UNIT –I CRYSTAL PHYSICS

IV. CONDENSED MATTER PHYSICS

Lecture - II

Dr. T. J. Shinde

Department of Physics

Smt. K. R. P. Kanya Mahavidyalaya, Islampur

Lecture - II

Simple Crystal Structures

� Simple cubic (SC)

� Face centered cubic (FCC)

� Body centered cubic (BCC)

� Hexagonal close packed (HCP)

� Diamond structure� Diamond structure

� Zinc blende structure

� NaCl structure

� CsCl structure

NUMBER OF ATOMS PER UNIT CELL (n)

• The total number of atoms present in an unit cell is known as number of atoms per

unit cell.

COORDINATION NUMBER (CN)

• It is the number of nearest neighboring atoms to a particular atom.

ATOMIC RADIUS (r)

• It is the radius of an atom. It is also defined as half the distance between two nearest

neighboring atoms in a crystal.

ATOMIC PACKING FACTOR (APF) ATOMIC PACKING FACTOR (APF)

• It is the ratio of volume occupied by the atoms or molecules in an unit cell (v) to the

total volume of the unit cell (V).

APF =

No. of atoms present in the unit cell x Volume of the atom

APF =Volume of the unit cell

Volume occupied by the atoms in an unit cell

Volume of the unit cell

•Each and every corner atom is shared by eight adjacent unit cells. The contribution of each and every corner atom to one unit cell is 1/8.

• The total number of atoms present in a unit cell =1/8 x 8 =1.

CORDINATION NUMBER (CN)

• For SCS atom, there are four nearest neighbours in its

Simple Cubic Structure (SC)

4

• For SCS atom, there are four nearest neighbours in its • own plane.

• There is another nearest neighbour in a plane which lies just above this atom and yet another nearest neighbour in another plane which lies just below this atom.

Therefore the co-ordination number is 6.

ATOMIC RADIUS (R)

Since the atoms touch along cube edges, the atomic radius for a simple cubic unit cell is,

r = a/2 ( 0.5a) (where a = 2r, is the lattice constant)ATOMIC PACKING FACTOR (APF)

APF =

4

3π (0.5a) 3

1

atomsunit cell atom

volume

∴ APF = 0.5236Thus 52 percent of the volume of the simple cubic unit cell is occupied by atoms and the remaining 48 percent volume of the unit cell is vacant or void space.

volumeunit cell

APF = a3

3

Face Centered Cubic structure (FCC)

• Atoms are arranged at the corners and center of each cube face of the cell.

• The atoms in a FCC unit cell touches along the face diagonal.

• Each and every corner atom is shared by eight • Each and every corner atom is shared by eight adjacent unit cells.

• Therefore each and every corner atom contributes 1/8 of its part to one unit cell. So the total number of atoms contributed by the corner atoms is 1/8 × 8 = 1.

•Two unit cells share each and every face centered atom.

• Therefore, the contribution of a face centered atom to unit cell is 1/2.

• So, the total number of atoms contributed by the face centered atoms = 1/2 × 6 = 3.

• The total number of atoms present in a FCC unit cell = 1+3 = 4.COORDINATION NUMBER

• In its own plane it touches four face centered atoms. The face • In its own plane it touches four face centered atoms. The face centered atoms are its nearest neighbors.

• In a plane, which lies just above this corner atom, it has four more face centered atoms as nearest neighbors.

• Similarly, in a plane, which lies just below this corner atom it has yet four more face centered atoms as its nearest neighbors.

•Therefore the number of nearest neighbours i.e., co-ordination number for FCC unit cell = 4+4+4 = 12

ATOMIC RADIUS (R)

• Consider the figure. AB = AC = a and AC = 4r.

From the triangle ABC, AC2 = AB2 + BC2From the triangle ABC, AC = AB + BC

AC2 = a2 + a2 ; AC2 = 2a2

AC =

i.e 4r =

Therefore atomic radius = 2a4

2a

2a

ATOMIC PACKING FACTOR (APF)

a

2

a

APF =

43

π ( 2a/4)34atoms

unit cell atomvolume

a3unit cellvolume

∴APF = 0.74

Thus 74 percent of the volume of the FCC unit cell is occupied by atoms and the remaining 26 percent volume of the unit cell is vacant or void space.

∴APF = 0.74

ex: Al, Cu, Au, Pb, Ni, Pt, AgExamples :-

� Aluminum (a = 0.405)

� Gold (a = 0.408)

BODY CENTERED CUBIC STRUCTURE•A body centred cubic structure has eight comer atoms and one body centred atom.

• The atom at the centre touches all the eight corner atoms.

• In BCC unit cell, each and every corner atom is shared by eight adjacent unit cells. So, the total number of atoms contributed by the corner atoms number of atoms contributed by the corner atoms is1/8 × 8 = 1.

• A BCC unit cell has one full atom at the centre of the unit cell.

• The total number of atoms present in a BCC unit cell = 1+1 = 2.

CO-ORDINATION NUMBER (CN)

Let us consider a body centred atom. The nearest neighbour for a body centred atom is a corner atom. A body centred atom is surrounded by eight corner atoms.

Therefore, the co-ordination number of a BCC unit cell = 8.

ATOMIC RADIUS (R)

For a BCC unit cell, the atomic radius can be calculated

From figure AB = BC = AD = ‘a’ and CD = 4r.From the triangle, ACD,From the triangle, ACD,

CD2 = AC2 + AD2 ; CD2 = 2a2 + a2

(4r)2 = 3a2 ; 16r2 = 3a2

i.e. r2 =

∴atomic radius r = a

23a16

34

ATOMIC PACKING FACTOR (APF)

APF =

4

3π ( 3a/4)32

atoms

unit cell atomvolume

a3

unit cell

volume

APF = 38

π = 0.688

Thus 68 percent of the volume of the BCC unit cell is occupied by atoms and the remaining 32 percent volume of the unit cell is vacant or void space.

Examples :-� Chromium (a=0.289 nm)

� Iron (a=0.287 nm)

� Sodium (a=0.429 nm)

HEXAGONAL CLOSED PACKED STRUCTURE

•It consists of three layers of atoms. The bottom layer has six corner atoms and one face centred atom. The middle layer has three full atoms. The upper layer has six corner atoms and one face centred atom.

• Each and every corner atom contributes 1/6 of its part to one unit cell.to one unit cell.

• The number of total atoms contributed by the corner atoms of both top and bottom layers is 1/6 × 12 = 2.•The face centred atom contributes 1/2 of its part to one unit cell.

a

a

ex: Cd, Mg, Ti, Zn

�Since there are 2 face centred atoms, one in the top and the other in the bottom layers, the number of atoms contributed by face centred atoms is 1/2× 2 = 1.

� Besides these atoms, there are 3 full atoms in the middle layer.� Total number of atoms present in an HCP unit cell is 2+1+3 = 6.

CO-ORDINATION NUMBER (CN)

� The face centered atom touches 6 corner atoms in its plane.� The face centered atom touches 6 corner atoms in its plane.� The middle layer has 3 atoms.There are three more atoms, which are in the middle layer of the unit cell.� Therefore the total number of nearest neighbours is 6+3+3=12.

ATOMIC RADIUS (R)

� Consider any two corner atoms. Each and every corner atom touches each other. Therefore a = 2r. i.e., The atomic radius, r = a/2

ATOMIC PACKING FACTOR (APF)

APF = Where, v = 6 × 4/3 πr3

Substitute r = , v = 6 × 4/3 π = πa3

AB = AC = BO = ‘a’. CX = c/2;where c → height of the hcp unit cell.Area of the base = 6 × area of the triangle – ABO

= 6 × 1/2 × AB × OO′

Area of the base = 6 × 1/2 × a × OO′

In triangle OBO′

vV

a2

3a8

O 'OB 30= °

Now, substituting the value of OO′,

Area of the base = 6 × × a × a =

V = Area of the base × height = × c

O 'OB 30= °

cos30º = OO ' OO 'BO a

= ∴ OO′ = a cos 30º = a 32

32

12

23 3a2

23 3a2

)c

a(

33

2π

2ca33

aπ

V

v APF

2

3

===

CALCULATION OF c/a RATIO

In the triangle ABA′, A 'AB 30= °

23

a30cos ABAA' 0 ==

a3

123

a32

AA'32

AXBut ===

In the triangle AXC, AC2 = AX2 + CX2 ; Substituting the values of AC, AX and CX,

+

=

23

22

2 caa

43

222 ca

a +=

32

34

222

2aa

ac

=−=

38

ac

2

2

=

38

=a

c0.74

23

π)

83

(33

2π APF ===∴

Atomic Packing Fraction =74%

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 arelocated using unit distances along theaxes.

3-15

Directions in Cubic Unit Cells

• 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, converted to

integers.integers.

3-16

Procedure to Find Direction Indices

(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)

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

The direction indices are [212]

x

y

NOAll areintegers?

Convert them to

smallest possibleinteger by multiplying

by an integer.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] )

YES

NO

YESNO

3-17

Direction Indices - Example

• Determine direction indices of the given vector.

Origin coordinates are (3/4 , 0 , 1/4).

Emergence coordinates are (1/4, 1/2, 1/2).

Subtracting origin coordinates

from emergence coordinates,

(1/4, 1/2, 1/2)-(3/4 , 0 , 1/4) (1/4, 1/2, 1/2)-(3/4 , 0 , 1/4)

= (-1/2, 1/2, 1/4)

Multiply by 4 to convert all

fractions to integers

4 x (-1/2, 1/2, 1/4) = (-2, 2, 1)

Therefore, the direction indices are [ 2 2 1 ]

3-18

Miller Indices

• Miller Indices 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 crystallographic x,y and z axes of three nonparallel edges of

the cubic unit cell. z

x

y

Miller Indices =(111)

3-19

Miller Indices - Procedure

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,lare miller indicesof cubic crystal plane

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

Place a ‘bar’ over theNegative indices

3-20

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).

*******************xx

y

z

(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).

x

3-21

Miller Indices - Examples

• 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.

3-22

Miller Indices - Example

• Plot the plane (110)

The reciprocals are (1,-1, ∞)

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

To show this plane a

single unit cell, the z(110)

single unit cell, the

origin is moved along

the positive direction

of y axis by 1 unit.

x

y

3-23

Miller Indices – Important Relationship

• Direction indices of a direction perpendicular to a crystal

plane are same as miller indices of the plane.

• Example:-

z

z

• Interplanar spacing between parallel closest planes with

same miller indices is given by

lkhd

a

hkl 222++

=

3-24

[110] (110)

x

y

Planes and Directions in Hexagonal Unit Cells

• Four indices are used (hkil) called as Miller-Bravais indices.

• Four axes are used (a1, a2, a3 and c).

• Reciprocal of the intercepts that a crystal plane makes with

the a1, a2, a3 and c axes give the h,k,I and l indices

respectively.

3-25

Hexagonal Unit Cell - Examples

• Basal Planes:-

Intercepts a1 = ∞

a2 = ∞

a3 = ∞

c = 1

(hkli) = (0001)

• Prism Planes :-• Prism Planes :-

For plane ABCD,

Intercepts a1 = 1

a2 = ∞

a3 = -1

c = ∞

(hkli) = (1010)

3-26

Directions in HCP Unit Cells

• Indicated by 4 indices [u v t w].

• u,v,t and w are lattice vectors in a1, a2, a3 and c directions respectively.

• Example:-

For a , a , a directions, the direction indices areFor a1, a2, a3 directions, the direction indices are

[ 2 1 1 0], [1 2 1 0] and [ 1 1 2 0] respectively.

3-27

Summary of Crystal Structures

Volume Density

1) Example:- Copper (FCC) has atomic mass of 63.54 g/mol and

atomic radius of 0.1278 nm. Find volume density.

vρMass/Unit cell

Volume/Unit cell=

a= =4 R

2

1278.04 nm×= 0.361 nm

Volume density of metal = = VNnA

Where, n = number of atoms/unit cell; A = atomic weight ; V = Volume of unit cell = a3

for cubic ; N = Avogadro’s number = 6.023 x 1023 atoms/mol

a= =2 2

Volume of unit cell = V= a3 = (0.361nm)3 = 4.7 x 10-29 m3

vρ

FCC unit cell has 4 atoms.

Mass of unit cell = m =

×

−

g

Mg

molatmos

molgatoms6

23

10/10023.6

)/54.63)(4(= 4.22 x 10-28 Mg

33329

28

98.898.8107.4

1022.4

cm

g

m

Mg

m

Mg

V

m==

×

×==

−

−

3-30

2) Chromium has a BCC structure and a density of 7.19 g/ml . What is the radius of its atom? (Atomic mass of Cr = 51.996 amu)

3) Copper has a FCC structure and a density of 8.96 g/ml. What is the volume of the unit cell? (Atomic mass of Cu = 63.546 amu)

4) Strontium has BCC structure and r =215 pm. What is the density? (Atomic mass of Sr = 86.62 amu)

5) Nickel has a density of 8.9 g/ml. Its atomic radius is 125 pm. What is the crystal structure? (Atomic mass of Ni = 58.6934 amu).

6)Polonium metal crystallizes in a simple cubic structure. Calculate the density of the polonium metal if the atom radius is 176 pm.(Atomic mass of Po=209 amu).

7) Based on a literature density of 9.196 g cm-3, what is the radius of Po?

8) The radius of the copper atom is 127.8 pm, and its’ density is 8.95 g/cm3. Which unit cell is consistent with these data: sc, bcc, or fcc?

9) Zinc has HCP structure. The height of the unit cell is 4.935Å. Find (i). How many atoms are there in a unit cell? and (ii). What is the volume of the unit cell ?

1) Height of the unit cell, c = 4.935Å = 4.935 × 10-10m

In HCP structure, the number of atoms present in the unit cell is 6.

We know that, the ratio

a = , a = 4.935 × a = 3.022 Å

c 8a 3

=

3c

8× 3

8

We also know that, volume of the unit cell for HCP structure is,

V = a2c or a3V = a2c or a3

V = (3.022 × 10-10)3

V = 1.17 × 10-28 m3

10. Sodium crystallises in a cubic lattice. The edge of the unit cell is 4.16 Å. The density of sodium is 975kg/m3 and its atomic weight is 23. What type of unit cell does sodium form ? (Take Avagadro number as 6.023×××× 1026 atoms (Kg mole)-1

i) Edge of the unit cell, a = 4.16Å = 4.16 × 10-10m

ii) Density of the sodium, ρ = 975 kg/m3

iii) Atomic weight of sodium, M = 23iv) Avogadro's number, N = 6.023 × 1026 atoms/kg mole

Density of the crystal material,

∴Number of atoms in the unit cell,

3

nM

Naρ =

3Nan

Mρ

=

= 2 atoms

Since the body centred cubic cell contains 2 atoms in it, sodium crystallises in a BCC cell.

nM

=

26 10 3975 6.023 10 (4.16 10 )n

23

−× × × ×=

11. A metallic element exists in a cubic lattice. Each side of the unit cell is 2.88 Å. The density of the metal is 7.20 gm/cm3. How many unit cells will be there in 100gm of the metal?

- Edge of the unit cell, a = 2.88Å = 2.88 × 10-10mDensity of the metal, ρ = 7.20 gm/cm3

= 7.2 ×103 kg/m3

Volume of the metal = 100gm = 0.1kgVolume of the unit cell = a3 = (2.88 × 10-10)3

= 23.9 × 10-30m3

Volume of 100gm of the metal = 3

Mass 0.1Density 7.2 10

=×

= 1.39 × 10-5m3

∴Number of unit cells in the volume =

= 5.8 × 1023

Density 7.2 10×

5

30

1.39 10

23.9 10

−

−

×

×