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?

Step by step solution

01

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#

02

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.

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