Question

The atomic radius of atom in FCC structure having a lattice parameter 'a' is

• A. $a\sqrt{2}/2$
• B. $a/2\sqrt{2}$
• C. $a\sqrt{3}/4$
• D. $a/2$

Hint:

FCC: $4r = \sqrt{2}a$; BCC: $4r = \sqrt{3}a$.

Answer 1

The Correct Option is B

Step 1: Concept
In a Face-Centered Cubic (FCC) lattice, atoms touch along the face diagonal of the unit cell.

Step 2: Meaning
Let $a$ be the edge length and $r$ be the atomic radius. The face diagonal length is $\sqrt{2}a$.

Step 3: Analysis
The face diagonal consists of 4 atomic radii ($4r$). Therefore, $4r = \sqrt{2}a \implies r = \frac{\sqrt{2}a}{4}$.

Step 4: Conclusion
Simplifying the expression: $r = \frac{a}{2\sqrt{2}}$. Final Answer: (B)