Chapter 12: Problem 111
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?
Short Answer
Expert verified
The wavelength of light that corresponds to the given photon energy of Greenockite (CdS) is \(477\, \text{nm}\).
Step by step solution
01
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.
02
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:
\[E = 2.6\,\text{eV} \times \dfrac{1.6\times10^{-19}\,\text{J}}{1\,\text{eV}} = 4.16\times10^{-19}\,\text{J}\]
03
Use the formula to find the wavelength
Substitute the given values and constants into the formula and solve for the wavelength:
\[\lambda = \dfrac{hc}{E} = \dfrac{(6.626\times10^{-34}\,\text{Js})(3.00\times10^8\,\text{m/s})}{4.16\times10^{-19}\,\text{J}}\]
04
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 \(477\, \text{nm}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
- 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.
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).
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.