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

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Octahedral void :

$\begin{figure} \includegraphics {fig2c.ps} \end{figure}$

Let a small circle squeezed between the spheres in the void and let its centre (0,0).

Distance of centre of one of spheres form (0,0) =

(R² + R²) = (

2)R

Also this is equal to R + r where r is the radius of small sphere in-between.

Hence R + r =(

2)R

r = (

2 - 1)R = 0.414R

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