Show that the atomic packing factor for BCC is 0.68.

In a body-centered cubic structure, the atoms are located at the corners and the center of the unit cell. The diagonal of the unit cell passes through the centers of the corner atom, the central atom, and the other corner atom. So, the length of the diagonal is equal to 4 times the atomic radius, and we get the equation: \[Diagonal=4r\] The diagonal can also be found using the Pythagorean theorem in three dimensions. The diagonal connects one corner of the unit cell to the opposite corner, so its length is given by: \[Diagonal = \sqrt{a^2 + a^2 + a^2}\] Now we can equate these two expressions for the diagonal to find the relationship between the lattice parameter (a) and the atomic radius (r): \[4r = \sqrt{a^2 + a^2 + a^2}\]

In a BCC unit cell, there are two atoms present. One atom is present at the center, and the other is the combined contribution from the 1/8th portions of the 8 corner atoms. The volume occupied by a single atom is given by the volume of a sphere: \[V_{atom} = \frac{4}{3}\pi r^3\] The total volume occupied by the atoms in the unit cell is the sum of the volumes occupied by all the 2 atoms: \[V_{occupied} = 2 \times V_{atom} = 2 \times \frac{4}{3}\pi r^3\]

The total volume of the BCC unit cell is given by the cube of its lattice parameter (a): \[V_{total} = a^3\] From Step 1, we have the relationship: \[4r = \sqrt{a^2 + a^2 + a^2}\] \[4r = \sqrt{3a^2}\] Now, we want to find \(a^3\), so we start by finding \(a^2\): \[(4r)^2 = 3a^2\] \[a^2 = \frac{16r^2}{3}\] Finally, we find \(a^3\): \[a^3 = \left(\frac{16r^2}{3}\right)^{\frac{3}{2}}= \frac{64r^6}{27}\] So, the total volume of the unit cell is: \[V_{total} = a^3 = \frac{64r^6}{27}\]

The atomic packing factor (APF) is the ratio of the volume occupied by atoms in the unit cell to the total volume of the unit cell. Using our results from Steps 2 and 3, we calculate the APF: \[APF = \frac{V_{occupied}}{V_{total}} = \frac{2 \times \frac{4}{3}\pi r^3}{\frac{64r^6}{27}} = \frac{8\pi r^3}{64r^6 / 27}\] \[APF = \frac{8\pi r^3 \times 27}{64r^6} = \frac{216\pi r^3}{64r^6} = \frac{27\pi}{32r^3}\] Now we can simplify the expression for APF: \[APF = \frac{27\pi}{32r^3}\times \frac{r^3}{r^3} = \frac{27\pi}{32} \approx 0.68\] Hence, the atomic packing factor for BCC is approximately 0.68.