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Chapter 12: Problem 39

Aluminum metal crystallizes in a face-centered cubic unit cell. (a) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, \(a\) , from the atomic radius of aluminum \((1.43 \hat{\mathrm{A}}) .\) (d) Calculate the density of aluminum metal.

Short Answer

Expert verified
In a face-centered cubic unit cell, there are 4 aluminum atoms, and each atom has a coordination number of 12. The unit cell edge length, $a$, is 4.04 Å, and the density of aluminum metal is 2.70 g/cm³.

Step by step solution

01

Part (a): Number of aluminum atoms in a unit cell

In a face-centered cubic unit cell, there is one atom located at each corner and one atom at the center of each of the six faces. There are 8 corner atoms and 6 face atoms. Remember that each corner atom is shared by 8 unit cells, and each face atom is shared by 2, so we can calculate the total number of aluminum atoms in a unit cell like this: Number of aluminum atoms = (8 corner atoms × 1/8) + (6 face atoms × 1/2)
02

Calculate the Number of Aluminum Atoms

Now let's plug in the values and calculate the number of aluminum atoms in the FCC unit cell: Number of aluminum atoms = (8 × 1/8) + (6 × 1/2) Number of aluminum atoms = 1 + 3 = 4 In a face-centered cubic unit cell, there are 4 aluminum atoms.
03

Part (b): Coordination number of each aluminum atom

The coordination number is defined as the number of nearest neighbors that an atom has. In a face-centered cubic unit cell, each aluminum atom has 12 nearest neighbors. Therefore, the coordination number of each aluminum atom is 12.
04

Part (c): Estimate the length of the unit cell edge, a, from the atomic radius of aluminum

In a face-centered cubic unit cell, the atomic radius (r) is related to the edge length (a) by the following equation: \(a = 2\sqrt{2}r\) We are given the atomic radius of aluminum, r = 1.43 Å. Let's plug this value into the equation to find the edge length a.
05

Calculate the Edge Length a

By substituting the atomic radius in the equation, we get: \(a = 2\sqrt{2}(1.43\ Å)\) \(a = 2(2.02)\ Å\) \(a = 4.04\ Å\) The unit cell edge length, a, for aluminum is 4.04 Å.
06

Part (d): Calculate the density of aluminum metal

To calculate the density (ρ) of aluminum, we will use the following equation: ρ = (mass of aluminum atoms in the unit cell) / (volume of the unit cell) The mass of each aluminum atom can be found using its molar mass, which is 26.98 g/mol, and Avogadro's number, N = 6.022 x 10^23 atoms/mol. The volume of the unit cell can be calculated as: Volume = a^3
07

Calculate the Density of Aluminum Metal

First, let's calculate the mass of 4 aluminum atoms in the unit cell: Mass of 4 Al atoms = (4 atoms x 26.98 g/mol) / (6.022 x 10^23 atoms/mol) Mass of 4 Al atoms = 1.79 x 10^-22 g Next, let's calculate the volume of the unit cell: Volume = a^3 Volume = (4.04 Å)^3 Volume = 65.78 Å^3 To convert Å^3 to cm^3, multiply by 10^-24: Volume = 65.78 x 10^-24 cm^3 Now we can calculate the density of aluminum metal: ρ = (1.79 x 10^-22 g) / (65.78 x 10^-24 cm^3) ρ = 2.70 g/cm^3 The density of aluminum metal is 2.70 g/cm^3.

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Most popular questions from this chapter

The coordination number for the \(\mathrm{Al}^{3+}\) ion is typically between four and six. Use the anion coordination number to determine the \(\mathrm{Al}^{3+}\) coordination number in the following compounds: (a) AlF_ \(_{3}\) where the fluoride ions are two coordinate, (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}\) where the oxygen ions are six coordinate, (c) AlN where the nitride ions are four coordinate.

If you want to dope GaAs to make an n-type semiconductor with an element to replace Ga, which element(s) would you pick?

For each of the following groups, which metal would you expect to have the highest melting point: (a) gold, rhenium, or cesium; (b) rubidium, molybdenum, or indium; (c) ruthenium, strontium, or cadmium?

Unlike metals, semiconductors increase their conductivity as you heat them (up to a point). Suggest an explanation.

If you want to make a polymer for plastic wrap, should you strive to make a polymer that has a high or low degree of crystallinity?
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