Introduction to Materials

Science and Selection

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Metals

Ceramic

combination

with a metal

Associated

with polymers

Semiconductor

materials

Representative strengths of various

categories of materials

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Atomic Bonding

Primary bonding: Metallic bond, Covalent bond, Ionic bond.

Secondary bonding: Van der Waals interactions (London forces, Debye interaction, Keesom interaction)

Intermetallic compound is a compound such as Al3V formed by two or more metallic atoms

Metallic Bonding

The metallic bond forms when atoms give up their valence electron, then forms an electron sea.

The positively charged atom cores are bonded by mutual attraction to the negatively charged electrons.

When voltage is applied, the sea of electrons can move freely to produce a current.

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Covalent Bonding

Covalent bonding

requires that

electrons be shared

between atoms in

such a way that

each atom has its

outer sp orbital filled.

In silicon, with a

valence of four, four

covalent bonds must

be formed.

Ionic Bonding

An ionic bond is the result of electron transfer from one atom to another.

When sodium donates its valence electron to chlorine, each becomes an ion; attraction occurs, and the ionic bond is formed.

It is important to knot that ionic bonds are nondirectional. A positively charged Na+ will attract any adjacent Cl- equally in all directions!

© 2003 Brooks/Cole Publishing / Thomson Learning™

Ionic Bonding

© 2003 Brooks/Cole Publishing /

Thomson Learning™

When voltage is

applied to an ionic

material, entire ions

must move to cause

a current to flow.

Ion movement is

slow and the

electrical

conductivity is poor.

Atomic Arrangement

Levels of atomic arrangements in materials:

(a) Inert monoatomic gases have no regular ordering of atoms

(b, c) Some materials, including water vapor, nitrogen gas, amorphous silicon and silicate glass have short-range order.

(d) Metals, alloys, many ceramics and some polymers have regular ordering of atoms/ions that extends through the material. (c) 2003 Brooks/Cole Publishing / Thomson

Learning™

Example: LCD

Liquid crystal display.

These materials are amorphous in one state and undergo localized crystallization in response to an external electrical field

Widely used in liquid crystal displays. (Courtesy of Nick Koudis/PhotoDisc/GettyImages.)

Lattice, Unit Cells, and Crystal

Structures

Lattice - A collection of points that divide space into smaller equally sized segments.

Unit cell – The simplest repeating unit of any structure that can be stacked to fill space. All atoms must be the same in every unit cell.

Atomic radius - The apparent radius of an atom, typically calculated from the dimensions of the unit cell, using close-packed directions (depends upon coordination number).

Packing factor - The fraction of space in a unit cell occupied by atoms.

Definition of the lattice parameters

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Their use in

cubic,

orthorhombic,

and hexagonal

crystal systems

Hexagonal close packed

structure (HCP)

The hexagonal close-packed (HCP) structure (left) and its unit cell.

(c) 2003 Brooks/Cole Publishing / Thomson

A

B

A

Remember Miller Indices?

For directions: Determine coordinates

for “head” and “tail” of the direction

“head”-”tail”

Clear fraction/reduce results to lowest integers.

Enclose numbers in [] and a bar over negative integers.

For planes: Identify points at which

the plane intercepts the x, y, z axis.

Take reciprocals of these intercepts.

Clear fractions and do NOT reduce to the lowest integers.

Enclose the numbers in parentheses () and a bar over negative integers.

Special note for directions…

For Miller Indices of directions: Since directions are vectors, a direction and its

negative are not identical! [100] ≠ [100] Same line, opposite directions!

A direction and its multiple are identical! [100] is the same direction as [200] ( need to reduce!)

[111] is the same direction as [222], [333]!

Certain groups of directions are equivalent; they have their particular indices because of the way we construct the coordinates. Family of directions: <111>=[111], [111],[111],[111],…

Special note for planes…

For Miller Indices of planes: Planes and their negatives are identical (not the

case for directions!) E.g. (020) = (020)

Planes and their multiples are not identical (Again, different from directions!) We can show this by defining planar densities and planar packing fractions. E.g. (010) ≠ (020) See example!

Each unit cell, equivalent planes have their particular indices because of the orientation of the coordinates. Family of planes: {110} = (110),(110),(110),(101), (101),…

In cubic systems, a direction that has the same indices as a plane is perpendicular to that plane.

Calculate the planar density and planar packing fraction for the (010) and (020) planes in simple cubic polonium, which has a lattice parameter of 0.334 nm.

Example: Calculating the Planar Density and Packing Fraction

(c) 2003 Brooks/Cole Publishing /

Thomson Learning™ a0a0

2142

2

atoms/cm 1096.8atoms/nm 96.8

)334.0(

faceper atom 1

face of area

faceper atom (010)density Planar

SOLUTIONThe total atoms on each face is one. The planar density is:

The planar packing fraction is given by:

79.0)2(

)(

)( atom) 1(

face of area

faceper atoms of area (010)fraction Packing

2

2

20

2

r

r

r

a

However, no atoms are centered on the (020) planes. Therefore, the planar density and the planar packing fraction are both zero. The (010) and (020) planes are not equivalent!

(a0)2

In-Class Exercise 1: Determine planar

density and packing fraction.

Determine the planar density and packing fraction for FCC nickel in the (100), (110), and (111) planes. Which, if any, of these planes is close-packed?

Remember when

visualizing the plane, only

count the atoms that the

plane passes through the

center of the atom. If the

plane does NOT pass

through the center of that

atom, we do not count it!

First, find atomic radius for Nickel from Appendix 2, page 797 of

Textbook (6th Ed. Shackleford) to calculate the lattice parameter:

Solution for plane (100)

536.33536.02

125.04

2

4

125.0

0 nmnmr

FCCa

nmrNi

a0 For (100):

7854.0

2/4

2_

/10600.110536.3

2_

2

2

215

28

r

rfractionpacking

cmatomscm

atomsdensityplanar

Solution for plane (110)

5554.0

2/42

2_

/10131.110536.32

2_

2

2

215

28

r

rfractionpacking

cmatomscm

atomsdensityplanar

For (110):

a0

02a

It is important to visualize how the plane is cutting

across the unit cell – as shown in the diagram!

Solution for plane (111)

02a

02a

For (111):

Again try to visualize the plane, count the number of

atoms in the plane:

02a

2

000 866.02

32

2

1

2

1_ aaabhareaplane

215

28/10847.1

10536.3866.0

2_ cmatoms

cm

atomsdensityplanar

9069.0

866.0

4/22_

2

0

2

0 a

afractionpacking

Therefore, plane (111) is close-packed!

In-Class Exercise 2: Determine planar

density and packing fraction.

Determine the planar density and packing fraction for BCC lithium in the (100), (110), and the (111) planes. Which, if any, of these planes is close packed?

Solution for plane (100)

510.33510.03

152.04

3

4

152.0

0 nmnmr

BCCa

nmrLi

First, find atomic radius for Nickel from Appendix 2, page 797 of

Textbook (6th Ed. Shackleford) to calculate the lattice parameter:

For (100):

5890.0

4/3_

/10115.810510.3

1_

2

0

2

0

214

28

a

afractionpacking

cmatomscm

atomdensityplanar

Solution for plane (110)For (110):

It is important to visualize how the plane is cutting

across the unit cell – as shown in the diagram!

8330.0

2

4/32_

/10148.110510.32

2_

2

0

2

0

215

28

a

afractionpacking

cmatomscm

atomsdensityplanar

Solution for plane (111)

02a

For (111):

Note: Since the (111) does NOT pass through the

center of the atom in the middle of the BCC unit cell,

we do not count it!

2

000 866.02

33

2

1

2

1_ aaabhareaplane

214

28/10686.4

10510.3866.0

2/1_ cmatoms

cm

atomdensityplanar

3401.0

866.0

4/32/1_

2

0

2

0 a

afractionpacking

Therefore, there is no close-pack plane in BCC!

There are only (3)(1/6)=1/2 atoms in the plane.

02a

X-Ray Diffraction (XRD)

Max von Laue (1879-1960) won the Nobel Prize in 1912 for his discovery related to the diffraction of x-rays by a crystal.

William Henry Bragg (1862-1942) and his son William Lawrence Bragg (1890-1971) won the 1915 Nobel Prize for their contributions to XRD.

Diffraction – The constructive interference, or reinforcement, of a beam of x-rays or electrons interacting with a material. The diffracted beam provides useful information concerning the structure of the material.

Used widely as an equipment to determine crystal structures of various materials.

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(a)Destructive (out of phase) x-ray beam gives a weak signal.

(b)Reinforcing (in phase) interactions between x-rays and the crystalline material. Reinforcement occurs at angles that satisfy Bragg’s law.

Bragg’s Law:

sin2 hkldn

Bragg’s Law:

222

0

lkh

adhkl

Interplanar

spacing:d

Miller Indices

Where is half the angle

between the diffracted

beam and the original

beam direction

is the wavelength of X-ray

d is the interplanar spacing

Interplanar spacing (d-spacing)

The distance between two adjacent parallel planes of atoms with the same Miller Indices (dhkl).

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

where a0 is the lattice parameter and h,k and lrepresent the Miller Indices of the adjacent planes being considered.

222

0

lkh

adhkl

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(a) Diagram of a diffractometer, showing powder sample, incident and diffracted beams. (b) The diffraction pattern obtained from a sample of gold powder.

The results of a x-ray diffraction experiment using x-

rays with λ = 0.7107 Å (a radiation obtained from

molybdenum (Mo) target) show that diffracted peaks occur at the following 2θ angles:

Example: Examining X-ray Diffraction

Determine the crystal structure, the indices of the plane producing each peak, and the lattice parameter of the material.

EXAMPLE SOLUTION

We can first determine the sin2 θ value for each peak,

then divide through by the lowest denominator, 0.0308.

EXAMPLE SOLUTION (Continued)

We could then use 2θ values for any of the peaks to calculate the interplanar spacing and thus the lattice parameter. Picking peak 8:

2θ = 59.42 or θ = 29.71

Å

Å868.2)4)(71699.0(

71699.0)71.29sin(2

7107.0

sin2

2224000

400

lkhda

d

This is the lattice parameter for body-centered cubic iron.

In-Class Exercise 3: Repeat Distance

In a FCC unit cell, how many d111 are present

between the 0,0,0 point and 1,1,1 point?

03a0

2 2 2hkl

ad

h k l

The distance between the 0,0,0 and 1,1,1 point is:

Answer:

The interplanar spacing is:

3111

0

222

0111

aad

Now, the number of interplanar spacings (d111) between the

specified points are:

End of Tutorial 2