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

$\begin{array}{1 1}r=R(0.414)\\r=R(1.414)\\r=R(2.414)\\r=R(0.214)\end{array} $

1 Answer

A sphere fitting into a octahedral void is shown by shaded circle.ABC is a right angle triangle.Applying Phythagoras theorem,
$BC^2=AB^2+AC^2$
$(2R)^2=(R+r)^2+(R+r)^2$
$\Rightarrow 2(R+r)^2$
(or) $\large\frac{(2R)^2}{2}=$$(R+r)^2$
$\sqrt{\large\frac{(2R)^2}{2}}=$$R+r$
(or) $\sqrt{\large\frac{4R^2}{2}}$$=R+r$
$\sqrt{2R^2}=R+r$
$\sqrt 2R=R+r$
$r=\sqrt 2R-R$
$r=R(\sqrt 2-1)$
$\;\;=R(1.414-1)$
$r=R(0.414)$
answered Jul 30, 2014 by sreemathi.v
 

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