Answer:
8.60 g/cm³
Explanation:
In the lattice structure of iron, there are two atoms per unit cell. So:
[tex]\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }[/tex] where [tex]V_{molar} = \frac{A}{\rho }[/tex] an and A is the atomic mass of iron.
Therefore:
[tex]\frac{2}{a^{3} } = \frac{N_{A} * p }{A}[/tex]
This implies that:
[tex]A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }[/tex]
= [tex]\frac{4}{\sqrt{3} }r[/tex]
Assuming that there is no phase change gives:
[tex]\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }[/tex]
= 8.60 g/m³
