Problem 80 - Chemistry The Central Science

Question

Red light-emitting diodes are made from GaAs and GaP solid solutions, GaP \(_{x} A s_{1-x}(\) see Exercise 12.79\() .\) The original red LEDs emitted light with a wavelength of 660 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

Answer

The composition used in red LEDs, GaP\(_x\)As\(_{1-x}\), with a wavelength of 660 nm is approximately \(x \approx 0.55\).

Step 1: Calculate the energy corresponding to the wavelength of light emitted by the LEDs

Using Planck's equation, we can find the photon energy related to the wavelength of the emitted light: \[E = h \nu\] where: \(E\) is the energy of the photon, \(h\) is the Planck constant, which is approximately \(6.63 \times 10^{-34}\, J s\), \(\nu\) is the frequency of the light. Since the frequency \(\nu\) can also be expressed as \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light (approximately \(3 \times 10^8\, m/s\)), and \(\lambda\) is the wavelength of the emitted light, 660 nm in this case. Now we can rewrite the equation as: \[E = h \frac{c}{\lambda}\] Plug in the given values: \[E = (6.63 \times 10^{-34} \, J s) \times \frac{3 \times 10^8 \, m/s}{660 \times 10^{-9} \, m}\]

Step 2: Calculate the photon energy

As we have the equation \(E = h \frac{c}{\lambda}\) and the values of \(h\), \(c\), and \(\lambda\), we can now calculate the photon energy: \[E = (6.63 \times 10^{-34} \, J s) \times \frac{3 \times 10^8 \, m/s}{660 \times 10^{-9} \, m} = 3.02 \times 10^{-19} \, J\] So, the energy of the emitted photons is approximately \(3.02 \times 10^{-19}\, J\).

Step 3: Apply the linear relation for band gap

We are given that the band gap energy for GaP\(_x\)As\(_{1-x}\) varies linearly with the composition. This can be expressed as: \[E = E_{0} + x(E_{1} - E_{0})\] where: \(E\) is the band gap energy of GaP\(_x\)As$_{1-x}\), \(E_{0}\) is the band gap energy of GaAs (at \(x=0\)), \(E_{1}\) is the band gap energy of GaP (at \(x=1\)), \(x\) is the composition we want to find. From the literature, the values of \(E_{0}\) and \(E_{1}\) are approximately \(1.43\, eV\) and \(2.26\, eV\), respectively. We must convert the calculated photon energy from Step 2 to electron-volts (eV). To do this, we divide the energy calculated in Step 2 by the elementary charge (\(1.6 \times 10^{-19}\, C\)): \[E = 3.02 \times 10^{-19} \, J \times \frac{1 \, eV}{1.6 \times 10^{-19} \, C} = 1.89\,eV\] Now we can insert the known values into the linear equation: \[1.89\,eV = 1.43\,eV + x(2.26\,eV - 1.43\,eV)\]

Step 4: Solve for the value of x

Now we have only one unknown value, \(x\), we can rearrange the equation and solve for \(x\). \[x = \frac{1.89\,eV - 1.43\,eV}{2.26\,eV - 1.43\,eV} = \frac{0.46\,eV}{0.83\,eV}\] \[x \approx 0.55\] So, the composition used in the red LEDs, GaP\(_x\)As\(_{1-x}\), is approximately \(x \approx 0.55\).

Key Concepts

GaAs

Gallium Arsenide (GaAs) is a compound made up of gallium and arsenic. It is a popular material in semiconductors because it offers a direct band gap. This means electrons can move between the valence and conduction bands without losing energy.
This direct band gap makes GaAs useful in optoelectronic devices like LEDs and laser diodes. It is efficient at converting electronic signals into light.
Compared to silicon, GaAs allows for faster electron mobility, which is great for high-frequency applications.

GaP

Gallium Phosphide (GaP) is another semiconductor material, consisting of gallium and phosphorus. Unlike GaAs, GaP has an indirect band gap. This means the process of electron transition between bands involves a change in momentum with a loss of energy, making it less efficient for light emission.
However, GaP is still widely used in optoelectronics, especially for LED technology. It is commonly used for green to red LEDs due to its robust properties.

• GaP is cheaper than GaAs.
• It provides good transparency to visible light.
• It’s often alloyed with GaAs to enhance performance.

Planck's equation

Planck's equation is vital in understanding how light interacts with materials. It describes the energy of photons, which are particles of light. The equation is represented as: \[ E = h u \]Here, \(E\) is the energy of a photon, \(h\) is the Planck constant, around \(6.63 \times 10^{-34} \, J \cdot s\), and \(u\) is the frequency of the light.
By using the speed of light \(c\) and the wavelength \(\lambda\), we can express frequency as \(u = \frac{c}{\lambda}\). This lets us calculate the energy of light emissions from semiconductors, crucial for devices like LEDs.

LEDs

Light Emitting Diodes (LEDs) are semiconductor devices that emit light when an electric current passes through them. They utilize materials like GaAs and GaP to produce light efficiently.
LEDs are highly valued for their energy efficiency and durability. They work by allowing electrons to recombine with holes in the material, releasing energy as photons, or light.

• LEDs consume less power than traditional bulbs.
• They have a longer lifespan.
• They're available in many colors, depending on the semiconductor material used.

Semiconductor

Semiconductors are materials that have electrical conductivity between conductors and insulators. They are the foundation for modern electronics, including computers, smartphones, and LEDs.
Their ability to conduct electric current can be altered by introducing impurities, a process known as doping. This allows semiconductors to have specific properties needed for different applications.

• They can be intrinsic or extrinsic based on purity.
• They efficiently control the flow of electricity.
• Widely used in diodes, transistors, and solar cells.

Overall, semiconductors' versatility makes them indispensable in technology today.