§2.4 Crystal Structure and
Complex Lattice
Ⅰ.Crystal structure is the real
arrangement of atom in crystals
Crystal structure ＝ Space lattice + Basis
or structure unit
+
=
The difference between space lattice and crystal structure
Fe
Fe : Al = 1 : 1
Al
2×3 atoms / cell
Ⅱ.Typical crystal structures of metals
1. BCC
 Example: α-Fe , V, Nb, Ta,
Cr, Mo, W, alkali metals
 n = 2 atoms/cell
 CN=8
 The number of nearest neighbours around
each atom is called — Coordination Number.
 Packing fraction
＝
Volume of atoms / cell
Volume of unit cell
To determineξ, The atom is looked as a hard
sphere, and the nearest neighbours touch each
other.
4
3 3
2  (
a)
3
4


 0.68
∴ For BCC,
3
a
2. FCC
 Example: γ-Fe , Al , Ni , Pb , Cu ,
Ag , Au , stainless steal
 n= 8×1/8+6×1/2=4 atoms/cell
 CN=12

4
2 3
4  (
)
3
4

 0.74
3
a
3. HCP
 Example: Be, Mg, Zn, Cd, Zr, Hf
Ti( low temperature)
1
1
 n＝  12   2  3  6
6
2
•
 CN＝12
•
•
•
 ξ＝0.74
c 2 2
3 
a ( )  
a
2
3 2 
•
2
2
c
8
 
 1.633
a
3
CN  12   0.74
••
•
•
•
•
••
•
•
••
4. Summary
a0 vs. r
Atoms
per cell
Coordination
Number
Packing
factor
Examples
SC
a0  2r
1
6
0.52
Polonium
(Po),α-Mn
BCC
4
a0 
r
3
2
8
0.68
Fe,Ti,W,Mo,
Nb,Ta,K,Na,
V,Zr,Cr
FCC
a0 
4
r
2
4
12
0.74
Fe,Cu,Au,Pt
,Ag,Pb,Ni
2
12
0.74
Ti,Mg,Zn,Be
,Co,Zr,Cd
Structure
HCP
a0  2r
c0  1.633a0
§2.5 Interstices in typical crystals of
metals
Definition:
In any of the crystal structures, there are small
holes between the usual atoms into which smaller
atoms may be placed. These locations are called
interstitial sites.
Ⅰ. Two types of Interstitials in
typical crystals
Octahedral interstitial
Tetrahedral interstitial
1. Octahedral interstitial
BCC
3
a
2
a
2
a
2
FCC
a
2
a
2
HCP
a
2
2. Tetrahedral interstitial
BCC
3
a
2
5
a
4
a
FCC
3
a
4
a
2
HCP

Ⅱ.Determination of the sizes of
interstitials
Definition:
By size of an interstitial we mean diameter of
the maximum hard sphere which can be
accommodated in the interstitial without distorting
the lattice.
di
diameter of interstitial atom
＝
da
diameter of atom in lattice point
1. Octahedral interstitial
condition for touching
di  d a  a
di
a

1
da da
For BCC
3
da 
a
2
di
a

 1  0.15
da
3
2
For FCC
2
da 
a
2
di
a

 1  0.41
da
2
2
2. Tetrahedral interstitial
A
B
L
interstitial
d a  di
2
H
D
C
host atom
d a  di 2
L 2 H 2
(
) ( ) ( )
2
2
2
di  d a  L  H
2
2
di
 
da
L2  H 2
1
da
For BCC
La
3
da 
a
2
a
H
2
di
  0.29
da
For FCC
2
L
a
2
a
H 
2
2
da 
a
2
di
 0.22
da
3. Summary
n
CN
interstices
ξ
oct.
tete.
di/da
oct.
tete.
BCC
2
8
0.68
6
6/2=3
12
12/2=6
0.15
0.29
FCC
4
12
0.74
4
4/4=1
8
8/4=2
0.41
0.22
HCP
6
12
0.74
6
6/6=1
12
12/6=2
0.41
0.22
Examples and Discussions
1. Both FCC and BCC are close-packed
structures while BCC is more open?
2. The interstitial atoms most likely occupy the
oct. interstitial position in FCC and HCP, while
in BCC two types of interstitial can be occupied
equally.
3. The solid solubility in BCC is much lower than
in FCC.
4. Diffusion of interstitial atoms in BCC diffusion is
much faster than in FCC or HCP at same
temperature.
5. Determine the relationship between the atomic
radius and the lattice parameter in SC, BCC,
and FCC structures when one atom is located
at each lattice point.
6. Determine the density of BCC iron, which has
a lattice parameter of 0.2866nm.
Solution:
For a BCC cell,
Atoms/cell = 2
a0 = 0.2866nm = 2.866×10－8cm
Atomic mass = 55.847g/mol
Volume of unit cell = a03 = 23.54×10 －24cm3/cell
(number of atoms / cell )(atomic mass of iron )
( volume of unit cell )( Avogadro' s number )
(2)(55.847)
3


7
.
882
g
/
cm
(23.54 10-24 )(6.02 10 23 )
Density  
Exercise
1. Determine the coordinates of centers of both
the octahedral and the tetrahedral interstitials
in HCP referred to a, b and c.
c
b
a
120o
4. Prove that the A-face-centered hexagonal lattice is
not a new type of lattice in addition to the 14 space
lattices.
5. Draw a primitive cell for BCC lattice.
Thanks for your attention !