 Solution

Interpret the sphere as an atom of iron, and assume neighbouring atoms touch without overlap. Its radius is $r=0.124\mathrm{nm}=1.24\times 10^{-10}\mathrm{m}$, and its mass is $m_{\mathrm{Fe}}=56u\simeq 9.30\times 10^{-26}\mathrm{kg}$, using atomic mass unit $u=1.66054\times 10^{-27}\mathrm{kg}$.

In the face-centered cubic lattice, atoms touch along a face diagonal: $\sqrt{2}{a}_F=4r$, so $a_F=2\sqrt{2}r=0.3507\mathrm{nm}$. Four atoms occupy a conventional cube. The hard-sphere atomic volume in cubic lattices therefore givesIn the body-centered cubic lattice, atoms touch along the body diagonal: $\sqrt{3}{a}_B=4r$, so $a_B=4r/\sqrt{3}=0.2864\mathrm{nm}$. There are two atoms per conventional cube, givingThese are primitive unit cell volumes, not the volumes of the conventional cubes or of the atomic spheres. The denser FCC contact packing gives the higher mass density in this fixed-radius model. Actual hot phases have temperature-dependent dimensions and bonding, so these numbers should not be treated as exact experimental phase densities.