Question

Crystalline Structures and Unit Cells An atom has a radius of 142 \(\mathrm{pm}\) and crystallizes in the face- centered cubic unit cell. What is the volume of the unit cell in \(\mathrm{cm}^{3} ?\)

Short answer

The volume of the FCC unit cell is approximately \(3.657 \times 10^{-23} \mathrm{cm}^3\).

Step-by-step solution

Understanding the Face-Centered Cubic Unit Cell

In a face-centered cubic (FCC) unit cell, atoms are located at each of the eight corners and the centers of each face of the cube. The cube's edge length, 'a', can be related to the atomic radius, 'r', through the geometrical relationship in FCC crystals: \( a = 2\sqrt{2}r \).

Calculate Cube Edge Length 'a'

Given the atomic radius of 142 pm, first convert the radius to centimeters: \( 142 \mathrm{pm} = 142 \times 10^{-12} \mathrm{cm} \). Now use the geometrical relationship to calculate 'a': \( a = 2\sqrt{2} \times 142 \times 10^{-12} \mathrm{cm} \).

Calculate the Volume of the Unit Cell

The volume 'V' of a cube with edge length 'a' is given by \( V = a^3 \). After obtaining the value of 'a' from Step 2, cube it to find the volume of the unit cell in cubic centimeters.

Key concepts

Understanding the Face-Centered Cubic Unit Cell

The face-centered cubic (FCC) unit cell is a common and important crystal structure found in metals such as aluminum, copper, and silver. Imagine a cube where each corner has an atom, and each face of the cube has another atom at its center. Each of these face-centered atoms is shared among adjacent unit cells, and the same applies to the corner atoms, which are shared among eight unit cells. This means that each FCC unit cell effectively contains only 4 whole atoms.

Geometrically, the relationship between the atomic radius and the edge length of the unit cell is a key concept in understanding FCC structures. If you visualize a diagonal plane cutting through the cube from one corner to the opposite corner, passing through the face-centered atoms, it includes four radii and forms an isosceles right triangle. Using the Pythagorean theorem, the relationship between the atomic radius (r) and the edge length (a) of the unit cell is derived as: \( a = 2\root{2}{r} \).

Grasping this relationship is crucial as it not only helps in calculating the edge length of the unit cell when the atomic radius is known, but also in understanding the dense packing and efficiency of the FCC structure.

Calculating the Atomic Radius

The atomic radius is a measure of the size of an atom from its center to the boundary of its electrons. When dealing with crystalline structures, the atomic radius plays a pivotal role because it directly influences the unit cell's dimensions. For a face-centered cubic unit cell, the atomic radius can be seen as half the length from the center of one atom to the center of an adjacent atom that it is touching along the face diagonal.

In the exercise, the given atomic radius of 142 pm (picometers) is essential for the calculation. To work within the standard SI unit system, we convert this measurement to centimeters: \( 142 \mathrm{pm} = 142 \times 10^{-12} \mathrm{cm} \).

Using compact and straightforward conversion factors helps students focus on the primary calculations without getting lost in complex unit manipulations. Conversion to centimeters is crucial because most chemical and physical constants are defined in terms of SI units, ensuring consistency and correctness in the subsequent calculations.

Unit Cell Volume Calculation

Once the edge length 'a' of the unit cell is found using the atomic radius and the geometrical relationship in FCC crystals, the next step is to determine the volume of the unit cell. For any cube, the volume 'V' is simply the cube of the edge length: \( V = a^3 \).

From the earlier calculation of 'a', one must cube this value to find the volume of the unit cell. Bear in mind that this volume represents the space occupied by four whole atoms in the FCC crystal structure. This calculation illustrates how the atomic radius influences the unit cell volume and hence the density and properties of the crystalline material.

Understanding how to perform these volume calculations is fundamental for students exploring material science and solid-state physics. It allows them to relate microscopic atomic parameters to the macroscopic properties of materials, providing a deeper comprehension of the material's characteristics. Moreover, by practicing unit cell volume calculations, students develop valuable skills in geometric reasoning and algebraic manipulation that are widely applicable in scientific problem-solving.