BCC unit cell: lattice constant and packing

In a body centred cubic (BCC) unit cell there is an atom at each of the 8 corners and one atom at the centre of the cube. The atoms touch each other along the body diagonal.

        o-----------o
       /|          /|
      / |         / |      o = corner atom
     o-----------o  |      @ = body-centre atom
     |  |   @    |  |
     |  o--------|--o      atoms touch along
     | /         | /       the body diagonal
     |/          |/
     o-----------o

Let $R$ = atomic radius and $a$ = lattice constant (edge of the cube).

Lattice constant

Along the body diagonal there is one corner atom radius, one full centre atom diameter and another corner radius:

$$\begin{aligned} \text{Body diagonal} &= \sqrt{3}\,a = R + 2R + R = 4R \\ a &= \frac{4R}{\sqrt{3}} = 2.309\,R \end{aligned}$$

Face diagonal

$$\text{Face diagonal} = \sqrt{2}\,a = \sqrt{2}\times\frac{4R}{\sqrt{3}} = 4\sqrt{\frac{2}{3}}\,R = 3.266\,R$$

Body diagonal

$$\text{Body diagonal} = \sqrt{3}\,a = 4R$$

Packing density

Number of atoms per cell: $8\times\frac{1}{8} + 1 = 2$.

$$\begin{aligned} \text{APF} &= \frac{\text{volume of atoms in cell}}{\text{volume of cell}} = \frac{2\times\frac{4}{3}\pi R^3}{a^3} \\ &= \frac{\frac{8}{3}\pi R^3}{\left(\frac{4R}{\sqrt{3}}\right)^3} = \frac{\frac{8}{3}\pi R^3\cdot3\sqrt{3}}{64R^3} = \frac{\sqrt{3}\,\pi}{8} \\ &= 0.68 \end{aligned}$$

Answer: $a = 4R/\sqrt{3} = 2.309R$; face diagonal $= 3.266R$; body diagonal $= 4R$; packing density $= \sqrt{3}\pi/8 = 0.68$ (68%). Coordination number is 8. Examples: Fe ($\alpha$), Cr, Na, W.