# 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:

**AC ^{2} = AB^{2} + BC^{2} = a^{2} + a^{2} = 2a^{2 }**

Now calculate the body diagonal of the cube:

**AD ^{2} = AC^{2} + CD^{2} = 2a^{2 }+ a^{2} = 3a^{2}**

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):

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

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

\( \frac{r}{R}=0.225~~~~or~~~~r=0.225R \)

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