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
4.4

X-ray Diffraction and Reciprocal Lattice

Determination of Crystal Structures
X-ray Diffraction
Neutron and Electron Diffraction

4.5
4.6
4.7

4.1

Summary
Terminal Questions
Solutions and Answers
Appendix A

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.
.

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

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

internal structure, it is important to understand how they interact with crystals. The
first aspect we must appreciate is that for achieving diffraction, any probe
(electromagnetic or otherwise) must have wavelength of the order of inter-atomic
distances (- 10-'~m).For X-rays to satisfy this condition, their energy must be

E = -hc
=

( ~ . ~ ~ X ~ O - ~ ~ J S ) X ( ~ X I O ~ ~ S - ~ )

(10-'Om) x (1.6 x lo-'' J ~ V - ' )

= 1 2 . 4 ~l o 3 e v .
That is, X-rays of a few keV will be required for probing microscopic structure of
solids using diffraction phenomenon. You have learnt the praperties of X-rays in
PHE-11 course on Modem Physics. However, for brevity, we state them here:
X-rays are produced by sudden deceleration of electrons in metals as well as by
excitation of core electrons in the atoms of a metallic target.
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

From PHE-09 course on

Optics* you may
that
when visible light is incident
on an arrangement of two or
more slits and their separation
is of the order of wavelength of
light, the diffraction pattern
may be observed. The
diffraction of electromagnetic
waves by atonis in a solid is an
analogous phenomenon; the
wavelength of electromagnetic
radiations used should be of the
order of inter-atomic distances
in solids (- ]A).A crystal
differs from a diffraction
grating in that the diffracting
centers here are not in one
plane! So, we can imagine a
as a 3-dimens,onal
space-grating.

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.)

.
. . . . . . . . . . . . O
*

In specular reflection, the

angle of incidence n ~ s be
t
equal to the angle of
reflection. Hence, here the
diffracted beams are just
like reflected ones in optics
term~nology.This leads to
the use of terms X-ray
d@ractron and reflection
intnchangeably in the texr

To obtain sharp diffraction spots, Bragg considered specular reflection by atoms in

any one plane and reflected waves from successive parallel planes separated through
distance d (Fig.4.2a) to interfere constructively. The path-difference between two
specylarly reflected waves (X-rays) is just AB + BC. In terms of 8 and d, we can writ
the total path-difference as
A = AB+BC=2dsin8.

(4.1)

As you know, for constructive interference, this path difference should be an integral
multiple of 1.This leads us to Bragg ~onditionfor constructive interference:
2dsin8=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.

Structure Unfolding

SAQ 1
Radiation of wavelength 4 x 10-~mis incident on a crystal with interplanar separation
distance of 2 x 1~ - ~ cinma solid. Shall we observe Bragg diffraction pattern? Justify
your answer.

Techniques

Spend
2 min.

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
waves, we will have to consider their phases. You will appreciate here that the
difference in phases arises because waves travel different distances due to spacing
between individual scatterers. We shall discuss diffraction of X-ravs in a more general
way by first considering the scattering of wave from a single atom and then collective
effect of all lattice sites in a crystal to obtain the condition for diffraction maxima.
d

a) Scattering from a n 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, kois 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

The main contribution to X-ray

scattering is made by core
electrons. The valence
electrons are a) much less in
m~mberand b) can be involved
in bonding and may not be
associated with the atom whose
lattice position is under
consideration.

"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 . 4we 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 ioand; are unit vectors along the incident and scattered directions,
respectively. For specular scattering, as shown in Fig.4.2bY8 , = O2 = 8 = half of the
scattering angle. Hence, net path difference between two scattered waves is given by

wAk

If k o and k are the wave vectors in the incident and scatter directions respectively,
we can write

28

2n
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 = 2 k s i n 8,
where 0 is half the scattering angle.

(4.7a)

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

Structure Unfolding
Techniques

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.

where N is the number of scatterers.

65

Crystal Structure

Note that electrons d o not have discrete positions in a n atom; they a r e 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.
The atomic scattering
factorf, is defined as the
ratio of radiation amplitude
scattered by the actual
electron distribution in an
atom Cf) to that scattered by
a single electron
This
is a dimensionless quantity.

we).

We then define an atomic scattering factor,

f, =

fa

i
i

,which is the integral in Eq. (4.16):

ir.Ak d v

(4.17)

v(r>e

For spherically symmetric charge distribution, as you will prove in TQ 3, the atomic
scattering factor has the form
R
fa =

Pnr2p(r)

sin rAk

dr ,

where R is the radius of the atom.

This result shows thatf,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

For the forward direction, i.e. 0 = 0, Ak

= 0 and

sin rAk
4 1. Then Eq. (4.18) reduces
rAk

Structure Unfolding
Techniques

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?
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

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 a2denote 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 ajvectors.
From the figure it is clear that we can express the position vector Ryof any atom as a
combination of R', and aj, i.e.

+ ai.

RY= R:

(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

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

Then Eq. (4.22) simplifies to

fc=FxSc.
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!

Structure Unfolding
Techniques

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

to obtain

sc = (

1 - eiNa.Ak
- e","

You can easily calculate the sum

The sum of a finite geometric

)1

I-r"

series is given by -,
I-r
where r is common ratio.

You can readily show that

N
sin -a.Ak
i(N-l)p,Ak
21
<Sc=
xe 2
sin L a . ~ k
2
SAQ 3

Spend
5 min.

Derive Eq. (4.30).

Since (e'J2 = I , we can write

lscl

N ,
sin 2 -a.Ak
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?
The plot of

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

2

Crystal Structure

I2 = N'.

between 0 and 2n. At a d k = 0 and a.Ak = 2n:; Isc

principal maximum.

This defines the

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,

l2 can be taken to be non-zero only in the immediate vicinity of the

(sc

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.
I

For a given h, the condition for constructive interference defines an, infinite number of
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
Techniques

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
~ the crystal scattering factor fc can
number of scattering atoms in the lattice, N , Mand
be-written as

IC(~,
= Fhkl

(4.35)

N~~~~~
9

where h, k and 1are integers. It may be noted that h, k, and 1are 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). Fhkldenotes 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.
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

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, Fhklis 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

(fa, = faZ = f a )

Eq. (4.37) becomes

From this result, we note that

Fhu= 0
= 2f,

forh+k+lodd
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, 1will 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

Structure Unfolding
Techniques

The concentration of atoms in this intermediate plane is equal to that of the top and
bottom planes defined by the faces the cube and so this plane will give reflected
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.

( k :i)

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

This implies that

F=fi

+h

=fi -h

for h + k + 1 even
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 1l), (200), etc. The
interplanar distances in these cases can be determined by using the equation (2.13).

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 =402 =

dOl2
= dO2,
= dZo1
etc.
Table 4.1

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.

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

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.

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 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.

Structure Unfolding
Techniques

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.

(a)

p,

(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 Pzand 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
resultant photograph can throw light on the symmetry, shape and size of the unit cell.
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
Transmitting beam

pi-.! i

--------

X-rays source Filter

Photographic film

Oriented differently

(b)

Exit hole

Entrance 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

Structure Unfolding
Techniques

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 . 1 6 ~ )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,

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
sinZ0
set of h, k, 1 values that result into the fraction
= 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.

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

Kinetic energy of a neutron is
given by
E =p2/2m,,
where m, is = Mass of neutron
(= 1.67 x 1 0 - ~ ' k ~ )

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,

Structure Unfolding
Techniques

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

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

Spend
2 min.

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 -inangstroms, 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

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?

a-+

2-

Crystal Structure

SUMMARY

4.5

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~istitutingit.
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.

TERMINAL QUESTIONS

4.6

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 sin28were 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-~mwhile 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.

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

-i-a

-e 2

-e

.Ak

" 1

i-a

.Ak

+e2

sin -a.Ak
ifia,Ak
2
e 2
,
1
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$,
=0

if h, k and 1 are all even or all odd

for any other even-odd combinations of h, k and I.

This means that reflections like (1 11), (200), (220) etc. will be present, whereas
those of the type (loo), (1 10) and (2 11) 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

l , l , - ! - (

4 4 4

( '),

. The fcc bases are (0,0, O), -,O,2

( O , ~ , ~ () l ,, ~ , O ) Hence
.
2 2
2 2

Structure Unfolding
Techniques

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 11), (220), (3 11), (400),
(33 1) etc.

d = - =nh
2 sin 8

lx0.2nm =0.15nm.
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

= Fnr2p(r)-

4.

sin rAk
dr .
r Ak

qwsine(ko = h (assuming all first order reflections)

:.

4d(,,2

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/Nsimilar to table 4.2, we get,

the common factor

= 0.061.
4a

For h = 1.932A,

a=

Structure Unfolding
Techniques

A-

= 3.91 A.

The common factor corresponds to N values of 3,4,8, 11, 12 and 16. Hence the
planes represented are (1 11), (200), (220), (3 1l), (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
Consider the sum over all the lattice vectors in the direct lattice

where T is an arbitrary vector,

unit cells in a direct lattice.

R, is the lattice vector, and N is the total number of

If we set

where G is the reciprocal lattice vector, we can write

where al ,laz, a 3 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 a;.az
C , = 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

Sn,,"= 1 for m = n
-0 f o r m # 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 dhkris interplanar spacing.

Since k

2x
-, 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

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

Structure Unfolding
Techniques

NOTES