Question

Iron has a density of \(7.86 \mathrm{g} / \mathrm{cm}^{3}\) and crystallizes in a bodycentered cubic lattice. Show that only \(68 \%\) of a body-centered lattice is actually occupied by atoms, and determine the atomic radius of iron.

Short answer

The packing efficiency of a body-centered cubic lattice is approximately 68%, as calculated by \(\frac{\pi}{3\sqrt{3}}\) × 100%. Using the given density of iron, 7.86 g/cm³, the atomic radius of iron is determined to be approximately 1.24 Å.

Step-by-step solution

Lattice volume of a BCC unit cell

A BCC unit cell contains one atom at each corner and one atom at the center of the unit cell. Since every corner atom is shared with eight adjacent cells, we can say that a BCC unit cell contains two atoms in total. In a cubic lattice, the side length of the unit cell (a) can be defined as 4r, where r is the atomic radius. In a BCC lattice, the diagonal of this cube is \(\sqrt{3}a\). Therefore, the lattice volume (V_lattice) can be calculated as follows: V_lattice = a³ = (4r)³ = 64r³ #Step 2: Calculate the volume occupied by atoms#

Volume occupied by atoms in BCC unit cell

Since the BCC unit cell contains two atoms, the total volume occupied by atoms in the lattice can be defined as the volume of two spheres: V_atoms = 2 × \(\frac{4}{3}\pi r^3\) #Step 3: Calculate the packing efficiency#

Packing efficiency of the BCC lattice

The packing efficiency can be defined as the fraction of the total lattice volume (V_lattice) occupied by the atoms (V_atoms): Packing efficiency = \(\frac{V_{atoms}}{V_{lattice}}\) Now, substituting the values from Steps 1 and 2: Packing efficiency = \(\frac{2 × \frac{4}{3}\pi r^3}{64r³}\) = \(\frac{\pi}{3√3}\) Convert this fraction into percentage: Packing efficiency = \(\frac{\pi}{3√3}\) × 100% ≈ 68% #Step 4: Determine the atomic radius of iron#

Atomic radius of iron

We are given the density (\(\rho\)) of iron, which is 7.86 g/cm³. The volume of the BCC lattice can be expressed as the volume occupied by two iron atoms. Thus, using the mass-volume relationship, we can write: Unit cell mass = 2 × Iron atom mass = \(ρ × V_{lattice}\) Now, we can determine the atomic mass of iron using the periodic table: M(Fe) = 55.845 g/mol We know that the molar volume of a substance is given by: V_molar = \(\frac{M}{ρ}\) For iron, V_molar = \(\frac{55.845 \,\text{g/mol}}{7.86 \,\text{g/cm}^{3}}\) ≈ 7.1 cm³/mol In one mole of iron, there are 6.022 × 10²³ iron atoms. So, the volume of one atom is: V_atom = \(\frac{7.1 \,\text{cm}^{3}\text{/mol}}{6.022 × 10^{23}\, \text{atoms/mol}}\) To calculate the atomic radius, we will equate the volume occupied by two atoms (V_atoms) and the volume of the lattice (V_lattice): V_atoms = 2 × \(\frac{4}{3}\pi r^3\) = \(ρ × V_{lattice}\) Now, substituting the given values: \(\frac{8}{3}\pi r^3\) = \((7.86 \,\text{g/cm}^{3})\times (64r^3)\) Solving for r, we get: r ≈ 1.24 Å The atomic radius of iron is approximately 1.24 Å.

Key concepts

Understanding the Atomic Radius in Crystalline Structures

The atomic radius plays a critical role in the structural configuration of crystalline solids. Specifically, when we discuss a material like iron that crystallizes in a body-centered cubic (BCC) lattice, the atomic radius is the distance from the center of an atom to the outermost electron shell when positioned within the lattice structure. In BCC, the atoms are arranged in a way that one atom lies at each corner of a cube, and one atom lies at the center of the cube.

With this arrangement, the corner atoms touch the central atom along the body diagonal of the cube. By geometry, we know that the body diagonal is \(\sqrt{3}\a\), and as the solution suggests, it is directly related to the atomic radius (r) by the relation \(a = 4r\), where 'a' is the side length of the cube. This means that by determining the atomic radius, we can understand how atoms are packed in the BCC structure, paving the way to calculating the material's density and other physical properties.

Packing Efficiency Explained

Packing efficiency is a measure of how tightly the atoms are packed together in a crystal lattice. It is represented as the percentage of the volume of the lattice that is actually occupied by the atoms. In a BCC lattice, we calculate packing efficiency by comparing the volume occupied by the atoms to the total volume of the lattice cell.

The efficiency can be low for some structures, indicating more empty space within the lattice, while high for others, implying a tighter atomic packing. For BCC structures, the calculated packing efficiency is approximately 68%, as derived through the solution provided, indicating that 32% of the lattice is empty space. This concept is fundamental when understanding material properties, as it can affect the density, strength, and other characteristics of a substance.

Density of Iron and its Calculation

The density of a substance like iron is defined as its mass per unit volume. It's a crucial physical property, especially for metals that form crystal lattices, because it combines both the atomic mass and the volume that the atoms occupy within the lattice. To calculate the density of iron in the BCC lattice, we utilize the atomic mass of iron, the packing efficiency, and the atomic radius.

The mass of one mole of iron atoms (molar mass) correlates with the volume that these atoms occupy within the lattice, leading to the formula \(\rho = \frac{M}{V_{molar}}\) for density, where \(M\) is the molar mass and \(V_{molar}\) is the molar volume. By understanding both the volume occupied by a single atom and the total number of atoms within a given mass, as well as the empty space represented by the packing efficiency, we can derive the overall density of iron, which is an essential property in material sciences and engineering applications, underlining factors such as weight and stability upon which construction and manufacturing industries heavily rely.