Consider the bcc unit cells of the solids $1$ and $2$ with the position of atoms as shown below. The radius of atom $B$ is twice that of atom $A$ . The unit cell edge length is $50 %$ more in solid $2$ than in $1$ . What is the approximate packing efficiency in solid $2$ ?

65 %

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90 %

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75 %

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45 %

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Solution

The correct option is B $90 %$
Packing efficiency = $\frac{V o l u m e o c c u p i e d b y s p h e r e}{V o l u m e o f c u b e} \times 100$
taking $r_{A}$ as the radius of atom $A$
$r_{B}$ as the radius of atom $B$
$a_{1}$ is the edge length of solid 1 and $a_{2}$ is the edge length of solid 2
Given, $r_{B} = 2 r_{A}$
$a_{2} = a_{1} + \frac{50}{100} = 1.5 a_{1}$
For bcc lattice:
$4 r_{A} = \sqrt{3} a_{1}$

$r_{A} = \frac{\sqrt{3} a_{1}}{4}$

$a_{1} = \frac{4 r_{A}}{\sqrt{3}}$

$a_{2} = 1.5 \frac{4 r_{A}}{\sqrt{3}}$

= $\frac{3}{2} (\frac{4 r_{A}}{\sqrt{3}}$ )
$a_{2} = 2 \sqrt{3} r_{A}$

Packing efficiency = $\frac{(\frac{4}{3} π r_{A}^{3} \times 1) + (\frac{4}{3} π r_{B}^{3} \times 1)}{a_{2}^{3}} \times 100$

= $\frac{(\frac{4}{3} π r_{A}^{3}) + (\frac{4}{3} π r_{B}^{3})}{(2 \sqrt{3} r_{A})^{3}} \times 100$

= $\frac{\frac{4}{3} π r_{A}^{3} \times 9}{8 \times 3 \sqrt{3} r_{A}^{3}} \times 100$

= $\frac{π}{2 \sqrt{3}} \times 100$

= $90.72 %$

= $90 %$

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