Q. An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is :
A) $\frac{\sqrt{2}}{4} \times 288$ pm
B) $\frac{4}{\sqrt{3}} \times 288$ pm
C) $\frac{4}{\sqrt{2}} \times 288$ pm
D) $\frac{\sqrt{3}}{4} \times 288$ pm
Explanation
[Solution] For bcc, $4r = \sqrt{3}a$, so $r = \frac{\sqrt{3}}{4} \times 288$ pm. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.