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CAP 331 CAP102 CAP 202 EAP 201 CAP 331 BAP 423 BAP 533 CEM 233 PCP 514 INS 5304 MLS 816 |
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Cubic System
The simple cubic space lattice is also the
primitive cell, whereas the conventional cells of fcc and bcc are not their
primitive unit cells.
Figure 10 shows the primitive translation vectors
and primitive unit cell of fcc lattice. The primitive translation vectors and primitive cell of bcc lattice are shown in figure 11 and 12 respectively.
Body-Centered Cubic (bcc) Crystal Structure: Figure13a shows the atomic-site unit cell for bcc crystal structure. In this unit cell the solid spheres represent the centers where atoms are located and clearly indicate their relative positions. If we represent the atoms in this cell as hard spheres, then the unit cell appears as shown in figure13b. In this unit cell we see that the center atom is surrounded by eight nearest neighbors and is said to have a coordination number of 8.
Number of atoms in bcc unit cell = 1 (at the
center) + 8 ´ = 2 atoms per unit cell.
The atoms in the bcc unit cell contact each other
across the cube diagonal as shown in figure14.
So the relationship between the length of the cube
side a and the atomic radius R is
For bcc lattice
Since there are two atoms in bcc, the volume of
atoms of radius R in one unit cell is
The volume of bcc unit cell is
Therefore,
Face-Centered Cubic (fcc) Crystal Structure: In this unit cell there is one lattice point at each corner of the cube and one at the center of cube face. The atomic-site unit cell, the hard sphere model and isolated unit cell of fcc lattice are shown in figure15.
Hexagonal Close-Packed (hcp) Crystal Structure:
There are an infinite number of ways of arranging
identical spheres in a regular array that maximizes the atomic packing factor.
One is fcc structure; another is the hexagonal close-packed structure shown in
fig.16.
The APF is 0.74 for both the structures. Both of these structures have same coordination number of 12.
Number of atoms in hcp unit cell = 3 (in the middle layer) + 2´
and bottom layer) + 2´6´
= 6 atoms per unit
cell.
Atomic Packing:
Spheres are arranged in a single closest-packed layer A (figure 17) by
placing each sphere in contact with six others. This layer may serve as the
either the basal plane for an hcp structure or the (111) plane of the fcc
structure.
A second layer B may be added by placing each
sphere of B in contact with three spheres of bottoms layer as shown in figure
17. A third layer C may be added in two ways.
We obtain the fcc structure if the spheres of the third layer are added over the holes in the first layer that are not occupied by B. We obtain the hcp structure if the spheres of the third layer are placed directly over the centers of the spheres in the first layer. Clearly seen in figure 18.
Directions in Cubic Unit Cells 
The orientation of a crystal plane is determined
by three points in the plane, provided they are not collinear. If each point lay
on a different crystal axes, the plane could be specified by giving the
coordinates of the points in terms of the lattice constant a1,
a2, and a3.
However, it turns out to be more useful for
structure analysis to specify the orientation of a plane by the indices, called
Miller indices, determined by the following rules (refer figure 19).
1.
Find the intercepts on the axes in terms of the lattice constants a1,
a2, and a3. The axes may be those of a
primitive or non-primitive cell.
2.
Take the reciprocals of these numbers and then reduce to three integers
having the same ratio, usually the smallest three integers. The result, enclosed
in parenthesis (hkl), is called the index of the plane.
3.
The indices (hkl) may denote a single plane or a set of parallel
planes.
4.
If a plane cuts an axis on the negative side of the origin, the
corresponding index is negative, indicated by the minus sign (or bar) above
index: (h
Let us now try to understand that for given Miller
indices how we can draw the corresponding planes in cubic unit cell. For this,
let us undertake the exercise of drawing (101), (1
a)
Plane with Miller indices (101).
·
Find the reciprocal of Miller
indices. These are 1,
¥ and 1 in this case.
·
This implies the corresponding
crystallographic plane passes through the unit cube with intercepts at a1=1,
a2=¥,
and a3=1. · Such plane is shown in following figure.
b)
Plane with Miller indices (1
·
Find the reciprocal of Miller
indices. These are 1, -1 and
¥
in this case. · T 
c)
Plane with Miller indices (421).
·
Find the reciprocal of Miller
indices. These are 1,
¥ and 1 in this case.
·
This implies the corresponding
crystallographic plane passes through the unit cube with intercepts at a1=1,
a2=¥,
and a3=1. · Such plane is shown in following figure.
An important relationship for the cubic system, and only the cubic system, is that the direction indices of a direction perpendicular to a crystal plane are the same as the Miller indices of that plane. For example, [100] direction is perpendicular to (100) crystal plane.
In cubic crystal structure the inter-planar
spacing between two closest parallel planes with same Miller indices is
designated as dhkl, where h, k and l are the Miller indices of
the planes. This spacing represents the distance from a selected origin
containing one plane and another parallel plane with same indices which is
closest to it.
For cubic crystal
where a = lattice constant, (edge of the unit cube). |
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or 
(at the center of top 
(at corners of top and bottom layer)
l).
0) and (421) crystallographic planes in cubic unit cell.
0). 
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Ó Rajdeep Singh Rawat, Natural Sciences and Science Education, National Institute of Education Updated on 3 Aug 2004 Contact │ rsrawat@nie.edu.sg |
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