A) 32%
B) 20%
C) 34%
D) 28%
Correct Answer: A
Solution :
For bcc unit cell, \[\text{Packing}\,\text{fraction}\,\text{=}\frac{\text{Total}\,\text{volume}\,\text{of}\,\text{spheres}}{\text{Volume}\,\text{of}\,\text{the}\,\text{unit}\,\text{cell}}\] \[\text{=}\frac{\frac{8}{3}\pi {{r}^{3}}}{\frac{64}{3\sqrt{3}}{{r}^{3}}}=0.68\]. \[\therefore \]% of occupied space \[=0.68\times 100\] \[=68%\] \[\therefore \]% of free space\[~=100-68=32%\]
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