Derivation Radius of Tetrahedral and Octahedral Voids.

Dear Student,

There is no derivation of radius of tetrahedral and octahedral void. Derivation of the relationship between radius of the octahedral void and tetrahedral void  with radius of atoms in close packing is shown below :
 
For octagedral :

$$\begin{gathered} \mathrm{Consider} \mathrm{triangle} \mathrm{ABC} \\ {\mathrm{BC}}^2={\mathrm{AB}}^2 + {\mathrm{AC}}^2 \\ (2\mathrm{R}{)}^2=(\mathrm{R}+\mathrm{r}{)}^2+(\mathrm{R}+\mathrm{r}{)}^2 \\ (2\mathrm{R}{)}^2=2(\mathrm{R}+\mathrm{r}{)}^2 \\ \mathrm{or}, \sqrt{2}{\mathrm{R}}^2=(\mathrm{R}+\mathrm{r}{)}^2 \\ \mathrm{or}, \sqrt{2}\mathrm{R}=\mathrm{R}+\mathrm{r} \\ \mathrm{or}, \mathrm{r}=\sqrt{2}\mathrm{R}-\mathrm{R}=\mathrm{R}(1.414-1)=0.414 \mathrm{R} \end{gathered}$$

For tetrahedral void:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube}=\mathrm{a} \\ \mathrm{From} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{ABC}, \mathrm{face} \mathrm{diagonal} \\ \mathrm{AB}=\sqrt{{\mathrm{AC}}^2+{\mathrm{BC}}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2}=\sqrt{2}\mathrm{a} \\ \mathrm{As} \mathrm{spheres} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} \mathrm{actually} \mathrm{touching} \mathrm{each} \mathrm{other}, \mathrm{face} \mathrm{diagonal} \mathrm{AB}=2\mathrm{R} \\ \mathrm{So}, 2\mathrm{R}=\sqrt{2}\mathrm{a} \\ \mathrm{or}, \mathrm{R}=\frac{1}{\sqrt{2}\mathrm{a}} ....\left(\mathrm{i}\right) \\ \mathrm{From} \mathrm{right} \mathrm{angled} \mathrm{trainage} \mathrm{ABD} \\ \mathrm{AD}\sqrt{{\mathrm{AB}}^2+{\mathrm{BD}}^2}=\sqrt{{\sqrt{2\mathrm{a}}}^2+{\mathrm{a}}^2}=\sqrt{3}\mathrm{a} \\ \mathrm{But} \mathrm{as} \mathrm{small} \mathrm{sphere} \mathrm{touches} \mathrm{other} \mathrm{spheres}, \mathrm{body} \mathrm{diagonal} \mathrm{AD}=2(\mathrm{R}+\mathrm{r}) \\ \mathrm{or}, 2(\mathrm{R}+\mathrm{r})=\sqrt{3}\mathrm{a} \\ \mathrm{or},\mathrm{R}+\mathrm{r}=\frac{\sqrt{3}}{2}\mathrm{a} ...\left(\mathrm{ii}\right) \\ \mathrm{Dividing} \left(\mathrm{ii}\right) \mathrm{by} \left(\mathrm{i}\right) \\ \mathrm{R}+\frac{\mathrm{r}}{\mathrm{R}}=\frac{\sqrt{3}}{2}\times \frac{\mathrm{a}}{{\displaystyle \raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}}=\frac{\sqrt{3}}{\sqrt{2}} \\ \mathrm{or}, 1+\frac{\mathrm{r}}{\mathrm{R}}=\frac{\sqrt{3}}{\sqrt{2}}=1.225 \\ \mathrm{or}, \frac{\mathrm{r}}{\mathrm{R}}=1.225-1=0.225 \\ \mathrm{or}, \mathrm{r}=0.225 \mathrm{R} \end{gathered}$$