PH0101 UNIT 4 LECTURE 6

PH 0101 UNIT 4 LECTURE 6 1

PH0101 UNIT 4 LECTURE 6RELATION BETWEEN LATTICE CONSTANT AND

DENSITY

DIAMOND CUBIC STRUCTURE

PROBLEMS


RELATION BETWEEN LATTICE CONSTANT

AND DENSITY

Consider a cubic crystal of lattice constant ‘a’.

Density of the crystal = ρ

Volume of the unit cell = V = a3

Number of atoms per unit cell = n

Atomic weight of the material = M

Avagadro number = N



AND DENSITY

In M gram of material there are ‘N’ atoms i.e., mass of N atoms is ‘M’ gram.

Mass of 1 atom =

Mass of ‘n’ molecules i.e., mass of an unit cell = (1) Density ‘ρ’=

i.e. ρ =

MN

nMN

massvolume

mV



AND DENSITY

ρ =

mass ‘m’ = ρ a3 (2)

Equating (1) and (2), we get,

ρ a3 =

ρ =

3

ma

nMN

3

nMNa



AND DENSITY

ρ = Number of atoms per unit cell Atomic weight

3Avagadro number (Lattice cons tan t)


PROBLEMS

Worked ExampleSodium 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

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

Density of the sodium, ρ = 975 kg/m3

Atomic weight of sodium, M=23


PROBLEMS

Avogadro's number, N = 6.023 × 1026 atoms/kg mole

Density of the crystal material,

Number of atoms in the unit cell,

3

nMNa

3NanM

26 10 3975 6.023 10 (4.16 10 )n23


PROBLEMS

= 2 atoms

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


PROBLEMS

Worked ExampleA 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


PROBLEMS

Volume of 100gm of the metal =

= 1.39 × 10-5m3

Number of unit cells in the volume =

= 5.8 × 1023

3

Mass 0.1Density 7.2 10

5

30

1.39 1023.9 10



It is formed by carbon atoms. Every carbon atom is surrounded by four other carbon atoms

situated at the corners of regular tetrahedral by the covalent linkages.

The diamond cubic structure is a combination of two interpenetrating FCC sub lattices displaced along the body diagonal of the cubic cell by 1/4th length of that diagonal.

Thus the origins of two FCC sub lattices lie at (0, 0, 0) and (1/4, 1/4,1/4)



a

a/4

Z

Y

2r

a/4W

a/4

X



The points at 0 and 1/2 are on the FCC lattice, those at 1/4 and 3/4 are on a similar FCC lattice displaced along the body diagonal by one-fourth of its length.

In the diamond cubic unit cell, there are eight corner atoms, six face centred atoms and four more atoms.

No. of atoms contributed by the corner atoms to an unit cell is 1/8×8 =1.

No. of atoms contributed by the face centred atoms to the unit cell is 1/2 × 6 = 3

There are four more atoms inside the structure.



No.of atoms present in a diamond cubic unit cell is 1 + 3 + 4 = 8. Since each carbon atom is surrounded by four more carbon atoms,

the co-ordination number is 4.

ATOMIC RADIUS(R)From the figure,in the triangle WXY,

XY2 = XW2 + WY2

=

2 2a a4 4



XY2 =

Also in the triangle XYZ,

XZ2 = XY2 + YZ2

=

XZ2 =

2a8

22a a8 4

23a16



But XZ = 2r

(2r)2 =

4r2 =

r2=

Atomic radius r =

23a16

23a16

23a64

3 a8



Atomic packing factor (APF)

APF =

v =

i.e. v =

APF =

vV

348 r3

34 3a83 8

34 3a83 8

3

3 3

8 4 3 3 a3 8 a



APF =

i.e. APF = 34%

Thus it is a loosely packed structure.

3 0.3416

PH0101 UNIT 4 LECTURE 6

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