ruman ansari
TRANSCRIPT
NaCl &DIAMOND STRUCTURE
NAME:-Ruman Ansari
Branch:-MEBatch:-2015
1
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
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
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
RELATION BETWEEN LATTICE CONSTANT AND DENSITY
RELATION BETWEEN LATTICE CONSTANT AND DENSITY
Number of atoms per unit cell Atomic weightAvagadro number (Lattice constant)
3
NaCl STRUCTURE
Nacl having fcc structure so, n=4 & the co- ordinate no. N=6Atomic weight of Nacl =23+35.45=58.45Density of Nacl =2180 kg/Lattice constant for Nacl structure
a=
NaCl STRUCTURE
3m
31
Nanm
NaCl STRUCTURE
a=
a=5.625Inter atomic distance is a/2 hence Inter atomic distance is = 5.625/2 = 2.8125A
31
2610023.6218045.584
DIAMOND CUBIC STRUCTURE
a
a/4
Z
YY
2r
a/4W
a/4
X
DIAMOND CUBIC STRUCTURE
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.
DIAMOND CUBIC STRUCTURE
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.
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
DIAMOND CUBIC STRUCTURE
XY2 =
Also in the triangle XYZ,
XZ2 = XY2 + YZ2
=
XZ2 =
2a8
22a a8 4
23a16
DIAMOND CUBIC STRUCTURE
But XZ = 2r
(2r)2 =
4r2 =
r2=
Atomic radius r =
23a16
23a16
23a64
3 a8
DIAMOND CUBIC STRUCTURE
Atomic packing factor (APF)
APF =
v =
i.e. v =
APF =
vV
348 r3
3
4 3a83 8
3
3 3
8 4 3 3 a3 8 a
DIAMOND CUBIC STRUCTURE
APF =
i.e. APF = 34%
Thus it is a loosely packed structure.
3 0.3416