Question

The diameter of a rubidium atom is \(4.95 \AA\). We will consider two different ways of placing the atoms on a surface. In arrangement \(\mathrm{A}\), all the atoms are lined up with one another. Arrangement \(\mathrm{B}\) is called a close-packed arrangement because the atoms sit in the "depressions" formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side? (b) How many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement \(\mathrm{B}\) ? (c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for \(\mathrm{Rb}\) metal?

Short answer

The total number of rubidium atoms that can be placed on a 1.0 cm square surface using arrangement A is approximately \(2.0 \times 10^{15}\) atoms, while in arrangement B it is approximately \(2.4 \times 10^{15}\) atoms. Going from arrangement A to B, the number of atoms has increased by a factor of around 1.20. When extended to three dimensions, arrangement B would lead to a greater density for rubidium metal due to its close-packed structure.

Step-by-step solution

Calculate the number of atoms that can be placed in a row for arrangement A

First, convert the diameter of a rubidium atom from angstroms to centimeters: 4.95 Å = \(4.95 \times 10^{-8}\) cm. Then, divide the length of the surface (1.0 cm) by the diameter of a rubidium atom to find the number of atoms that can be placed in a row in arrangement A: Number of atoms in a row (A) = (Surface length) / (Diameter of a rubidium atom) = \(\frac{1.0}{4.95 \times 10^{-8}}\) Calculate the value.

Calculate the number of rows of atoms for arrangement A

In arrangement A, all atoms are lined up with one another. Therefore, the number of rows will be the same as the number of atoms in a row (calculated in step 1). Number of rows (A) = Number of atoms in a row (A)

Calculate the total number of atoms that can be placed on a surface for arrangement A

Multiply the number of rows by the number of atoms in a row to find the total number of atoms: Total number of atoms (A) = (Number of rows) x (Number of atoms in a row) Calculate the value.

Calculate the number of atoms that can be placed in a row for arrangement B

In arrangement B, the atoms sit in the "depressions" formed by the previous row of atoms, which means that each row is shifted by half the diameter of an atom. Therefore, the number of atoms that can be placed in a row in arrangement B is equal to the number of atoms in a row for arrangement A, plus 0.5: Number of atoms in a row (B) = Number of atoms in a row (A) + 0.5 Calculate the value.

Calculate the number of rows of atoms for arrangement B

Due to the close-packed arrangement, the rows are shifted and closer together. The distance between the rows is now equal to the radius of the atom times √3. Divide the length of the surface (1.0 cm) by this distance to find the number of rows in arrangement B: Number of rows (B) = (Surface length) / (Radius of a rubidium atom x √3) = \(\frac{1.0}{(4.95 \times 10^{-8}) / 2 \times \sqrt{3}}\) Calculate the value.

Calculate the total number of atoms that can be placed on a surface for arrangement B

Multiply the number of rows by the number of atoms in a row to find the total number of atoms for arrangement B: Total number of atoms (B) = (Number of rows) x (Number of atoms in a row) Calculate the value.

Compare the number of atoms in arrangement A and B, and find the factor by which the number of atoms has increased

Divide the total number of atoms in arrangement B by the total number of atoms in arrangement A to find the factor by which the number of atoms has increased: Factor = Total number of atoms (B) / Total number of atoms (A) Calculate the value.

Determine which arrangement leads to a greater density when extended to three dimensions

For arrangement A, the atoms are simply stacked one on top of the other, so the overall density would not change. In arrangement B, however, the close-packed arrangement not only exists in two dimensions but also in the third dimension. This means that when extended to three dimensions, arrangement B has a higher density due to its close-packed structure. So, the greater density for Rb metal is achieved in arrangement B.

Key concepts

Rubidium Atom Diameter

Understanding the size of atoms is crucial in the realm of chemistry and materials science. The diameter of a rubidium atom, as given, is approximately 4.95 angstroms or \(4.95 \times 10^{-8}\) centimeters. Rubidium, like other alkali metals, has a relatively large atomic size due to its single valence electron which is located far from its nucleus.

Let's put this scale into perspective: A single human hair is roughly 1 million angstroms thick. If we were to line up the rubidium atoms side by side, we would need around 202,020 atoms just to match the width of a thin human hair. This atomic size is also important when considering how atoms will pack together, as different arrangements will result in different densities and properties of the material. Knowing the atom's diameter is the first step in exploring these packing arrangements.

Atomic Surface Packing

Atomic surface packing refers to how atoms organize themselves on a given surface. This is influenced by the shape of the surface and the nature of the atomic interactions. There are several ways atoms can pack on a surface, but two primary examples include the simple aligned arrangement, referred to as arrangement A in the exercise, and the close-packed arrangement, known as arrangement B.

In arrangement A, rubidium atoms are lined up precisely one after another, similar to how oranges might be arranged in a box. Each row directly mirrors the one before it, creating an orderly grid. In contrast, the close-packed arrangement B optimizes the use of space. Here, each atom sits in the depressions formed by the preceding row, similar to the way cannonballs might be tightly stacked. This method reduces the amount of unused space, allowing for a greater number of atoms to be packed together, which can significantly impact the material's density.

Calculating Atom Density

Calculating atom density, or the number of atoms per unit area (or volume), provides insight into the properties of a substance. In the exercise, we compared two different arrangements on a surface, and the calculations illustrated how arrangement B can hold more atoms than arrangement A.

To calculate the density of atoms for arrangement A, we divided the square surface area by the area occupied by a single atom (using its diameter). However, the process is slightly more complex for arrangement B, where the hexagonal close-packing comes into play. By shifting every other row, more atoms can fit within the same space. The exercise showed that the number of atoms in arrangement B, when stacked in three dimensions, substantially increases, resulting in higher material density. This theoretical approach can be applied to understand the structure of real-world materials, such as metals, where atomic packing significantly impacts mechanical and electrical properties.