Question

Distance between body centred atom & a corner atom in sodium(a = 4.225$\stackrel{\circ }{A}$) is:

• A. 3.66$\stackrel{\circ }{A}$
• B. 3.17$\stackrel{\circ }{A}$
• C. 2.99$\stackrel{\circ }{A}$
• D. 2.54$\stackrel{\circ }{A}$

Answer 1

Answer: (A)

3.66$\stackrel{\circ }{A}$

For B.C.C cell $r=\frac{\sqrt{3}}{4}a$ so distance between body centered atom and a corner atom is $2r=\frac{\sqrt{3}}{4}\times a\times 2=3.66\stackrel{\circ }{A}$