Question

The band gap in aluminum phosphide (AlP) is \(2.5\) electron-volts \(\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\). What wavelength of light is emitted by an AlP diode?

Short answer

The wavelength of light emitted by an AlP diode is calculated using the energy-photon relationship and Planck's equation. First, convert the band gap energy from electron-volts to joules: \(E = 2.5 \ \text{eV} \cdot (1.6 × 10^{-19} \text{ J/eV})\). Then, find the wavelength using the equation \(\lambda = \dfrac{h * c}{E}\), where h is Planck's constant and c is the speed of light. Plugging the values, we obtain \(\lambda ≈ 1.98 × 10^{-7} \text{ m}\), or approximately \(198 \text{ nm}\) when expressed in nanometers.

Step-by-step solution

Convert the energy from electron-volts to joules

The energy of the band gap is given in electron-volts (eV). We need to convert this to joules (J) using the provided conversion factor. \( 1 eV = 1.6 × 10^{-19} J \) So, \(E = 2.5 \ \text{eV} \cdot (1.6 × 10^{-19} \text{ J/eV})\)

Calculate the energy using Planck's equation

Planck's equation relates the energy of a photon with its frequency and wavelength. It should be mentioned that we ultimately need to calculate the wavelength of the emitted light, but to find that, we use the energy of emitted photon. The energy (E) of a photon is given by: \(E = h * f\) where, h is the Planck's constant (\(6.63 × 10^{-34} \text{ J s}\)) and f is the frequency. The relationship between frequency (f) and wavelength (\(\lambda\)) is given by: \(f = \dfrac{c}{\lambda}\) where, c is the speed of light in vacuum (\(3 × 10^{8} \text{ m/s}\)). Combining the two equations: \(E = h * \dfrac{c}{\lambda}\) Now we should have enough information to calculate the wavelength (\(\lambda\)).

Calculate the wavelength of the emitted light

From the above equation, the wavelength (\(\lambda\)) can be calculated as: \(\lambda = \dfrac{h * c}{E}\) Now plug in the relevant values: \(\lambda = \dfrac{6.63 × 10^{-34} \text{ J s} * 3 × 10^{8} \text{ m/s}}{2.5 * 1.6 × 10^{-19} \text{ J}}\)

Evaluate and simplify the expression

Evaluate the expression to find the final value for the wavelength: \(\lambda ≈ 1.98 × 10^{-7} \text{ m}\) This is the wavelength of the light emitted by the AlP diode. Since it's given in meters, you may choose to express the result in nanometers for convenience, by multiplying it by \(10^{9}\): \(\lambda ≈ 198 \text{ nm}\)

Key concepts

Photon Energy

Photon energy is an important concept in understanding how light and other forms of electromagnetic radiation carry energy. Each photon carries a discrete amount of energy, which is directly proportional to its frequency—the higher the frequency, the higher the energy.
One way to calculate the energy of a photon is by using Planck's equation, where the energy (\( E \)) is equal to the Planck's constant (\( h \)) multiplied by the frequency (\( f \)) of the photon:

• \( E = h \times f \)

In practical applications such as the one involving the band gap in aluminum phosphide (AlP), understanding photon energy helps us determine the characteristics of light emitted when electrons transition across the band gap.
Knowing the energy allows us to calculate other important properties like the wavelength of the emitted light, which is crucial in designing optoelectronic devices.

Electron-volts (eV)

An electron-volt is a unit of energy that's often used in physics, especially when dealing with atomic and subatomic particles. Unlike joules, electron-volts provide a convenient way to express very small quantities of energy typical of individual atoms or particles.

• 1 eV equals \(1.6 \times 10^{-19}\) joules.

This conversion is crucial when calculating energy-related properties, like those involving the band gap of a semiconductor like AlP. The band gap gives us the energy required to excite an electron from the valence band to the conduction band in a semiconductor. When expressed in electron-volts, it provides insight into the energy states of electrons, allowing us to predict behaviors like the emission of specific light wavelengths when an electron returns to a lower energy state.

Wavelength Calculation

Calculating the wavelength of emitted light requires using the relationship between energy and wavelength. Knowing one property's value, such as energy, allows us to solve for the wavelength.
Using the formula \( \lambda = \frac{h \times c}{E} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, \( c \) is the speed of light, and \( E \) is the energy of the photon:

• By inserting known values, you can determine the exact wavelength of light emitted.
• This calculation helps us understand the color or type of light a material emits.

For the specific example of aluminum phosphide, knowing the band gap in eV allows us to convert this energy into joules and use the formula for an accurate calculation. Wavelength is usually expressed in nanometers for ease of use, relating directly to visible light properties.

Planck's Constant

Planck's constant is a fundamental value in quantum mechanics, which establishes a proportional relationship between the energy and frequency of a photon. It is a tiny constant with a value of \(6.63 \times 10^{-34}\) J·s.

• It acts as a bridge linking the particle and wave nature of light.
• Helps determine the energy carried by photons for given frequencies.

Planck's constant is essential in calculations involving photon energy, such as determining the wavelength of light emitted from a diode.
It provides a foundation for understanding the quantization of energy levels and is indispensable for making accurate predictions in quantum physics. In practical exercises, like the one involving aluminum phosphide, Planck's constant allows us to connect energy values in electron-volts to measurable physical properties like wavelength.