Question

Calculate the total volume occupied by all particles in fcc unit cell if volume of unit cell is $6.4 \times 10^{-23}$ cm$^3$.

• A. $3.321 \times 10^{-23}$ cm$^3$
• B. $4.350 \times 10^{-23}$ cm$^3$
• C. $5.126 \times 10^{-23}$ cm$^3$
• D. $4.736 \times 10^{-23}$ cm$^3$

Hint:

Memorize the packing efficiencies for instant calculations:
- Simple Cubic (sc) = 52.4%
- Body-Centered Cubic (bcc) = 68%
- Face-Centered Cubic (fcc) ccp hcp = 74%

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the total volume of a face-centered cubic (fcc) unit cell. We need to find the actual volume occupied by the spherical atoms (particles) inside it.

Step 2: Key Formula or Approach:
The packing efficiency of a crystal lattice is the percentage of the total unit cell volume that is occupied by the particles.
For an fcc (or ccp) lattice, the packing efficiency is a standard known value of 74%.
$$\text{Occupied Volume} = \text{Total Volume} \times \text{Packing Fraction}$$

Step 3: Detailed Explanation:
Given:
Total volume of unit cell = $6.4 \times 10^{-23} \text{ cm}^3$
Packing fraction for fcc = $0.74$
Multiply the total volume by the packing fraction:
$$\text{Occupied Volume} = 0.74 \times (6.4 \times 10^{-23})$$
$$\text{Occupied Volume} = 4.736 \times 10^{-23} \text{ cm}^3$$

Step 4: Final Answer:
The volume occupied by the particles is $4.736 \times 10^{-23} \text{ cm}^3$, matching option (d).