Problem 38 - Chemistry The Central Science In Si Units

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 \((\mathbf{b})\) face-centered cubic.

Step-by-Step Solution

Answer

48.21 \(\AA^3\) for primitive, 135.44 \(\AA^3\) for face-centered cubic.

Step 1: Understand the Problem

We are asked to calculate the volume of cubic unit cells composed of atoms with an atomic radius of \(182 \text{ pm}\). We need to find the volume for two types of cubic unit cells: primitive and face-centered cubic.

Step 2: Convert Units

Convert the atomic radius from picometers to angstroms since the volume is required in \(\AA^3\). \(1 \text{ pm} = 0.01 \text{ Å}\), so \(182 \text{ pm} = 1.82 \text{ Å}\).

Step 3: Calculate Edge Length for Primitive Cell

For a primitive cubic unit cell, the atoms touch along the edge. Therefore, the edge length \(a\) is twice the atomic radius: \(a = 2 \times 1.82 = 3.64 \text{ Å}\).

Step 4: Calculate Volume of Primitive Cell

The volume \(V\) of a cubic unit cell with edge length \(a\) is given by \(V = a^3\). For the primitive cell, \(V = 3.64^3 \). Calculate this to find \(V = 48.21 \AA^3\).

Step 5: Calculate Edge Length for Face-Centered Cubic Cell

In a face-centered cubic (FCC) unit cell, the atoms touch along the face diagonal. The face diagonal is \(4\) times the atomic radius, giving the equation \(a\sqrt{2} = 4 \times 1.82\). Solve for \(a\) to get \(a = \frac{4 \times 1.82}{\sqrt{2}} \approx 5.14 \text{ Å}\).

Step 6: Calculate Volume of Face-Centered Cubic Cell

Using the edge length \(a\) for FCC, calculate the volume using \(V = a^3\). Thus, \(V = 5.14^3\). Calculate this to get \(V = 135.44 \AA^3\).

Key Concepts

Atomic Radius

The atomic radius is a fundamental attribute of an atom, defined as the distance from the nucleus to the outermost shell of an electron. It essentially gives us a measure of the size of an atom. In our exercise, the atomic radius is given as 182 picometers (pm). To effectively use this measurement for volume calculations, we need to convert it to angstroms (Å). Atomic radii are often expressed in different units, and knowing these conversions is crucial for accurate calculations.
For example, in our problem, 1 pm equals 0.01 Å, thus converting 182 pm to angstroms yields 1.82 Å. This conversion is essential because volume calculations for cubic cells are typically in cubic angstroms ().
Understanding the atomic radius helps predict how atoms pack in a unit cell, influencing the calculation of edge lengths for different cubic structures.

Primitive Cubic Cell

The primitive cubic cell is the simplest type of cubic lattice structure. In this arrangement, the atoms are located at each of the eight corners of the cube. The primitive cell contains only one atom per unit cell, as each corner atom is shared among eight neighboring unit cells.
For a primitive cubic cell, the atoms touch each other along the edge of the cube. Therefore, the edge length, denoted as 'a', is simply twice the atomic radius. In our case, with an atomic radius of 1.82 Å, the edge length is calculated as:
\[ a = 2 \times 1.82 \text{ Å} = 3.64 \text{ Å} \]Once the edge length is known, it can be used to compute the volume of the unit cell, following the cubic volume formula.

Face-Centered Cubic Cell

The face-centered cubic (FCC) unit cell is a more efficient packing structure compared to the primitive cubic cell. In an FCC lattice, atoms are located at each corner and in the center of each face of the cube. This arrangement results in four atoms per unit cell.
In the FCC structure, atoms touch along the face diagonal rather than along the edge. The face diagonal becomes key in calculating the edge length, using the relation:
\[ a\sqrt{2} = 4\times r \]where \(r\) is the atomic radius. Solving for \(a\), the formula is:
\[ a = \frac{4 \times 1.82}{\sqrt{2}} \approx 5.14 \text{ Å} \]This formula provides the edge length necessary for volume calculations.
The FCC packing pattern leads to a high density, influencing properties like strength and durability in materials.

Volume Calculation

Volume calculation is essential for determining the spatial occupancy of atoms within unit cells. In cubic unit cells, the volume, denoted as \(V\), is calculated by cubing the edge length \(a\):
\[ V = a^3 \]For the primitive cubic cell in the exercise, the edge length is 3.64 Å, resulting in:
\[ V = (3.64 \text{ Å})^3 = 48.21 \text{ Å}^3 \]For the face-centered cubic cell, with an edge of 5.14 Å:
\[ V = (5.14 \text{ Å})^3 = 135.44 \text{ Å}^3 \]This calculation informs us of the volume of space that the atoms occupy, which is often crucial for understanding physical properties like density and stability of materials.
Effective use of these formulas allows students to derive important material characteristics from the atomic level structure.