Distance between body centred atom & a corner atom in sodium (a = 4.225 Å) is
Answer & explanation
Correct answer: option 1
[Hint ⇒ for B.C.C. cell $r=\frac{\sqrt{3}}{4}a$
So distance between body centered atom and a corner atom is $2r=\frac{\sqrt{3}}{4}×a×2=3.66Å$]