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 to form a square grid. 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 \(\mathrm{Rb}\) atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement \(\mathrm{B} ?(\mathrm{c})\) 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?

To find the number of rubidium atoms that can be placed in arrangement A, let's first convert 1.0 cm to angstroms: \[ 1.0 \, \text{cm} = 1.0 \times 10^{-2} \, \text{m} = 1.0 \times 10^{8} \, \text{\AA} . \] Now, divide the total length of the side by the diameter of a rubidium atom to find the number of atoms that can fit along each side of the surface: \[ \text{Number of atoms per side} = \frac{1.0 \times 10^{8} \, \text{\AA}}{4.95 \, \text{\AA}} \approx 2.02 \times 10^{7} . \] Since the atoms form a square grid, we can compute the total number of atoms by squaring the number of atoms per side: \[ \text{Total number of atoms} = (2.02 \times 10^{7})^2 \approx 4.08 \times 10^{14} . \]

In arrangement B, the atoms are close-packed and form a hexagonal arrangement. For this arrangement, the width of each hexagon along the square surface is given by the diameter of a rubidium atom multiplied by the square root of 2 (since it is the diagonal of a square whose side length is the diameter): \[ \text{Width of each hexagon} = 4.95 \, \text{\AA} \cdot \sqrt{2} \approx 7.00 \, \text{\AA} . \] Now, divide the total length of the side by the width of each hexagon to find the number of hexagons that can fit along each side of the surface: \[ \text{Number of hexagons per side} = \frac{1.0 \times 10^{8} \, \text{\AA}}{7.00 \, \text{\AA}} \approx 1.43 \times 10^{7} . \] To compute the total number of atoms, we need to multiply the number of hexagons per side by the number of atoms per side calculated in part (a) and then multiply this value by a correction factor to account for the arrangement: \[ \text{Total number of atoms} = 1.43 \times 10^{7} \cdot 2.02 \times 10^{7} \cdot \frac{4\sqrt{2}}{\pi} \approx 5.66 \times 10^{14} . \]

To find the factor by which the number of atoms increases when moving from arrangement A to arrangement B, divide the number of atoms found in arrangement B by the number of atoms found in arrangement A: \[ \text{Increase factor} = \frac{5.66 \times 10^{14}}{4.08 \times 10^{14}} \approx 1.39 . \] Thus, the number of atoms on the surface increased by a factor of approximately 1.39 in going from arrangement A to arrangement B. As for which arrangement would lead to greater density in three dimensions, arrangement B would likely result in a greater volume density for rubidium metal, since the close-packed arrangement is more efficient in filling space compared to the square grid arrangement in arrangement A.