Question

Chromium metal crystallizes with a body centred cubic lattice. The length of the unit cell is found to be 287 pm. Calculate the atomic radius, the number of atoms per unit cell.

Answer 1

To solve this question, we must first derive a relation between the edge length of the unit cell and the atomic radius. From this we will get the value of the atomic radius. After this, we must individually count the atoms present in the unit cell and find their sum to get the total number of atoms present in the unit cell.

Complete step by step answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
In a body centred cubic lattice, atoms are placed at all the vertices of the unit cell. Along with this, there is one atom that is placed at the centre of the cube.
In order to understand the relation between the unit cell edge length and the radius of the atom:
Let the edge length of the unit cell be ‘a’ units. Hence, using Pythagoras theorem, the diagonal formed on the face of the unit cell can be given as:
Length of face diagonal = $\sqrt {{a^2} + {a^2}}$ = $a\sqrt 2$ units.
Using the Pythagoras theorem again, the length of the body diagonal can be calculated as:
Length of body diagonal = $\sqrt {{a^2} + {{(a\sqrt 2 )}^2}}$ = $a\sqrt 3$ units.
We can observe that there are 3 atoms present on the body diagonal of the unit cell. Considering the radius of the atom to be ‘r’ units, the length of the body diagonal can be equated as:
$a\sqrt 3$ = 4r
r = $\dfrac{{\sqrt 3 }}{4}$ a
Using this relation, we can calculate the atomic radius of chromium as:
r = $\dfrac{{\sqrt 3 }}{4}$ a
r = $\dfrac{{\sqrt 3 }}{4}$ (287 pm) = $\dfrac{{\sqrt 3 }}{4}$ ( $287 \times {10^{ - 12}}$ ) = $124,27 \times {10^{ - 12}}$ m.
We can observe that there are $\dfrac{1}{8}$ th of atoms present on all the vertices of the unit cell. Along with this, there is one complete atom present at the centre of the unit cell. Hence, the total number of atoms present in the unit cell of chromium is:
Total number of atoms present = (number of atoms present at one vertex) (number of vertices) + (number of atoms present at the centre of the unit cell)
$= \left( {\dfrac{1}{8}} \right)\left( 8 \right) + \left( 1 \right)$
= 2 atoms

Note:
The bcc arrangement does not allow the atoms to pack together as closely as the fcc or hcp arrangements. The bcc structure is often the high temperature form of metals that are close - packed at lower temperatures.