Q. Atomic radius of $fcc$ is
A) $\frac{a}{2}$
B) $\frac{a}{2\sqrt{2}}$
C) $\frac{\sqrt{3}}{4}a$
D) $\frac{\sqrt{3}}{2}a$
Explanation
(b) For the $fcc$ structure
$4r = \left( a^{2} + a^{2} \right)^{1/2} = a\sqrt{2}$
$\Rightarrow r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.