PH 0101 UNIT 4 LECTURE 6 1
PH0101 UNIT 4 LECTURE 6RELATION BETWEEN LATTICE CONSTANT AND
DENSITY
DIAMOND CUBIC STRUCTURE
PROBLEMS
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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.
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
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RELATION BETWEEN LATTICE CONSTANT
AND DENSITY
ρ =
mass ‘m’ = ρ a3 (2)
Equating (1) and (2), we get,
ρ a3 =
ρ =
3
ma
nMN
3
nMNa
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RELATION BETWEEN LATTICE CONSTANT
AND DENSITY
ρ = Number of atoms per unit cell Atomic weight
3Avagadro number (Lattice cons tan t)
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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
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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
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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 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
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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
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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)
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DIAMOND CUBIC STRUCTURE
a
a/4
Z
Y
2r
a/4W
a/4
X
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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.
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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
=
2 2a a4 4
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DIAMOND CUBIC STRUCTURE
XY2 =
Also in the triangle XYZ,
XZ2 = XY2 + YZ2
=
XZ2 =
2a8
22a a8 4
23a16
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DIAMOND CUBIC STRUCTURE
But XZ = 2r
(2r)2 =
4r2 =
r2=
Atomic radius r =
23a16
23a16
23a64
3 a8
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DIAMOND CUBIC STRUCTURE
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
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DIAMOND CUBIC STRUCTURE
APF =
i.e. APF = 34%
Thus it is a loosely packed structure.
3 0.3416
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