Chapter 9, Problem 51E

Interpretation Introduction

Interpretation:

        The atomic radius of calcium in its cubic close packing structure is given and its density has to be determined.

Concept introduction:

        In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.

       In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

       Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                  $\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\\ \end{array}$