Calculate the volume in \(\AA^{3}\) of each of the following types of cubic unit cells if it is composed of atoms with an atomic radius of \(182 \mathrm{pm}\). (a) primitive (b) face-centered cubic.

In a primitive cubic unit cell, there is one atom at each corner of the cube. The atoms are touching along the edge of the cube. The atomic radius (r) is half the distance between the centers of two adjacent atoms. Therefore, the edge length (a) of the cube will be twice the atomic radius. Given atomic radius \(r = 182 \mathrm{pm}\). Calculate the edge length: \[a = 2r = 2(182 \mathrm{pm}) = 364 \mathrm{pm}\] Now, we can calculate the volume of the primitive unit cell (V): \[V = a^{3} = (364 \mathrm{pm})^{3} = 48,298,304 \mathrm{pm^{3}}\] As \(1 \mathring{A} = 100 \mathrm{pm}\), the volume in \(\mathring{A}^{3}\) is \[V = 48,298,304 \mathrm{pm^{3}} \times \left(\frac{1 \mathring{A}}{100 \mathrm{pm}}\right)^{3} = 48.30 \mathring{A}^{3}\]

In a face-centered cubic (FCC) unit cell, there are corner atoms like the primitive unit cell and one atom located at the center of each face. The atoms touch along the face diagonals of the cube. Since we have a face diagonal, we can use Pythagorean theorem in the plane of the face to find the edge length. Let the edge length be 'a'. In a square, the face diagonal is \(\sqrt{2a^{2}}\). The face diagonal of the cube can be represented as: \[\sqrt{2a^{2}} = 4r\] Again, the atomic radius \(r = 182 \mathrm{pm}\). Calculate the edge length: \[a = \frac{1}{\sqrt{2}} \times 4r = \frac{1}{\sqrt{2}} \times 4(182 \mathrm{pm}) = 364\sqrt{2} \mathrm{pm}\] Now, we can calculate the volume of the FCC unit cell (V): \[V = a^{3} = (364\sqrt{2} \mathrm{pm})^{3} = 100,530,148 \mathrm{pm^{3}}\] As \(1 \mathring{A} = 100 \mathrm{pm}\), the volume in \(\mathring{A}^{3}\) is \[V = 100,530,148 \mathrm{pm^{3}} \times \left(\frac{1 \mathring{A}}{100 \mathrm{pm}}\right)^{3} = 100.53 \mathring{A}^{3}\]