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

Step 1: Understanding the Concept:
For a Face-Centered Cubic (FCC) lattice, the atoms touch along the face diagonal of the cubic unit cell. The relationship between the edge length (\(a\)) and the atomic radius (\(r\)) is essential to find the unit cell volume (\(V = a^3\)).
Step 2: Key Formula or Approach:
For an FCC unit cell: \[ a = 2\sqrt{2} \cdot r \] Volume, \(V = a^3\)
Step 3: Detailed Explanation:
Given radius, \(r = 141.4\text{ pm}\). Let's convert this to \(\text{cm}\) to match the options. \[ 1\text{ pm} = 10^{-12}\text{ m} = 10^{-10}\text{ cm} \] \[ r = 141.4 \times 10^{-10}\text{ cm} \] Notice that \(141.4 \approx 100 \times \sqrt{2}\) (since \(\sqrt{2} \approx 1.414\)). This approximation makes calculation easier: \[ r \approx 100\sqrt{2} \times 10^{-10}\text{ cm} = \sqrt{2} \times 10^{-8}\text{ cm} \] Now, calculate the edge length '\(a\)': \[ a = 2\sqrt{2} \cdot r = 2\sqrt{2} \cdot (\sqrt{2} \times 10^{-8}\text{ cm}) \] \[ a = 2 \cdot (\sqrt{2} \cdot \sqrt{2}) \times 10^{-8}\text{ cm} \] \[ a = 2 \cdot 2 \times 10^{-8}\text{ cm} = 4 \times 10^{-8}\text{ cm} \] Now, calculate the volume '\(V\)': \[ V = a^3 = (4 \times 10^{-8}\text{ cm})^3 \] \[ V = 4^3 \times (10^{-8})^3\text{ cm}^3 \] \[ V = 64 \times 10^{-24}\text{ cm}^3 \] To match the scientific notation format in the options: \[ V = 6.4 \times 10^{-23}\text{ cm}^3 \] Step 4: Final Answer:
The calculated volume perfectly matches option (C).