Q. The edge of unit cell of fcc Xe crystal is 620 pm. The radius of Xe atom is
A) 189.37 pm
B) 209.87 pm
C) 219.25 pm
D) 235.16 pm
Explanation
(c)
$r = \frac{a}{2\sqrt{2}} = \frac{620}{2\sqrt{2}} = 219.25\ pm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.