PH0101 UNIT 4 LECTURE 6
RELATION BETWEEN LATTICE CONSTANT AND
DENSITY
DIAMOND CUBIC STRUCTURE
PROBLEMS
PH 0101 UNIT 4 LECTURE 6
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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
=
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RELATION BETWEEN LATTICE CONSTANT
AND DENSITY
In M gram of material there are ‘N’ atoms i.e., mass of N
atoms is ‘M’ gram.
M
Mass of 1 atom =
N
nM
Mass of ‘n’ molecules i.e., mass of an unit cell =
N
mass
Density ‘ρ’=
(1)
volume
m
i.e. ρ =
V
PH 0101 UNIT 4 LECTURE 6
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RELATION BETWEEN LATTICE CONSTANT
AND DENSITY
ρ= m
a3
mass ‘m’ = ρ a3
(2)
Equating (1) and (2), we get,
ρ a3 = nM
N
ρ
nM
=
Na 3
PH 0101 UNIT 4 LECTURE 6
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RELATION BETWEEN LATTICE CONSTANT
AND DENSITY
Number of atoms per unit cell  Atomic weight
ρ =
3
Avagadro number  (Lattice cons tan t)
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PROBLEMS
Worked Example
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
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
PH 0101 UNIT 4 LECTURE 6
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PROBLEMS
Avogadro's number, N
= 6.023 × 1026 atoms/kg mole
Density of the crystal material,   nM3
Na
3

Na
Number of atoms in the unit cell, n 
M
975  6.023  1026  (4.16  1010 )3
n
23
PH 0101 UNIT 4 LECTURE 6
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PROBLEMS
= 2 atoms
Since the body centred cubic cell contains 2
atoms in it, sodium crystallises in a BCC cell.
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PROBLEMS
Worked Example
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 × 10-10m
Density of the metal, ρ = 7.20 gm/cm3
= 7.2 ×103 kg/m3
Volume of the metal = 100gm = 0.1kg
Volume of the unit cell = a3 = (2.88 × 10-10)3
= 23.9 × 10-30m3
PH 0101 UNIT 4 LECTURE 6
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PROBLEMS
Volume of 100gm of the metal =
Mass
0.1

Density 7.2  103
= 1.39 × 10-5m3
1.39  105
Number of unit cells in the volume =
23.9  1030
= 5.8 × 1023
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DIAMOND CUBIC STRUCTURE
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)
PH 0101 UNIT 4 LECTURE 6
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DIAMOND CUBIC STRUCTURE
a
Z
2r
a/4
X
Y
a/4
W
a/4
PH 0101 UNIT 4 LECTURE 6
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DIAMOND CUBIC STRUCTURE
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.
PH 0101 UNIT 4 LECTURE 6
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DIAMOND CUBIC 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
a
=  
4
2
a
  
4
2
PH 0101 UNIT 4 LECTURE 6
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DIAMOND CUBIC STRUCTURE
a2
XY2 =
8
Also in the triangle XYZ,
XZ2 = XY2 + YZ2
= a
2
8
a
 
4
2
2
3a
XZ2 =
16
PH 0101 UNIT 4 LECTURE 6
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DIAMOND CUBIC STRUCTURE
But XZ = 2r
2
3a
(2r)2 =
16
2
3a
2
4r =
16
2
3a
r2=
64
Atomic radius r =
3 a
8
PH 0101 UNIT 4 LECTURE 6
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4  3a 
8 


3  8 
3
DIAMOND CUBIC STRUCTURE
Atomic packing factor (APF)
APF
v
=
v
V
=
8 
i.e. v =
APF
=
4 3
r
3
4  3a 
8 


3  8 
3
8  4  3 3 a 3
3  83  a 3
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DIAMOND CUBIC STRUCTURE
APF
=
 3
 0.34
16
i.e. APF = 34%
Thus it is a loosely packed structure.
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