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