Phy351 ch 3

PHY351

Chapter 3

CRYSTAL
STRUCTURE

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.
2

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
3

The space lattice and unit cells
Properties of SOLIDS depends upon:

bonding force
crystal structure

4

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.

Space Lattice

Unit Cell
5

Question 2:
a.

6

Define the following terms:
i. Lattice point
ii. Motif
iii. Lattice constant

Crystal Systems and Bravais Lattice
Only SEVEN different types of unit cells are necessary to
create all point lattices which:







7

Cubic
Tetragonal
Orthorhombic
Rhombohedral
Hexagonal
Monoclinic
Triclinic

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


8

i.

Cubic Unit Cell
 a=b=c

 α = β = γ = 900

Simple

9

Body Centered

Face centered

ii. Tetragonal
 a =b ≠ c

 α = β = γ = 900

Simple

10

Body Centered

iii. Orthorhombic
 a≠ b≠ c

 α = β = γ = 900

Simple

11

Body Centered

Face Centered

Base Centered

iv. Rhombohedral

a= b≠ c

 a =b = c

 α=β=γ≠

Simple

12

v. Hexagonal

900

 α = β = 900
γ = 1200

Simple

vi. Monoclinic

vii. Triclinic

 a≠ b≠ c

 a≠ b≠ c

 α = γ = 900 ≠ β

 α ≠ β ≠ γ ≠ 900

Simple

13

Base
Centered

Simple

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.

14

BCC Structure

FCC Structure

HCP Structure

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)

15

(a) Atomic-site unit cell

Figure 3.4:
BCC crystal structure

(b) Hard-sphere unit cell (c) Isolated unit cell

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;

4R
a
3
16

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)

17

(a) Atomic-site unit cell

Figure 3.6:
FCC crystal structure

(b) Hard-sphere unit cell

(c) Isolated unit cell

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;

4R
a
2
18

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)

19

Figure 3.8: HCP crystal structure:
(a) Schematic of the crystal structure (larger cell)
(b) Hard-sphere model

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:
(2 x 6 x 1/6) + (2 x ½) + 3 =
Ideal HCP c/a ratio is 1.633.

20

6 atoms

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

Figure 3.8: HCP crystal structure:
(c) Isolated unit cell schematic

Therefore each HCP unit cell (primitive cell) has

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

21

2 atoms

Atomic Packing Factor (APF)
Formula:

APF =

Volume of atoms in unit cell
Volume of unit cell

22

APF for BCC structure:

APF =

Volume of atoms in unit cell
Volume of unit cell

But;
Vatoms =

 4R
2.
 3


3






V unit cell = a3 =

= 8.373R3

3
Therefore; APF = 8.723 R3 = 0.68
12.32 R
23

 4R 




3


3

= 12.32 R3

Question 3
3.1 Determine the atomic packing factor (APF) for:
a.
b.

Face centered cubic structure
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.
24

Volume density
Formula volume density of metal:

v

25

= Mass per Unit cell
Volume per Unit cell

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

26

Solution:

v

= Mass per Unit cell

Volume per Unit cell

Volume of unit cell = V
= a3

Known FCC unit cell has 4 atoms.
Therefore;

But;
a=

Mass of unit cell = m

4R

-9

=

4 x 0.1278nm
4  0.1278 x 10

2

= 4 (63.54)
6.02 x 1023

2

Therefore;
Volume of unit cell = (0.361nm)3
= 4.7 x 10-29 m3

= 4.22 x 10-22g

27

= 0.361 nm

m
10
 v  V  4.22xx10
4.7

-22

-29

=8.95 x 106 g/m3

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.
28

Figure3.9 Atomic position in a BCC unit cell.

29

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.

30

Fig 3.1: Some directions in cubic unit cells

PROCEDURE TO FIND DIRECTION INDICES
Produce the direction vector till it
emerges from surface of cubic cell

z
(1,1/2,1)

Determine the coordinates of point
of emergence and origin

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

(0,0,0)

Subtract coordinates of point of
Emergence by that of origin
Are all are
integers?
YES
Are any of the direction
vectors negative?
YES

31

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

NO

x

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

The direction indices are [212]
Convert them to
smallest possible
integer by multiplying

by an integer.

NO
Represent the indices in a square
bracket without comas (Eg: [212] )

Example:
Determine the direction indices of the cubic direction shown
below:

32

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.

ii.

33

Draw the following direction vectors in cubic unit cells:
a. [100] and [110]
b. [112]
c. [1 1 0]
d. [3 2 1]
Determine the direction indices of the cubic direction
between the position coordinates (3/4 , 0 , ¼) and (1/4 ,
½ , ½)

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

Miller Indices = (111)

y
34

x

MILLER INDICES - PROCEDURE
Choose a plane that does not pass
through origin

z

y

Determine the x,y and z intercepts
x

of the plane
Find the reciprocals of the intercepts
Fractions?
Place a ‘bar’ over the
Negative indices

Yes

Clear fractions by
multiplying 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)
35

Miller Indices = (111)

Miller Indices - Examples
z

 Intercepts of the plane at x,y & z

axes are 1, ∞ and ∞

(100)
y

 Taking reciprocals we get (1,0,0).
 Miller indices are (100).

x
x

 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).
36

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.
Figure EP3.7 a

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.

37

Figure EP3.7 c

Plot the plane (110)
• The reciprocals are (1,-1, ∞)
• The intercepts are x=1, y=-1 and z= ∞ (parallel to z axis)

Figure EP3.7 b

38

Note:
To show this plane in a single unit cell, the origin is moved along the positive
direction of y axis by 1 unit.

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.

39

(a)

(b)

Planar Atomic Density
Formula planar atomic density:



40

p

=

Equivalent number of atoms whose
centers are intersected by selected area
Selected area

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.

41

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

- Area of 110 plane =

2a  a  2a 2

Therefore;



p

=


42

2
2 0.287 2

17.2atoms
nm2

1.72  101319
atoms

 1.72 x 10 atoms/m2
mm2

Linear Atomic Density
Formula linear atomic density:



l

=

Number of atomic diameters intersected by
selected length of line in direction of interest
Selected length of line

43

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

44

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:

2  0.361nm
Therefore;



l



2atoms
2  0.361nm

45



3.92atoms
nm

3.92  106 atoms

 3.92 x 109 atoms/m
mm

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:

46

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.

47

Figure 3.22
Schematic diagram of the cross section of a sealed-off filament Xray tube.

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

48

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.

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.

49

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,…..
50

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.

51

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.

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 110 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
52

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.

53

Phy351 ch 3

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