Class 12 Chemistry

Solid State

Efficiency of Close Packed Structure

Packing efficiency in Simple Cubic Lattice:

A unit cell of simple cubic lattice contains one atom. Let the side of a simple cubic lattice is ‘a’ and radius of atom present in it is ‘r’.

Since, edges of atoms touch each other, therefore, a = 2r

Volume of cube =Side3 =a^3=(2r)^3

Volume of one atom =4/3πr^3

Packing Efficiency =text(Volume of 1 sphere in unit cell)/text(Total volume of unit cell)xx100

=(4/3πr^3)/((2r)^3)xx100

=(4πr^3)/(3xx8xx\r^3)xx100

=(4xx3.14xx100)/(24)=52.4%

Thus,packing efficiency of bcc structure=52.4%

Calculation of dimensions of a unit cell:

Let
The edge of a unit cell is ‘a’.
The density of unit cell is ‘d’
Molar mass of unit cell is ‘M’.
Number of atoms present in unit cell is ‘z’.
Mass of each atoms present in unit cell is ‘m’.

Mass of one atom in unit cell m=M/(N_A)

Therefore, mass of unit cell = Number of atoms present in unit cell xx Mass of one atom

Or, Mass of unit cell =zm

Or, Mass of unit cell =z\xx(M)/(N_A)

We know that density d=text(Mass)/text(Volume)

Therefore, density d of unit cell =text(Mass of unit cell)/text(Volume of unit cell)

Or, d=(z\xx(M)/(N_A))/(a^3)

Or, d=(zm)/(a^3N_A)

Where, d is density, z is number of atoms present in unit cell, a is length of edge, and NA is Avogadro constant.

Above expression has five parameters, d, z, a, m and NA . By knowing any four of them fifth can be calculated.

Solid State - Introduction

Solid State - Crystal Lattice

Solid State - Efficiency of Close Packing Structure - 1

Solid State - Efficiency of Close Packing Structure - 2

Solid State - Imperfections in Solids - Crystal Defects

Solid State - Electrical and Magnetic Properties