Phy351 ch 3

CRYSTAL STRUCTURE

Chapter 3

PHY351

Long-Range-Order (LRO)

Most metals and alloys, semiconductors, ceramics and some polymers have a crystalline structure (where the atoms or ions display in a long-range-order (LRO).

The atoms or ions in these material form a regular repetitive, grid like pattern in three dimension.

These materials are referred as a crystalline structure.

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Question 1:

a. Define the following terms:

i. SRO atom/ion arrangement

ii. no order atom/ion arrangement

iii. crystalline solid

iv. crystal structure

v. amorphous

b. Give 2 example material of

i. SRO atom/ion arrangement

ii. no order atom/ion arrangement

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Properties of SOLIDS depends upon:

bonding force

crystal structure

The space lattice and unit cells

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Space lattice

An IMAGINARY NETWORK of lines with atoms at lines intersection that representing the arrangement of atoms is called space lattice.

Unit cells

Unit cell is that block of atoms which REPEATS itself to form space lattice.

Unit Cell

Space Lattice

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Question 2:

a. Define the following terms:

i. Lattice point

ii. Motif

iii. Lattice constant

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Crystal Systems and Bravais Lattice

Only SEVEN different types of unit cells are necessary to create all point lattices which:

Cubic

Tetragonal

Orthorhombic

Rhombohedral

Hexagonal

Monoclinic

Triclinic

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According to Bravais (1811-1863), fourteen standard unit cells can describe all possible lattice networks based on FOUR basic types of unit cells which are:

Simple

Body Centered

Face Centered

Base Centered

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i. Cubic Unit Cell a = b = c

α = β = γ = 900

Simple Body Centered Face centered

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ii. Tetragonal a =b ≠ c

α = β = γ = 900

Simple Body Centered

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iii. Orthorhombic a ≠ b ≠ c

α = β = γ = 900

Simple Base CenteredFace CenteredBody Centered

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iv. Rhombohedral a =b = c

α = β = γ ≠ 900

Simple

v. Hexagonal a = b ≠ c α = β = 900

γ = 1200

Simple

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vi. Monoclinic a ≠ b ≠ c

α = γ = 900 ≠ β

vii. Triclinic a ≠ b ≠ c

α ≠ β ≠ γ ≠ 900

SimpleSimple

Base

Centered

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Principal Metallic Crystal Structures

90% of the metals have either Body Centered Cubic (BCC), Face Centered Cubic (FCC) or Hexagonal Close Packed (HCP) crystal structure.

HCP is denser version of simple hexagonal crystal structure.

BCC Structure FCC Structure HCP Structure14

Body Centered Cubic (BCC) Crystal Structure

Represented as:

one atom at each corner of cube

one at the center of cube.

Examples :-

Chromium (a=0.289 nm)

Iron (a=0.287 nm)

Sodium (a=0.429 nm)

(b) Hard-sphere unit cell

Figure 3.4: BCC crystal structure

(a) Atomic-site unit cell (c) Isolated unit cell15

Each unit cell has:

eight 1/8 atom at corners

1 full atom at the center.

Therefore each unit cell has (8 x 1/8) + 1 = 2 atoms

Atoms contact each other at cube diagonal. Therefore;

3

4Ra

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Face Centered Cubic (FCC) Crystal Structure

FCC structure is represented as one atom each

at the corner of cube

at the center of each cube face.

Examples :- Aluminum (a = 0.405) Gold (a = 0.408)

(b) Hard-sphere unit cell

Figure 3.6: FCC crystal structure

(a) Atomic-site unit cell (c) Isolated unit cell17

Each unit cell has:

eight 1/8 atom at corners

six ½ atoms at the center of six faces

Therefore each unit cell has (8 x 1/8) + (6 x ½) = 4 atoms

Atoms contact each other across cubic face diagonal. Therefore;

2

4Ra

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Hexagonal Close-Packed Structure (HCP)

The HCP structure is represented as

1 atom at each of 12 corners of a hexagonal prism

2 atoms at top and bottom face

3 atoms in between top and bottom face.

Examples:- Zinc (a = 0.2665 nm, c/a = 1.85) Cobalt (a = 0.2507 nm, c/a = 1.62)

Figure 3.8: HCP crystal structure:(a) Schematic of the crystal structure (larger cell)

(b) Hard-sphere model19

Each atom has:

six 1/6 atoms at each of top and bottom layer

two ½ atoms at top and bottom layer

3 full atoms at the middle layer

Therefore each HCP unit cell has:

Ideal HCP c/a ratio is 1.633.

(2 x 6 x 1/6) + (2 x ½) + 3 = 6 atoms

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Isolated HCP unit cell also called the primitive cell.

The atoms at locations marked:

‘1’ contribute 1/6 of an atom

‘2’ contribute 1/12 of an atom

‘3’ contribute 1 atom

Therefore each HCP unit cell (primitive cell) has

(4 x1/6) + (4 x 1/12) + 1 = 2 atoms

Figure 3.8: HCP crystal structure:

(c) Isolated unit cell schematic

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APF = Volume of atoms in unit cell

Volume of unit cell

Atomic Packing Factor (APF)

Formula:

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APF = 8.723 R3

12.32 R3 = 0.68

Vatoms =

= 8.373R3

3

3

4

R

= 12.32 R3

V unit cell = a3 =

3

4.2

3R

APF for BCC structure:

But;

Therefore;

APF = Volume of atoms in unit cell

Volume of unit cell

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Question 3

3.1 Determine the atomic packing factor (APF) for:a. Face centered cubic structure

b. Hexagonal Close-Packed Structure (HCP)

3.2 Calculate the volume of the zinc crystal structure unit cell by using the following data: pure zinc has HCP crystal structure with lattice constant a = 0.2665 nm and c = 0.4947nm.

3.3 By using data in question 2.2, find the volume of the larger cell.

3.4 What are the three most common metal crystal structures? List five metals that have each of these crystal structures.

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Volume density

Formula volume density of metal:

vMass per Unit cell

Volume per Unit cell=

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Example:-

Copper (FCC) has atomic mass of 63.54 g/mol and atomic radius of 0.1278 nm. Calculate the volume density.

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v Mass per Unit cell Volume per Unit cell

=

Solution:

Known FCC unit cell has 4 atoms.

Therefore;

Mass of unit cell = m

= 4 (63.54) 6.02 x 1023

= 4.22 x 10-22g

2

4Ra = =

2

1278.04 nm= 0.361 nm

Volume of unit cell = V= a3

But;

Therefore;Volume of unit cell = (0.361nm)3

= 4.7 x 10-29 m3

v V

m 4.22 x 10-22 =8.95 x 106 g/m3

4.7 x 10-29

4 x 0.1278 x 10-9

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Atom Positions in Cubic Unit Cells

Cartesian coordinate system is use to locate atoms.

In a cubic unit cell

y axis is the direction to the right.

x axis is the direction coming out of the paper.

z axis is the direction towards top.

Negative directions are to the opposite of positive directions.

Atom positions are located using unit distances along the axes.

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Figure3.9 Atomic position in a BCC unit cell.

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Direction

In cubic crystals, direction indices are vector components of directions resolved along each axes, resolved to smallest integers.

Direction indices are position coordinates of unit cell where the direction vector emerges from cell surface, after converted to integers.

Fig 3.1: Some directions in cubic unit cells30

(0,0,0)

(1,1/2,1)

zProduce the direction vector till it emerges from surface of cubic cell

Determine the coordinates of pointof emergence and origin

Subtract coordinates of point of Emergence by that of origin

(1,1/2,1) - (0,0,0)= (1,1/2,1)

Are all areintegers?

Convert them to

smallest possible

integer by multiplying

by an integer.

2 x (1,1/2,1)= (2,1,2)

Are any of the directionvectors negative?

Represent the indices in a square bracket without comas with a over negative index (Eg: [121])

Represent the indices in a square bracket without comas (Eg: [212] )

The direction indices are [212]

x

y

YES

NO

YESNO

PROCEDURE TO FIND DIRECTION INDICES

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Example:

Determine the direction indices of the cubic direction shown below:

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Tips:The vector representing EF can be found by subtracting the coordinate of the tips of the vector F from the coordinate of the tail E.

Exercise 3.4

i. Draw the following direction vectors in cubic unit cells:

a. [100] and [110]

b. [112]

c. [1 1 0]

d. [3 2 1]

ii. Determine the direction indices of the cubic direction between the position coordinates (3/4 , 0 , ¼) and (1/4 , ½ , ½)

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Lattice plane

Miller Indices are are used to refer to specific lattice planes of atoms.

They are reciprocals of the fractional intercepts (with fractions cleared) that the plane makes with the crystallographic x,y and z axes of three nonparallel edges of the cubic unit cell.

z

x

y

Miller Indices = (111)

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Choose a plane that does not pass through origin

Determine the x,y and z intercepts

of the plane

Find the reciprocals of the intercepts

Fractions? Clear fractions bymultiplying by an integer

to determine smallest set of whole numbers

Enclose in parenthesis (hkl)where h,k,l are miller indicesof cubic crystal plane

forx,y and z axes. Eg: (111)

Place a ‘bar’ over theNegative indices

Yes

MILLER INDICES - PROCEDURE

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z

x

y

Miller Indices = (111)

Miller Indices - Examples

Intercepts of the plane at x,y & z axes are 1, ∞ and ∞

Taking reciprocals we get (1,0,0).

Miller indices are (100).

Intercepts are 1/3, 2/3 & 1.

taking reciprocals we get (3, 3/2, 1).

Multiplying by 2 to clear fractions, we get (6,3,2).

Miller indices are (632).

xx

y

z

(100)

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Plot the plane (101)

Taking reciprocals of the indices we get (1 ∞ 1).

The intercepts of the plane are x=1, y= ∞ (parallel to y) and z=1.

Plot the plane (2 2 1)

Taking reciprocals of the indices we get (1/2 1/2 1).

The intercepts of the plane are x=1/2, y= 1/2 and z=1.

Figure EP3.7 a

Figure EP3.7 c37

Plot the plane (110)

• The reciprocals are (1,-1, ∞)

• The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis)

Note:

To show this plane in a single unit cell, the origin is moved along the positive

direction of y axis by 1 unit.

Figure EP3.7 b

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Exercise 3.7

1. Draw the following crystallographic planes in cubic unit cells:

a. (101)

b. (110)

c. (221)

d. (110)

2. Determine the miller index for the plane given cubic unit cell.

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Planar Atomic Density

Formula planar atomic density:

p =

Equivalent number of atoms whose

centers are intersected by selected area

Selected area

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Example:-

In Iron (BCC, a=0.287 nm), the (110) plane intersects center of 5 atoms (four ¼ and 1 full atom). Calculate the planar atomic density.

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Solution:

- Equivalent number of atoms = (4 x ¼ ) + 1 = 2 atoms

- Area of 110 plane =

Therefore;

222 aaa

p 2287.02

2=

2

13

2

1072.12.17

mm

atoms

nm

atoms

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1.72 x 1019 atoms/m2

Linear Atomic Density

Formula linear atomic density:

=

Number of atomic diameters intersected by

selected length of line in direction of interest

Selected length of line

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l

Example:-

For a FCC copper crystal (a=0.361 nm), the [110] direction intersects 2 half diameters and 1 full diameter.

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Solution:

- The number of atomic diameters intersected by this length of line are:

½ + ½ + 1 = 2 atomic diameters

- The length of the line is:

Therefore;

mm

atoms

nm

atoms

nm

atoms 61092.392.3

361.02

2

l

nm361.02

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3.92 x 109 atoms/m

Interplanar Spacing

The distance between two adjacent parallel planes of atoms with the same Miller indices is called the interplanar spacing, dhkl .

The interplanar spacing in cubic materials is given by the general equation:

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Crystal Structure Analysis –X-ray Diffraction

Information about crystal structure are obtained using X-Rays diffraction technique.

The X-rays source:

- The wavelength (0.05-0.25 nm) which same as distance between crystal lattice planes.

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Figure 3.22 Schematic diagram of the cross section of a sealed-off filament X-

ray tube.

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Crystal planes of target metal act as mirrors reflecting X-ray beam.

If rays leaving a set of planes are out of phase (as in case of arbitrary angle of incidence) no reinforced beam is produced.If rays leaving are in phase, reinforced beams are produced.

Figure 3.25 The reflection of an X-ray beam by the (hkl) planes of a crystal

a) No reflected beam is produced at an arbitrary angle of incidence.

b) At the Bragg angle , the reflected rays are in phase and reinforce one another.

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For rays reflected from different planes to be in phase, the extra distance traveled by a ray should be a integral multiple of wave length λ .

c) Similar to b) except that the wave representation has been omitted.

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Let us consider incident X-rays 1 and 2 as figure 3.25c.

For these rays to be in phase, the extra distance of travel of ray 2 is equal to MP + PN which must be an integral number of wavelength.

Thus;

nλ = MP + PN

where;

n = order of the diffraction

= 1,2,3,…..

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Since both MP and PN equal to dhklsin, the condition for constructive interference (the production of a diffraction peak of intense radiation) must be:

nλ = 2 dhkl sinθ

This equation known as Bragg’s law.

Bragg’s law gives the relationship among the angular positions of the reinforced diffracted beams in terms of the wavelength of the incoming X-ray radiation and of the interplanar spacing dhkl of the crystal planes.

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Example 3.15:

A sample of BCC iron was placed in an X-ray diffractometer using incoming X-rays with a wavelength of 0.1541 nm. Diffraction from the planes was obtained at 2 = 44.70. Calculate a value for the lattice constant, a of BCC iron. (Assume first-order diffraction with n = 1).

Answer: 0.287 nm

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References

A.G. Guy (1972) Introduction to Material Science, McGraw Hill.

J.F. Shackelford (2000). Introduction to Material Science for Engineers, (5th Edition), Prentice Hall.

W.F. Smith (1996). Priciple to Material Science and Engineering, (3rd Edition), McGraw Hill.

W.D. Callister Jr. (1997) Material Science and Engineering: An Introduction, (4th Edition) John Wiley.

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