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Question

# First three nearest neighbour distances for primitive cubic unit cell will be, respectively: (Here, edge length of unit cell =l )

A
l,2 l,3 l
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B
3 l,2 l,l
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C
l,2 l,2 l
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D
l,2 l, l
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Solution

## The correct option is A l,√2 l,√3 lThe edge length of the unit cell is ′l′. The structure of cubic unit cell is shown below. The first nearest atom for any atom in a cubic unit cell is the atom located at adjacent corner of it. Hence, it will have 6 nearest atom to it in simple cubic. ∴ Coordination Number = 6 Thus, the length of first nearest atom is, (1) →l (First nearest neighbour distance) The second nearest atom will be at the face diagonal 'C'. Thus, in △ABC AC2=AB2+BC2⇒AC2=l2+l2⇒AC2=2l2⇒AC=l√2 Thus, (2) → l√2 (Second nearest neighbour distance) The third nearest atom is the one at axial diagonal 'E' In △AEF AE2=AF2+EF2⇒AE2=l2+2l2⇒AE2=3l2⇒AE=l√3 Thus, (3)→ l√3 (Third nearest neighbour distance)

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