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Crystal Structure 1 Lattice Sites in Cubic Unit Cell

Lecture 2 SSP

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Page 1: Lecture 2 SSP

Crystal Structure 1

Lattice Sites in Cubic Unit Cell

Page 2: Lecture 2 SSP

Crystal Structure 2

Crystal Directions

Fig. Shows [111] direction

We choose one lattice point on the line as an origin, say the point O. Choice of origin is completely arbitrary, since every lattice point is identical.

Then we choose the lattice vector joining O to any point on the line, say point T. This vector can be written as;

R = n1 a + n2 b + n3c

To distinguish a lattice direction from a lattice point, the triple is enclosed in square brackets [ ...] is used.[n1n2n3]

[n1n2n3] is the smallest integer of the same relative ratios.

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Crystal Structure 3

210

X = 1 , Y = ½ , Z = 0[1 ½ 0] [2 1 0]

X = ½ , Y = ½ , Z = 1[½ ½ 1] [1 1 2]

Examples

Page 4: Lecture 2 SSP

Crystal Structure 4

Negative directions

When we write the

direction [n1n2n3] depend on the origin, negative directions can be written as

R = n1 a + n2 b + n3c

Direction must be

smallest integers.

Y direction

(origin) O

- Y direction

X direction

- X direction

Z direction

- Z direction

][ 321 nnn

][ 321 nnn

Page 5: Lecture 2 SSP

Crystal Structure 5

X = -1 , Y = -1 , Z = 0 [110]

Examples of crystal directions

X = 1 , Y = 0 , Z = 0 [1 0 0]

Page 6: Lecture 2 SSP

Crystal Structure 6

Crystal Planes

Within a crystal lattice it is possible to identify sets of equally spaced parallel planes. These are called lattice planes.

In the figure density of lattice points on each plane of a set is the same and all lattice points are contained on each set of planes.

b

a

b

a

The set of planes in 2D lattice.

Page 7: Lecture 2 SSP

Crystal Structure 7

Miller Indices

Miller Indices are a symbolic vector representation for the orientation of an atomic plane in a crystal lattice and are defined as the reciprocals of the fractional intercepts which the plane makes with the crystallographic axes.

To determine Miller indices of a plane, take the following steps;

1) Determine the intercepts of the plane along each of the three crystallographic directions

2) Take the reciprocals of the intercepts

3) If fractions result, multiply each by the denominator of the smallest fraction

Page 8: Lecture 2 SSP

Crystal Structure 8

Axis X Y Z

Intercept points 1 ∞ ∞

Reciprocals 1/1 1/ ∞ 1/ ∞Smallest

Ratio 1 0 0

Miller İndices (100)

Example-1

(1,0,0)

Page 9: Lecture 2 SSP

Crystal Structure 9

Axis X Y Z

Intercept points 1 1 ∞

Reciprocals 1/1 1/ 1 1/ ∞Smallest

Ratio 1 1 0

Miller İndices (110)

Example-2

(1,0,0)

(0,1,0)

Page 10: Lecture 2 SSP

Crystal Structure 10

Axis X Y Z

Intercept points 1 1 1

Reciprocals 1/1 1/ 1 1/ 1Smallest

Ratio 1 1 1

Miller İndices (111)(1,0,0)

(0,1,0)

(0,0,1)

Example-3

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Crystal Structure 11

Axis X Y Z

Intercept points 1/2 1 ∞

Reciprocals 1/(½) 1/ 1 1/ ∞Smallest

Ratio 2 1 0

Miller İndices (210)(1/2, 0, 0)

(0,1,0)

Example-4

Page 12: Lecture 2 SSP

Crystal Structure 12

Axis a b c

Intercept points 1 ∞ ½

Reciprocals 1/1 1/ ∞ 1/(½)

Smallest Ratio 1 0 2

Miller İndices (102)

Example-5

Page 13: Lecture 2 SSP

Crystal Structure 13

Miller Indices

Reciprocal numbers are:

2

1 ,

2

1 ,

3

1

Plane intercepts axes at cba 2 ,2 ,3

Indices of the plane (Miller): (2,3,3)

(100)

(200)(110)

(111) (100)

Indices of the direction: [2,3,3]a

3

2

2

bc

[2,3,3]

Page 14: Lecture 2 SSP

Crystal Structure 14

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Crystal Structure 15

Indices of a Family or Form

Sometimes when the unit cell has rotational symmetry, several nonparallel planes may be equivalent by virtue of this symmetry, in which case it is convenient to lump all these planes in the same Miller Indices, but with curly brackets.

Thus indices {h,k,l} represent all the planes equivalent to the plane (hkl) through rotational symmetry.

)111(),111(),111(),111(),111(),111(),111(),111(}111{

)001(),100(),010(),001(),010(),100(}100{

Page 16: Lecture 2 SSP

Crystal Structure 16

There are only seven different shapes of unit cell which can be stacked together to completely fill all space (in 3 dimensions) without overlapping. This gives the seven crystal systems, in which all crystal structures can be classified.

Cubic Crystal System (SC, BCC,FCC) Hexagonal Crystal System (S) Triclinic Crystal System (S) Monoclinic Crystal System (S, Base-C) Orthorhombic Crystal System (S, Base-C, BC, FC) Tetragonal Crystal System (S, BC) Trigonal (Rhombohedral) Crystal System (S)

3D – 14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM

TYPICAL CRYSTAL STRUCTURES

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Crystal Structure 17

Page 18: Lecture 2 SSP

triclinic monoclinic

orthorhombic

tetragonal rhombohedral

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hexagonal

cubic

Page 20: Lecture 2 SSP

Packing FractionPacking Fraction

In crystallography, atomic packing factor (APF) or packing fraction is the fraction of volume in a crystal structure that is occupied by atoms.

•Natoms is the number of atoms in the unit cell

•Vatom is the volume of an atom

•Vunit cell is the volume occupied by the unit cell

Page 21: Lecture 2 SSP

Unit Cell:

The smallest repeating unit in a three dimensionalstructure (lattice)

Characterized by a coordination number (number of nearest neighbors)

and a lattice parameter, a (edge length)

Bonding

Atomic radius -- Bulk density

Different geometric arrangements depending on the crystal type

Page 22: Lecture 2 SSP

Packing Fraction ofSimple cubic, Body centered, face centeredPacking Fraction ofSimple cubic, Body centered, face centered

Page 23: Lecture 2 SSP

Packing in Crystals

Simple Cubic Crystal

packing can be determined exactly

If these were atoms then there would be 8(1/8) atoms per cell or 1 atom per cell.

Page 24: Lecture 2 SSP

Packing Fraction of Simple Cubic Lattice

The packing fraction would be

(4/3)πr3/d3

r is related to d, r = d/2

Therefore, the packing is (4/3) π(d/2)3/d3 = 4π/24 =π /6 = 0.52

Page 25: Lecture 2 SSP

Simple Cubic (SC)Simple Cubic (SC)

• SC lattice and crystal structure

a

a = 2R

Where: R = atomic radius a = lattice parameter

a = 2R

Where: R = atomic radius a = lattice parameter

SC

Packing fraction = 54%

Coordination number = 6

Page 26: Lecture 2 SSP

Body Centered Cubic (BCC)

BCC

a = 4R

3Where: R = atomic radius a = lattice parameter

a

Packing fraction = 68%

Coordination number = 8

Where does this come from?

Page 27: Lecture 2 SSP

Cubic Packing - BCC

a

a

√2 a √2 a

a√3a

√3a=4R

a=4R/√3

Page 28: Lecture 2 SSP

Face Centered Cubic (FCC)

FCC

a = 2R 2

a = 4R

2

Where: R = atomic radius a = lattice parameter

A close-packed structure with a packing fraction of 74%

Coordination number = 12

Page 29: Lecture 2 SSP

Some Observations

Since a crystal structure is a lattice + basis the packing fraction of the simple cubic lattice can go beyond one atom bases.

However, some crystal structures that appear simple cubic are in fact not: The sodium chloride structure is actually face centered cubic with a basis of two atoms.

Crystal structure, in itself, is a course.

Page 30: Lecture 2 SSP

Calculating packing fraction for body centeredCalculating packing fraction for body centered

Body-centered cubic crystal structureBody-centered cubic crystal structure

Simple cubic: 0.52 Body-centered cubic: 0.68 Hexagonal close-packed: 0.74 Face-centered cubic: 0.74 Diamond cubic: 0.34