91Ó°ÊÓ

First, we need to find the volume of one atom. We know that atoms in HCP structures can be approximated as closely packed spheres. The volume V of a sphere is given by the formula: \(V_\textrm{sphere} = \frac{4}{3}\pi r^3\) where \(r\) is the atomic radius.

Next, we need to know how many atoms are in the HCP unit cell. In an HCP structure, there are: - 6 atoms making up the hexagonal faces, shared by 6 adjacent unit cells, so 1 atom per unit cell - 12 atoms shared by 2 unit cells, so 6 atoms per unit cell - 3 atoms wholly contained in the unit cell, not shared by any other unit cell So, there are a total of \(1+6+3=10\) atoms per unit cell in HCP.

We now multiply the volume of one atom by the number of atoms in the unit cell to find the total volume occupied by atoms: \(V_\textrm{atoms} = 10V_\textrm{sphere} = 10\left(\frac{4}{3}\pi r^3\right)\)

Next, we will determine the volume V of the HCP unit cell. The height h and base edge length a are given in terms of the atomic radius: - h = \(2\sqrt{\frac{2}{3}}r\) - a = \(2r\) The area A of a regular hexagon with side-length a is: \(A_\textrm{hexagon} = \frac{3}{2}\cdot\sqrt{3}\cdot a^2 = \frac{3}{2}\cdot\sqrt{3}\cdot (2r)^2\) The volume V of the HCP unit cell is then the product of the area A of the hexagon and the height h: \(V_\textrm{unit cell} = A_\textrm{hexagon}\times h = \frac{3}{2}\cdot\sqrt{3}\cdot (2r)^2\times 2\sqrt{\frac{2}{3}}r\)

Now we can find the APF by dividing the total volume occupied by atoms in the unit cell \(V_\textrm{atoms}\) by the volume of the unit cell \(V_\textrm{unit cell}\): \(\textrm{APF} = \frac{V_\textrm{atoms}}{V_\textrm{unit cell}}\) \(\textrm{APF} = \frac{10\left(\frac{4}{3}\pi r^3\right)}{\frac{3}{2}\cdot\sqrt{3}\cdot (2r)^2\times 2\sqrt{\frac{2}{3}}r}\) After simplifying this expression, we get: \(\textrm{APF} = \frac{10\left(\frac{4}{3}\pi r^3\right)}{12\sqrt{2}\pi r^3}\) \(\textrm{APF} = \frac{\cancel{10\left(\frac{4}{\cancel{3}}\pi \cancel{r^3}\right)}}{\cancel{12{}_{=4\times3}}\cancel{4}\sqrt{\cancel{2}}\cancel{\pi} \cancel{r^3}}}\) \(\textrm{APF} = \frac{5}{6} \approx 0.74\) Therefore, the atomic packing factor for HCP is \(\boxed{0.74}\).