The relationship between **edge length $a$** and **atomic radius $r$** in a **body-centred cubic (bcc)** unit cell is:
- $a = 2r$
- $a = 4r/\sqrt 3$
- $a = 4r/\sqrt 2$
- $a = r\sqrt 3$
Show answer and solution
Correct answer: B. $a = 4r/\sqrt 3$
In bcc, body diagonal $= a\sqrt 3 = 4r$ (along it: corner atom + central atom + corner atom = 4r in radii).
**So $a = 4r/\sqrt 3$.**