kuliah xrd mr

Analisis Difraksi Sinar-X

(X-Ray Diffraction Analysis)

Crystal Structure

LATTICE = Kisi susunan titik dalam ruang yang memiliki lingkungan identik antara

satu dengan lainnya

CRYSTAL STRUCTURE = Susunan atom (kelompok atom) yang berulang .

It can be described by associating with each lattice point

a group of atoms called the MOTIF (BASIS)

Primitive Cell: simplest cell, contain one lattice point

Not necessary have the crystal symmetry

UNIT CELL = The smallest component of the crystal, which when

stacked together with pure translational repetition reproduces the

whole crystal

5 Kisi Bravais dalam 2D

P P NP

Definition:

Bravais Lattice: an infinite array of discrete points with an

arrangement and orientation that appears exactly the same from

whichever of the points the array is viewed.

Name

Number of Bravais lattices

Conditions

Triclinic

1 (P)

a1 a2 a3

Monoclinic

2 (P, C)

a1 a2 a3

= = 90°

Orthorhombic

4 (P, F, I, A)

a1 a2 a3

= = = 90°

Tetragonal

2 (P, I)

a1 = a2 a3

= = = 90°

Cubic

3 (P, F, I)

a1 = a2 = a3

= = = 90°

Trigonal

1 (P)

a1 = a2 = a3

= = < 120° 90°

Hexagonal

1 (P)

a1 = a2 a3

= = 90°

= 120°

3D: 14 Bravais Lattice, 7 Crystal System

Kisi FCC

Logam Cu memiliki kisi face-centered cubic Atom-atom identik terletak pada sudut dan pada bagian muka kisi

Jenis Kisi adalah type F

also Ag, Au, Al, Ni...

BCC Lattice

-Fe merupakan sebuah kisi body-centered cubic

Atom-atom Identik terletak pada sudut dan body center (nothing at face centers)

Lattice type I

Also Nb, Ta, Ba, Mo...

FCC Lattices

Sodium Chloride (NaCl) - Na is much smaller than Cs

Face Centered Cubic

Rocksalt structure

Lattice type F

Also NaF, KBr, MgO….

Unit cell contents Counting the number of atoms within the unit cell

Many atoms are shared between unit cells

Atoms Shared Between: Each atom counts:

corner 8 cells 1/8

face center 2 cells 1/2

body center 1 cell 1

edge center 4 cells 1/4

lattice type cell contents

P 1 [=8 x 1/8]

I 2 [=(8 x 1/8) + (1 x 1)]

F 4 [=(8 x 1/8) + (6 x 1/2)]

C

2 [=(8 x 1/8) + (2 x 1/2)]

e.g. NaCl

Na at corners: (8 1/8) = 1 Na at face centres (6 1/2) = 3

Cl at edge centres (12 1/4) = 3 Cl at body centre = 1

Unit cell contents are 4(Na+Cl-)

CRYSTAL STRUCTURE

DETERMINATION BY X-RAY DIFRACTION (XRD)

Bravais lattices Crystals possess a regular, repetitive

internal structure. The concept of symmetry

describes the repetition of structural features.

Symmetries are most frequently used to classify the

different crystal structures. In general one can generate 14

basic crystal structures through symmetries. These are called

Bravais lattices. Any crystal structures can be reduced to one

of these 14 Bravias lattices.

CRYSTAL STRUCTURE DETERMINATION

The determination of an unknown structure

proceeds in three major steps:

1. The shape and size of the unit cell are deduced

from the angular positions of the diffraction

lines.

2. The number of atoms per unit cell is then

computed from the shape and size of the unit cell,

the chemical composition of the specimen, and

its measured density.

3. Finally, the position of the atoms within the unit

cell are deduced from the relative intensities of

the diffraction

Size and Shape of the Unit Cell This step is same as the indexing of power diffraction pattern. It involves: • The accurate determination of peak positions. • Determination of the unit cell parameters from the peak positions.

Bragg’s Law Interplanar d-spacing Systematic Absences

Bragg's Law When x-rays are scattered from a crystal lattice, peaks of scattered intensity are observed which correspond to the following conditions: 1. The angle of incidence = angle of scattering.

2. The pathlength difference is equal to an integer number of

wavelengths. The condition for maximum intensity contained in Bragg's

law above allow us to calculate details about the crystal structure, or if the crystal structure is known, to determine the wavelength of the x-rays incident upon the crystal.

Indexing Tetragonal system:

2

2

2

22

2

1

c

l

a

kh

dhkl

hkld2/sin

222

2

2

2

2222

2 )(42

sin ClkhAc

l

a

kh

dhkl

2222 4/;4/ cCaA

The value of A is obtained from hk0 lines.

When l = 0,

Possible values for h2 + k2 are 1, 2, 4, 5, 8, 9, 10, …

The rest of lines are

Possible values for l2 are 1, 4, 9, 16, …

Indexing Hexagonal system:

2

2

2

22

2 3

41

c

l

a

khkh

dhkl

222

2

22

2

2222 )(

43sin ClkhkhA

c

l

a

khkh

2222 4/;3/ cCaA

Possible values for h2 + hk + k2 are 1, 3, 4, 7, 9, 12, …

The rest of lines are

Possible values for l2 are 1, 4, 9, 16, …

2222

2222

)(sin

)(sin

ClkhkhA

atau

ClkhkhA

yCxAClkhkhA 2222 )(sin

l h,k

0,0 1,0 1,1 2,0 2,1 2,2 3,0 3,1

0 0 x 3x 4x 7x 12x 9x 13x

1 y x + y 3x + y 4x + y 7x + y 12x + y 9x + y 13x + y

2 4y x + 4y 3x + 4y 4x + 4y 7x + 4y 12x + 4y 9x + 4y 13x + 4y

3 9y x + 9y 3x + 9y 4x + 9y 7x + 9y 12x + 9y 9x + 9y 13x + 9y

4 16y x + 16y 3x + 16y 4x + 16y 7x + 16y 12x + 16y 9x + 16y 13x + 16y

h,k

1,0 1,1 2,0 2,1 2,2 3,0 3,1

Sin2 Sin2 /3

Sin2 /4

Sin2 /7 Sin2 /12 Sin2 /9 Sin2 /13

1 0,097 0,032 0,024

2 0,112 0,037 0,028

3 0,136 0,045 0,034

4 0,209 0,070 0,052

5 0,332 0,111 0,083

6 0,390 0,130 0,098 7 0,434 0,145 0,109 8 0,472 0,157 0,118 9 0,547 0,182 0,137

10 0,668 0,223 0,167 11 0,722 0,241 0,180

A = 0,112

Example:

sin2 sin2/3 sin2/4 sin2/7

0.097 0.032

0.112 0.037

0.136 0.045

0.209 0.070

0.332 0.111

0.390 0.130

0.434 0.145

0.472 0.157

0.547 0.182

0.668 0.223

0.722 0.241

0.024

0.028

0.034

0.052

0.083

0.098

0.109

0.118

0.137

0.167

0.180

0.014

0.016

0.019

0.030

0.047

0.056

0.062

0.067

0.078

0.095

0.103

100

110

Let’s say A = 0.112

1

2

3

4

5

6

7

8

9

10

11

sin2 sin2-A sin2-3A

0.097

0.112

0.136

0.209

0.332

0.390

0.434

0.472

0.547

0.668

0.722

0.806

0.879

0

0.024

0.097

0.220

0.278

0.322

0.360

0.435

0.556

0.610

0.694

0.767

0.054

0.098

0.136

0.211

0.332

0.386

0.470

0.543

0.097 belongs to Cl2.

What is the l?

There are two lines

between 100 and 110.

Probably, 10l1 and

10l2

0.024 and 0.097

are different ls.

0.097/0.024 ~ 4

0.220/0.024 ~ 9

0.390/0.024 ~ 16

C = 0.02441

1

2

3

4

5

6

7

8

9

10

11

12

13

100

110

101

102

002

112

0 2 4 6 8 10 12 14 16 18

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

Cl2

l2

Slope = C = 0.02441

Indexing orthorhombic system:

222

2

2

2

2

2

222

4sin ClBkAh

c

l

b

k

a

h

More difficult! Consider any two lines having indices

hk0 and hkl Cl2; get a guesses value C then put it back

and get A and B. If the guess is right, the number will be

consistent. If not try another guesses value of C.

Indexing Monoclinic and Triclinic system:

Even more complex, 6 variables, must have enough

diffraction lines for the computer to indexing.

Effect of Cell distortion on the powder Pattern:

Determination of the number of atoms in a unit cell:

After establishing shape and size we find the number

of atoms in that unit cell.

M

NVn C 0

VC: unit cell volume; : density

N0: Avogodro’s number;

M: molecular weight;

n: number of molecules in a unit cell

We need to index the powder pattern in order to obtain

the unit cell parameters unit cell volume

66054.1)10022.6()()10( 23

03

338 C

C

VN

cm

gcmVnM

in Å3

in g/cm3

A

MENENTUKAN INTENSITAS PUNCAK

DIFRAKSI SINAR X

Multiplicity factor

Lattice Index Multiplicity Planes

Cubic

(with highest

symmetry)

(100) 6 [(100) (010) (001)] ( 2 for negatives)

(110) 12 [(110) (101) (011), (110) (101) (011)] ( 2 for negatives)

(111) 12 [(111) (111) (111) (111)] ( 2 for negatives)

(210) 24* (210) 3! Ways, (210) 3! Ways,

(210) 3! Ways, (210) 3! Ways

(211) 24 (211) 3 ways, (211) 3! ways,

(211) 3 ways

(321) 48*

Tetragonal

(with highest

symmetry)

(100) 4 [(100) (010)] ( 2 for negatives)

(110) 4 [(110) (110)] ( 2 for negatives)

(111) 8 [(111) (111) (111) (111)] ( 2 for negatives)

(210) 8* (210) = 2 Ways, (210) = 2 Ways,

(210) = 2 Ways, (210) = 2 Ways

(211) 16 [Same as for (210) = 8] 2 (as l can be +1 or 1)

(321) 16* Same as above (as last digit is anyhow not permuted)

* Altered in crystals with lower symmetry

Cubic hkl hhl hk0 hh0 hhh h00

48* 24 24* 12 8 6

Hexagonal hk.l hh.l h0.l hk.0 hh.0 h0.0 00.l

24* 12* 12* 12* 6 6 2

Tetragonal hkl hhl h0l hk0 hh0 h00 00l

16* 8 8 8* 4 4 2

Orthorhombic hkl hk0 h0l 0kl h00 0k0 00l

8 4 4 4 2 2 2

Monoclinic hkl h0l 0k0

4 2 2

Triclinic hkl

2

* Altered in crystals with lower symmetry (of the same crystal class)

Multiplicity factor

If atom B is different from atom A the amplitudes must be weighed by the

respective atomic scattering factors (f)

The resultant amplitude of all the waves scattered by all the atoms in the UC gives

the scattering factor for the unit cell

The unit cell scattering factor is called the Structure Factor (F)

Scattering by an unit cell = f(position of the atoms, atomic scattering factors)

electronan by scattered waveof Amplitude

ucin atoms allby scattered waveof AmplitudeFactor StructureF

[2 ( )]i i h x k y l zE Ae fe 2 ( )h x k y l z In complex notation

2FI

[2 ( )]

1 1

j j j j

n ni i h x k y l zhkl

n j j

j j

F f e f e

Structure factor is independent of the shape and size of the unit cell

For n atoms in the UC

If the UC distorts so do the planes in it!!

nnie )1(

)(2

Cosee ii

Structure factor calculations

A Atom at (0,0,0) and equivalent positions

[2 ( )]j j j ji i h x k y l z

j jF f e f e

[2 ( 0 0 0)] 0i h k lF f e f e f

22 fF F is independent of the scattering plane (h k l)

nini ee

Simple Cubic

1) ( inodde

1) ( inevene

B Atom at (0,0,0) & (½, ½, 0) and equivalent positions

[2 ( )]j j j ji i h x k y l z

j jF f e f e

1 1[2 ( 0)]

[2 ( 0 0 0)] 2 2

[ 2 ( )]0 ( )2 [1 ]

i h k li h k l

h ki

i h k

F f e f e

f e f e f e

F is independent of the ‘l’ index

C- centred Orthorhombic

Real

]1[ )( khiefF

fF 2

0F

22 4 fF

02 F

e.g. (001), (110), (112); (021), (022), (023)

e.g. (100), (101), (102); (031), (032), (033)

X-Ray Fluorescence Analysis

Analisis X-ray Fluoresensi

Pendahuluan

Prinsip Kerja

Skema Cara Kerja Alat

Preparasi Sampel

Instrumen XRF

Contoh spektra

Radiasi Elektromagnetik

1Hz - 1kHz 1kHz - 1014Hz

1014Hz - 1015Hz

1015Hz - 1021Hz

Extra-Low Frequency

(ELF)

Radio Microwave Infrared

Visible Light

X-Rays,

Gamma Rays

Low energy High energy

Pendahuluan

Teknik fluoresensi sinar x (XRF) merupakan suatu teknik analisis yang dapat menganalisa unsur-unsur yang membangun suatu material, dengan dasar interaksi sinar-X dengan materi analit.

Teknik ini juga dapat digunakan untuk menentukan konsentrasi unsur berdasarkan pada panjang gelombang dan jumlah sinar x yang dipancarkan kembali setelah suatu material ditembaki sinar X.

Fitur XRF

• Akurasi yang relative tinggi

• Dapat menentukan unsur dalam material tanpa adanya standar (bandingkan dg. AAS)

• Dapat menentukan kandungan mineral dalam bahan biologis maupun dalam tubuh secara langsung.

Kelebihan

• tidak dapat mengetahui senyawa apa yang dibentuk oleh unsur-unsur yang terkandung dalam material yang akan kita teliti.

• tidak dapat menentukan struktur dari atom yang membentuk material itu.

Kelemahan

Prinsip Kerja

Menembakkan radiasi foton elektromagnetik

ke material yang diteliti.

Radiasi elektromagnetik yang

dipancarkan akan berinteraksi dengan

elektron yang berada di kulit K suatu unsur.

Elektron yang berada di kulit K akan

memiliki energi kinetik yang cukup untuk

melepaskan diri dari ikatan inti, sehingga

elektron itu akan terpental keluar.

Peristiwa pada tabung sinar-X.

Prinsip Kerja XRF

Pada teknik XRF,

• dienggunakan sinar-X dari tabung pembangkit sinar-X untuk mengeluarkan electron dari kulit bagian dalam untuk menghasilkan sinar-X baru dari sample yang di analisis.

Prinsip Kerja XRF

• Untuk setiap atom di dalam sample,

intensitas dari sinar-X karakteristik tersebut

sebanding dengan jumlah (konsentrasi)

atom di dalam sample.

• Intensitas sinar–X karakteristik dari setiap

unsur, dibandingkan dengan suatu standar

yang diketahui konsentrasinya, sehingga

konsentrasi unsur dalam sample bisa

ditentukan.

Skema Cara Kerja Alat

Instrumen XRF

Instrumen XRF terdiri dari :

• Sumber cahaya

• Optik

• Detektor

Syarat Sampel

Serbuk Ukuran serbuk < 4 00 mesh

Padatan Permukaan yang dilapisi akan meminimalisir efek

penghamburan Sampel harus datar untuk menghasilkan analisis kuantitatif

yang optimal

Cairan Sampel harus segar ketika dianalisis dan analisis dilakukan

secara cepat jika sampel mudah menguap Sampel tidak boleh mengandung endapan

Sumber Cahaya

• Tabung Sinar X

• End Window

• Side Window

• Radioisotop

Tabung sinar x

• End Window

Side Window

Be W indow

Silicone Insulation

Glass En v elope

Filament

Electron beam

Target (Ti, Ag,

Rh, etc.)

Copper Anode

H V Lead

Radioisotop

Isotope

Fe-5 5

Cm-244

Cd-109

Am-241

Co-57

Energy (keV)

5.9

14.3, 18.3

22, 88

59.5

122

Elements (K-lines)

Al – V

Ti-Br

Fe-Mo

Ru-Er

Ba - U

Elements (L-lines)

Br-I

I- Pb

Yb-Pu

None

none

Optik

Source Detector

Filter

Detector

X-Ray

Source

Source Filter

Contoh Spektra

Contoh Spektra

Detektor

• Si(L i)

• P N Diode

• Silicon Drift Detectors

• Proportional Counters

• Scintillation Detectors

Si ( Li) Detektor

WIndow

Si(Li)

crystal

Dewar

filled with

LN 2

Super-Cooled Cryostat

Cooling: LN2 or Peltier

Window: Beryllium or Polymer

Counts Rates: 3,000 – 50,000 cps

Resolution: 120-170 eV at Mn K-alpha

FET

Pre-Amplifier

PIN Diode

• Cooling: Thermoelectrically cooled (Peltier)

• Window: Beryllium

• Count Rates: 3,000 – 20,000 cps

• Resolution: 170-240 eV at Mn k-alpha

Silicon Drift Detector

Packaging: Similar to PIN Detector Cooling: Peltier Count Rates; 10,000 – 300,000 cps Resolution: 140-180 eV at Mn K-alpha

Proportional Counter

Anode Filament

Fill Gases: Neon, Argon, Xenon, Krypton

Pressure: 0.5- 2 ATM

Windows: Be or Polymer

Sealed or Gas Flow Versions

Count Rates EDX: 10,000-40,000 cps WDX: 1,000,000+

Resolution: 500-1000+ eV

Window

Scintillation Detector

PMT (Photo-multiplier tube) Sodium Iodide Disk Electronics

Connector Window: Be or Al Count Rates: 10,000 to 1,000,000+ cps Resolution: >1000 eV

(0,0,0)

(0, ½, ½)

(½, ½, 0)

(½, 0, ½)

Fractional Coordinates

Cs (0,0,0)

Cl (½, ½, ½)

Density

Calculation

AC NV

nA

n: number of atoms/unit cell

A: atomic mass

VC: volume of the unit cell

NA: Avogadro’s number

(6.023x1023 atoms/mole)

Calculate the density of copper.

RCu =0.128nm, Crystal structure: FCC, ACu= 63.5 g/mole

n = 4 atoms/cell, 333 216)22( RRaVC

3

2338/89.8

]10023.6)1028.1(216[

)5.63)(4(cmg

8.94 g/cm3 in the literature

(100) (111)

(200) (110)

2

222

2

1

a

lkh

dhkl

For cubic system

Lattice spacing

Bragg’s Law

For cubic system:

But no all planes have the

diffraction !!!

sin2

sinsin

hkl

hklhkl

d

dd

QTSQn

222 lkh

adhkl

• X-ray diffraction from a crystal: Bragg’s Law

sin2 hkldn

222 lkh

adhkl

X-Ray Diffraction

n: order of

diffraction peak

dhkl: interplanar

spacing

(hkl): Miller

indices of plane

Crystal Structures [OGN 21.2] • Body-centered

cubic (BCC)

(200) (211)

Powder diffraction

X-Ray

Phase purity. In a mixture of compounds each crystalline phase present will contribute to

the overall powder X-ray diffraction pattern. In preparative materials

chemistry this may be used to identify the level of reaction and purity of the

product. The reaction between two solids Al2O3 and MgO to form MgAl2O4

may be monitored by powder X-ray diffraction.

•At the start of the reaction a mixture of Al2O3 and MgO will produce an X-

ray pattern combining those of the pure phases. As the reaction proceeds,

patterns (a) and (b), a new set of reflections corresponding to the product

MgAl2O4, emerges and grows in intensity at the expense of the reflection

from Al2O3 and MgO. On completion of the reaction the powder diffraction

pattern will be that of pure MgAl2O4.

•A materials chemist will often use PXRD to monitor the progress of a

reaction.

•The PXRD method is widely employed to identify impurities in materials

whether it be residual reactant in a product, or an undesired by-product.

•However the impurity must be crystalline.

Observable diffraction

peaks

222 lkh Ratio

Simple

cubic

SC: 1,2,3,4,5,6,8,9,10,11,12..

BCC: 2,4,6,8,10, 12….

FCC: 3,4,8,11,12,16,24….

222 lkh

adhkl

nd sin2

Ex: An element, BCC or FCC, shows diffraction peaks

at 2: 40, 58, 73, 86.8,100.4 and 114.7.

Determine:(a) Crystal structure?(b) Lattice constant?

(c) What is the element?

2theta theta (hkl)

40 20 0.117 1 (110)

58 29 0.235 2 (200)

73 36.5 0.3538 3 (211)

86.8 43.4 0.4721 4 (220)

100.4 50.2 0.5903 5 (310)

114.7 57.35 0.7090 6 (222)

2sin222 lkh

a =3.18 A, BCC, W

0

5

10

15

20

25

30

0 20 40 60 80

Bragg Angle (, degrees)

Lo

ren

tz-P

ola

riza

tio

n f

act

or

Polarization factor Lorentz factor

2

1

2

1

SinCos

SinfactorLorentz 21 2CosIP

CosSin

CosfactoronPolarizatiLorentz

2

2 21

Intensity of powder pattern lines (ignoring Temperature & Absorption factors)

CosSin

CospFI

2

22 21

Valid for Debye-Scherrer geometry

I → Relative Integrated “Intensity”

F → Structure factor

p → Multiplicity factor

POINTS

As one is interested in relative (integrated) intensities of the lines constant

factors are omitted

Volume of specimen me , e (1/dectector radius)

Random orientation of crystals in a with Texture intensities are

modified

I is really diffracted energy (as Intensity is Energy/area/time)

Ignoring Temperature & Absorption factors valid for lines close-by in

pattern