Atomic radius of fcc is. | PHYSICS - CLASSIFICATION OF METALS, CONDUCTORS AND SEMICONDUCTORS | SolveeIt | Solveeit

Today, August 19

Question

Atomic radius of fcc is.

A

$\frac{a}{2}$

B

$\frac{a}{2\sqrt{2}}$

C

$\frac{\sqrt{3}}{4}a$

D

$\frac{\sqrt{3}}{2}a$

Answer

$\frac{a}{2\sqrt{2}}$

Explanation

Solution

For the fcc structure

$4r = (a^{2} + a^{2})^{1/2} = a\sqrt{2}$

⇒ $r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$