Question

A solid XY has bcc structure. If the distance of closest approach between the two atoms is 173 pm, the edge length of the cell is :
A. 200 pm
B. \[\dfrac{{\sqrt 3 }}{{\sqrt 2 }}\]Pm
C. 142.2 pm
D. \[\sqrt 2 \]pm

Answer 1

Hint: Body-centered cubic unit cell is one kind of cubic lattice. In this cubic lattice each unit cell contains one atom at its body center. And contains one atom at the eight corners of the unit cell.

Complete step by step solution:
In the B.C.C the closest distance between two atoms in the unit lattice is \[\dfrac{{\sqrt {\text{3}} a}}{2}\]. Where a is the edge length of the unit lattice.
Now according to this formula \[\dfrac{{\sqrt {\text{3}} a}}{{\text{2}}}{\text{ = 173pm}}\]
Therefore, the edge length is,
\[
\dfrac{{\sqrt {{\text{3a}}} }}{{\text{2}}}{\text{ = 173pm}} \\
{\text{a = }}\dfrac{{{\text{2 \times 173}}}}{{\sqrt {\text{3}} }} \\
{\text{a = 200pm}} \\
\]

So, the correct option is A.

Additional information:
The contribution of each corner atoms is \[{\dfrac{{\text{1}}}{{\text{8}}}^{{\text{th}}}}\] of a total atom. So, the contribution of all eight atoms is, \[\dfrac{{{1}}}{{{8}}} x 8 = 1\].
For the body center atoms, the contribution of that atom is a total of a total atom.
Total number of atoms is 2 in a B.C.C lattice.

Note: Face-centered cubic unit cell is one kind of cubic lattice. In this unit cell each face of the unit cell contains one atom at its center. And contains one atom at the eight corners of the unit cell. For example , NaCl is a solid crystal with an F.C.C unit cell. Now the number of atoms in a unit cell can be calculated by calculation summation of the contribution of the atoms present in the unit cell.
The contribution of each corner atoms is \[{\dfrac{{\text{1}}}{{\text{8}}}^{{\text{th}}}}\] of a total atom. So, the contribution of all eight atoms is, \[\dfrac{{\text{1}}}{{{8}}} \times 8 = 1\].
For the face center atoms, the contribution of each atoms \[{\dfrac{{\text{1}}}{{\text{2}}}^{{\text{th}}}}\] of a total atom. So, the total contribution of the total six atoms is \[\dfrac{{{1}}}{2} \times 6 = 3\].
Total number of atoms is 4 in a F.C.C lattice.