Question

What fraction of the total volume of a cubic closest packed structure is occupied by atoms? (Hint: \(V_{\text { sphere }}=\frac{4}{3} \pi r^{3} .\) ) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers.

Short answer

The fraction of the total volume occupied by atoms in a cubic closest packed structure is approximately 74% (\(\frac{\sqrt{2}}{1}\)), while in a simple cubic structure, it is 25% (\(\frac{1}{4}\)). A cubic closest packed structure occupies a higher volume compared to a simple cubic structure.

Step-by-step solution

Determine the number of atoms per unit cell in each structure.

For the cubic closest packed (ccp) structure, there are 4 atoms per unit cell. In the simple cubic structure, there is 1 atom per unit cell.

Calculate the total volume occupied by atoms.

In the ccp structure, the volume occupied by atoms is given by: \[V_{ccp} = 4 \times \frac{4}{3} \pi r^{3}\] In the simple cubic structure, the volume occupied by atoms is given by: \[V_{simple} = 1 \times \frac{4}{3} \pi r^{3}\]

Find the volume of the unit cell in each structure.

In the ccp structure, the edge length of the unit cell is \(2\sqrt{2}r\), and the volume of the unit cell is given by: \[V_{ccp\_cell}=(2\sqrt{2}r)^{3}\] In the simple cubic structure, the edge length of the unit cell is \(2r\), and the volume of the unit cell is given by: \[V_{simple\_cell}=(2r)^{3}\]

Calculate the fraction of the total volume occupied by atoms in each structure.

In the ccp structure, the fraction of the total volume occupied by atoms is given by: \[\frac{V_{ccp}}{V_{ccp\_cell}} = \frac{4\times\left( \frac{4}{3} \pi r^{3}\right)}{(2\sqrt{2}r)^{3}}\] In the simple cubic structure, the fraction of the total volume occupied by atoms is given by: \[\frac{V_{simple}}{V_{simple\_cell}} = \frac{1\times\left( \frac{4}{3} \pi r^{3}\right)}{(2r)^{3}}\]

Compare the answers and reduce the fractions.

The ccp and simple cubic structures have the following fractions of the total volume occupied by atoms: \[\frac{V_{ccp}}{V_{ccp\_cell}} = \frac{32\pi r^3}{(2\sqrt{2}r)^3} = \frac{32}{16\sqrt{2}} = \frac{2}{\sqrt{2}}= \frac{\sqrt{2}}{1} \approx 0.74 \] \[\frac{V_{simple}}{V_{simple\_cell}} = \frac{4\pi r^3}{(2r)^3} = \frac{1}{4} = 0.25\] Comparing the two answers, we can see that a cubic closest packed structure has a higher fraction of its total volume occupied by atoms (\(\approx\) 74%) than a simple cubic structure (25%).

Key concepts

Cubic Closest Packed Structure

The cubic closest packed (ccp) structure is a highly efficient type of atomic arrangement, where atoms are packed together as tightly as possible. This arrangement maximizes the use of available space, resulting in a high packing density. In the ccp structure, each unit cell contains 4 atoms.

One of the key characteristics of this structure is how the atoms touch each other. Each atom is surrounded by 12 others, forming an arrangement similar to hexagonal closest packing, but laid out differently. This results in efficient use of space, as indicated by the high packing factor.

For calculations, the volume of atoms per unit cell is given by the formula \(V_{ccp} = 4 \times \frac{4}{3} \pi r^{3}\). The edge length of the ccp unit cell is \(2\sqrt{2}r\), making the total unit cell volume \(V_{ccp\_cell}=(2\sqrt{2}r)^{3}\). Ultimately, about 74% of the unit cell's volume is filled with atoms.

Simple Cubic Structure

The simple cubic structure is one of the least efficient ways to pack atoms. In this arrangement, atoms are positioned at each corner of a cube, resulting in a less dense packing. Only one atom is effectively present per unit cell due to the sharing of corner atoms among adjacent cells.

In this structure, each atom directly touches only 6 others, one on each face of the cube that it is part of. This gives the simple cubic structure the smallest coordination number among cubic structures, contributing to its lower packing efficiency.

To determine the volume of atoms in a simple cubic unit cell, use the formula \(V_{simple} = 1 \times \frac{4}{3} \pi r^{3}\). The unit cell edge length is \(2r\), and the volume of the unit cell is \(V_{simple\_cell}=(2r)^{3}\). The packing fraction here is approximately 25%, indicating that much of the space within a simple cubic lattice is unoccupied.

Atomic Packing Factor

The atomic packing factor (APF) is a measure of the efficiency of space usage in a crystal structure. It is defined as the fraction of volume in a unit cell that is occupied by atoms. This concept helps us understand how densely packed the atoms are in a given structure.

For the cubic closest packed (ccp) structure, the APF is about 0.74 (or 74%), indicating a very efficient packing. This is due to the overlapping arrangement of atoms, which minimizes empty space.

In contrast, the simple cubic structure has an APF of just 0.25 (or 25%), reflecting its less efficient packing with large amounts of unoccupied space. Understanding the APF helps in comparing different structures and assessing their potential properties, such as density and stability.

Volume Occupied by Atoms

To determine how much space within a unit cell is filled by atoms, we calculate the volume occupied by these atoms. This volume is essential for assessing the density and structural properties of any crystalline material.

In the context of cubic structured crystals, this occupied volume is derived from the count of atoms per unit cell and the individual volume of each atom. The formula \(V = n \times \frac{4}{3} \pi r^{3}\) is used, where \(n\) is the number of atoms per unit cell. In the ccp structure, this volume results in a high packing fraction due to overlap between layers.

Comparatively, the simple cubic structure yields a smaller occupied volume since only one atom effectively fills each cell. Knowing the volume occupied by atoms not only allows for the computation of the packing factor but also provides insights into the material’s potential properties and stability across different phases.