Iron has a density of 7.86 \(\mathrm{g} / \mathrm{cm}^{3}\) and crystallizes in a body-centered 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 body-centered cubic (BCC) lattice, there are 8 atoms at the corners of the unit cell and 1 atom at the center of the unit cell. Each corner atom is shared by 8 adjacent unit cells, so the total contribution of corner atoms to the overall lattice count is equal to 8 * (1/8) = 1 atom. The center atom, clearly, contributes 1 atom to the lattice. Therefore, the total number of atoms in a BCC lattice is 1 + 1 = 2 atoms. A BCC lattice has a side length 'a'. As the BCC lattice has a central atom, we form a right-angled triangle with the side length 'a' and a diagonal, which is 4 times the atomic radius of the lattice element, i.e., 4r. Using Pythagorean theorem: \(a^2 + a^2 = (4r)^2\) We know that the volume of the unit cell is a^3. Therefore: \(a^3\)= \(\frac{(4r)^2}{2} * a\) The volume of 2 atoms in a BCC lattice is: \(Volume_{atoms}\) = \(2 * \frac{4}{3} * \pi * r^3\) Now we can find the packing efficiency as: \(Packing \thinspace Efficiency\)= \(\frac{Volume_{atoms}}{a^3}\) * 100

We know that the density of a substance can be given by the formula: \(Density\)= \(\frac{Mass}{Volume}\) For the BCC lattice of iron, we have: \(Density\)= \(\frac{2 * Molar \thinspace Mass_{Iron}}{a^{3} * Avogadro's \thinspace Number}\) Given the density of iron as 7.86 g/cm³, we can rewrite the equation as: \(7.86\)= \(\frac{2 * 55.847 \thinspace g/mol}{a^{3} * 6.022 \times 10^{23} atoms/mol}\) From Step 1, we know that: \(a^3\)= \(\frac{(4r)^2}{2} * a\) Substituting the value of a³, we get: \(7.86\)= \(\frac{2 * 55.847 \thinspace g/mol}{(\frac{(4r)^2}{2} * a) * 6.022 \times 10^{23} atoms/mol}\) Now substitute the value of a from Step 1: \(7.86\)= \(\frac{2 * 55.847 \thinspace g/mol}{(\frac{(4r)^2}{2} * (\sqrt{2}\times 4r) ) * 6.022 \times 10^{23} atoms/mol}\) Solving for the atomic radius 'r', we get: \(r \approx 1.24 \thinspace Å \) Now we can substitute this value in the Packing Efficiency formula we derived earlier: \(Packing \thinspace Efficiency \approx \frac{2 * \frac{4}{3} * \pi * (1.24)^3}{(\sqrt{2}\times 4* 1.24)^3}\) * 100 \(Packing \thinspace Efficiency \approx\) 68 % Thus, it's shown that only 68% of a body-centered lattice is actually occupied by atoms, and the atomic radius of iron is approximately 1.24 Å.