question_answer

Percentage of free space in a body centre cubic unit cell is

A)  30%                                      

B)     32%                      

C)     34%                                      

D)     28%

Correct Answer: B

Solution :

No. of effective atom in a unit cell in b.c.c. = 2 Let side length of the cube\[=a\] from b.c.c. structure \[4r=\sqrt{3a}\] or,          \[a=\frac{4r}{\sqrt{3}}\] Packing fraction\[=\frac{2\times \frac{4}{3}\pi {{(r)}^{3}}}{{{\left( \frac{4r}{\sqrt{3}} \right)}^{3}}}\times 100\] \[=\frac{\sqrt{3\pi }}{8}\times 100=68%\] Percentage of free space = 32%.