Crystal Structures

Crystal structures

1 Md Imran Momtaz, Dept. of EEE, BUET

The Crystalline State

• Lattice• Basis

• Unit Cell


Crystal = Lattice + Basis at each lattice pointCrystal = Lattice + Basis at each lattice point

Molecules and general bonding principles


(a) Force vs. interatomic separation(b) Energy vs. interatomic separation

Molecules and general bonding principles


• In previous slide, only electron scenario has been shown. Many electron case can be analyzed in the same manner.

Crystal structure of some well known material


• Copper, Cu:

FCC unit cellCrystal structure of Cu An FCC unit cell Reduced sphere representation

4 atoms per unit cell4 atoms per unit cell2 2

Ra=

3 2APF

π=



• Iron, Fe:

BCC structure of Fe Reduced sphere representation

2 atoms per unit cell2 atoms per unit cell3

4R

a= 3

8APF

π=



• Zinc, Zn:

HCP Crystal structure of Zn Unit cellSmallest unit cell

(Hexagonal unit cell)

74%APF =



• Silicon, Si & Germanium, Ge:

Diamond Cubic Crystal Structure of Si (Same for Ge as well)

8 atoms per unit cell8 atoms per unit cell

Note: A diamond cubic crystal structure can be formed by superimposing two FCC structures by (a/4, a/4, a/4) distance



• ZnS & GaAs (and some other well known III-V semiconductor as well):

Zinc Blend Cubic Crystal Structure of Si (Same for Ge as well)

8 atoms per unit cell8 atoms per unit cell

Note: A Zinc Blend cubic crystal structure can be formed by superimposing two FCC structures by (a/4, a/4, a/4) distance



• NaCl & CsCl:

FCC Crystal of NaCl BCC Crystal of CsCl

Numerical Problem


• Consider FCC structure of Cu crystal.

a) How many atoms are there per unit cell?

b) Prove that,

c) Determine APF.

d) Determine atomic concentration and density of crystal. Given, atomic mass of Cu is 63.55 gm mol-1 and Rcu = 0.128 nm.

2 2R a=

Crystal Direction & Plane


a, b, c, α, β and γ are known as lattice parameters.

For Cubic Crystal structure, a = b = c andα = β = γ = 90°

For Cubic Crystal structure, a = b = c andα = β = γ = 90°

For Hexagonal Close structure, a = b ≠ c andα = β = 90°, γ = 120°

For Hexagonal Close structure, a = b ≠ c andα = β = 90°, γ = 120°

Crystal Direction


1. Take the x, y and z axis intercepts. a/2, b, c/2

Result

2. Normalize them w.r.t. lattice constants. 1/2, 1, 1/2

3. Put them within [] without comma. [½ 1 ½]

4. Multiply with suitable number, if needed. [1 2 1]

a/3, b/2, c

Another example

1/3, 1/2, 1

[1/3 ½ 1]

[2 3 6]

Crystal Direction


Note: If the intercept is negative, place a bar over the number.

-a, b, c

Another example

-1, 1, 1

[-1 1 1]

[111]

Note: If the direction does not start from origin, shift that direction to some unit cell so that its initial point starts from origin.

Family of planes:It is represented by <> .

<100> means [100],[010],[001],[100],[010] [001]and

<110> means

[110],[011],[101],[110],[110] .etc

Crystal Planes: Miller Indices


1. Take the x, y and z axis intercepts. a/2, b, ∞

2. Normalize them w.r.t. lattice constants. 1/2, 1, ∞

3. Invert them, put them within () without comma.

(2 1 0)

4. Multiply with suitable number, if needed. (2 1 0)

Result

This is Miller Indices.

Crystal Planes: Miller Indices


Crystal Plane


Note: If the intercept is negative, place a bar over the number.

-a, b, c

Another example

-1, 1, 1

(-1 1 1)

(111)

Note: If the direction starts from origin, shift that direction to a unit cell.

Family of planes:It is represented by {} .

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

{110} means

(110), (011), (101), (110), (110) .etc

Important Point


[hkl] direction is perpendicular to (hkl) plane. This is applicable for cubic crystal

system ONLY.

[hkl] direction is perpendicular to (hkl) plane. This is applicable for cubic crystal

system ONLY.

Planer Concentration


• Significance:

It can justify different properties of the materials.

e.g. It can justify the oxide growth probability. It also can justify the required amount of force required to penetrate the layer.

•Planer Concentrationatoms ina certain plane

areaof that plane=

Numerical Problem


• Consider FCC structure of Cu crystal.

a) Determine planer concentration at (100) plane

b) Determine planer concentration at (110) plane

Engineering

Crystal Structures