Question

Tungsten crystallizes in a body-centered cubic unit cell. If the edge of the unit cell is $316.5\,{\text{pm}}$ , What is the atomic radius of the tungsten atom?

Answer 1

Hint: In a body-centered cubic unit cell, the radius is one-fourth of diagonal length. The diagonal edge length of the body-centered cubic unit cell is $a\sqrt 3 $.

Formula used: $r\, = \dfrac{{a\sqrt 3 }}{4}$

Complete step by step answer:
The formula to calculate the atomic radius of body-centered cubic unit cell is as follows:
$r\, = \dfrac{{a\sqrt 3 }}{4}$
Where,
r is the atomic radius.
a is the edge length of the unit cell.
Substitute $316.5\,{\text{pm}}$ for the edge length of the unit cell.
$r\, = \dfrac{{{\text{316}}{\text{.5}}\,\,{\text{pm}} \times \sqrt 3 }}{4}$
$r\, = 137.05\,{\text{pm}}$
So, the atomic radius of the body-centered cubic unit cell of the tungsten atom is $137.05\,{\text{pm}}$.

Therefore, the atomic radius of the tungsten atom is $137.05\,{\text{pm}}$.

Additional Information: The formula of radius depends upon the type of unit cell. In face-centered cubic lattice, the diagonal edge length is $a\sqrt 2 $ so, the radius is the one-fourth of the edge length $a\sqrt 2 $ . The formula to determine the radius of face-centered cubic lattice is as follows:
$r\, = \dfrac{{a\sqrt 2 }}{4}$
In a simple cubic unit cell, the edge length is $a$ and the relation between edge length and atomic radius is $a\, = \,2r$.

Note: The atomic radius is one-fourth of the edge length for the BCC unit cell. The edge length depends upon the type of unit cell, so the relationship between edge length and atomic radius also depends upon the type of unit cell. In the different unit cells, the arrangement of atoms is different. So, the edge length is different.