Question

A solid has a ‘BCC’ structure. If the distance of nearest approach between two atoms is 1.73${{A}^{\circ }}$, the edge length of the cell is:
(A) 314.20 pm
(B) 216 pm
(C) 200 pm
(D) 141 pm

Answer 1

The length of body diagonal is 4r and length of face diagonal is $\sqrt{3}a$. Every lattice have their own different relation of radius and edge length. The distance of the nearest approach between two atoms is twice the radius of the atom.

Complete step by step answer:
There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them. Each unit cell has six sides, and each side is a parallelogram. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called a simple cubic part. If the unit cell also contains an identical component in the center of the cube, then it is body centered cubic(BCC). If there are components in the center of each face in addition to those at the corners of the cube,then the unit cell is face-centered cubic (FCC).
If a is the edge length of the cube and r is the radius of each atom. Distance of nearest approach between two atoms = 2r = 1.73${{A}^{\circ }}$.
Therefore, $r=\dfrac{1.73}{2}=0.865{{A}^{\circ }}$
For BCC lattice, the relation of edge length and radius of the atom is,
$\sqrt{3}a=4r$
$a=\dfrac{4r}{\sqrt{3}}$
$a=\dfrac{4\times 0.865}{\sqrt{3}}$
$a=1.99=2{{A}^{\circ }}=200 pm$
The edge length of the BCC lattice is $200 pm$.
The correct answer is the (C) option.

Note: Each unit cell has six sides, and each side is a parallelogram. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called a simple cubic part.