Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Semiconductors and Electronic Devices

Question:

Distance between body centred atom & a corner atom in sodium (a = 4.225 Å) is

Options:

3.66 Å

3.17 Å

2.99 Å

2.54 Å

Correct Answer:

3.66 Å

Explanation:

[Hint ⇒ for B.C.C. cell $r=\frac{\sqrt{3}}{4}a$

So distance between body centered atom and a corner atom is $2r=\frac{\sqrt{3}}{4}×a×2=3.66Å$]