Question

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

• A. $l,\sqrt{2}l,\sqrt{3}l$
• B. $\sqrt{3}l,\sqrt{2}l,l$
• C. $l,\sqrt{2}l,2l$
• D. $l,\sqrt{2}l,l$

Answer 1

The correct option is A $l,\sqrt{2}l,\sqrt{3}l$

The edge length of the unit cell is ${}^{\prime }{l}^{\prime }$. 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.
$\therefore$ Coordination Number = $6$
Thus, the length of first nearest atom is,
(1) $\to l$ (First nearest neighbour distance)
The second nearest atom will be at the face diagonal 'C'.
Thus, in $△ABC$
$A{C}^2=A{B}^2+B{C}^2\Rightarrow A{C}^2=l^2+l^2\Rightarrow A{C}^2=2l^2\Rightarrow AC=l\sqrt{2}$
Thus,
(2) $\to l\sqrt{2}$ (Second nearest neighbour distance)
The third nearest atom is the one at axial diagonal 'E'
In $△AEF$
$A{E}^2=A{F}^2+E{F}^2\Rightarrow A{E}^2=l^2+2l^2\Rightarrow A{E}^2=3l^2\Rightarrow AE=l\sqrt{3}$
Thus,
(3)$\to l\sqrt{3}$ (Third nearest neighbour distance)