Questions & Answers

Question

Answers

(A) 0.134a

(B) 0.027a

(C) 0.067a

(D) 0.047a

Answer

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To solve this question we have to find the relation between the radius (r) and the edge of the body centered unit cell (a). We know that the total number of atoms present in the body centered cubic cell is = 2.

In body centered unit cells, atoms in the body diagonals are in contact with each other. So the length of the body diagonal is = $R+2R+R=4R$

The body diagonal = $\sqrt{3}a$

So, 4R = $\sqrt{3}a$---1

According to the question

\[\dfrac{a}{2}=\left( R+r \right)\]---2

From equation 1 and 2

\[\dfrac{a}{2}=\dfrac{a\sqrt{3}}{4}+r\]

\[a(\dfrac{2-\sqrt{3}}{4})=r\]

r = 0.067a

1 atom at body center $+\text{ (}\dfrac{1}{8})(8)$ at the corner = 2 atoms

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