PowerPoint ????

Chapter 7



Introduction to

Crystallography

Crystalline Substances: e.g., Diamond and Table Salt









Diamonds

7.1 periodicity and lattices of crystal structure

7.1.1 The characteristics of crystal structure

1. A few definitions:

• Solids can be divided into two primary categories,

crystalline and amorphous.

• Crystalline Solids are built from atoms or molecules

arranged in a periodic manner in space, e.g., rock salt and

diamond.

• Amorphous Solids possess short-range order only. They

are not related through symmetry, e.g. glass, rosin, amber

glass.



• Short-Range Order: Fixed bond lengths and angles

• Long-Range Order: Associated with a lattice point

Crystals:

Crystals are solids that are built from atoms or

molecules arranged in a periodic manner in space.

2. Fundamental characteristics of crystal

a) Spontaneous formation of polyhedral shapes

F+V=E+2









b) Uniformity: periodic distribution of

atoms/molecules

Single crystal gold bead with

naturally formed facets

HRTEM images of hollow beads

•Anisotropy

Different periodicity and density for different direction.





2150 g/mm2



Conductivity









1150 g/mm2



570 g/mm2

Graphite

NaCl

•Definite sharp melting points

T







t

• Symmetry: crystal shape (macroscopic)

lattice arrangement (microscopic)



•X-ray diffraction by crystals:

atomic distances match the wavelength of x-ray.

7.1.2 The lattice and unit cell



Lattice:

• A periodic pattern of points in space, such

that each lattice point has identical

surroundings.

• Can be reproduced by translational motion

along the vector between any two points.

a. The lattice and its unit in 1D:

Tm = ma (m = 0, 1,  2，) a: basic vector.

 Each motif in this

1D pattern can be

represented by a

point.

A one-dimensional lattice  This 1D system is

represented by a

lattice of repeating

points.







Different choice of unit cell for a 1D lattice

Note: the translation vectors connecting any two lattice

points constitute a translation group.

Examples of 1D lattice

b. Lattice and its unit in 2D:

Tmn = ma + nb（m, n = 0,1,  2， )

a & b: independent basic vectors









•Crystal structure = lattice + structural motif

(basis)

Lattice:

•A periodic pattern of points in space, such that

each lattice point has identical surroundings.

•Can be reproduced by translational motion

along the vector between any two points.









this 2D pattern 2D lattice. 2D lattice.

itself is not a lattice,

but can be

represented by a

2D Lattice.

2D Primitive Cell

Unit Cell Choice



 There is always more than one possible

choice of unit cell

 By convention the unit cell is chosen so

that it is as small as possible while

reflecting the full symmetry of the lattice



1) The highest symmetry

2) The smallest area (or volume)

Five 2D lattices









ab   90, a=b =90

  120 ab =90







ab =90 a=b =120



Primitive

unit cell

Centered

There are literally thousands of crystalline

materials, there are only 5 distinct planar lattices

c. Lattices and its unit in 3D:



T = ma + nb + pc (m, n, p = 0, 1,  2, …)









What’s the smallest structure motif of a graphene sheet?

The Choice of a Unit Cell: Have the highest symmetry

and minimal size





c















b

a

c



a b

The Choice of a Primitive Cell



1) The axial system consisting of the basis

vectors should be right handed.

2) The basis vectors should coincide as much as possible

with directions of the highest symmetry.

3) Should be the smallest volume that satisfies condition 2.

4) Of all lattice vectors none is shorter than a.

5) Of those not directed along a none is shorter than b.

6) Of those not lying in the a, b plane none is shorter than c.

7) The three angles between the basis vectors a,b,c are

either all acute or obtuse.

Crystal structure = lattice + structural motif

(basis)

Atomic Coordinates: Fractional coordinates



Fractional coordinates:

The positions of atoms inside a

unit cell are specified using 0.5

fractional coordinates(x,y,z). i

These coordinates specify the

position as fractions of the unit 0.6



cell edge lengths

i: (1.0, 0.6, 0.5)

Example:

Cubic unit cell of CsCl,

a=b=c

===90

Cs:(0,0,0)

Cl: (1/2,1/2,1/2)









Single Crystal: Composed of only one particular type of space lattice.

Polycrystalline matter: Clusters of multiple crystals.

Chapter 7 Introduction to Crystallography

a. The lattice and its unit in 1D:

T = ma (m = 0, 1,  2，)









unit cell and its choice for one-dimensional lattice

b. Lattice and its unit in 2D:

T = ma + nb（m, n = 0,1,  2， )









•Crystal structure = lattice + structural motif

(basis)

Five 2D lattices









ab   90, a=b =90

  120 ab =90







ab =90 a=b =120



Primitive

unit cell

Centered

There are literally thousands of crystalline materials,

there are only 5 distinct planar lattices

c. Lattices and its unit in 3D:



T = ma + nb + pc (m, n, p = 0, 1,  2, …)





c















b

a

c



a b

Crystal structure = lattice + structural motif

(basis)

Atomic Coordinates: Fractional coordinates



Fractional coordinates:

The positions of atoms inside a

unit cell are specified using 0.5

fractional coordinates(x,y,z). i

These coordinates specify the

position as fractions of the unit 0.6



cell edge lengths

i: (1.0, 0.6, 0.5)

7.1.3 Crystal systems and Bravais Lattices

a. Crystal systems

A total of seven types of crystal

systems exist with different symmetry

elements.

Crystal Characteristic Unit cell Choice of axis Lattic

systems symmetry parameters Point

elements Group

Triclinic Nil abc Ci



Monoclinic C2, h abc b // 2-fold axis C2h

==90

Orthorhombic 3C2, 2 h abc a, b, c // 2-fold D2h

===90 axes



Unit cell is chosen in such a way that it contains as many

symmetry elements of the lattice as possible and, meanwhile,

has the smallest volume.

Trigonal 3-fold rotation Rhombohedr D3d

axes al

a=b=c

== are used to specify a set of symmetry-

equivalent directions.

[uvw] zone axis Direction Vector = ua + vb + wc

d. d-spacing dhkl





(010)





(110)

d010



d110





( 2 10)

(210)



( 1 20)

The spacing between adjacent planes in a family is

referred to as a “d-spacing”



Cubic : 1/d2 = (h2+k2+l2)/a2 or d2 = a2/(h2+k2+l2)

Tetragonal: 1/d2 = (h2+k2)/a2 + l2/c2

Orthorhombic: 1/d2 = h2/a2+k2/b2 + l2/c2

Hexagonal: 1/d2 = (4/3)(h2+hk+k2)/a2 + l2/c2

Monoclinic: 1/d2 = [(h/a )2 + (k/b )2sin2 + (l/c )2-

(2hl/ac)cos]/sin2

Triclinic:

7.1.5 Real crystals and Crystal defects：

Real crystals are only close approximations of space lattices





Edge dislocation

Screw Dislocation

• Formed by shear stress

• Also linear and along a

dislocation line

7.2 Symmetry in crystal structures.



7.2.1 Symmetry elements and symmetry

operations

 Crystallographers make use of all the symmetry in a

crystal to minimize the number of independent

coordinates





a. Lattice symmetry

b. Point symmetry

c. Other translational symmetry elements: screw axes

and glide planes



a. Lattice symmetry --- translation operation

m 

Tmnp=ma+nb+pc

Tmnp=

n

 

 p

 

b. Point symmetry elements compatible with 3D translations



• Point symmetry operation does not alter at least one

point that it operates on: rotation axes, mirror planes,

rotation-inversion axes



Reflection Mirror Plane m

Rotation operation Rotation axis n = 1, 2, 3, 4, 6

Inversion Center of symmetry 1

Rotatory inversion Inversion axis 3, 4, 6





Translational symmetry of lattice prohibits the presence

of 5-fold axis!

Rotation axes, 1,2,3,4,6 only!! Why ???



ma

B1  B2

 Limitation induced by

Translation symmetry!





 

a a a

A1 A2 A3 A4

Lattice points A1, A2, A3, A4 (m-1)/2 1

Through n-fold operation m-1 2

A1  B1 m= 3, 2, 1, 0, -1

A4  B2 cos =1, 1/2, 0, -1/2, -1

A1A4 // B1B2  = 0º, 60º, 90º, 120º, 180º

B1B2 ＝a +2acos = ma n= 1, 6, 4, 3, 2

cos  = (m-1)/2 rotation axes, 1,2,3,4,6 only!!

The symmetry elements of a cube





Twofold axis

Threefold axis

Fourfold axis









2 3 4 6

Rotation axis

1 0 0 General equivalent positions:

  (x,y,z); (-x, y, -z)

R(2)  0 1 0

0 0 1 2-fold axis // b axis

 



0 1 0

 

R(3)  1 1 0

0 0 1

 

general equivalent positions:

(x,y,z), (-y, x-y, z) (-x+y, -x, z)

3-fold axis // c axis

0 1 0 general equivalent positions: (x,y,z),

  (-y, x, z), (-x,-y,z), (y,-x,z)

R(4)  1 0 0

0 0 1 4 fold axis // c axis

 





1 1 0 general equivalent positions: (x,y,z), (x-

  y, x, z), (-y, x-y, z), (-x,-y,z), (y-x, -x, z),

R(6)  1 1 0 (y, y-x,z)

0 0 1 6 fold axis // c axis

 

c. Screw axes nm (m < n) and glide planes:

2 m

A two-fold screw 21 n  C T (t );

1

m

1

n C  R( ),

1

n T (t )  a

n n

21 // a axis



(x,y,z)(x, –y, -z)

(x+1/2,-y, -z)



Helical

structure









The direction of such an axis is usually along a unit cell edge, and the

translation must be a subintegral fraction of the unit translation in that direction.

Higher order screw axes









Screw 31 Screw 32

41-3, 61-5

Glide operations: An a glide

• a, b, c, n and d glides occur

• The a glide has translational component

of 1/2a

• n glide has translational component a/2

+b/2 or b/2+c/2 or …

• d glide has translational component of the

type a/4 + b/4 + c/4

• e glide (double glide plane).



Mirror: M

Glide operation: G = MT(t)

t= a/2, b/2, c/2, (a+b)/2, (b+c)/2, (a+c)/2, (a+b+c)/4 etc.

GG = [MT(t)]2 = T(2t) , t = 2t ( a basic vector of lattice)



Zig-zag structure

Summary of symmetry elements and

symmetry operations in crystal structure



Symm. operation | Symm. Element

• Rotation operation | Cn rotation axis

• Reflection operation | mirror plane

• Inversion operation | center of symmetry

• Rotation inversion operation | inversion axis

• Translation operation | lattice

• Screw operation | screw axis

• Glide operation | glide plane

(n=1, 2, 3, 4, 6)

Combinations of the 8 point symm. elements (5 n-axis,

i, m, S4) result in 32 crystallographic point groups.

Combining symmetry elements

When a crystal possesses more than one of the above

symmetry elements, these macroscopic symmetry elements

must all pass through a common point. There are 32

possible combinations of the above symmetry elements that

pass through a point and these are the 32 crystallographic

point groups.





7 Crystal systems

14 Bravais lattices

32 crystallographic point groups

230 space groups

7.2.2 Crystallographic point group and space group

1.Crystallographic point group

Combinations of the 8 point symm. elements (5 n-axis, i, m,

S4) result in 32 crystallographic point groups.

e.g., Monoclinic system: point symmetry element -- 2, m.

2 – {E, C2} -- C2 point group ; m -- {E,} --Cs point group.

2,m -- {E, C2, , i} -- C2h point group.

Two notations of crystallographic point group

schonflies notation vs. international notation

C2 2

Cs m

C2h 2/m

• 7 crystal systems.

• Each crystal system contains a set of

distinctive crystallographic point groups.

• A total of 32 crystallographic point groups, e.g.

Crystal Schonflies International Symmetry examples

system notation notation elements

monoclinic C2 2 C2 BiPO4



Cs m  KNO2



C2h 2/m C2, h, i, KAlSi3O8



orthorhombic D2 222 3C2 HIO3



C2v mm2 C2, 2 NaNO2



D2h mmm 3C2, 3, i, Mg2SO4

2. Crystallographic space group

Point groups (32) + translational symmetries = Space groups (230)

Schonflies notation and International notation for space group

e.g, D2h16 - P21/n 21/m 21/a C2h5 – P21/c

system directions

1 2 3

Cubic a a+b+c a+b

hexagonal c a 2a+b

Tetragonal c a a+b

Trigonal a+b+c a-b -

Trigonal* c a

Orthorhombic a b c

monoclinic b - -

Example: monoclinic point group C2h-2/m

Six space groups belong to C2h point group, denoted:

C2h1-P2/m, C2h2-P21/m, C2h3-C2/m,

C2h4-P2/c, C2h5-P21/c, C2h6-C2/c,

Equivalent positions: e.g., C2h5 – P21/c





c glide

plane

at b/4.





A set of equivalent positions located within a unit cell!

General equivalent positions :

4 e 1 (x, y, z), (-x, y  1 , 1  z), (-x,  y,  z), (x,

2 2

1

2  y, 1  z)

2

e.g., C2h5 – P21/c









Special equivalent positions :

2 d 1 ( 1 ,0, 1 ), ( 1 , 1 ,0);

2 2 2 2 2 c 1 (0,0, 1 ), ( 1 ,0,0);

2 2



2 b 1 ( 1 ,0,0),( 1 , 1 , 1 );

2 2 2 2 2 a 1 (0,0,0),(0, 1 , 1 )

2 2





Inversion center

Diamond: face-centred cubic Oh7- Fd3m







41





d glide

plane

1/8,3/8,

5/8,8/7 Sideview topview

Structure motif – 2C

• Lattice points: (0,0,0)+, (1/2,1/2,0)+, (0,1/2,1/2)+, (1/2,0,1/2) +

7.2.3 The description and application of crystal structure

Example 1. Crystal of iodine

Crystal System orthorhombic

Space group D182h-Cmca (or C 2/m 2/c 21/a)

Cell parameters a=713.6 pm b= 468.6 pm c = 987.4 pm

Number of molecules per unit cell Z = 4

Atomic coordinate for I x y z

0 0.15434 0.11741

Lattice points within a unit cell: (0,0,0)+, (1/2, 1/2, 0)+ (C-center).

General equivalent positions:

(x,y,z); (-x, -y, -z); (-x, -y+1/2, z+1/2); (x, y+1/2, -z+1/2)

(0, .15434, .11741) (1/2, .65434, .11741)

(0, -.15434, -.11741) (1/2, .34566, -.11741)

(0, .34566, .61741) (1/2, .84566, .61741)

(0, .65434, 38259) (1/2, .15434, 38259)

a) Bond length (Bond distance)

r1-2= [(x1-x2)2a2+(y1-y2)2b2+(z1-z2)2c2]1/2 = 2.715 A

c) Density of crystal

V = a x b x c = 3.27 x 10 8 pm 3

D = 8 x 127.0 /( 6.02 x 10 23 x 327.0 x 10 –24 ) g cm-3 =5.16 g cm-3

7.3 X-ray diffraction of crystals

7.3.1 The source and property of X-ray









X-ray tube

the wavelengths of X-ray are in the range

of 100-0.01Å

• 1-0.01Å: hard x-ray

• 100～1Å：soft x-ray

• 2.5-0.5Å: used in crystal structure

analysis

• 1-0.05Å: used in medical perspective,

detection of materials wound

X-rays produced by electronic transition between

atomic energy levels

High energy A part of the

electrons are

electron beam

blocked; their

kinetic energies

M giving rise to

L L “white” x-ray.

radiation

As for Cu:

K Cu+ 1s12s22p6…

2S  1s22s22p5…

1/2

e

2P

3/2

K1=1.540594Å

2P K2=1.544422Å

1/2

e

IK1  2IK2

K1







K2









Notice: K2 can not be striped by the monochromator.

Synchrotron Radiation X ray Source

Benefits of Synchrotron radiation X-ray :

• Narrow range of x-ray wave-length– high

monochromicity

• High intensity of x-ray.

• High intensity and high quality of diffraction data

• High resolution – characterization microcrystals

Toooooooo Expensive facility!

SPring-8, at Osaka, Japan. www.spring8.or.jp

ESRF - European Synchrotron Radiation Facility , Polygone

Scientifique Louis Néel - 6, rue Jules Horowitz - 38000 Grenoble

- France , http://www.esrf.fr

The Advanced Photon Source (APS) at Argonne National

Laboratory, http://www.aps.anl.gov/aps.php

7.3.2 Laue equation and Bragg’s Law



1. Laue equations

Laue first mathematically

described diffraction from

crystals

• consider X-rays scattered

from every atom in every

unit cell in the crystal and

how they interfere with each

other

Max Von Laue

• to get a diffraction spot you

must have constructive

interference

Laue equation (Based on diffraction by 1D lattice)

Interference condition:

the difference in path lengths of

adjacent lattice points must be a

multiple integral of the wavelength.

AD-CB = h

AD = a·S = acos

CB = a·S0 = acos0

For each h value, the

a(cos-cos0) = h, h=0, 1, 2 …. diffraction rays from a 1D

lattice make a cone.

Where, 

a— lattice parameter 0



0—angle between a and s0

— angle between a and s

Expanded to 3D lattice case

a·(S-S0) = a(cos-cos0) = h

b·(S-S0) = b(cos-cos0) = k

c·(S-S0) = c(cos-cos0) = l

This is the Laue Equation!



where,

a,b,c—lattice parameter

0,0,0—angle between a,b,c and s0

,, —angle between a,b,c and s

h,k,l — indices of diffraction, integers

which may not be prime to each other and are

different from Miller indices for crystal plane!

In the diffraction direction, the difference between the

incident and the diffracted beam through any two

lattice points must be an integral number of

wavelengths.

The vector from (000) to (mnp):

Tmnp = ma + nb +pc

The differences in wavelengths:

 =Tmnp · (S-S0)

=(ma + nb +pc) ·(S-S0)

= ma ·(S-S0)+nb ·(S-S0)+pc ·(S-S0)

=mh+ nk+pl

=(mh+nk+pl)

2. The Bragg’s Law

Bragg discovered that you could consider the diffraction

arising from reflection from lattice planes







s0 s

 O P 

dhkl





Conditions to obtain constructive inferences,

a. the scattered x-ray must be coplanar with the

incident x-ray and the normal of the lattice plane.

b. S=S0

=AD+DB = 2d(hkl)sinn

s0 s

Condition for diffraction:

2d(hkl) sinn = n (n=1, 2, 3, … )  O 

dhkl

A

n: the angle of reflection D

n: the order of the reflection B

2dnhnknlsinnh,nk,nl=(dnhnknl = dhkl/n)

However, the reflection order n is

not measurable!

Reformulated Bragg equations: reflection indices



2dhkl sin = 

n=2

Virtual reflection plane





2d(hkl) sinn = n (n=1, 2, 3, … )

Lattice plane 2dhkl sin =  (dnhnknl = dhkl/n)

indices



Diffraction indices or virtual reflection plane indices.

The Bragg equation defines the direction of diffraction

beams from a given set of lattice planes!

(100)

diffraction

crystal planes -

(100), (200), … (100) (200) (300)

Families of planes (200)



Lattice plane

directions-(100)



The Bragg Law vs. Laue equation:

•Both equations define the relationship between the direction of

diffraction beams, the x-ray wave-length and the parameters of

a crystal lattice.

•The Bragg law: simple, easier to derive lattice parameters from

the direction of diffraction beams.

2dhkl sin = 

3. Reciprocal lattice --- Why we need it?

     

 * b  c  * c  a c *  a  b Basis vectors of a



a  b  reciprocal lattice

V V V

* * * * *

r  ha  kb  lc r  1 / d hkl

102 112 A reciprocal lattice

101 point corresponds to

111 a diffraction (lattice)

100 110 plane of its original

a* 001 lattice!!!!!!!!!!!!

c* A reciprocal vector r*

b*

000 010 is perpendicular to a

r* lattice plane with the

1 22 indices (hkl).

1 01 1 11

1 00 1 10

4. Ewald sphere (reflection sphere) O1: origin of

crystal lattice

Radius = 1/

G O: origin of

reciprocal lattice

S r* G: a reciprocal

1/dhkl lattice point / a

 2 diffraction point!

A S0 O1 1/ O

Only when the

reciprocal lattice

point is located on

the Ewald sphere

can constructive

OG  2 O1O * sin   ( 2 ) sin   1 interference occur

 d hkl (diffraction) !

 2d hkl sin θ  λ

Note: very few diffraction data can be

observed when a single crystal is fixed.

How to get more diffraction data?

• Rotate the crystal to enable more reciprocal points

(diffractions) dynamically located on the Ewald

sphere.

• While the crystal is fixed, use x-rays with varying

wavelengths (e.g., white x-ray). By doing so, the

Ewald sphere becomes filled.



The first technique is preferred and has been

widely used in practice!

7.3.3. The intensity of diffraction beam

1. The principle of X-ray scattering

To same your time,

Let’s neglect part 1 and

QR move directly to part 2!

r If you are interested in

s-s0 P s this part, just go

s0 O through it by yourself.



For elastic scattering, each electron scatters the plane wave

causing a spherical wave (exp[2i(kr)]).

The phase difference is: =(r•s - r•so)/λ

The scattered x-ray: exp{2i[r(s-s0)/]} or exp{2i[rq/]}

The contribution of the scattering of all electrons of a given atom:



  (r) exp(2iq  r /  )d

3

r

Rn

For the crystal structure ：

 (r )    cell (r  Rn )

b

n

c



A     cell (r  Rn ) exp(2iq  r /  )d 3 r a

n





A   cell 

( r ) exp(2iq  r /  )d 3r  exp(2iq  Rn /  )

n



 F ( q) exp(2iq  Rn /  )

n



F(q) --- structure factor



F ( q)    cell ( r ) exp(2iq  r /  )d 3 r

Supposed that there are N1，N2，N3 periods along a，

b，c, and all the atoms locate on the position of lattice

points, F(q) can be replace with a constant ‘f’. f is

scattering factor of atoms.

N1 1 N 2 1N3 1

Amnp  f  e 2i /  ( n1an2bn3c )q

n1 0 n2 0 n3 0





For the case of 1D and f=1,



2iN aq / 

N 1

1 e

AN   e 2inaq / 

 2iaq / 

n 0 1 e

The intensity:



 N 

sin  2

a q

   sin 2 Nh 

I  AN  AN AN 

2 *



  sin 2 h 

sin 2  a  q 

 

30 2

|A| (N=5) 250 |A|2(N=15)

200

20

150

100

10

50

0 h 0 h

-1.0 -0.6 -0.2 0.2 0.6 1.0 -1.0 -0.6 -0.2 0.2 0.6 1.0

In the case of 3-D:

 N 1  2  N 2  2  N 3 

sin 

2

a  q  sin  b  q  sin  c q

I  Amnp 

2

 f

2         

     

sin 2  a  q  sin 2  b  q  sin 2  c  q 

     

Therefore,

aq/=h，bq/=k，cq/=l (h，k，l should be integer)

or aq=h，bq=k，cq=l

----- Laue conditions for X-ray diffraction



I f N N N

2

I  Fhkl N N N

2 2 2 2 2 2 2

1 2 3 1 2 3



Only those scatterings fulfilling these conditions give

rise to measurable diffraction beams.

2. The intensity of diffraction beam



• The directions of the diffraction beams



Bragg Equation

2dhkl sin = 





The directions of the diffraction beams are

determined by the cell parameters.

• The intensity of diffraction beam I  K F 2

hkl hkl

Structure factor:



Fhkl     ρ( x , y , z )e 2 πi ( hx  ky lz )

dxdydz

Electron density distribution in a unit cell.

n

  f je

2 πi ( hx j  ky j  lz j )



j 1 atomic coordinates: x,y,z



Sum over all atoms fj: atomic scattering factor defined by

within a unit cell. atomic electron density distribution.

• The intensity of the diffraction beams are determined by

types of atoms and the arrangement of atoms in the cell!

• By measuring the cell parameters and the intensities of

diffraction points, the atomic arrangement within a unit cell

can be derived.

3. systematic absence

Calculation of structure factor

Example A, Body center crystal

Simplified case: Each lattice point is a

metal atom.

n

Fhkl   f j e

2i ( hx j  ky j  lz j )



j 1



1 1 1

2i ( h  k  l )

 f1e 2i ( h 0 k 0l 0 )  f1 e 2 2 2



i ( h  k  l )

 f1 (1  e )

While h+k+l =2n+1,

e πi ( h  k l )  e( 2 n 1 )πi  1  Fhkl  0

systematic absence !

Body-center crystal –

A general case: each lattice point contains n atoms

• The total number of atoms within a unit cell is 2n;

• For jth atom in a structure motif (a lattice point): (xj,yj,zj)

• Its body-center equivalent point is: (0.5+xj, 0.5+yj, 0.5+zj)

2n

Fhkl   f i e 2πi ( hxi  kyi lzi )

i 1



n 1 1 1

2 πi [ h (  x j ) k (  y j )l (  z j )]

 { f j e

2 πi ( hx j  ky j lz j )

 fj e 2 2 2

}

j 1

n

  f je

2 πi ( hx j  ky j lz j )

[ 1  eπi( h k l ) ]

j 1



While h+k+l =2n+1,

System

πi ( h  k  l ) ( 2 n 1 ) πi

e e  1  Fhkl  0 absence

Example B. Unit cell has a 21 screw axis along the c axis at

x=y=0

Equivalent position (x,y,z) and ( -x , -y , z+1/2)

N /2

Fhkl   f j {exp[i 2 (hx j  ky j  lz j )] (x,y,z)

Sum over j 1

all atoms of

a molecule. 1

 exp[i 2 (hx j  ky j  l ( z j  ))]}

2

N /2

F00l  f

j 1

j exp( i 2πlz j )[( 1  exp( iπl )]

(x , y , z+1/2)



N /2

Note:

F00l  2  f j exp[i 2 lz j ] l = 2n •A structuremotif (lattice

j 1

point) consists of two

F00l  0 l = 2n+1 molecules.

•The two molecules are

correlated by the 21 screw

systematic absence

axis.

systematic absence

• Crystals of the same lattice type exhibit similar behavior

in system absence!



• Crystal structure which contain centering, glide plane

and screw axis will have systematic absences.



• Namely, some reflections/diffractions will be

systematically absent in case the lattice have centering,

glide plane and screw axis.



Please derive the conditions of system absence due to

3-,4-, 6-fold screw axes and d-glide plane by yourself!

systematic absence and sysmmetry

Types of Conditions for extinction Cause of extinction Centering and

reflection symmetry

elements

hkl h+k+l = odd In-centred (bodycentred) I

h+k = odd C-centred C

h+l = odd B-centred B

k+l = odd A-centred A

h,k,l not all even and not all Face-centred F

odd

-h+k+l not multiples of 3 R-centred R(hexagonal)

0kl k =odd Translation in b/2 b

l =odd (100) c/2 c

k+l =odd glide (b+c)/2 n

k+l not multiples of 4 Planes (b+c)/4 d

00l l =odd Translation c/2 21, 42, 63

l not multiples of 3 Along c/3 31, 32, 62, 64

l not multiples of 4 (001) c/4 41, 43

l not multiples of 6 Screw axis c/6 61, 65

7.3.4 Applications of X-ray diffraction

1. Methods

* Single crystal diffraction



Monochromatic camera method -- Monochromatic X-ray

Rotation, Oscillation, Weissenberg …

Laue photography --- white X-ray

Diffractometer -- Monochromatic X-ray





Crystal Diffraction beam

2

Incident beam

* Powder diffraction



Monochromatic X-ray Diffraction beam



powder

2

Incident beam 2









P

Diffraction beam



Incident beam 2

O



sample R







Powder Diffractometer

Radiation sources

X-ray tubes Monochromator – e.g.HOPG



Synchrotron radiation Filter – e.g. Ni for CuK



Detectors

•Film

- poor sensitivity, high background, low dynamic range

•Scintillation counters

- good sensitivity, low background, high dynamic range

•Imaging plates

- good sensitivity, low background, good dynamic range, very

efficient data collection

•CCDs and Multiwire detectors

- fast readout, good sensitivity, low background, good dynamic range,

very efficient data collection

Automated diffractometer method

2. The applications



a. crystal structure determination



Intensity data collection



Indexing Crystal system and

Cell parameters

I hkl  K Fhkl

2









Phase problem



F (hkl)      ( x, y, z )e 2i ( hx kylz ) dxdydz



 ( x, y , z )  V 1  F ( hkl)e 2i ( hx kylz )

h k l

Indexing of the cubic system:

a0

d hkl  2dhklsin= 

h2  k 2  l 2

sin 2   ( / 2a) 2 (h 2  k 2  l 2 )

sin   h  k  l

2 2 2 2 sin 2 θ1 : sin 2 θ2 : sin 2 θ3 : sin 2 θ4 :



 Get the directions of the diffraction beams, i

 Get the sin 2 θ1 : sin 2 θ2 : sin 2 θ3 : sin 2 θ4 :

 Get the ( h 2  k 2  l 2 )1 : ( h 2  k 2  l 2 )2 : ( h 2  k 2  l 2 )3 ....



 Get the lattice type (space group) and unit cell

parameters.

Indexing of the cubic system:

sin 2   ( / 2a) 2 (h 2  k 2  l 2 )



sin   h  k  l

2 2 2 2 sin 2 θ1 : sin 2 θ2 : sin 2 θ3 : sin 2 θ4 :



Characteristic line sequence in cubic system:

P: (hkl) 100, 110, 111, 200, 210, 211, 220, 221, 222, 300, ….

Systematic absence

（h2＋k2＋l2 ) 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, …

I: (hkl) 100, 110, 111, 200, 210, 211, 220, 221, 222, 300, ….



（h2＋k2＋l2 ) 2, 4, 6, 8, 10, 12, 14, 16 … (1: 2: 3: 4: 5: 6: 7: 8:…)

F: (hkl) 100, 110, 111, 200, 210, 211, 220, 221, 222, 300, ….

（h2＋k2＋l2 ) 3, 4, 8, 11, 12, 16, 19, 20 …

Observed diffractions: hkl all odd or all

Example for the indexing of cubic system and its applications



Sample: NaCl Condition: Cu K, =1.5418Å, R=50 mm

(1) Measure sample and relative intensity

(2) Calculate the position of diffraction lines (usually 2 in Ewald

sphere)

(3) Calculate 

(4) Calculate sin2

(5) Calculate sin21: sin22 : sin23 : sin24 :…=3:4:8:11:…

(6) Identify Bravais lattice → face cubic

(7) Indexing and calculate h2+k2+l2, calculate dhkl and a.



FC: (hkl) 100, 110, 111, 200, 210, 211, 220, 221, 222, 300, ….

（h2＋k2＋l2 ) 3, 4, 8, 11, 12, 16, 19, 20 …

(7) Index and calculate h2+k2+l2

No. I 2  sin2 h2+k2+l2 hkl

1 W 27.46 13.73 0.05631 3 111

2 S 31.80 15.90 0.07508 4 200

3 S 45.60 22.80 0.15016 8 220

4 W 54.06 27.03 0.20647 11 311

5 S 57.50 28.75 0.22524 12 222

6 S 66.44 33.22 0.30032 16 400

7 W 73.30 36.65 0.35663 19 331

8 S 77.56 38.78 0.37540 20 420

9 S 84.30 42.15 0.45045 24 422







Measured in Ewald sphere Indexing

by x-ray diffraction

(8) Calculate lattice parameter

1.54182 Why use high angle

sin 2 θ   ( h2  k 2  l 2 ) values?

4a 2

1.5418

a   h2  k 2  l 2

2 sin θ



Least-square method, plot

method, high angle values,…

a=5.628 Å 0° 90 

(9) = 2.165g/cm3 for NaCl °



ρV 2.165  ( 5.628  10 8 )3

n  4

M 23  35 .5

N0 6.022  10 23

One unit cell contains 4 NaCl

Example. Index cubic pattern and calculation lattice parameter

Line 2  sin2 sin2i h2+k2+l2 hkl

/sin21

1 40.26 20.13 0.1184 1 2 110



2 58.26 29.13 0.2370 2 4 200



3 73.20 36.60 0.3555 3 6 211



4 87.02 43.51 0.4740 4 8 220



5 100.64 50.32 0.5923 5 10 310



6 114.92 57.46 0.7109 6 12 222



7 131.16 65.58 0.8290 7 14 321



8 153.58 76.79 0.9470 8 16 400





If =1.5418 Å, body-centred cubic



 1.5418

a  h k l 

2 2 2

 42  02  02  3.16 Å

2 sin  2 sin 76.79

b. Applications of powder diffractions









Peak Positions Peak Intensities Peak Shapes and Widths

b. Applications of powder diffractions



Information contained in a Diffraction Pattern



Peak Positions

Crystal system, cell parameters, qualitative phase identification



Peak Intensities

Unit cell contents, quantitative phase fractions



Peak Shapes and Widths

Crystallite size, Non-uniform microstrain

(i) Peak Positions and

Intensities

Qualitative Analysis:

One crystal phase correspond to

a set of diffraction peaks.

( being different from the

spectroscopic analysis)

Phase analysis

Quantitative Analysis:

The peak intensities are

proportional to the weight

percentage of the corresponding

phase.

(ii) Changes of lattice parameters――Solid solution、

doping

Using high angle diffraction data or applying least square method.





Why use high angle

2dsin=n

values?





Maximal 

 minimal d/





0° 90 

°

(ii) Changes of lattice spacing along specific

directions――residue stress





stress

d





d d d’

(iii) The width of diffraction peaks ―― Crystallite

size

30 2

|A| (N=5) 250 |A|2(N=15)

200

20

150

100

10

50

0 h 0 h

-1.0 -0.6 -0.2 0.2 0.6 1.0 -1.0 -0.6 -0.2 0.2 0.6 1.0







I  Fhkl N N N  Fhkl A

2 2 2 2 2 2

1 2 3





N1, N2, N3 periods along the lattice axes within a microcrystal

(iii) The width of diffraction peaks ―― Crystallite

size

Scherrer formula: (1nm to 100nm) Widely exploited in

research of nanoparticles!

K  K 

Dhkl  

  cos ( B  B0 )  cos Why use small

angle values?

Dhkl average size along the direction perpendicular to (hkl) plane.

B measured peak width

B0 Instrumental width, using standard sample (e.g. -SiO2 with

crystallite size of 25－44m）









Instrumental width（B0）  B0 （B）

[100]





[001]









XRD pattern of ZnO nanowires

(iii) The width of diffraction peaks ―― Lattice

Distortion

晶格的畸变（不均匀应变、微观应变、内应力）



d1 d4 d5 ＝d /d

d2 d3 2(d+d)sin(+)=

or 2d(1+)sin(+)=

d  and  are very small,

d1 d2

and hence ,

2＝－2tg or

’＝ 2tg

Separation of the effects of Crystallite size and

Lattice Distortion





K 

  i    '

 2tg

D  cos

i







 cos sin  K

 2 

  D



Measuring two or more diffraction peaks.

(iv) The profile of diffraction peaks ―― Crystallite

size distribution

 N 

sin  2

a  S  Derived from

I  AN

2

 AN AN 

*    Bragg equation

2 

sin  a  S 

 

m

sin 2 ns 

f p ( s )  K  P ( n)

n 1 sin (s )

2







fp(s) is the line profile of diffraction peak

P(n) is Crystallite size distribution function

Instrumental（B0） Convolution









 B0 （B）





h( s )   g (t ) f (s  t )dt



or h( s)  g ( s)  f ( s)



F (h( s))  F ( g ( s))  F ( f ( s))

(vi) Texture (002)

30.0k





25.0k





20.0k









Intensity a.u.

15.0k





10.0k





5.0k



(103) (004)

0.0

30 40 50 60 70 80 90

2-Theta degree









ZnO nano-arrays

140.0k (201)



120.0k





100.0k









Intensity

80.0k





60.0k





40.0k

(002)

20.0k





0.0

30 40 50 60 70 80

2-Theta degree









(vii) Small angle scattering --- Particle size

b. Applications of powder diffractions

Applications

Qualitative Analysis

Quantitative Analysis

Lattice Parameter Determination

Crystallite size / size distribution & Lattice Distortion

Analysis (Non-uniform microstrain)

Crystallinity Analysis

Residue Stress Analysis

Texture analysis

Structure Solution and Refinement

Radical distribution function (for amorphous materials)

7.2.5 Electron Diffraction and Neutron Diffraction

1. Electron Diffraction

de Brogli wave length

of electron in a field V:

h

＝

2meeV



100 kV ----  ~ 0.00370 nm



Atom-level resolution!

a) TEM image of the tip part of one TeO2 nanorod. (b)

Enlarged TEM image. (c) The corresponding electron

2. Neutron Diffraction diffraction pattern.



----- Scatterring of atomic nuclear

~ higher atomic resolution

7.4 Quasi-crystal, liquid crystal and amorphous

Quasi-crystal





Liquid crystal





Amorphous

Quasi-crystal

Crystal









There is no translation

symmetry.