Question

Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of \(4 \mathrm{~nm}\). Calculate the mass of the quantum dot, using the density of silicon \(\left(2.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\). (b) How many silicon atoms are in the quantum dot? (c) The density of germanium is \(5.325 \mathrm{~g} / \mathrm{cm}^{3}\). If you made a 4-nm quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical.

Short answer

(a) The mass of the silicon quantum dot is \(7.70674 \times 10^{-20} \mathrm{~g}\). (b) There are \(1.65 \times 10^3\) silicon atoms in the quantum dot. (c) In a 4-nm germanium quantum dot, there would be \(1.478 \times 10^3\) germanium atoms.

Step-by-step solution

a) Mass of the quantum dot

First, find the volume of the quantum dot using the sphere formula, \(V = \frac{4}{3} \pi r^{3}\). Since the diameter is given, the radius is \(r = \frac{d}{2} = \frac{4 \mathrm{~nm}}{2} = 2 \mathrm{~nm}\). However, we need to convert the radius from nanometers to centimeters, as density is given in grams per cubic centimeter. So, \(\frac{2 \mathrm{~nm}}{1 \times 10^{7} \mathrm{~nm/cm}} = 2 \times 10^{-7} \mathrm{~cm}\). Now, we can find the volume: \(V = \frac{4}{3} \pi (2 \times 10^{-7})^{3} = 3.35103 \times 10^{-20} ~\mathrm{cm}^3\). Next, use the density formula to find the mass: \(m = \rho V = 2.3 \mathrm{~g/cm}^3 \times 3.35103 \times 10^{-20} ~\mathrm{cm}^3 = 7.70674 \times 10^{-20} \mathrm{~g}\).

b) Number of silicon atoms

Now, we need to find out how many silicon atoms are in the quantum dot. First, find the mass of one silicon atom by using the atomic mass: Silicon has an atomic mass of 28.0855 g/mol, so we can find the mass of a single silicon atom: \(\frac{28.0855 \mathrm{~g}}{6.022 \times 10^{23}\ \mathrm{atoms/mol}} = 4.6637 \times 10^{-23} \mathrm{~g/atom}\). Now, divide the total mass of the quantum dot by the mass of one silicon atom to find the number of silicon atoms: \(\frac{7.70674 \times 10^{-20} \mathrm{~g}}{4.6637 \times 10^{-23} \mathrm{~g/atom}} = 1.65 \times 10^3 \ \text{atoms}\).

c) Number of germanium atoms

For the germanium quantum dot, follow the same procedure as above. First, the volume of the germanium quantum dot remains the same as the one for silicon: \(V = 3.35103 \times 10^{-20} ~\mathrm{cm}^3\). Next, find the mass of the germanium quantum dot: \(m = \rho V = 5.325 \mathrm{~g/cm}^3 \times 3.35103 \times 10^{-20} ~\mathrm{cm}^3 = 1.78329 \times 10^{-19} \mathrm{~g}\). Now, find the mass of one germanium atom by using its atomic mass: Germanium has an atomic mass of 72.630 g/mol, so the mass of a single germanium atom is \(\frac{72.630 \mathrm{~g}}{6.022 \times 10^{23}\ \mathrm{atoms/mol}} = 1.206 \times 10^{-22} \mathrm{~g/atom}\). Finally, divide the total mass of the germanium quantum dot by the mass of one germanium atom to find the number of germanium atoms: \(\frac{1.78329 \times 10^{-19} \mathrm{~g}}{1.206 \times 10^{-22} \mathrm{~g/atom}} = 1.478 \times 10^3 \ \text{atoms}\).

Key concepts

Silicon Density

Silicon, a fundamental material in the semiconductor industry, is noted for its relatively low density compared to other semiconductor materials. The density of silicon is given as 2.3 grams per cubic centimeter (g/cm³). This means that for every cubic centimeter of silicon, it weighs 2.3 grams.
Understanding the density of silicon is crucial when calculating properties of objects made from it, like quantum dots. Density is a key factor in determining the mass of a material based on its volume.

• Density is used in the formula for mass: \( m = \rho V \), where \( m \) is mass, \( \rho \) is density, and \( V \) is volume.
• Having a lower density compared to similar materials makes silicon suitable for applications where minimizing weight is beneficial.

Therefore, knowing the density of silicon allows us to calculate the mass of any silicon-based object once we know its volume.

Germanium Density

Germanium, another key material used in electronics, has a higher density than silicon. The density of germanium is 5.325 g/cm³. In many contexts, germanium is favored for its electrical properties but is heavier, affecting design and application decisions.
This density value means that for every cubic centimeter of germanium, it weighs 5.325 grams. When considering its use in quantum dots, this information is critical in determining the physical characteristics, such as mass, of germanium-based structures.

• The formula for mass applies here too: \( m = \rho V \), demonstrating how increased density leads to an increased mass for a given volume.
• Germanium's comparatively high density can influence its suitability for certain technical applications due to the additional weight.

Understanding germanium's density aids in identifying the number of atoms and their arrangement within a specific volume, enhancing the overall grasp of its physical and electronic properties.

Atomic Mass Calculation

To determine how many atoms are in a given sample of a material, such as a quantum dot, you must understand how to calculate atomic mass. Each element has a standard atomic mass based on the average mass of its atoms, measured in grams per mole (g/mol). This average takes isotopic abundances into account.
For instance, silicon's atomic mass is approximately 28.0855 g/mol, while germanium's atomic mass is about 72.630 g/mol. The atomic mass helps estimate the number of atoms by using Avogadro's number, which is \(6.022 \times 10^{23}\) atoms/mol.

• The mass of a single atom is calculated by dividing the atomic mass by Avogadro's number: for silicon, it’s \( \frac{28.0855}{6.022 \times 10^{23}} = 4.6637 \times 10^{-23} \text{ g/atom} \).
• Germanium follows a similar approach: \( \frac{72.630}{6.022 \times 10^{23}} = 1.206 \times 10^{-22} \text{ g/atom} \).

Knowing how to calculate atomic mass from these constants allows scientists and engineers to calculate the number of atoms for any substance, bridging the gap between macroscopic measurements and microscopic quantities.

Volume of a Sphere

The volume of a sphere is a critical geometric property that helps in calculating the mass and number of atoms in spherical quantum dots. The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.
For a better insight, when given a diameter, you first find the radius (half of the diameter) before calculating the volume. In the provided example, a quantum dot of diameter 4 nm has a radius of 2 nm, which must be converted to centimeters (2 \( \times 10^{-7} \text{ cm} \)) to match density units.

• Using the formula: if \( r = 2 \times 10^{-7} \text{ cm} \), then \( V = \frac{4}{3} \pi (2 \times 10^{-7})^3 = 3.35103 \times 10^{-20} \text{ cm}^3 \).
• This uniform method can be used for any sphere, provided the radius is known.

Calculating the volume is the foundational step in obtaining the mass of spherical objects using their density, facilitating subsequent computations for various physical properties.