Problem 111 - Chemistry: The Central Science in SI Units, Global Edition

Question

Greenockite (CdS) was utilized as a pigment known as vermillion. It has a band gap of \(2.6 \mathrm{eV}\) near room temperature for the bulk solid. What wavelength of light (in \(\mathrm{nm}\) ) would a photon of this energy correspond to?

Step-by-Step Solution

Answer

The wavelength of light that corresponds to the given photon energy of Greenockite (CdS) is \(477\, \text{nm}\).

Step 1: Write down the given information and the formula we will use

We are given: - Bandgap energy: \(E = 2.6 eV\) - We will need the speed of light constant: \(c = 3.00\times10^8 m/s\) We want to find: - Wavelength of light: \(\lambda\) The formula relating energy, wavelength, and the speed of light is: \[E = \dfrac{hc}{\lambda}\] where \(h\) is the Planck's constant \(h = 6.626\times10^{-34} Js\), \(c\) is the speed of light, and \(\lambda\) is the wavelength.

Step 2: Convert energy from electron-volt to joules

To find the wavelength, we first need to convert the given bandgap energy from eV to Joules. The conversion factor is: \[1\,\text{eV} = 1.6\times10^{-19}\,\text{J}\] Thus, converting the energy: \[E = 2.6\,\text{eV} \times \dfrac{1.6\times10^{-19}\,\text{J}}{1\,\text{eV}} = 4.16\times10^{-19}\,\text{J}\]

Step 3: Use the formula to find the wavelength

Substitute the given values and constants into the formula and solve for the wavelength: \[\lambda = \dfrac{hc}{E} = \dfrac{(6.626\times10^{-34}\,\text{Js})(3.00\times10^8\,\text{m/s})}{4.16\times10^{-19}\,\text{J}}\]

Step 4: Calculate the wavelength and convert it to nanometers

Calculate the wavelength and convert it to nanometers: \[\lambda = \dfrac{(6.626\times10^{-34}\,\text{Js})(3.00\times10^8\,\text{m/s})}{4.16\times10^{-19}\,\text{J}} = 4.77 \times 10^{-7} \,\text{m}\] To convert the wavelength to nanometers, multiply by \(10^9\) : \[\lambda = 4.77\times10^{-7}\,\text{m} \times \dfrac{10^9\,\text{nm}}{1\,\text{m}} = 477\,\text{nm}\] So, the wavelength of light that corresponds to the given photon energy of Greenockite (CdS) is \(477\, \text{nm}\).

Key Concepts

Bandgap Energy

Bandgap energy is a fundamental property of semiconductors, like Greenockite (CdS), which was used as a pigment. It measures the energy difference between the valence band and the conduction band in a solid material.

• Valence Band: This is where the electrons are before they gain enough energy to jump.
• Conduction Band: This is where the electrons move to after gaining energy.

Understanding bandgap energy is important because it determines a material's electrical conductivity and the type of light it absorbs. When the bandgap is large, the material is usually a good insulator. For semiconductors, like CdS, the bandgap is moderate, allowing control over their conductive properties.
CdS has a bandgap of 2.6 eV, which we can use to calculate the energy of a photon that can traverse this bandgap. This concept is crucial in photovoltaics and electronics, where bandgap energies dictate the efficiency and functionality of devices.

Photon Energy

Photon energy refers to the energy carried by a single photon, the smallest unit or "packet" of light energy. In physics, photon energy can be calculated using the formula:\[ E = hf \]where:

• \(E\) is the photon energy in Joules (J).
• \(h\) is Planck's constant, \(6.626 \times 10^{-34} \text{ Js}\).
• \(f\) is the frequency of the light in Hertz (Hz).

To find the energy of a photon needed to match the bandgap energy of Greenockite, we used the conversion from electron-volts (eV) to Joules, enhancing comprehension by using the mathematical formula:\[ E = 2.6 \, \text{eV} = 4.16 \times 10^{-19} \, \text{J} \]This calculation helps determine the wavelength of light absorbed or emitted by materials, especially in studying the interaction between light and matter. Light with energy matching the bandgap will get absorbed, allowing electrons to jump from one band to another.

Planck's Constant

Planck's constant \(h\) is a vital concept in quantum mechanics and physics in general. It plays a key role in the energy-frequency relationship for photons. Planck introduced this constant, \(6.626 \times 10^{-34} \text{ Js}\), which is used to describe how much energy a photon has based on its frequency. It's a fundamental bridge between the energy and the properties of electromagnetic waves.In calculations involving bandgap energy or photon energy, Planck's constant is indispensable because it allows us to connect energy to wavelength. The formula linking these is:\[ E = \dfrac{hc}{\lambda} \]where:

• \(h\) is the Planck's constant.
• \(c\) is the speed of light, \(3.00 \times 10^{8} \text{ m/s}\).
• \(\lambda\) is the wavelength of light in meters.

Understanding Planck's constant is essential for anyone studying quantum mechanics or fields related to the quantum description of light. It helps to comprehend how electrons behave in semiconductors and plays a significant role in the development of various technologies, from quantum computing to lasers.