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Chapter 12: Problem 99

Gold adopts a face-centered cubic structure with a unit cell edge of 408 pm (Figure 12.11). How many gold atoms are there in a sphere that is $20 \mathrm{nm}\( in diameter? Recall that the volume of a sphere is \)\frac{4}{3} \pi r^{3}$.

Short Answer

Expert verified
There are approximately 246,004 gold atoms in the sphere with a diameter of 20 nm.

Step by step solution

01

Calculate the volume of the sphere

To calculate the volume of the sphere with a diameter of 20 nm, first, convert the diameter to radius (r) by dividing it by 2. The radius is 10 nm. Now, convert the radius from nm to pm by multiplying by 1000. The radius becomes 10,000 pm. Using the formula for the volume of a sphere, \(V = \frac{4}{3}\pi r^{3}\), we can find the volume: V = \frac{4}{3}\pi (10,000)^{3} = 4.19 \times 10^{12} \text{ pm}^{3}
02

Calculate the volume of the unit cell

The edge of the unit cell is given as 408 pm. In a cubic lattice, the volume of the unit cell is the cube of the edge length: V_\text{unit cell} = a^3 = (408 \text{ pm})^3 = 6.813 \times 10^{7} \text{ pm}^{3}
03

Calculate the number of unit cells in the sphere

To obtain the number of unit cells within the sphere, divide the volume of the sphere by the volume of the unit cell: \text{Number of unit cells} = \frac{V_\text{sphere}}{V_\text{unit cell}} = \frac{4.19 \times 10^{12} \text{ pm}^{3}}{6.813 \times 10^{7} \text{ pm}^{3}} = 61500.5 Since the number of unit cells must be an integer, we'll round up to the nearest whole number: \text{Number of unit cells} = 61501
04

Determine the number of gold atoms in the sphere

In a face-centered cubic lattice, there are 4 gold atoms per unit cell. To find the total number of gold atoms in the sphere, multiply the number of unit cells by the number of atoms in each unit cell: \text{Number of gold atoms} = (\text{Number of unit cells}) \times (\text{Atoms per unit cell}) = 61501 \times 4 = 246004 So, there are approximately 246,004 gold atoms in the sphere with a diameter of 20 nm.

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Most popular questions from this chapter

Calcium crystallizes in a face-centered cubic unit cell at room temperature that has an edge length of \(558.8 \mathrm{pm}\). (a) Calculate the atomic radius of a calcium atom. (b) Calculate the density of Ca metal at this temperature.

Arrange the following metals in increasing order of expected melting point: La, W, Ta, Hf. Explain this trend in melting points.

Energy bands are considered continuous due to the large number of closely spaced energy levels. The range of energy levels in a crystal of copper is approximately \(1 \times 10^{-19} \mathrm{~J}\). Assuming equal spacing between levels, the spacing between energy levels may be approximated by dividing the range of energies by the number of atoms in the crystal. (a) How many copper atoms are in a piece of copper metal in the shape of a cube with edge length \(0.5 \mathrm{~mm}\) ? The density of copper is $8.96 \mathrm{~g} / \mathrm{cm}^{3}$. (b) Determine the average spacing in J between energy levels in the copper metal in part (a). (c) Is this spacing larger, substantially smaller, or about the same as the \(1 \times 10^{-18}\) J separation between energy levels in a hydrogen atom?

You are given a gray substance that melts at \(700^{\circ} \mathrm{C} ;\) the solid is a conductor of electricity and is insoluble in water. Which type of solid (molecular, metallic, covalent-network, or ionic) might this substance be?

Write the chemical equation that represents the formation of (a) polychloroprene from chloroprene (polychloroprene is used in highway- pavement seals, expansion joints, conveyor belts, and wire and cable jackets) (b) polyacrylonitrile from acrylonitrile (polyacrylonitrile is used in home furnishings, craft yarns, clothing, and many other items).
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