Question

The diameter of a rubidium atom is $4.95\text{AA}$. 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 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 Rb metal?

Short answer

(a) Total atoms (arrangement A) ≈ 4.08*10^{14} (b) Total atoms (arrangement B) ≈ 4.12*10^{14} (c) Factor by which the number of atoms increased from A to B ≈ 1.01 Arrangement B would lead to a greater density for Rb metal if extended to three dimensions.

Step-by-step solution

Convert units

Convert the diameter of the rubidium atom from ångströms (Å) to centimeters (cm), as the surface area is given in cm^2. Diameter (in Å) = 4.95 Å Conversion factor: 1 Å = 1*10^{-8} cm Diameter (in cm) = 4.95 Å * (1*10^{-8} cm/Å) = 4.95*10^{-8} cm

Calculate the number of atoms for arrangement A

Calculate the number of atoms that can fit in each row and column of a 1 cm x 1 cm square grid, and then calculate the total number of atoms for arrangement A. Number of atoms in a row = 1cm / Diameter (in cm) Number of atoms in a row = 1 cm / 4.95*10^{-8} cm ≈ 2.02*10^7 Total atoms (arrangement A) = (Number of atoms in a row)^2 Total atoms (arrangement A) = (2.02*10^7)^2 ≈ 4.08*10^{14}

Calculate the number of atoms for arrangement B

In a close-packed arrangement, every other row is shifted by half an atom's diameter and has one less atom than the other rows. Calculate the number of atoms in both even and odd rows and then calculate the total number of atoms for arrangement B. Number of atoms in an even row = 2.02*10^7 (Same as arrangement A) Number of atoms in an odd row = 2.02*10^7 - 1 (One less atom per odd row) Number of atoms in an odd row ≈ 2.02*10^7 - 1 Number of rows = Number of atoms in a row (As square surface) Number of rows = 2.02*10^7 Total atoms (arrangement B) = [Even rows * (Number of atoms in an even row)] + [Odd rows * (Number of atoms in an odd row)] Total atoms (arrangement B) ≈ (1.01*10^7) * (2.02*10^7) + (1.01*10^7) * [(2.02*10^7) - 1] ≈ 4.12*10^{14}

Calculate the factor by which the number of atoms increased in arrangement B compared to arrangement A

Compare the number of atoms on the surface for both arrangements to find what factor the number of atoms increased from arrangement A to B. Factor = Total atoms (arrangement B) / Total atoms (arrangement A) Factor ≈ 4.12*10^{14} / 4.08*10^{14} ≈ 1.01

Determine which arrangement would lead to a greater density for Rubidium metal if extended to three dimensions

Since the close-packed arrangement (arrangement B) has a higher number of atoms on the surface compared to arrangement A, the density would be greater if extended to three dimensions for arrangement B. So, the arrangement B would lead to a greater density for Rubidium metal if it is extended to three dimensions. To summarize: (a) Total atoms (arrangement A) ≈ 4.08*10^{14} (b) Total atoms (arrangement B) ≈ 4.12*10^{14} (c) Factor by which the number of atoms increased from A to B ≈ 1.01 Arrangement B would lead to a greater density for Rb metal if extended to three dimensions.

Key concepts

Square Grid Arrangement

The square grid arrangement is a simple and straightforward way of arranging atoms. Imagine placing atoms on a flat surface, lining them up in neat, straight rows and columns, very much like a chessboard.
This arrangement offers a clear view of each atom's position. One can easily count how many atoms fit into a given area.
For example, if we have a square surface of 1.0 cm x 1.0 cm and each atom has a diameter of 4.95 Å, we can calculate how many atoms fit across this surface.

• First, convert the atom diameter into centimeters by using the conversion rate: 1 Å = 1×10^{-8} cm.
• This gives a diameter of 4.95 × 10^{-8} cm for each rubidium atom.

The structure of this arrangement leads to each atom occupying a specific spot with equal spacing all around. While simple, this arrangement is not the most space-efficient.

Close-Packed Arrangement

A close-packed arrangement is more efficient than the square grid arrangement. It is designed to fit more atoms in a given space by placing each atom into the gaps or depressions created by atoms in the previous row.
Think of how oranges are often stacked in a grocery store, each row nestled into the grooves of the row beneath it. This method uses every available space to maximize the number of atoms.
In our example, the rubidium atoms can be placed closer together using this arrangement, thereby increasing the total number of atoms that fit into the 1.0 cm x 1.0 cm area.

• For close-packed arrangement B, rows are staggered, with every other row shifted by half an atom’s diameter, allowing a tighter packing.
• The calculation shows that more atoms (approximately 4.12×10^{14}) fit onto the surface compared to the square grid's arrangement.

Hence, the close-packed arrangement allows more particles to occupy the same space, which could lead to increased material density when extended into three dimensions.

Atomic Density

Atomic density refers to how densely packed atoms are in a given volume or area. It is a crucial concept in determining how materials behave. When atoms are closely packed, like in the close-packed arrangement, the atomic density is higher.
This increased density affects properties like the strength, conductivity, and thermal expansion of the material.
For rubidium, extending to three dimensions, the close-packed arrangement would result in a higher density than the square grid arrangement.
The concept of density is essential in materials science and engineering.

• A close-packed arrangement allows more atoms within the same space, increasing the atomic density.
• This higher density often correlates with greater material strength and stability.

Understanding atomic density helps in creating materials with desired properties for various applications, from electronics to structural components.

Rubidium Atom Size

The size of a rubidium atom is a critical factor when considering atomic arrangements. With a diameter of approximately 4.95 Å, rubidium atoms are relatively large compared to some other alkali metals.
Knowing the size of the atom is essential when calculating how many can fit within a given space.
The diameter directly impacts the ability to pack atoms more tightly or loosely.
When converting the atom’s diameter into centimeters, we use the conversion factor: 1 Å = 1×10^{-8} cm.
This conversion helps align the measurement units to compute the total number of atoms that could fit onto a specific area, like the 1.0 cm x 1.0 cm surface in our example.

• The knowledge of rubidium atom size aids in predicting how they will behave in both square grid and close-packed arrangements.
• It also allows for better forecasting of material properties when rubidium is part of a compound or alloy.

Rubidium's relatively large atomic size makes it an interesting case for study when considering various atomic arrangements.