Packing Efficiency in BCC Structures

In BCC structure, the atom at the centre is in touch with other two atoms which are diagonally arranged as shown in figure.
In △EFD,
b² = a² + a² = 2a²
or, b = √2 a

In △AFD,
c² = a² + b² = a² + 2a² = 3a²
or, c = √3 a

The length of the body diagonal c is equal to 4r, r is the radius of the sphere (atom). As all the three spheres along the diagonal touch each other,
so, c = 4r
or, c = 4r = √3 a
or, a = 4r/√3
or, r = √3 a/4

We know that the total number of atoms associated with BCC unit cell is 2 , so, the volume (v) is
v = 2 × (4/3) πr³ = 8/3 πr³
And the volume (V) of the unit cell = a³ = (4r/√3)³ = 64r³/3√3

Now, the packing efficiency = (100 × v)/V = [(8/3 πr³)/(64/3√3) × r³ ] × 100
= (√3 π × 100)/8 = 68%

Therefore, 68% of unit cell is occupied by atoms and the rest 32% is empty space.

---

Also read...

Packing Efficiency in Simple Cubic Structures

Packing Efficiency in CCP and HCP Structures