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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 l
The 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+BC2AC2=l2+l2AC2=2l2AC=l2
Thus,
(2) l2 (Second nearest neighbour distance)
The third nearest atom is the one at axial diagonal 'E'
In AEF
AE2=AF2+EF2AE2=l2+2l2AE2=3l2AE=l3
Thus,
(3) l3 (Third nearest neighbour distance)

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