Q. The distance between the body centred atom and a corner atom in sodium $(a = 4.225\ \mathring{\mathrm{A}})$ is
A) 3.66 $\mathring{\mathrm{A}}$
B) 3.17 $\mathring{\mathrm{A}}$
C) 2.99 $\mathring{\mathrm{A}}$
D) 2.54 $\mathring{\mathrm{A}}$
Explanation
(a) Sodium has $bcc$ structure. The distance between body centre and a corner $= \frac{\sqrt{3}\ a}{2}$
$= \frac{\sqrt{3} \times 4.225}{2} = 3.66\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.