Q. A solid metal has ccp or fcc structure. The relation of side of cube $(a)$and radius of atom $(r)$ will be
A) $a = 2r$
B) $a = 2\sqrt{2}r$
C) $a = \frac{4}{\sqrt{3}}r$
D) $a = \sqrt{\frac{3}{2}}r$
Explanation
(b)
In case of ccp or fcc structure
$\ \ \ 4r = \sqrt{2}a\ \ \ \ \Rightarrow \ \ a = \frac{4r}{\sqrt{2}}$
$\therefore\ \ \ \ \ \ a = 2\sqrt{2}r\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.