KCET · Chemistry · Solid State
Vacant space in body centered cubic lattice unit cell is about
- A \(10 \%\)
- B \(23 \%\)
- C \(46 \%\)
- D \(32 \%\)
Answer & Solution
Correct Answer
(D) \(32 \%\)
Step-by-step Solution
Detailed explanation
For bcc, number of atoms \(=2\)
\[
\begin{aligned}
\text { Percentage packing fraction } & =\frac{V_{\text {sphere }}}{V_{\text {unit cell }}} \times 100 \\
& =\frac{2 \times \frac{4}{3} \times \pi r^2}{\left(\frac{4}{\sqrt{3}} r\right)^3} \times 100=68 \%
\end{aligned}
\]
\(\therefore\) Fraction of free space in bcc in \(100-68=32 \%\).
\[
\begin{aligned}
\text { Percentage packing fraction } & =\frac{V_{\text {sphere }}}{V_{\text {unit cell }}} \times 100 \\
& =\frac{2 \times \frac{4}{3} \times \pi r^2}{\left(\frac{4}{\sqrt{3}} r\right)^3} \times 100=68 \%
\end{aligned}
\]
\(\therefore\) Fraction of free space in bcc in \(100-68=32 \%\).
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