Question

Iridium crystallizes in a face-centered cubic unit cell that has an edge length of \(3.833 \AA\). The atom in the center of the face is in contact with the corner atoms, as shown in the drawing. (a) Calculate the atomic radius of an iridium atom. (b) Calculate the density of iridium metal.

Short answer

The atomic radius of an iridium atom is \(1.358 \AA\) and the density of iridium metal is \(22.4 g/cm^3\).

Step-by-step solution

Calculate the atomic radius

In a face-centered cubic unit cell, the diagonal of a face can be expressed in terms of the atomic radius and the edge length. The relation is: \(\sqrt{2} \times edge\,length = 4 \times atomic\,radius\) Given the edge length of iridium \(3.833 \AA\), we can calculate the atomic radius: \(atomic\,radius = \frac{\sqrt{2} \times edge\,length}{4}\) \(atomic\,radius = \frac{\sqrt{2} \times 3.833}{4}\) \(atomic\,radius = 1.358 \AA\)

Calculate the number of atoms per unit cell

In a face-centered cubic unit cell, there are: - 8 corner atoms, each contributing 1/8 to the unit cell - 6 face-centered atoms, each contributing 1/2 to the unit cell Therefore, the total number of atoms per unit cell is: \(n_{atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4\)

Calculate the density

To calculate the density, we will use the following formula: \(density = \frac{n_{atoms} \times molar\,mass \times N_{A}}{edge\,length^3}\) Where: - \(n_{atoms}\) is the number of atoms per unit cell - \(molar\,mass\) is the molar mass of iridium (\(192.22 g/mol\)) - \(N_{A}\) is Avogadro's number (\(6.022 \times 10^{23} mol^{-1}\)) - \(edge\,length\) is given in cm (convert from Å to cm) First, convert the edge length from Å to cm: \(edge\,length = 3.833 \times 10^{-8} cm\) Now, calculate the density: \(density = \frac{4 \times 192.22 \times 6.022 \times 10^{23}}{(3.833 \times 10^{-8})^3}\) \(density = 22.4 g/cm^3\) The atomic radius of an iridium atom is \(1.358 \AA\), and the density of iridium metal is \(22.4 g/cm^3\)

Key concepts

Understanding Atomic Radius in Face-Centered Cubic Crystals

The atomic radius is a critical factor in the crystal structure of a metal. In a face-centered cubic (fcc) unit cell, like that of iridium, the atoms are arranged in a way that optimizes space and is prevalent in the structure of many metals. To calculate the atomic radius, we rely on the geometrical arrangement of these atoms.

In the case of iridium, this calculation hinges on the relationship between the face diagonal and the atomic radii since the atoms touch along the face diagonal. Mathematically, this is expressed as \(\sqrt{2} \times edge\,length = 4 \times atomic\,radius\). By rearranging the equation, students can find the atomic radius. This concept's understanding is crucial for students exploring solid-state physics or chemistry, as it provides insights into the molecular scale that defines the properties of materials.

Calculating Density of Metals with Crystal Lattice Considerations

Density is an essential property of materials and understanding how to calculate it in metals with a crystalline structure is fundamental in material science. The density of a metal like iridium is calculated by taking into account the molar mass, the number of atoms within a unit cell, and the cell's volume.

To determine the density, we use the formula \(density = \frac{n_{atoms} \times molar\,mass \times N_{A}}{edge\,length^3}\), where \(N_{A}\) is Avogadro's number. This illustrates how macroscopic properties (like density) are deeply connected to the microscopic world, emphasizing the importance of unit cell structure in determining physical characteristics. Knowing how to perform this calculation is crucial for any student who wishes to progress in fields like metallurgy, chemistry, or material engineering.

Exploring Crystal Lattice Structures

The crystal lattice structure determines many of a material's properties, including its density, melting point, and how it interacts with light. The fcc lattice is a tight, efficient packing arrangement where each corner atom is shared by eight unit cells, and each face-centered atom is shared by two. This structure impacts how we calculate the number of atoms per unit cell and subsequently properties such as density.

In the fcc lattice, the contribution of corner atoms multiplied by eight and face-centered atoms multiplied by six yields a total of four atoms per unit cell. This intricate yet orderly arrangement bestows fcc metals with distinctive mechanical and thermal characteristics, making the study of crystal lattices intriguing and significant for students in physics, materials science, and engineering.