Question

Calculate the volume occupied by all particles in fcc unit cell if volume of unit cell is \(1.6 \times 10^{-23} \text{ cm}^3\).

• A. \(4.088 \times 10^{-23} \text{ cm}^3\)
• B. \(2.156 \times 10^{-23} \text{ cm}^3\)
• C. \(1.184 \times 10^{-23} \text{ cm}^3\)
• D. \(3.226 \times 10^{-23} \text{ cm}^3\)

Hint:

For quick questions on cubic unit cells:
• fcc packing efficiency = \(74%\)
• bcc packing efficiency = \(68%\)
• simple cubic packing efficiency = \(52.4%\)

Answer 1

The Correct Option is C

Concept:
For a face-centered cubic (fcc) unit cell, the packing efficiency is: \[ 74% = \frac{74}{100} = 0.74 \] This means \(74%\) of the total unit cell volume is occupied by particles.

Step 1: Write the given unit cell volume.
\[ V_{\text{cell}} = 1.6 \times 10^{-23}\ \text{cm}^3 \]

Step 2: Use packing efficiency of fcc.
\[ V_{\text{occupied}} = 0.74 \times V_{\text{cell}} \] \[ V_{\text{occupied}} = 0.74 \times 1.6 \times 10^{-23} \]

Step 3: Calculate the value.
\[ V_{\text{occupied}} = 1.184 \times 10^{-23}\ \text{cm}^3 \] Hence, the correct answer is:
\[ \boxed{(C)\ 1.184 \times 10^{-23}\ \text{cm}^3} \]