Question

A metal forms fcc structure. Calculate the volume of fcc unit cell in \(\text{cm}^3\) if void volume is \(1.66\times 10^{-23} \text{cm}^3\).

• A. \(4.912\times 10^{-23} \text{cm}^3\)
• B. \(8.151\times 10^{-23} \text{cm}^3\)
• C. \(9.346\times 10^{-23} \text{cm}^3\)
• D. \(6.385\times 10^{-23} \text{cm}^3\)

Hint:

fcc packing efficiency is 74 percent, so voids occupy 26 percent of the cell volume.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The packing efficiency of an fcc lattice is 74 percent, so 26 percent of the unit cell volume is void space.

Step 2: Key Formula or Approach:
\[ V_{\text{void}} = 0.26\,V_{\text{cell}} \;\Rightarrow\; V_{\text{cell}} = \frac{V_{\text{void}}}{0.26} \]

Step 3: Detailed Explanation:
Given void volume \(= 1.66\times10^{-23} \text{ cm}^3\).
\[ V_{\text{cell}} = \frac{1.66\times10^{-23}}{0.26} = 6.385\times10^{-23} \text{ cm}^3 \]
Check: the occupied volume is \(0.74\times6.385\times10^{-23} = 4.72\times10^{-23}\) and the void plus occupied volumes add to \(6.385\times10^{-23}\).
The other options correspond to dividing by wrong fractions, none of which is the void fraction of fcc.

Final Answer:
The volume of the fcc unit cell is \(6.385\times10^{-23} \text{ cm}^3\), option (D). \[ \boxed{6.385\times10^{-23}\text{ cm}^3 \text{ (D)}} \]