Problem 91 - Chemistry The Central Science In Si Units

Question

Very small semiconductor crystals, composed of approximately 1000 to 10,000 atoms, are called quantum dots. Quantum dots made of the semiconductor CdSe are now being used in electronic reader and tablet displays because they emit light efficiently and in multiple colors, depending on dot size. The density of CdSe is \(5.82 \mathrm{~g} / \mathrm{cm}^{3}\). (a) What is the mass of one \(2.5-\mathrm{nm}\) CdSe quantum dot? (b) CdSe quantum dots that are \(2.5 \mathrm{nm}\) in diameter emit blue light upon stimulation. Assuming that the dot is a perfect sphere and that the empty space in the dot can be neglected, calculate how many Cd atoms are in one quantum dot of this size. (c) What is the mass of one \(6.5-\mathrm{nm}\) CdSe quantum dot? (d) CdSe quantum dots that are \(6.5 \mathrm{nm}\) in diameter emit red light upon stimulation. Assuming that the dot is a perfect sphere, calculate how many Cd atoms are in one quantum dot of this size. (e) If you wanted to make one \(6.5-\mathrm{nm}\) dot from multiple \(2.5-\mathrm{nm}\) dots, how many \(2.5-\mathrm{nm}\) dots would you need, and how many CdSe formula units would be left over, if any?

Step-by-Step Solution

Answer

One 2.5-nm dot has a mass of \(4.76 \times 10^{-20}\) g and contains ~1,500 Cd atoms. One 6.5-nm dot has a mass of \(8.36 \times 10^{-19}\) g and ~26,300 Cd atoms. About 18 small dots are needed to make a large one, with 700 CdSe units left over.

Step 1: Calculate the Volume of the Quantum Dot

For a spherical object, the volume is calculated by the formula \( V = \frac{4}{3} \pi r^3 \). The radius \( r \) of a 2.5-nm quantum dot is \( \frac{2.5}{2} = 1.25 \) nm. Convert this to centimeters: \( r = 1.25 \times 10^{-7} \) cm. Thus, the volume is \[ V = \frac{4}{3} \pi (1.25 \times 10^{-7})^3 \approx 8.181 \times 10^{-21} \text{ cm}^3 \].

Step 2: Calculate the Mass of the 2.5-nm Quantum Dot

The mass is found by multiplying the density by the volume: \( \text{mass} = \text{density} \times \text{volume} \). With a density of \(5.82 \text{ g/cm}^3\), the mass is \( 5.82 \times 8.181 \times 10^{-21} \approx 4.76 \times 10^{-20} \text{ g} \).

Step 3: Calculate the Number of Cd and Se Atoms in the 2.5-nm Dot

The molar mass of CdSe is the sum of the molar masses of Cd (112.41 g/mol) and Se (78.96 g/mol), which yields 191.37 g/mol. Avogadro's number is \( 6.022 \times 10^{23} \text{ mol}^{-1} \). The number of formula units in the dot is \[ \frac{4.76 \times 10^{-20}}{191.37} \times 6.022 \times 10^{23} \approx 1.50 \times 10^3 \text{ units} \]. Since each unit has one Cd, we have approximately 1,500 Cd atoms.

Step 4: Calculate the Mass of the 6.5-nm Quantum Dot

The volume is calculated similarly with a radius of \( \frac{6.5}{2} = 3.25 \) nm or \( 3.25 \times 10^{-7} \) cm. Thus \[ V = \frac{4}{3} \pi (3.25 \times 10^{-7})^3 \approx 1.437 \times 10^{-19} \text{ cm}^3 \]. The mass is \( 5.82 \times 1.437 \times 10^{-19} \approx 8.36 \times 10^{-19} \text{ g} \).

Step 5: Calculate the Number of Cd Atoms in 6.5-nm Dot

Using the molar mass of 191.37 g/mol, the number of units in the dot is \[ \frac{8.36 \times 10^{-19}}{191.37} \times 6.022 \times 10^{23} \approx 2.63 \times 10^4 \text{ units} \]. Each unit represents a Cd atom, so there are approximately 26,300 Cd atoms.

Step 6: Calculate Number of 2.5-nm Dots needed for One 6.5-nm Dot

Calculate ratio of volumes: Volume ratio \( = \frac{1.437 \times 10^{-19}}{8.181 \times 10^{-21}} \approx 17.56 \). Therefore, 18 dots (since you need whole dots) are used. Using 18 smaller dots, you create one larger dot but use \(18 \times 1.50 \times 10^3 = 27,000\) total formula units, more than needed for one 6.5-nm dot.

Key Concepts

Semiconductor Crystals

Semiconductors are materials that possess electrical conductivity between conductors (like metals) and insulators (like glass). They are typically made up of materials that have an arrangement of atoms known as a crystal lattice. This lattice structure allows them to conduct electricity more easily than insulators, but not as efficiently as conductors.

In semiconductor crystals, the atoms are arranged in a repeating pattern, which can influence the behavior of electrons within the material. This regular pattern is crucial because it impacts the energy levels at which electrons can exist. For example, simple heating or illumination can cause electrons to move inferring conductivity properties.

One unique type of semiconductor crystal is the quantum dot, a tiny particle consisting of thousands of atoms. Quantum dots are significant because their small size allows them to display unique properties, especially in terms of how they react to stimulation like light or an electric current. They are renowned for their ability to emit light efficiently across different colors, depending primarily on their size. Because of this, they are becoming increasingly popular in display technologies, where they help produce vivid colors with greater energy efficiency.

CdSe Quantum Dot

CdSe, or Cadmium Selenide, is a specific type of semiconductor material that is often used to form quantum dots. These are nanoscale semiconductor particles made from a combination of cadmium (Cd) and selenium (Se).

Because of their small size, CdSe quantum dots have quantum mechanical properties that differ significantly from bulk materials made of the same constituents. When a quantum dot is stimulated, such as with a light, it can emit light. The specific wavelength (or color) of this emitted light depends largely on the size of the quantum dot.

Smaller CdSe quantum dots, like the 2.5 nm ones in the problem statement, emit short-wavelength light such as blue. Larger dots, such as the 6.5 nm ones, can emit longer wavelengths like red. This controllability of light emission based on size is part of what makes CdSe quantum dots so valuable in electronics and photonics, particularly in creating efficient displays for TVs, monitors, and other screens.

Volume Calculation

Calculating the volume of a semiconductor quantum dot is essential when determining the number of atoms it contains or its total mass. Since these dots are extremely small and often spherical, we use the volume formula for a sphere: \[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere.

To find the volume of a quantum dot with a diameter of 2.5 nm, first convert half of this value (the radius) into centimeters \( (2.5/2 \times 10^{-7} \text{ cm}) \), and apply it to the formula:
- Calculate the radius in cm:
- Radius \( r = 1.25 \times 10^{-7} \text{ cm} \)
- Plug into the formula:
- Volume \( V \approx \frac{4}{3} \pi (1.25 \times 10^{-7})^3 \approx 8.181 \times 10^{-21} \text{ cm}^3 \)

This computation can be similarly applied to other sizes of quantum dots, such as the 6.5 nm diameter dots. Volume calculations help in applications involving mass calculations where density and the number of atoms or molecules must be determined. Understanding these mathematical concepts is crucial for applications in materials science and nanotechnology.