A body-centered cubic lattice is made up of hollow spheres of B. Spheres of solid A are present in hollow spheres of B. Radius A is half of radius of B. What is the ratio of total volume of spheres of B unoccupied by A in a unit cell and volume of unit cell?

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$\frac{7 \sqrt{3 π}}{64}$

$\frac{7 \sqrt{3}}{128}$

$\frac{7 . π}{24}$

$\frac{7 π}{64 \sqrt{3}}$

answer is D.

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Detailed Solution

Number of atoms of B in unit cell = 2
Total volume of B unoccupied by A in a unit cell

$2 \times \frac{4}{3} (R^{3} - r^{3}) \times π = \frac{7 π R^{3}}{3}$

$Vol. of unit cell = a^{3} = \frac{64}{3 \sqrt{3}} R^{3}$

$For bcc, \sqrt{3} a = 4 R$

$∴ ρ_{ratio} = \frac{7 π R^{3} / 3}{\frac{64}{3 \sqrt{3}} R^{3}} = \frac{7 π}{64 \sqrt{3}}$

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