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Prove that the void space percentage in zinc blende structure is 25%

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Anions are in fcc positions and half of the tetrahedral holes are occupied by cations.There are H anions and 8 tetrahedral hols per unit cell
Here,face diagonal =$\sqrt 2a$
$\therefore 4r=\sqrt 2a$
$a=2\sqrt 2r$
$r_-$=radius of anion.
$r_+$=radius of cation.
$a$=edge length of the cell
Volume of unit cell =$a^3$
$\Rightarrow (2\sqrt 2 r_-)^3=16\sqrt 2r_-^3$
$\therefore$ Packing fraction =$\large\frac{4\times (\Large\frac{4}{3}\normalsize \pi r_-^3)+\Large\frac{1}{2}\normalsize \times 8\times (\Large\frac{4}{3}\normalsize \pi r_+^3)}{16\sqrt 2 r_-^3}$
$\Rightarrow \large\frac{\pi}{3\sqrt 2}$$\left\{1+(\large\frac{r_+}{r_-})^3\right\}$
Since,for tetrahedral holes
$\large\frac{r_+}{r_-}$$=0.225$
$\therefore PF=\large\frac{\pi}{3\sqrt 2}$$[1+(0.225)^3]$
$\Rightarrow 0.7485$ /unit volume of unit cell
$\therefore$ Void space =1-0.7485
$\Rightarrow 0.2515$ /unit volume of unit cell
$\therefore$ Percentage of void space =25.15% $\simeq 25\%$
answered Aug 1, 2014 by sreemathi.v
 

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