Consider a metallic element that crystallizes in a cubic close-packed lattice. The edge length of the unit cell is \(408 \mathrm{pm}\). If close-packed layers are deposited on a flat surface to a depth (of metal) of \(0.125 \mathrm{~mm}\), how many close-packed layers are present?

Understanding the atomic radius is essential when exploring the cubic close-packed (ccp) structure of metals. The atomic radius is half the distance between the centers of two adjacent atoms in a crystal lattice. In the ccp structure, which is also referred to as face-centered cubic (fcc), the atoms are packed in such a way to occupy the least amount of space, resulting in a highly efficient use of space.

To determine the atomic radius in a ccp structure, you can use the face diagonal of the cube. Since atoms touch along the face diagonal in the ccp structure, it can be calculated from the edge length, represented by the mathematical relationship derived from the Pythagorean theorem, for the face diagonal of a square as \( a\sqrt{2} \)

Improving your comprehension

The relationship between the face diagonal and the atomic radius is crucial. In the ccp structure, the face diagonal is equivalent to four times the atomic radius (\(4r\)), enabling us to solve for \(r\) when given the edge length of the unit cell. This conversion step from the given edge length to the atomic radius is necessary to understand the geometric properties of the crystal lattice and the spatial arrangement of atoms within the unit cell.

The unit cell geometry of a cubic close-packed structure is fascinating for its symmetry and efficiency. A unit cell is the basic repeating unit that defines the crystal structure. In the case of the ccp structure, the unit cell is face-centered cubic, meaning that atoms are located at each of the corners and the centers of all the cube faces.

Each corner atom is shared by eight neighboring unit cells, and each face-centered atom is shared by two unit cells, reducing the overall number of atoms per unit cell to an equivalent of four whole atoms.

Connection to the Exercise

For our exercise, knowing the unit cell geometry helps to visualize how the atoms stack upon each other and interact to create a whole layer. The ccp structure allows for tightly packed atomic layers, and this close packing results in the highest possible atom density. To apply this to practical problems, like determining the number of layers within a given depth, it is first essential to understand how atoms fill the unit cell and to be able to calculate the implications of the atomic radius on the overall crystal structure.

Crystal lattice layers refer to the planes of atoms stacked over one another in a repeating pattern. In a cubic close-packed structure, these layers follow a specific sequence often denoted by the letters A, B, and C, to indicate the different orientations of atoms in consecutive layers.

In a ccp structure, each layer is offset from the previous one, so the atoms in one layer nest in the gaps of the previous layer, allowing for the close packing. This arrangement leads to a three-dimensional repeating pattern where no two same letters are adjacent, typically following an ABCABC... sequence. Understanding the stacking of these layers is critical for determining properties like density, stability, and the behavior of the material under various conditions.

Practical Calculation in Context

Relating back to the exercise, once the atomic radius is known, the height of these continuous layers can be calculated because each new layer adds an atomic diameter's worth of height, which is twice the atomic radius (\(2r\)). By converting the material's depth to the same dimensional units as the atomic radius and height, it is then possible to calculate precisely how many of these atomic layers fit within a given depth of material.