Question

The diameter of a rubidium atom is 4.95 A. 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? ( b) How many Rb atoms could be placed on a square surface that is 1.0 \(\mathrm{cm}\) on a side, using arrangement \(\mathrm{B} ?(\mathbf{c})\) 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

The number of Rb atoms that can be placed on a 1.0 cm x 1.0 cm surface using arrangement A is approximately \(4.08 \times 10^{14}\), while using arrangement B, it is 460. The number of atoms on the surface increased by a factor of \(1.13 \times 10^{-12}\) in going to arrangement B from arrangement A. In three dimensions, arrangement B (close-packed) would lead to a greater density for Rb metal because it allows the atoms to sit more closely, maximizing the number of atoms per unit volume.

Step-by-step solution

(Step 1: Conversion of units)

To perform the calculations, we need to convert the given diameter of the Rb atom and the surface dimensions to the same units. Here, we convert the diameter of the Rb atom from Angstroms to centimeters and surface dimensions from centimeters to Angstroms. 1 Angstrom (A) = \(10^{-8}\) cm Surface dimensions: 1.0 cm x 1.0 cm = 100 A x 100 A (as 1 cm = 10 A) Diameter of Rb atom: 4.95 A = 4.95 x \(10^{-8}\) cm.

(Step 2: Arrangement A - Square grid)

In arrangement A, all Rb atoms are lined up in a square grid. To find out how many Rb atoms can fit in a 1.0 cm x 1.0 cm square surface, first, we need to calculate the number of atoms along one side of the surface. Number of atoms along one side = Surface side length (cm) ÷ Diameter of one atom (cm). Number of atoms along one side = 1.0 cm ÷ (4.95 x \(10^{-8}\) cm) = \(2.02 \times 10^7\). Now, we calculate the total number of atoms on the surface: Total atoms (A) = (Number of atoms along one side)² = \((2.02 \times 10^7)^2\) = \(4.08 \times 10^{14}\).

(Step 3: Arrangement B - Close-packed)

In arrangement B, the Rb atoms sit in the "depressions" formed by the previous row of atoms, resulting in a slight increase in packing density. As it's a close-packed arrangement, the atoms will form layers of equilateral triangles, and thus, the distance between adjacent atoms in a row will be equal to their diameter, which is 4.95 A. The distance between rows will be shorter than in arrangement A, due to the atoms sitting in the depressions. Using the equilateral triangle, we find the height to be: Height = Diameter x sin(60°) = 4.95 x \(\frac{\sqrt{3}}{2}\) = 4.29 A. Now, we calculate the number of Rb atoms along one side: Number of atoms along one side = Surface side length (A) ÷ Diameter of one atom (A) = 100 A ÷ 4.95 A = 20.2 ≈ 20. Similarly, we calculate the number of rows of atoms: Number of rows = Surface side length (A) ÷ Height of one row (A) = 100 A ÷ 4.29 A = 23.3 ≈ 23. Total atoms (B) = Atoms along one side × Rows of atoms = 20 × 23 = 460.

(Step 4: Comparing both arrangements and calculating density)

We now compare the number of atoms in both arrangements: Total atoms (A) = \(4.08 \times 10^{14}\) Total atoms (B) = 460 Increase in atoms = Total atoms (B) ÷ Total atoms (A) = 460 ÷ \(4.08 \times 10^{14}\) = \(1.13 \times 10^{-12}\) In three dimensions, arrangement B (close-packed) would lead to greater density because the atoms are more closely packed, and the arrangement allows the atoms to sit in the depressions of previous rows, maximizing the number of atoms per unit volume.

Key concepts

Atomic Diameter Measurement

Understanding atomic diameter measurement is essential when analyzing the structures of materials at a microscopic scale. The diameter of an atom is the distance between the farthest points of an atom's electron cloud. Measuring the atomic diameter can be more complex than it might seem because atoms are not solid objects with well-defined edges.

To overcome this difficulty, scientists often determine atomic diameter with reference to the concept of an average bond length found in crystal structures. For instance, the diameter of a rubidium (Rb) atom is given as 4.95 Angstroms (A), which is a common unit used in atomic scale measurements. One angstrom is equivalent to one ten-billionth (or 10^-10) of a meter. To be useful in practical calculations, such as packing atoms on a surface, the atomic diameter needs to be converted into the same units as the surface dimensions. In this case, the diameter in Angstroms was converted to centimeters for compatibility with the surface area given in centimeters.

Square Grid Atomic Arrangement

The square grid atomic arrangement represents a two-dimensional structure where each atom is positioned at the corner of a square. This arrangement allows atoms to line up in rows and columns, creating a grid pattern. The simplicity of this structure makes it a useful model for visualization and calculation.

In context to our exercise, the square grid arrangement assists in determining how many rubidium atoms could fit on a surface with specified measurements. By dividing the length of one side of the surface (\(1.0 \text{ cm}\) or 100 A) by the diameter of an Rb atom, the number of atoms that can fit along that side is found. Multiplying the number of atoms along one side by itself gives the total number of atoms that the surface can accommodate in this arrangement. However, this type of arrangement is not the most efficient in terms of space utilization as there are gaps at each corner of the squares that are not filled by atoms.

Close-Packed Atomic Arrangement

When transitioning to the close-packed atomic arrangement, the efficiency of space utilization is improved. In such an arrangement, each atom is surrounded by as many atoms as possible. In a two-dimensional setting, this commonly results in a hexagonal pattern where each atom is in contact with six others.

In a close-packed arrangement, atoms of the subsequent row occupy the 'depressions' left between atoms of the previous row, which resembles a formation akin to stacked oranges at a market. The exercise involved calculating the new layout based on the diameter of an Rb atom and the height of each row formed by the atoms sitting in the depressions of the row below. For rubidium and similar materials, this arrangement results in higher packing density compared to a square grid, implying that more atoms can occupy the same surface area. This is verified by the exercise where the close-packed arrangement allows a significant increase in the number of Rb atoms per square centimeter of surface area. If extrapolated into three dimensions, this principle means that close-packed arrangements would produce a material with greater atomic density, which is a cornerstone in material science and solid-state physics.