Question

Gold adopts a face-centered cubic structure with a unit cell edge of \(4.08 \AA\) (Figure 12.11 ). How many gold atoms are there in a sphere that is \(20 \mathrm{nm}\) in diameter? Recall that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\).

Short answer

Approximately 246,716 gold atoms are present in the spherical gold structure with a diameter of 20 nm.

Step-by-step solution

Convert Diameter to Radius and Units

Since the diameter of the sphere is given in nm, and the unit cell edge is given in Å, let's first convert the diameter to a radius and then convert the units to Å (1 nm = 10 Å). Radius (r) = Diameter/2 = 20 nm/2 = 10 nm Now, convert the radius to Å: Radius (r) = 10 nm * 10 Å/nm = 100 Å

Calculate Volume of the Sphere

Using the formula for the volume of a sphere, V_sphere = (4/3) * π * r³: V_sphere = (4/3) * π * (100 Å)³ = (4/3) * π * (1,000,000 ų) ≈ 4,188,790 ų

Calculate Volume of the Unit Cell

The unit cell of gold has a cubic shape with an edge length of 4.08 Å. The volume of a cube is given by V_cube = a³, where a is the length of the edge. V_unit_cell = (4.08 Å)³ ≈ 67.9 ų

Determine the Number of Unit Cells in the Sphere

Now that we have the volume of the gold sphere and the unit cell, we can find the number of unit cells that fit inside the sphere: Number of unit cells = V_sphere / V_unit_cell = 4,188,790 ų / 67.9 ų ≈ 61,679 unit cells Since we'll need the number of gold atoms inside the unit cells, we should round this value to the nearest whole number. Therefore, there are approximately 61,679 unit cells inside the gold sphere.

Calculate the Number of Gold Atoms per Unit Cell

For a face-centered cubic structure, each corner atom is shared by 8 unit cells, and each face atom is shared by 2 unit cells. Each unit cell contains: - 1/8 of each of the 8 corner atoms - 1/2 of each of the 6 face atoms Number of atoms per unit cell = (8 * 1/8) + (6 * 1/2) = 1 + 3 = 4 atoms

Calculate the Number of Gold Atoms in the Sphere

Finally, we can find the total number of gold atoms in the sphere by multiplying the number of unit cells by the number of atoms per unit cell: Number of gold atoms = 61,679 unit cells * 4 atoms/unit cell ≈ 246,716 gold atoms Approximately 246,716 gold atoms are present in the spherical gold structure.

Key concepts

Crystal Lattice

When studying materials at the atomic level, one fundamental concept is the crystal lattice. Think of it as a three-dimensional grid that extends in all directions, made up of repeated units called unit cells. These unit cells are the smallest distinct section that, by repeating over and over, create the entire crystal of a material. Imagine a brick wall, where each brick is identical and placed in a specific pattern; similarly, in a crystal lattice, each 'brick' or unit cell contains a particular arrangement of atoms or molecules.

A crystal lattice isn't just a useful theoretical construct; it actually represents the regular, repeating pattern in which atoms are arranged in solid materials. The arrangement within the unit cells determines many of the material's properties, including melting point, hardness, and density. For various elements and compounds, these unit cells can have different shapes, sizes, and symmetries, leading to a diversity of crystal structures found in nature.

Face-Centered Cubic Structure

One important type of unit cell is the face-centered cubic (FCC) structure. This structure is one where atoms are located at each of the corners of a cube, as well as at the centers of each face of the cube. Because of this arrangement, the face-centered cubic structure is extremely efficient in filling space, which is why many metals, including gold, silver, and aluminum, crystallize in this form.

In an FCC crystal, atoms on the corners are shared with adjacent unit cells and hence are not 'fully' part of one single unit cell. The calculations account for this by considering that only 1/8th of each corner atom is within a specific unit cell. Similarly, an atom on a face is shared with one other unit cell, meaning only half of it is within one unit cell. Therefore, the contribution of these atoms to a single unit cell needs to be accounted for differently – which is where students often need clarification. It is the understanding of this structure that helps in determining the number of atoms in a substance and predicting physical properties.

Volume Calculation

The ability to calculate the volume of different shapes is an essential skill in chemistry, especially when dealing with problems related to unit cells in a crystal lattice. The volume can tell us a lot about the substance we are studying, like how many atoms or unit cells can fit into a given space, or what the density of a substance might be.

For regular shapes like the unit cell in a face-centered cubic structure, which is a cube, the volume is simply calculated by cubing the edge length. However, for irregular shapes, such as spheres, one must use the appropriate mathematical formula, such as \( V_{sphere} = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. In our example, with a sphere of gold atoms, the volume determines how much space is taken up by the sphere and, consequently, how many unit cells fit into that space. This type of volume calculation is foundational to understanding more complex topics in materials science and solid-state physics.