

(b) The value of first parenthesis is non-zero for h + k + I = 4n ; n being integer or for h + k + 1 = odd; and 0 otherwise.

Hence when conditions (a) and (b) are satisfied simultaneously, then and then only the geometric structure factor is non-zero.

i.e. F is non-zero for all h, k, 1 odd or even and h + k + 1 = odd or integer multiple of 4.

Hence the (hko planes giving diffraction maxima are ( I 1 1), (220), (3 1 1), (400), (33 1) etc.

nh lx0.2nm d = - = =0.15nm. 2 sin 8 2sin42"

For spherical symmetric charge distribution the volume element

dV= 2x ? sine dQ dr. (4.54)

Since sine d9 = d (cose) and r.Ak = rAk cose, we can write

Integrating over cose in the limit -1 to +I, we obtain

sin rAk = Fnr2p(r)- dr .

r Ak

4. qw sine(ko = h (assuming all first order reflections) 2 a :. 4d(,, sin2e,hko = h2 where d(,,, = Jm

By determining the values of h, k, I , we can determine the type of lattice. Constructing the table of sin28/N similar to table 4.2, we get,

A- the common factor - = 0.061.

4a

Structure Unfolding Techniques

I For h = 1.932A, a = = 3.91 A.

The common factor corresponds to N values of 3,4,8, 1 1, 12 and 16. Hence the planes represented are (1 1 1), (200), (220), (3 1 l), (222) and (400) respectively. Occurance of these planes is characteristic of a fcc lattice. Hence the given crystal is having fcc structure with lattice constant of 3.91 A.

85

Crystal Structure - --

APPENDIX A

Sn,," = 1 for m = n

Consider the sum over all the lattice vectors in the direct lattice

where T is an arbitrary vector, R, is the lattice vector, and N is the total number of unit cells in a direct lattice.

If we set

where G is the reciprocal lattice vector, we can write

* * *

where al ,laz, a3 are the set of reciprocal lattice vectors and a,, a2, a3 are the crystal basis in a direct lattice. h, k, l and n,, n,, n3 are integers. But a;.a, = ~ T C , a;.az = 0 = a;.a3 and so on. Therefore,

This shows that when T = G, the sum in Eq. (A.2) has the value N. When T # G, the situation will be analogous to what we obtained while evaluating the sum Sc in Eq. (4.27). And if N is large, this sum vanishes except for some values of Ak. Actually, these values turn out to be simply those for which T = G. Therefore, we can write

-0 form#n This means that lattice sum will vanish whenever vector T is not equal to some reciprocal lattice vector G. You are familiar with the properties of delta function from the PHE-14 course. If you compare this expression with that for lattice structure factor Sc (Eq. (4.27)), you will find that Sc vanishes for every value of Ak except when Ak = G.

This is another form of Bragg condition (or Laue equation). You will recall that magnitude of reciprocal lattice vector is inversely proportional to the interplanar distance dhkl. Hence

implies that

where dhkr is interplanar spacing.

2x Since k = - , for n = 1 we can write h

This is Bragg condition.

Further, Ak = G in Eq. (4.7) implies that

Hence

For elastic scattering, the magnitudes of k and k, are equal and this result simplifies to

Structure Unfolding Techniques

This is yet another form of Bragg diffraction condition for X-rays in terms of reciprocal lattice vectors.

NOTES