Iron has a density of \(7.86 \mathrm{~g} / \mathrm{cm}^{3}\) and crystallizes in a bodycentered cubic lattice. Show that only \(68 \%\) of a body-centered lattice is actually occupied by atoms, and determine the atomic radius of iron.

In a BCC lattice, the atoms are located at the corners of the cubic unit cell and at the center of the cube. To calculate the volume of the unit cell, we can use the formula: \[V_{\text{cell}} = a^3\] where \(a\) is the lattice parameter (length of one edge of the cubic unit cell). Since we don't know the value of \(a\) yet, we will rewrite the volume of the unit cell in terms of the atomic radius (\(r\)) of iron. In a BCC lattice, the length of a body diagonal is equal to \(\sqrt{3}a\), and it also passes through two atoms (one at a cube corner and one at the center) whose radii add up to \(\sqrt{3}a\): \[\sqrt{3}a = 4r\] Now we can rewrite the volume of the unit cell as: \[V_{\text{cell}} = \left(\frac{4r}{\sqrt{3}}\right)^3\]

The volume of a single atom can be calculated using the formula for the volume of a sphere: \[V_{\text{atom}} = \frac{4}{3}\pi r^3\] In a BCC lattice, there are 2 atoms per unit cell (1 from the cube corners shared by 8 unit cells and 1 located at the center). Therefore, the volume occupied by atoms in a unit cell is: \[V_{\text{atoms}} = 2 \times V_{\text{atom}} = 2 \times \frac{4}{3}\pi r^3\]

To calculate the percentage of volume occupied by atoms in the unit cell, we will divide the volume of atoms by the volume of the unit cell: \[\text{% occupied} = \frac{V_{\text{atoms}}}{V_{\text{cell}}} \times 100\] Substituting the expressions for \(V_{\text{atoms}}\) and \(V_{\text{cell}}\) from Steps 1 and 2, we get: \[\text{% occupied} = \frac{2 \times \frac{4}{3}\pi r^3}{\left(\frac{4r}{\sqrt{3}}\right)^3} \times 100\] Simplifying this expression results in: \[\text{% occupied} = \frac{3\pi}{4\sqrt{3}}\times 100 \approx 68\%\] So, only 68% of a body-centered lattice is actually occupied by atoms.

Now we will determine the atomic radius of iron using the given density and the lattice parameter of the BCC lattice. First, we can rewrite the mass of the unit cell in terms of the density and volume: \[\text{mass of unit cell} = \rho \times V_{\text{cell}}\] where \(\rho\) is the density of the substance. Since there are 2 atoms in each unit cell of a BCC lattice, we can rewrite the atomic mass of iron (\(M_{\text{Fe}}\)) as: \[M_{\text{Fe}} = \frac{\text{mass of unit cell}}{2}\] Now, we will find the atomic radius \(r\) from the density and lattice parameter: \[7.86 \mathrm{~g}/\mathrm{cm}^{3} \times \left(\frac{4r}{\sqrt{3}}\right)^{3} = 2 \times M_{\text{Fe}}\] Finally, we can solve for the atomic radius \(r\): \[r = \frac{\sqrt{3}}{16\pi} \times \frac{2 \times M_{\text{Fe}}}{7.86 \mathrm{~g}/\mathrm{cm}^{3}}\] By substituting the value of \(M_{\text{Fe}}\) (the atomic mass of iron) into the equation, the atomic radius of iron can be calculated.