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Chapter 10: Problem 65

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

Expert verified
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

01

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, \(E = 2.5 \ \text{eV} \cdot (1.6 × 10^{-19} \text{ J/eV})\)
02

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 (\(\lambda\)).
03

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: \(\lambda = \dfrac{6.63 × 10^{-34} \text{ J s} * 3 × 10^{8} \text{ m/s}}{2.5 * 1.6 × 10^{-19} \text{ J}}\)
04

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 \(10^{9}\): \(\lambda ≈ 198 \text{ nm}\)

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