Problem 74
Question
Red light-emitting diodes are made from GaAs and GaP solid solutions, \(\mathrm{GaP}_{x}\) As \(_{1-x}\) (see Exercise 12.73 ). The original red LEDs emitted light with a wavelength of \(660 \mathrm{nm}\). If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1\), estimate the composition (the value of \(x\) ) that is used in these LEDs.
Step-by-Step Solution
Verified Answer
The composition of the red LED made from \(\mathrm{GaP}_{x}\) As \(_{1-x}\) solid solution is approximately \(x \approx 0.597\).
1Step 1: Convert the wavelength to energy
First, we need to convert the wavelength of emitted light (\(660 \mathrm{nm}\)) to energy. This conversion is done using the speed of light (\(c\)), Planck's constant (\(h\)), and the wavelength (\(\lambda\)) using the following equation:
\[ E = \frac{hc}{\lambda} \]
Where \(E\) is the energy (in Joules), \(c = 3 \times 10^{8} \: \mathrm{m/s}\) (speed of light), and \(h = 6.63 \times 10^{-34} \: \mathrm{Js}\) (Planck's constant).
2Step 2: Calculate the energy emitted by LED
Now, we will plug the values into the energy equation:
\[ E = \frac{ (6.63 \times 10^{-34}\: \mathrm{Js}) \times (3 \times 10^{8} \: \mathrm{m/s}) }{ 660 \times 10^{-9} \: \mathrm{m} } \]
Which simplifies to:
\[ E = 3.01 \times 10^{-19} \: \mathrm{J} \]
3Step 3: Convert energy to electron-volts
For simplicity and easy comparison, it is common to convert energy to electron-volts (eV). One electron-volt is equal to \(1.6 \times 10^{-19} \: \mathrm{J}\). Thus, to convert the energy to eV, we can use the following equation:
\[ E = 3.01 \times 10^{-19} \: \mathrm{J} \times \frac{1 \: \mathrm{eV}}{1.6 \times 10^{-19} \: \mathrm{J}} \]
This simplifies to:
\[ E = 1.88 \: \mathrm{eV} \]
4Step 4: Consider the band gap's relationship with the composition
We have the band gap energy value for the LED as \(1.88 \: \mathrm{eV}\). We know that the band gap varies linearly between gallium arsenide (GaAs) at \(x=0\) and gallium phosphide (GaP) at \(x=1\). The band gaps for GaAs and GaP are \(1.44 \: \mathrm{eV}\) and \(2.29 \: \mathrm{eV}\), respectively.
The linear relationship between the band gap (\(E_g\)) and the composition (\(x\)) is given by the equation:
\[ E_g = x \times E_g(\mathrm{GaP}) + (1-x) \times E_g(\mathrm{GaAs}) \]
5Step 5: Calculate the composition
Now, we can plug in the values and solve for \(x\):
\[ 1.88 = x \times 2.29 + (1 - x) \times 1.44 \]
Rearrange the equation:
\[ x = \frac{1.88 - 1.44}{2.29 - 1.44} \]
Solving for \(x\):
\[ x \approx 0.597 \]
Therefore, the composition of the red LED made from \(\mathrm{GaP}_{x}\) As \(_{1-x}\) solid solution is approximately \(x \approx 0.597\).
Key Concepts
LED composition calculationEnergy-wavelength conversionElectron-volt energy unit
LED composition calculation
Understanding the composition of light-emitting diodes (LEDs) is crucial when designing these devices to emit specific colors of light. The composition directly affects the LED's semiconductor band gap, which in turn determines the color of light emitted. For instance, red LEDs typically require a band gap close to the energy associated with red light.
In our exercise, we're looking at a compound composed of gallium arsenide (GaAs) and gallium phosphide (GaP), denoted as \( \mathrm{GaP}_{x} \) As \( _{1-x} \). The variable 'x' here represents the mole fraction of GaP in the compound. The goal is to calculate this value to produce red light with a wavelength of 660 nm. The band gap energies for pure GaAs and GaP are given, and we are informed that the overall band gap energy varies linearly between these two values.
To calculate the proper composition 'x', we use the relationship that the band gap energy of the LED corresponds to the energy of the emitted light. After computing this energy, we use the known band gap energies of pure GaAs and GaP to establish a linear equation. Solving this gives us the value of 'x', and thus the exact composition required for emitting red light at 660 nm wavelength.
In our exercise, we're looking at a compound composed of gallium arsenide (GaAs) and gallium phosphide (GaP), denoted as \( \mathrm{GaP}_{x} \) As \( _{1-x} \). The variable 'x' here represents the mole fraction of GaP in the compound. The goal is to calculate this value to produce red light with a wavelength of 660 nm. The band gap energies for pure GaAs and GaP are given, and we are informed that the overall band gap energy varies linearly between these two values.
To calculate the proper composition 'x', we use the relationship that the band gap energy of the LED corresponds to the energy of the emitted light. After computing this energy, we use the known band gap energies of pure GaAs and GaP to establish a linear equation. Solving this gives us the value of 'x', and thus the exact composition required for emitting red light at 660 nm wavelength.
Energy-wavelength conversion
The conversion between the energy of a photon and its wavelength is a fundamental concept in understanding how LEDs work. This process is described by the equation \( E = \frac{hc}{\lambda} \), where 'E' is energy in Joules, 'h' is Planck's constant (\(6.63 \times 10^{-34} \: \mathrm{Js}\)), 'c' is the speed of light (\(3 \times 10^{8} \: \mathrm{m/s}\)), and '\(\lambda\)' is the wavelength of light.
This equation shows the inverse relationship between energy and wavelength: shorter wavelengths correspond to higher energy photons and vice versa. When dealing with LEDs, understanding this relationship is critical because the energy of the photons they emit (and thus the color of the light) is directly related to the properties of the semiconductor material used, such as its band gap energy.
For the red LED we discussed, emitting light at 660 nm requires a photon with a certain energy calculated using the above equation. The ability to perform this calculation is vital for engineers and physicists working with optical technologies, as it provides a direct link between the physical properties of the material and the characteristics of the light produced.
This equation shows the inverse relationship between energy and wavelength: shorter wavelengths correspond to higher energy photons and vice versa. When dealing with LEDs, understanding this relationship is critical because the energy of the photons they emit (and thus the color of the light) is directly related to the properties of the semiconductor material used, such as its band gap energy.
For the red LED we discussed, emitting light at 660 nm requires a photon with a certain energy calculated using the above equation. The ability to perform this calculation is vital for engineers and physicists working with optical technologies, as it provides a direct link between the physical properties of the material and the characteristics of the light produced.
Electron-volt energy unit
The electron-volt (eV) is a unit of energy that is particularly convenient and widely used in the field of semiconductors and photonics. It is defined as the amount of kinetic energy gained or lost by a single electron moving through an electric potential difference of one volt. One electron-volt is equivalent to \(1.6 \times 10^{-19} \: \mathrm{J}\).
Since the energies involved in photon emission and electronic transitions in semiconductors are typically very small, the electron-volt is much more practical than using joules. For example, the LED's energy calculated earlier can be converted into electron-volts, resulting in a more manageable number that better illustrates the minute energy changes within the semiconductor.
Understanding how to convert between joules and electron-volts is important when studying semiconductor devices, as technical literature and datasheets often provide energy levels and band gaps in eV. Moreover, it simplifies comparisons of energy levels across different materials and permits ease of calculations, especially when coupled with the knowledge of how to translate between energy and wavelength for photons.
Since the energies involved in photon emission and electronic transitions in semiconductors are typically very small, the electron-volt is much more practical than using joules. For example, the LED's energy calculated earlier can be converted into electron-volts, resulting in a more manageable number that better illustrates the minute energy changes within the semiconductor.
Understanding how to convert between joules and electron-volts is important when studying semiconductor devices, as technical literature and datasheets often provide energy levels and band gaps in eV. Moreover, it simplifies comparisons of energy levels across different materials and permits ease of calculations, especially when coupled with the knowledge of how to translate between energy and wavelength for photons.
Other exercises in this chapter
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