Question

Calculate the packing efficiency in case of a metal Crystal for
1. simple cubic
2. body -centred cubic
3.Face centred cubic

Answer 1

Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cellThe smallest repeating unit of a crystal lattice.. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells.

Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in Figure 8.2.1. Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in part (d) in Figure 8.2.1  is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure 8.2.2 .

Figure 8.2.1 Unit Cells in Two Dimensions (a–c) Three two-dimensional lattices illustrate the possible choices of the unit cell. The unit cells differ in their relative locations or orientations within the lattice, but they are all valid choices because repeating them in any direction fills the overall pattern of dots. (d) The triangle is not a valid unit cell because repeating it in space fills only half of the space in the pattern.

Figure 8.2.2 Unit Cells in Three Dimensions  These images show (a) a three-dimensional unit cell and (b) the resulting regular three-dimensional lattice.

Answer 2

In a simple cubic lattice the atoms are located only on the corners of the cube.
Let take edge length or side of the cube = a,
Let take radius of each particles = r
The relation between radius and edge a
a = 2r
The volume of the cubic unit cell = side3
= a3
= (2r)3
= 8r3
Number of atoms in unit cell = 8 x 1 /8
= 1
The volume of the occupied space = (4/3)πr



(ii) In body centered cubic two atoms diagonally





Let take edge length or side of the cube = a,

Let take radius of each particles               = r

The diagonal of a cube is always a√3

The relation between radius and edge a will

 a√3 = 4r

divide by root 3 we get

                        a          = 4r/√3

total number of atoms in body centered cubic

number of atoms at the corner = 8 x 1/8 = 1

number of atoms at the center = 1

total number of atoms               = 2

The volume of the cubic unit cell    = side3

                                                            = a3

                                                            = (4r/√3)3

The volume of the occupied space = (4/3)πr3




(iii)

Let take edge length or side of the cube = a,

Let take radius of each particles               = r


The diagonal of a square  is always a√2

The relation between radius and edge a will

 a√2 = 4r

divide by root 3 we get

                        a          = 4r/√2

total number of atoms in body centered cubic

number of atoms at the corner = 8 x 1/8 = 1

number of atoms at the face    = 6 x1/2    = 3

total number of atoms                                 = 4

The volume of the cubic unit cell             = side3

                                                                       = a3

                                                                       = (4r/√2)3

                                                                        = (2√2 r)3


The volume of the occupied space = (4/3)πr3