Question

Calculate the volume in \(\AA^{3}\) of each of the following types of cubic unit cells if it is composed of atoms with an atomic radius of \(182 \mathrm{pm}\). (a) primitive (b) face-centered cubic.

Short answer

The volume of the given cubic unit cells with an atomic radius of \(182 \mathrm{pm}\) is: (a) For primitive unit cell: \(V = 48.30 \mathring{A}^{3}\) (b) For face-centered cubic unit cell: \(V = 100.53 \mathring{A}^{3}\)

Step-by-step solution

(a) Primitive Unit Cell)

In a primitive cubic unit cell, there is one atom at each corner of the cube. The atoms are touching along the edge of the cube. The atomic radius (r) is half the distance between the centers of two adjacent atoms. Therefore, the edge length (a) of the cube will be twice the atomic radius. Given atomic radius \(r = 182 \mathrm{pm}\). Calculate the edge length: \[a = 2r = 2(182 \mathrm{pm}) = 364 \mathrm{pm}\] Now, we can calculate the volume of the primitive unit cell (V): \[V = a^{3} = (364 \mathrm{pm})^{3} = 48,298,304 \mathrm{pm^{3}}\] As \(1 \mathring{A} = 100 \mathrm{pm}\), the volume in \(\mathring{A}^{3}\) is \[V = 48,298,304 \mathrm{pm^{3}} \times \left(\frac{1 \mathring{A}}{100 \mathrm{pm}}\right)^{3} = 48.30 \mathring{A}^{3}\]

(b) Face-Centered Cubic Unit Cell)

In a face-centered cubic (FCC) unit cell, there are corner atoms like the primitive unit cell and one atom located at the center of each face. The atoms touch along the face diagonals of the cube. Since we have a face diagonal, we can use Pythagorean theorem in the plane of the face to find the edge length. Let the edge length be 'a'. In a square, the face diagonal is \(\sqrt{2a^{2}}\). The face diagonal of the cube can be represented as: \[\sqrt{2a^{2}} = 4r\] Again, the atomic radius \(r = 182 \mathrm{pm}\). Calculate the edge length: \[a = \frac{1}{\sqrt{2}} \times 4r = \frac{1}{\sqrt{2}} \times 4(182 \mathrm{pm}) = 364\sqrt{2} \mathrm{pm}\] Now, we can calculate the volume of the FCC unit cell (V): \[V = a^{3} = (364\sqrt{2} \mathrm{pm})^{3} = 100,530,148 \mathrm{pm^{3}}\] As \(1 \mathring{A} = 100 \mathrm{pm}\), the volume in \(\mathring{A}^{3}\) is \[V = 100,530,148 \mathrm{pm^{3}} \times \left(\frac{1 \mathring{A}}{100 \mathrm{pm}}\right)^{3} = 100.53 \mathring{A}^{3}\]

Key concepts

Atomic Radius

Understanding the concept of atomic radius is key in calculating properties of solid materials. Atomic radius is defined as half the distance between the nuclei of two adjacent atoms that are just touching each other. This concept is useful primarily in the realm of crystalline solids where atoms are arranged in repeating structures. In cubic unit cells, the atomic radius helps determine the arrangement and dimensions of these cells. For example, if you know the atomic radius, you can calculate the edge length of various types of cubic cells, which is vital for figuring out their volume. In our case study, we have an atomic radius of 182 pm (picometers), which serves as a cornerstone measurement for further calculations.

Primitive Unit Cell

The primitive unit cell represents the simplest form of a crystal lattice. It comprises atoms positioned only at the corners of the cube. In this structure, the edge length of the cube is equal to twice the atomic radius because the atoms are in contact along the cube's edges.

To calculate the volume of this type of unit cell, once the edge length is known, we cube the edge length. Given an atomic radius of 182 pm, the edge length becomes \( 2 \times 182 = 364 \) pm. The volume can be calculated as:

• \( V = a^3 = (364 \text{ pm})^3 \)

Converting it to \(\mathring{A}^3\), where 1 \(\mathring{A} = 100 \text{ pm}\), results in a volume of approximately 48.30 \(\mathring{A}^3\). This highlights the influence of atomic radius on unit cell dimensions.

Face-Centered Cubic

The face-centered cubic (FCC) unit cell is a slightly more complex structure than the primitive unit cell. Besides atoms at the corners, there is also an additional atom at the center of each face. This leads to a larger coordination number and a more compact arrangement.

In the FCC lattice, the atoms touch each other along the face diagonals rather than along the edges. Therefore, the calculation for the edge length involves a bit more geometry. Using the Pythagorean theorem, we find that the face diagonal is \( \sqrt{2a^2} \), which simplifies the relation with the atomic radius as \( 4r = \sqrt{2a^2} \). Solving for \( a \), we can express the edge length in terms of the atomic radius. With our given atomic radius of 182 pm, the edge length becomes \( 364\sqrt{2} \) pm, leading to a cell volume of approximately 100.53 \(\mathring{A}^3\).

Edge Length Calculation

Calculating the edge length is crucial for evaluating the volume of cubic unit cells. The approach varies slightly between primitive and face-centered cubic structures due to their different atomic arrangements.

For primitive cubic cells, the calculation is straightforward: the edge length is simply twice the atomic radius, as atoms are only touching each other along the edges. Therefore, with an atomic radius of 182 pm, the edge length is \( 364 \text{ pm} \).

For face-centered cubic cells, the process involves more geometry. Here, the atoms are in contact along the face diagonals, requiring us to apply the Pythagorean theorem to find the edge length. We derive \( a = \frac{1}{\sqrt{2}} \times 4r \), which for an atomic radius of 182 pm results in an edge length of \( 364\sqrt{2} \text{ pm} \). Knowing the edge length allows for the subsequent calculation of the cell's volume, which is crucial for understanding the material's physical properties.