Question

A metal has a fcc structure if the edge length of unit cell is 200 pm, calculate the volume occupied by particles in a unit cell.

• A. \(2.08\times 10^{-24} \text{cm}^3\)
• B. \(5.92\times 10^{-24} \text{cm}^3\)
• C. \(8\times 10^{-24} \text{cm}^3\)
• D. \(3.14\times 10^{-24} \text{cm}^3\)

Hint:

Use r = a root 2 over 4 for fcc, then multiply the volume of one sphere by 4 atoms per cell.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a face-centred cubic (fcc) cell the atoms touch along the face diagonal. There are 4 atoms per unit cell. The volume occupied equals 4 times the volume of one sphere.

Step 2: Key Formula or Approach
\[ \sqrt2\,a=4r\Rightarrow r=\frac{\sqrt2\,a}{4} \]

Step 3: Radius
\[ r=\frac{1.414\times200}{4}=70.7\ \text{pm}=7.07\times10^{-9}\ \text{cm} \]

Step 4: Volume of one atom
\[ V_1=\frac43\pi r^3=\frac43(3.1416)(7.07\times10^{-9})^3=1.48\times10^{-24}\ \text{cm}^3 \]

Step 5: Volume of four atoms
\[ V=4\times1.48\times10^{-24}=5.92\times10^{-24}\ \text{cm}^3 \]

Step 6: Check
The cell volume is \(a^3=(2\times10^{-8})^3=8\times10^{-24}\ \text{cm}^3\), which is option (C). The fraction occupied is \(5.92/8=0.74\), the known packing efficiency of fcc, so the result is consistent.

Final Answer:
The four atoms in the fcc cell fill \(5.92\times10^{-24}\ \text{cm}^3\), option (B). \[ \boxed{5.92\times10^{-24}\ \text{cm}^3} \]