Question

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

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

Step-by-step solution

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}

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}

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

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.