Question

Percentage of free space in body centred cubic (bcc) unit cell is

• A. 0.3
• B. 0.32
• C. 0.34
• D. 0.28

Answer 1

Answer: (B)

0.32

Answer 2

In bcc unit cell, the number of atoms = 2 Thus, volume of atoms in unit cell $(v)=2\times\frac{4}{3}\pi r^3$ For bcc structure$(r)=\frac{\sqrt{}3}{4}a$ $(V)=2\times\frac{4}{3}\pi\Bigg(\frac{\sqrt{3}}{4}a\Bigg)^3=\frac{\sqrt{3}}{8}\pi a^3$ Volume of unit cell $(V)=a^3$ Percentage of volume occupied by unit cell $=\frac{Volume \, of\, the\, atoms \, in\, unit\, cell}{Volume \, of \,unit\, cell}$ $=\frac{\frac{\sqrt{3}}{8}\pi a^3}{a^3}\times100=\frac{\sqrt{3}}{8}\pi \times 100=68\%$ Hence, the free space in bcc unit cell = 100 -68=32%.