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):
\( 1+\frac{r}{R}=\frac{\sqrt{3}}{\sqrt{2}}~~~,~~~\frac{r}{R}=\frac{\sqrt{3}}{\sqrt{2}}-1 \)
\( \frac{r}{R}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}=\frac{1.732-1.414}{1.414} \)
\( \frac{r}{R}=0.225~~~~or~~~~r=0.225R \)
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