Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
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}$$\left\{1+(\large\frac{r_+}{r_-})^3\right\}$
Since,for tetrahedral holes
$\therefore PF=\large\frac{\pi}{3\sqrt 2}$$[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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App