Problem 84 - Chemistry A Molecular Approach

Question

When spheres of radius \(r\) are packed in a body-centered cubic arrangement, they occupy \(68.0 \%\) of the available volume. Use the fraction of occupied volume to calculate the value of \(a,\) the length of the edge of the cube, in terms of \(r\).

Step-by-Step Solution

Answer

The edge length \(a\) of the cube in a body-centered cubic arrangement can be calculated in terms of the sphere's radius \(r\) as: \(a = r \sqrt[3]{\frac{\frac{8}{3}\pi}{0.68}}\).

Step 1: Understanding the body-centered cubic arrangement

A body-centered cubic (BCC) unit cell has a sphere at each of the eight corners of a cube and one sphere in the center. Each corner sphere is shared among eight unit cells, and the central sphere belongs entirely to one unit cell.

Step 2: Calculating the total volume occupied by spheres

The volume of one sphere with radius r is given by the formula for the volume of a sphere: \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). In a BCC unit cell, 1/8 of each of the 8 corner spheres and the whole central sphere contribute to the unit cell volume, making the total volume contributed by spheres \(V_{\text{total spheres}} = 8 \cdot \frac{1}{8}V_{\text{sphere}} + 1\cdot V_{\text{sphere}} = 2V_{\text{sphere}} = \frac{8}{3}\pi r^3\).

Step 3: Calculating the volume of the cube

Since spheres occupy 68.0% of the volume of the cube in a BCC arrangement, the volume of the cube \(V_{\text{cube}}\) can be calculated using the occupied volume fraction: \(V_{\text{cube}} = \frac{V_{\text{total spheres}}}{0.68} = \frac{\frac{8}{3}\pi r^3}{0.68}\).

Step 4: Determining the edge length of the cube

The volume of a cube with edge length \(a\) is \(V_{\text{cube}} = a^3\). Setting this equal to the volume found in the previous step, we get \(a^3 = \frac{\frac{8}{3}\pi r^3}{0.68}\). To solve for \(a\), take the cube root of both sides of the equation: \(a = \sqrt[3]{\frac{\frac{8}{3}\pi r^3}{0.68}}\).

Step 5: Simplifying the expression for the cube's edge length

To simplify the expression for \(a\), first cancel \(r^3\) in the numerator and the cube root, then take the constant terms outside the cube root: \(a = r \sqrt[3]{\frac{\frac{8}{3}\pi}{0.68}}\). This gives the edge length \(a\) of the cube as a function of the sphere's radius \(r\).

Key Concepts

Unit Cell Geometry

Understanding the geometry of a unit cell is fundamental when diving into the microscopic world of crystal structures. A unit cell is the smallest repeating unit in a crystal lattice - think of it as the basic building block from which the entire lattice is constructed. In a body-centered cubic (BCC) unit cell, we see a distinctive arrangement: there are spheres (atoms or molecules) at each of the cube's eight corners, as well as a single sphere positioned smack dab in the center.

This arrangement is important because it affects the unit cell's packing efficiency – how tightly the spheres are packed within the volume of the cube. As we delve into calculations, we see that while these corner spheres seem to belong to one unit cell, they are actually shared between eight adjacent unit cells, so only 1/8th of each corner sphere's volume counts towards the volume of a single unit cell. The central sphere, however, is entirely within the unit cell and is not shared.

Through this understanding, one can begin to visualize how crystal structures are formed and how the dimensions of the unit cell relate to the properties of the substance. When you're tackling problems involving unit cells, it's always helpful to sketch the 3D structure to aid in visualizing the arrangement and its geometry.

Volume of a Sphere

Let's take a closer look at the volume of a sphere, since this is an integral part of calculations involving particles in different unit cells. The formula for the volume of a sphere, \( V_{\text{sphere}} = \frac{4}{3}\pi r^3 \), is derived from the sphere's radius, represented by \( r \). This equation tells us that the volume increases rapidly as the radius grows – it's a cubic relationship, so even small changes in radius have a big impact on volume.

Why is this relevant in our BCC unit cell problem? Well, because we need to calculate how much of the unit cell's space these spheres actually take up. We're given that spheres pack to occupy 68% of the cell's volume, and by understanding the volume formula, we can use it to work out important features of the unit cell, like the edge length of the cubic arrangement.

It's useful to memorize this formula and understand its components as it will recur in a multitude of scientific and mathematical contexts, from calculating the charge distribution in atoms to determining the amount of liquid a spherical tank can hold.

Crystal Lattice Structures

Crystal lattice structures are fascinating and varied, with each type of structure offering unique properties that are essential to the nature of the material. In solid-state physics and materials science, crystal lattices are categorized based on the geometrical arrangement of atoms, ions, or molecules. The body-centered cubic (BCC) structure is one of several types of unit cell geometries, along with face-centered cubic (FCC) and hexagonal close-packed (HCP), among others.

In BCC structures, the atoms are positioned in a way that each central atom is equidistant from its neighboring corner atoms, forming a cube. This arrangement is not as densely packed as FCC structures, which is why it occupies 68% of the available volume with spheres, contrasting with FCC's higher packing efficiency. Each type of crystal lattice has its own implications in terms of the material's density, melting point, and other physical properties. Understanding these structures assists scientists and engineers not only in identifying materials, but also in designing new materials with desired properties for applications in technology and manufacturing. When working with these structures, always consider how the atomic arrangement impacts the material's broader characteristics.