Question

Calculate the void volume in fcc unit cell if total volume of a unit cell is \(6.4\times 10^{-23} \text{cm}^3\) ?

• A. \(3.172\times 10^{-23} \text{cm}^3\)
• B. \(3.911\times 10^{-23} \text{cm}^3\)
• C. \(1.664\times 10^{-23} \text{cm}^3\)
• D. \(2.051\times 10^{-23} \text{cm}^3\)

Hint:

fcc packing efficiency is 74 percent, so voids are 26 percent.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Packing efficiency is the fraction of the unit cell volume actually filled by atoms. For a face centred cubic (fcc) lattice, it is \(74\%\).

Step 2: Key Formula or Approach:
Void percentage \(= 100 - 74 = 26\%\). Void volume \(= 0.26 \times V_{cell}\).

Step 3: Detailed Explanation:
\[ V_{void} = 0.26 \times 6.4 \times 10^{-23} = 1.664 \times 10^{-23}\ \text{cm}^3 \]
The occupied volume is \(0.74 \times 6.4 \times 10^{-23} = 4.736 \times 10^{-23}\) cm\(^3\), and this plus the void volume gives back the total \(6.4 \times 10^{-23}\) cm\(^3\).
The other options do not match the \(26\%\) empty space of a close packed fcc lattice.

Final Answer:
The void volume is \(1.664 \times 10^{-23}\) cm\(^3\), option (C). \[ \boxed{1.664 \times 10^{-23}\ \text{cm}^3} \]