Rhodium crystallizes in a face-centered cubic unit cell that has an edge length of \(0.381 \mathrm{nm}\). (a) Calculate the atomic radius of a rhodium atom. (b) Calculate the density of rhodium metal.

The edge length of the unit cell is equal to the lattice constant, which we know to be 0.381 nm. We convert this value to meters: \[a = 0.381 * 10^{-9} \mathrm{m}\]

In an fcc unit cell, we can form an isosceles right triangle using the atomic radius (r), edge length (a), and two similar triangular faces along a diagonal in one of the faces. Using the Pythagorean Theorem, we can find the relationship between a and r for an fcc unit cell: \[a^2 + a^2 = (4r)^2\]

We will solve the relationship equation for r and then substitute the value of a that we have: \[r = \frac{a\sqrt{2}}{4}\] Now substitute the value of a, \[r = \frac{0.381 * 10^{-9}\sqrt{2}}{4} = 0.13493 * 10^{-9} \mathrm{m}\] Thus, the atomic radius of a rhodium atom is approximately 0.13493 nm.

To find the density of Rhodium, we will use the formula for density: \[\rho = \frac{mass}{volume}\] Volume of the Rhodium unit cell can be calculated by \(V = a^3\). Furthermore, there are four Rhodium atoms per fcc unit cell and there are \(N_A = 6.022 * 10^{23}\) atoms per mole. The molar mass of Rhodium is \(M_R = 102.91 \frac{g}{mol}\). We can calculate the mass of the Rhodium atoms in the unit cell using the molar mass and the number of atoms. First, let's calculate the volume of one Rhodium unit cell: \[V = (0.381*10^{-9})^3 = 5.5289*10^{-29}\,m^3\] Now, find the mass of four Rhodium atoms: \[mass = \frac{4 \,atoms * 102.91\frac{g}{mol}}{6.022 * 10^{23}\,atoms/mol} = 6.8473 * 10^{-22}\,g\] Now we can calculate the density: \[\rho = \frac{6.8473 * 10^{-22} \,g}{5.5289*10^{-29}\,m^3} = 12.40 \frac{g}{cm^3}\] The density of Rhodium metal is approximately 12.40 g/cm³.