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What is the minimum number of atoms that could be contained in the unit cell of an element with a face-centered cubic lattice? \((\mathbf{a}) 1,(\mathbf{b}) 2,(\) (c) 3 (d) 4 (e) 5.

Short Answer

Expert verified
In a face-centered cubic (FCC) lattice, there are 8 corner atoms and 6 face atoms. The corner atoms contribute 1 atom to the unit cell, while the face atoms contribute 3 atoms to the unit cell. Therefore, the minimum number of atoms in the unit cell of an element with a face-centered cubic lattice is \(1 + 3 = 4\). The correct answer is (d) 4.

Step by step solution

01

The face-centered cubic lattice structure

An FCC lattice has atoms positioned at every corner of the cube and at the center of each face. Each unit cell in an FCC lattice has 8 corner atoms and 6 face atoms. However, since the corner atoms are shared by 8 adjacent unit cells, and the face atoms are shared by 2 adjacent unit cells, we need to determine their contributions to the unit cell.
02

Calculate the contribution of the corner atoms

Each corner atom is shared by 8 unit cells. So, the contribution of a corner atom to a single unit cell is \( \frac{1}{8}\). Since there are 8 corner atoms, their total contribution to the unit cell is: \[8 \times \frac{1}{8} = 1\]
03

Calculate the contribution of the face atoms

Each face atom is shared by 2 unit cells. So, the contribution of a face atom to a single unit cell is \( \frac{1}{2}\). Since there are 6 face atoms, their total contribution to the unit cell is: \[6 \times \frac{1}{2} = 3\]
04

Determine the total number of atoms in the unit cell

Now we need to add the contributions from both the corner atoms and the face atoms to find the total number of atoms in the unit cell: \[1 + 3 = 4\] So, the minimum number of atoms that could be contained in the unit cell of an element with a face-centered cubic lattice is 4. The correct answer is (d) 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Unit Cell
The unit cell is the smallest structural component of a crystal lattice. It acts as the building block, repeating in three-dimensional space to form the entire crystal. In a face-centered cubic (FCC) lattice, the unit cell is a cube where the atoms are organized at specific points:
  • At each of the eight corners
  • At the center of each of the six faces
This arrangement is characteristic of the FCC lattice and is crucial for calculating how many atoms are actually within the unit cell. By understanding how these atoms are shared between neighboring unit cells, we gain insight into the makeup of materials at the atomic level.
Corner Atoms and Their Contribution
In a crystal structure, corner atoms are located at the vertices of the unit cell. A key understanding is that these atoms are not exclusive to a single cell but are shared by eight adjacent unit cells. Consequently, the contribution of a single corner atom to one unit cell is only \( \frac{1}{8} \) of an atom.
Given there are eight such corner atoms in a face-centered cubic unit cell:
  • The cumulative contribution of all corner atoms is calculated by multiplying:\[ 8 \times \frac{1}{8} = 1 \]atom per unit cell.
With this understanding, we see that these corner atoms collectively contribute one whole atom to the structure of the unit cell.
The Role of Face Atoms
Face atoms are positioned at the centers of each cube face in a face-centered cubic lattice. In contrast to corner atoms, each face atom is shared between only two unit cells. This means the contribution of each face atom is half an atom to a single unit cell. There are six face atoms in an FCC unit cell leading to:
  • The overall contribution calculated by:\[ 6 \times \frac{1}{2} = 3 \]atoms to the unit cell.
Thus, these face atoms contribute significantly to the total atom count, playing a vital role in the stability and properties of the crystal.
Comprehending the Crystal Lattice Structure
A crystal lattice structure refers to the orderly arrangement of atoms in a crystalline solid. The face-centered cubic (FCC) lattice is a specific type of structure where cubes repeat in a symmetrical pattern, providing both stability and high packing density. Each unit cell connects seamlessly to its neighbors, forming a continuous three-dimensional grid.
In the FCC structure:
  • Atoms are arranged efficiently, reducing empty space.
  • It contributes to the material's properties, such as ductility and strength.
Understanding the way units like FCC lattices compose a crystal helps explain the behavior and characteristics of different crystalline materials, which is crucial in materials science and engineering.

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Most popular questions from this chapter

Sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) adopts a cubic structure with \(\mathrm{Na}\) atoms represented by green spheres and \(\mathrm{O}\) atoms by red spheres. (a) How many atoms of each type are there in the unit cell? (b) Determine the coordination number and describe the shape of the coordination environment for the sodium ion. (c) The unit cell edge length is \(555 \mathrm{pm}\). Determine the density of \(\mathrm{Na}_{2} \mathrm{O}\)

Indicate whether this statement is true or false: If you want a semiconductor that emits blue light, you could either use a material that has a band gap corresponding to the energy of a blue photon or you could use a material that has a smaller band gap but make an appropriately sized nanoparticle of the same material.

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Indicate the type of solid (molecular, metallic, ionic, or covalent-network) for each compound: (a) SiC, (b) \(\mathrm{Ni}\) (c) \(\mathrm{CaCl}_{2}\) (d) camphor \(\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)\), (e) \(\mathrm{SiO}_{2}\).

At room temperature and pressure RbCl crystallizes with the NaCl-type structure. (a) Use ionic radii to predict the length of the cubic unit cell edge. (b) Use this value to estimate the density. (c) At high temperature and pressure, the structure transforms to one with a CsCl-type structure. Use ionic radii to predict the length of the cubic unit cell edge for the high- pressure form of RbCl. (d) Use this value to estimate the density. How does this density compare with the density you calculated in part (b)?

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