Question

Sodium metal crystallizes in a body-centered cubic lattice with the cell edge \[\left( a \right) = 4.29A^\circ \]. The radius of sodium atom is\[\left( {{\rm{in }}A^\circ } \right)\]:

Answer 1

Hint: For a body centered cubic unit cell the relationship between the radius of the atom and the edge length of the unit cell is as shown below.
\[r = \dfrac{{\sqrt 3 }}{4}a\]
Substituting the value of the edge length of the bcc unit cell of sodium, in the above equation, and calculate the radius of the sodium atom.

Complete step by step answer:
The body centered cubic unit cell is represented as bcc.
For a body centered cubic unit cell the relationship between the radius of the atom and the edge length of the unit cell is as shown below.
\[4r = \sqrt 3 a\]
The above relationship can also be rearranged as shown below:

\[r = \dfrac{{\sqrt 3 }}{4}a\]

Here r is the radius of the sodium atom and a is the edge length of the bcc unit cell of sodium.
The cell edge \[\left( a \right) = 4.29A^\circ \]
Substitute the value of the edge length of the bcc unit cell in the above equation.
$r = \dfrac{{\sqrt 3 }}{4}a$

$r = \dfrac{{\sqrt 3 }}{4} \times 4.29A^\circ$
$r = 1.857A^\circ$

Hence, the radius of the sodium atom is \[1.857A^\circ \].

Note: In the body centered cubic unit cell, one atom is present at each corner and one atom is present at the body centre. In the body diagonal, the atoms touch each other.