Question

Calculate the volume of unit cell having atomic radius \(141.4\text{ pm}\) forming fcc unit cell.

• A. \(9.3 \times 10^{-23}\text{ cm}^3\)
• B. \(8.1 \times 10^{-23}\text{ cm}^3\)
• C. \(6.4 \times 10^{-23}\text{ cm}^3\)
• D. \(4.7 \times 10^{-23}\text{ cm}^3\)

Hint:

In fcc: atoms touch along face diagonal $\Rightarrow$ always use \(a = 2\sqrt{2}r\).

Answer 1

The Correct Option is C

Concept:
In a face-centered cubic (fcc) unit cell, atoms touch along the face diagonal. Thus, relation between edge length \(a\) and atomic radius \(r\) is: \[ a = 2\sqrt{2}\,r \] Step 1: Substitute atomic radius. \[ r = 141.4 \text{ pm} \] \[ a = 2\sqrt{2} \times 141.4 \] Since \(2\sqrt{2} \approx 2.828\), \[ a = 2.828 \times 141.4 \approx 400 \text{ pm} \]
Step 2: Convert into cm. \[ 1 \text{ pm} = 10^{-10} \text{ cm} \] \[ a = 400 \text{ pm} = 400 \times 10^{-10} = 4 \times 10^{-8} \text{ cm} \]
Step 3: Calculate volume of unit cell. \[ V = a^3 = (4 \times 10^{-8})^3 \] \[ V = 64 \times 10^{-24} = 6.4 \times 10^{-23} \text{ cm}^3 \]
Step 4: Conclusion. \[ {V = 6.4 \times 10^{-23} \text{ cm}^3} \]