If the radius of the octachedral void is r and radius of the atoms in close packing is R,

A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled

∠POQ = 900

Now, applying Pythagoras theorem, we can write:

$\mathrm{PQ}^{2}=\mathrm{PO}^{2}+\mathrm{OQ}^{2}$

$\Rightarrow(2 R)^{2}=(R+r)^{2}+(R+r)^{2}$

$\Rightarrow(2 R)^{2}=2(R+r)^{2}$

$\Rightarrow 2 \mathrm{R}^{2}=(\mathrm{R}+\mathrm{r})^{2}$

$\Rightarrow \sqrt{2} \mathrm{R}=\mathrm{R}+\mathrm{r}$

$\Rightarrow \mathrm{r}=\sqrt{2} \mathrm{R}-\mathrm{R}$

$\Rightarrow \mathrm{r}=(\sqrt{2}-1) \mathrm{R}$

$\Rightarrow \mathrm{r}=0.414 \mathrm{R}$