Question

The diameter of a rubidium atom is \(495 \mathrm{pm}\) We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement 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? \((\mathbf{b})\) How many \(\mathrm{Rb}\) atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement B? (c) By what factor has the number of atoms on the surface increased in going to arrangement \(\mathrm{B}\) from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?

Short answer

In arrangement A, the total number of Rubidium atoms on the surface is \( \left( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm}} \right)^2 \). In arrangement B, the total number of Rubidium atoms on the surface is \( \left( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm} \times \frac{\sqrt{3}}{2}} \right)^2 \times \left(1 + \frac{1}{2}\right) \). The increase factor is calculated by dividing the total atoms in arrangement B by the total atoms in arrangement A. The greater density arrangement would be the one with a higher number of atoms on a 1.0 cm square surface.

Step-by-step solution

Since the diameter of a rubidium atom is given in pm, we need to convert it to cm to make it consistent with the surface dimensions: 1 pm = \(10^{-12}\) m = \(10^{-10}\) cm So, the side length of a Rubidium atom is: \(495 \times 10^{-10}\) cm #Step 2: Calculate the number of atoms in arrangement A#

In arrangement A, the atoms are lined up in a square grid. To find the number of atoms that can fit in one row, we'll divide the side length of the surface by the side length of an atom: Number of atoms in a row = \( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm}} \) To find the total number of atoms on the surface, we need to multiply the number of atoms in a row by the number of rows: Total atoms in arrangement A = \( \left( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm}} \right)^2 \) #Step 3: Calculate the number of atoms in arrangement B#

In arrangement B, the atoms are in a close-packed arrangement. This means they occupy more space, so we'll multiply the side length of an atom by \( \frac{\sqrt{3}}{2}\) to account for the additional space required. Modified side length of an atom in arrangement B = \(495 \times 10^{-10} \text{ cm} \times \frac{\sqrt{3}}{2}\) To find the number of atoms that can fit in one row, we'll divide the side length of the surface by the modified side length of an atom: Number of atoms in a row (arrangement B) = \( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm} \times \frac{\sqrt{3}}{2}} \) Since the atoms in arrangement B also occupy the spaces between rows of arrangement A, we need to multiply the number of atoms in a row by the number of rows and add an extra half-row for every row: Total atoms in arrangement B = \( \left( \frac{1.0 \text{ cm}}{495 \times 10^{-10}\text{ cm} \times \frac{\sqrt{3}}{2}} \right)^2 \times \left(1 + \frac{1}{2}\right) \) #Step 4: Calculate the increase factor and determine the greater density arrangement#

To find the increase factor, we need to divide the number of atoms in arrangement B by the number of atoms in arrangement A: Increase factor = \( \frac{\text{Total atoms in arrangement B}}{\text{Total atoms in arrangement A}} \) Since the density of an arrangement in three dimensions is determined by how close the atoms are to each other, the arrangement with the higher number of atoms on the surface will lead to a greater density. Therefore, the greater density arrangement is the one with a higher number of atoms on a 1.0 cm square surface.

Key concepts

Rubidium Atom

The rubidium atom is an alkali metal element with the chemical symbol Rb. It is known for its relatively large atomic size compared to other alkali metals. The diameter of a rubidium atom is given as 495 picometers (pm). To put that into perspective, 1 pm is one trillionth of a meter, or \(10^{-12}\) meters. When working with problems involving spatial arrangements, converting this diameter into more convenient units like centimeters (cm) becomes necessary, especially when considering macroscopic dimensions like surface areas, which are measured in centimeters. Therefore, the diameter of a rubidium atom in cm is \(495 \times 10^{-10} \text{ cm}\). This conversion is crucial because it allows us to calculate how many rubidium atoms can fit within a given surface area.

Square Grid

A square grid arrangement is a simple and straightforward way to organize spherical objects like atoms on a surface. In this setup, each atom is placed right next to another, forming rows and columns that are perpendicular to one another, just like squares on a chessboard. In this context, where rubidium atoms are positioned in a square grid, the arrangement allows for a systematic filling of the surface. To calculate how many atoms fit on a surface when using a square grid, we divide the side length of the surface by the side length of an atom. For example, on a 1.0 cm side, the number of rubidium atoms in a single line would be \( \frac{1.0 \text{ cm}}{495 \times 10^{-10} \text{ cm}} \). Since the atoms are arranged in a grid, the total number on the surface is the square of this value, providing a neat and organized structure.

Close-Packed Arrangement

In a close-packed arrangement, atoms are packed more efficiently on a surface compared to the more regular square grid. Each atom in this setup sits in the gaps or depressions left by the previous row, maximizing the number of atoms that can fit on the surface. This arrangement is inspired by nature's tendency to form structures that occupy the least space while accommodating the most particles.
The close-packed arrangement means adjusting the effective size of the atoms to account for this additional space utilization. Since the atoms fit into the depressions of preceding rows, they effectively occupy less space. Hence, they require modifying the particle's length by \( \frac{\sqrt{3}}{2} \). Therefore, to determine the number of atoms in this efficient layout, we divide the side length of the surface by this new effective length: \( \frac{1.0 \text{ cm}}{495 \times 10^{-10} \text{ cm} \times \frac{\sqrt{3}}{2}} \). This results demonstrates how close-packed structures can significantly increase the number of atoms on a surface.

Surface Density

Surface density relates to how many atoms occupy a surface area, and it is a determinant of the efficiency of space usage. The arrangement (whether square grid or close-packed) greatly impacts surface density. To compare the surface densities of two arrangements, one can calculate the increase factor of the number of atoms. For instance, in our problem, the increase factor is calculated by dividing the total number of atoms in a close-packed arrangement by the number of atoms in a square grid. This ratio reveals the improvement in space utilization:

• The more atoms per unit area, the greater the surface density.
• A higher density in three dimensions translates to a more compact metal structure.
• In general, close-packed arrangements contribute to a higher density in the material.

This not only implies a better arrangement for maximizing the surface but also hints at which structure might yield a higher density when extended into three dimensions.