analisis difraksi sinar-x (x-ray diffraction analysis)
DESCRIPTION
Analisis Difraksi Sinar-X (X-Ray Diffraction Analysis). Crystal Structure. Reading: Ashcroft 4-7. Ideal Crystal: Mengandung susunan atom/ion secara periodik Direpresentasikan oleh titik kisi Sekelompok atoms yang membentuk titik kisi. Basis. - PowerPoint PPT PresentationTRANSCRIPT
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Analisis Difraksi Sinar-X
(X-Ray Diffraction Analysis)
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Crystal Structure
Ideal Crystal: Mengandung susunan atom/ion secara periodikDirepresentasikan oleh titik kisiSekelompok atoms yang membentuk titik kisi
Basis
LATTICE = Kisi susunan titik dalam ruang yang memiliki lingkungan identik antara satu dengan lainnyaCRYSTAL 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)
Reading: Ashcroft 4-7
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{R = n1 a1 + n2 a2 + n3 a3}
Translationalvector
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Primitive Cell: simplest cell, contain one lattice pointNot 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
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Conventional cell vs. Primitive CellReflecting the symmetry
Different Basis
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5 Kisi Bravais dalam 2D
P P NP
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Square a=b =90
Rectangular a b =90
Centered Rectangular
a b =90
Hexagonal a=b =120
Oblique a b 90
5 Kisi Bravais dalam 2D
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Translationalvector
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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
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Kisi FCC
Logam Cu memiliki kisi face-centered cubic
Atom-atom identik terletak pada sudut dan pada bagian muka kisiJenis Kisi adalah type F
also Ag, Au, Al, Ni...
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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...
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Simple Cubic LatticeCaesium Chloride (CsCl) is primitive cubic
Different atoms at corners and body center. NOT body centered, therefore.
Lattice type P
Also CuZn, CsBr, LiAg
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FCC Lattices
Sodium Chloride (NaCl) - Na is much smaller than Cs
Face Centered Cubic
Rocksalt structure
Lattice type F
Also NaF, KBr, MgO….
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Diamond Structure: two sets of FCC Lattices
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One 4-fold axes
Why not F tetragonal?
Tetragonal: P, I
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Example
CaC2 - has a rocksalt-like structure but with non-spherical carbides
2-C
C
Carbide ions are aligned parallel to c
c > a,b tetragonal symmetry
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Orthorhombic: P, I, F, C
C F
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Another type of centering
Side centered unit cell
Notation:
A-centered if atom in bc plane
B-centered if atom in ac plane
C-centered if atom in ab plane
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Unit cell contents Counting the number of atoms within the unit cell
Many atoms are shared between unit cells
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Atoms Shared Between: Each atom counts:corner 8 cells 1/8face center 2 cells 1/2body center 1 cell 1edge center 4 cells 1/4
lattice type cell contentsP 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)]
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e.g. NaCl Na at corners: (8 1/8) = 1 Na at face centres (6 1/2) = 3Cl at edge centres (12 1/4) = 3 Cl at body centre = 1
Unit cell contents are 4(Na+Cl-)
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(0,0,0)(0, ½, ½)(½, ½, 0)(½, 0, ½)
Fractional Coordinates
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Cs (0,0,0)Cl (½, ½, ½)
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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
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Crystallographic Directions And Planes
Lattice DirectionsIndividual directions: [uvw][uvw]Symmetry-related directions: <uvw><uvw>
Miller Indices:1. Find the intercepts on the axes in terms of the lattice
constant a, b, c2. Take the reciprocals of these numbers, reduce to the
three integers having the same ratio(hkl)(hkl)
Set of symmetry-related planes: {hkl}{hkl}
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Crystal Structures [OGN 21.2]• Body-centered cubic (BCC)
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(100) (111)
(200) (110)
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2
222
2
1
a
lkh
dhkl
For cubic system
Lattice spacing
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Crystal Structure Analysis
X-ray diffraction
Essence of diffraction: Bragg Diffraction
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LightInterference fringes Constructive
Destructive
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Bragg’s Law
For cubic system: But no all planes have the diffraction !!!
sin2
sinsin
hkl
hklhkl
d
dd
QTSQn
222 lkh
adhkl
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• 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
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Crystal Structures [OGN 21.2]• Body-centered cubic (BCC)
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/hchE
35KeV ~ 0.1-1.4ACu K 1.54 A
Mo:
X-Ray Diffraction
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(200)(211)
Powder diffraction
X-Ray
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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.
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•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.
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The powder diffraction patterns and the systematic absences of three versions of a cubic cell. Comparison of the observed pattern with patterns like these enables the unit cell to be identified. The locations of the lines give the cell dimensions.
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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
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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)
2sin 222 lkh
a =3.18 A, BCC, W
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