An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is :
- 24 288 pm
- 4 3 288 pm
- 4 2 288 pm
- 34 288 pm
Answer
(4) 34 288 pm
Solution
In a bcc unit cell, the atoms touch each other along the body diagonal.
∴ Body diagonal = 4r
⇒ 3 a = 4r
⇒ r = 34 a
= 34 288 pm
Answer: (4)