Q. If lattice parameter for a crystalline structure is $3.6\ \mathring{\mathrm{A}}$, then atomic radius is fcc crystal in $\mathring{\mathrm{A}}$ is
A) 7.20
B) 1.80
C) 1.27
D) 2.90
Explanation
(c) Atomic radius for fcc crystal is
$r = \frac{a}{2\sqrt{2}} = \frac{3.6\mathring{\mathrm{A}}}{2\sqrt{2}} = 1.27\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.