MHT CET · Chemistry · Solid State
Find the void volume of fcc unit cell in \(\mathrm{cm}^3\) if the volume of fcc unit cell \(1.25 \times 10^{-22} \mathrm{~cm}^3\).
- A \(3.25 \times 10^{-23}\)
- B \(2.16 \times 10^{-23}\)
- C \(1.34 \times 10^{-23}\)
- D \(4.20 \times 10^{-23}\)
Answer & Solution
Correct Answer
(A) \(3.25 \times 10^{-23}\)
Step-by-step Solution
Detailed explanation
For fcc unit cell, packing efficiency \(=74 \%\)
\(\therefore \quad\) Percentage of unoccupied space (void volume)
\(\begin{aligned}
& =100-74=26 \% \\
& =1.25 \times 10^{-22} \mathrm{~cm}^3 \times \frac{26}{100} \\
& =3.25 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}\)
\(\therefore \quad\) Percentage of unoccupied space (void volume)
\(\begin{aligned}
& =100-74=26 \% \\
& =1.25 \times 10^{-22} \mathrm{~cm}^3 \times \frac{26}{100} \\
& =3.25 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}\)
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