If the radius of the octahedral void is $r$ and the radius of the atoms in close packing is $R$,derive the relation between $r$ and $R$.

(N/A) Consider a square plane of four atoms in close packing with an octahedral void at the center $O$. Let the centers of two adjacent atoms be $P$ and $Q$.
From the geometry of the octahedral void,we can form a right-angled triangle $\triangle POQ$ where the hypotenuse $PQ = 2R$ and the sides $PO = OQ = R + r$.
Applying the Pythagoras theorem in $\triangle POQ$:
$PQ^{2} = PO^{2} + OQ^{2}$
$(2R)^{2} = (R + r)^{2} + (R + r)^{2}$
$4R^{2} = 2(R + r)^{2}$
$2R^{2} = (R + r)^{2}$
Taking the square root on both sides:
$\sqrt{2}R = R + r$
$r = \sqrt{2}R - R$
$r = (\sqrt{2} - 1)R$
Since $\sqrt{2} \approx 1.414$,we get:
$r = 0.414R$