# Size of Tetrahedral void

### Size of Tetrahedral void

#### Relationship between Radius of Tetrahedral Void (r) and Radius of Sphere (R) Let us consider an FCC unit cell, if we divide this unit cell into 8 equal units by passing an imaginary plane through the center of the X, Y and Z axis. Then every cubic unit has 1 tetrahedral void at the body center. Consider edge length of cubic unit ‘a’ cm, then:

AB = a   ,   BC = a   ,   CD = a

First calculate the face diagonal of the cube:

AC2  =  AB2 + BC2   =   a2 + a2   =  2a2

Now calculate the body diagonal of the cube:

AD2  =  AC2 + CD2  =  2a2 + a2   =  3a2

Now the ratio of body diagonal to face diagonal will be:

$$\frac{AD}{AC}=\frac{\sqrt{3}.a}{\sqrt{2}.a}=\frac{\sqrt{3}}{\sqrt{2}}$$  ……….(i)

In term of Radius of atom (R) and Radius of tetrahedral void (r):

AC = 2R    ,    AD = 2R +2r

Now the ratio of body diagonal to face diagonal will be:

$$\frac{AD}{AC}=\frac{2R+2r}{2R}=1+\frac{r}{R}$$ ………..(ii)

Compare equation (ii) with (i):

$$1+\frac{r}{R}=\frac{\sqrt{3}}{\sqrt{2}}~~~,~~~\frac{r}{R}=\frac{\sqrt{3}}{\sqrt{2}}-1$$

$$\frac{r}{R}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}=\frac{1.732-1.414}{1.414}$$

$$\frac{r}{R}=0.225~~~~or~~~~r=0.225R$$