Diffraction Techniques2007 00 L4 rev1einstein.sc.mahidol.ac.th/~jose/pdf/L4.pdf · 2008-06-16 ·...

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Crystallography II Outline: 1- Classification of Bravais lattices: 1.1- The seven crystal systems an fourteen Bravais lattices 1.2- Crystallographic point groups and space groups 1.3- Symmetry operations in crystallographic point groups 1.4- Nomenclature 1.5- Space groups 2- Real and reciprocal lattices 2.1-Lattice directions and planes 2.2- Miller indices 2.3-Interplanar spacing 2.4-The reciprocal lattice 3- Laue diffraction conditions

Transcript of Diffraction Techniques2007 00 L4 rev1einstein.sc.mahidol.ac.th/~jose/pdf/L4.pdf · 2008-06-16 ·...

Page 1: Diffraction Techniques2007 00 L4 rev1einstein.sc.mahidol.ac.th/~jose/pdf/L4.pdf · 2008-06-16 · 1.1-Lattice directions and planes 1.2- Miller indices 1.3-Interplanar spacing 1.4-The

Crystallography IIOutline:1- Classification of Bravais lattices:

1.1- The seven crystal systems an fourteen Bravais lattices1.2- Crystallographic point groups and space groups1.3- Symmetry operations in crystallographic point groups1.4- Nomenclature1.5- Space groups

2- Real and reciprocal lattices2.1-Lattice directions and planes2.2- Miller indices2.3-Interplanar spacing2.4-The reciprocal lattice

3- Laue diffraction conditions

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Some Definitions

Symmetry Operation: An operation that leads to superimposition of an object on itself.

Symmetry Element: An imaginary geometric entity (line, point, plane) about which a symmetry operation takes place.

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Symmetry

“A body is said to be symmetrical when it can be divided into parts that are related to each other in certain ways. The operation oftransferring one part to the position of a symmetrically related part is termed a symmetry operation, the result of which is to leave the final state of the body indistinguishable from its original state. In general, successive application of the symmetry operation must ultimately bring the body actually into its original state again.”

From: Lipson/Cochran, The Determination of Crystal StructuresLipson, H. and W. Cochran. The Determination of Crystal Structures.

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Symmetry Operations

Some symmetry operations are compatible with translation, some are not.

Compatible are:•mirror•glide•twofold rotation•threefold rotation•fourfold rotation•sixfoldrotation

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Symmetry operations of all Bravais lattices:

1-Translations trough lattice vectors

2-any operation that leaves at least one fixed point

3-combinations of 1 and 2

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Symmetry operations in the crystallographic point groups

1) Rotations through integral multiples of 2π/n about some axis

2) Rotation reflections even if a lattice lacks of a 2π/n symmetry, it may have a combination of such rotation followed by a reflectionin a plane perpendicular to that axis.

3) Rotation-Inversions a rotation through 2π/n followed by an inversionnone of these operations alone are symmetry elements of the lattice.

4) Reflections (mirror planes)

5) Inversion (maps a point with position r into another with position –r)

6) Identity

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Along the c axis of the Hexagonal lattice, we have alternately, points of type 1 and type 2 .They are related by a a screw axis parallel to the c-axis:Shift c/2, and rotate 60 deg..

Screw axes, and glide planes

Glide plane:Translate c/2 parallelto the c axis, and reflecton the indicated plane

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Hexagonal Close-packed structure Screw axis

Two Hexagonal Bravais lattices interpenetrated and displaced by c/2

Screw axis

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Symmetry operations of all Bravais lattices:

1-Translations trough lattice vectors

2-any operation that leaves at least one fixed point

3-combinations of 1 and 2

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Simple Tetragonal(seen along thec-axis)

Simple Tetragonal(seen along thec-axis)

SimpleOrthorhombic

Base-centeredOrthorhombic“All points are in the same plane”

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Simple orthorhombicBase centered orthorhombic

SimpleMonoclinic

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Triclinic: a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90°Lattice: P

Monoclinic: a ≠ b ≠ c, α = γ = 90° ≠ βLattice: P, C

Orthorhombic: a ≠ b ≠ c, α = β = γ = 90°Lattice: P, C, I, F

Tetragonal: a = b ≠ c, α = β = γ = 90°Lattice: P, I

Trigonal/Hexagonal: a = b ≠ c, α = 90°, γ = 120°Lattice: P, R

Cubic: a = b = c, α = β = γ = 90°Lattice: P, I, F

The 14 Bravais Lattices

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Cubic

Tetragonal

Orthorhombic

Monoclinic

Triclinic

Trigonal

Hexagonal

Sym

metry decreases

•Groups are constructedby reducing the symmetryof an object of a particularcrystal system

•When all the symmetryoperations are found in aless symmetric group thenew object is considered tobelong to the less symmetricgroup

The 32 crystallographic point groups

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Crystallography IIIOutline:

1- Real and reciprocal lattices1.1-Lattice directions and planes1.2- Miller indices1.3-Interplanar spacing1.4-The reciprocal lattice

2- Laue diffraction conditions2.1-diffraction by a row of atoms2.2-diffraction by net2.3-diffracrion by a lattice2.4-Interference conditions in terms of the reciprocal lattice

3-Crystallography problem set hand out. (due on the 25th of July, room K410)

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2.1-Lattice directions and planes

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2.1-Lattice directions and planes

For lattice points:

u,v,w

For points u’,v’w’ that are not lattice points:

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2.1-Lattice directions and planes

Let a line pass through the origin of the unit cell, and through a point u,v,w(not necessarily integers).The line also passes through 2u,2v,2w, 3u,3v,3w…etc..We call [uvw] the indices of the direction of the line.

They are also the indices of any parallel to the given line(because the origin is arbitrary)

We convert the values of u,v,w to the set of smallest integers and write themIn square brackets:

[1/2 1/2 1] [112] and [224] all represent the same direction but the preferredform is [112]

Negative indices denoted with a bar on top [uvw]

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Directions of a form

All the directions that are related by symmetry, are called directions of a form

Example: The four diagonals of a cube: [111], [111], [111] and [111] are allrepresented by the directions of the form <111> (with angular brackets)

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Crystallographic planes:We might describe the orientation of a plane by specifying the distancesmeasured from the origin where the plane intersects the three axes.

problem with planes that are parallel to the axis!

To avoid the problem, we can specify the inverse of the distances ofintersection with the axes. These are called Miller indices of the plane

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Crystallographic planes:

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Crystallographic planes:

The Miller indices hkl refer to a system of equidistant parallel planes, one of whichpasses through the origin and the plane nearest to the origin intersects the threecrystallographic axes on a/h, b/k, c/l.

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The miller indices: some examples

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Miller indices for Hexagonal system

Because of the symmetry of the hexagonal lattice, a fourth (i) index is added.The intercept on each of the three axes related by a C3 axis and the intercepton the c axis.Because the three axes are coplanar, one of the indices is related to the other two:

h+k=-i

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Miller indices for Hexagonal system

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Inter-planar spacing dhkl

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Page 39: Diffraction Techniques2007 00 L4 rev1einstein.sc.mahidol.ac.th/~jose/pdf/L4.pdf · 2008-06-16 · 1.1-Lattice directions and planes 1.2- Miller indices 1.3-Interplanar spacing 1.4-The

The reciprocal lattice

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The reciprocal lattice

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Page 42: Diffraction Techniques2007 00 L4 rev1einstein.sc.mahidol.ac.th/~jose/pdf/L4.pdf · 2008-06-16 · 1.1-Lattice directions and planes 1.2- Miller indices 1.3-Interplanar spacing 1.4-The

Diffraction by a row of atoms

1- each atom acts as a secondary source of spherical waves

2-constructive interference occurs only in the directions along whichthe wave-fronts are in phase.

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Diffraction by a row of atoms

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Diffraction by a row of atoms

(cos cos )p hν μ λ− = ±We arrive to the condition

And for the particular case that the angle of incidence is μ = 90°we have a very simple expression:

cosp hν λ= ±

And the zero order cone reduces to a disk.

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Diffraction by a net or a plane of atoms

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Diffraction by a net or a plane of atoms

1 1(cos cos )p hν μ λ− = ±

2 2(cos cos )p kν μ λ− = ±

Both conditions must be satisfied: this happensAt the intersection of thetwo cones

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Diffraction by a lattice or a 3D array of atoms

1 1(cos cos )p hν μ λ− = ±

2 2(cos cos )p kν μ λ− = ±

All three conditions must be satisfied: this happensAt the intersection of thethree cones

3 3(cos cos )p lν μ λ− = ±

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The Ewald construction