/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A light-emitting diode (LED) has... [FREE SOLUTION] | 91影视

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Chapter 28: Problem 48

A light-emitting diode (LED) has the property that electrons can be excited into the conduction band by the electrical energy from a battery; a photon is emitted when the electron drops back to the valence band. (a) If the band gap for this diode is \(2.36 \mathrm{eV},\) what is the wavelength of the light emitted by the LED? (b) What color is the light emitted? (c) What is the minimum battery voltage required in the electrical circuit containing the diode?

Short Answer

Expert verified
Answer: The wavelength of the emitted light is approximately 526 nm, and the color is green. The minimum battery voltage required to power the LED is approximately 2.36 V.

Step by step solution

01

Determine the energy of the emitted photon

The energy of the emitted photon can be calculated using the given band gap energy. The band gap energy is provided in electron volts (eV). We need to convert it to joules (J) before proceeding. We can do this by multiplying the band gap energy by the charge of an electron (\(1.6 \times 10^{-19} \mathrm{C}\)). $$E = 2.36 \mathrm{eV} \times (1.6 \times 10^{-19} \mathrm{C})= 3.776 \times 10^{-19} \mathrm{J}$$
02

Calculate the wavelength of the emitted light

Since we now have the energy of the emitted photon, we can use the energy-wavelength relationship to find the wavelength of the light emitted. The relationship between energy and wavelength is given as: $$E = \dfrac{hc}{\lambda}$$ where \(E\) is the energy of the photon, \(h = 6.626 \times 10^{-34} \mathrm{Js}\) is Planck's constant, \(c = 2.998 \times 10^8 \mathrm{m/s}\) is the speed of light, and \(\lambda\) is the wavelength. We will solve for the wavelength: $$\lambda = \dfrac{hc}{E} = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^8 \mathrm{m/s})}{3.776 \times 10^{-19} \mathrm{J}} \approx 526 \times 10^{-9} \mathrm{m}$$
03

Determine the color of the emitted light

Now that we have the wavelength of the light emitted, let's determine the color of the light. The wavelength (in nanometers) falls within the visible light spectrum. The approximate ranges for colors in the visible spectrum are: - Violet: 380 nm - 450 nm - Blue: 450 nm - 495 nm - Green: 495 nm - 570 nm - Yellow: 570 nm - 590 nm - Orange: 590 nm - 620 nm - Red: 620 nm - 750 nm The calculated wavelength (~526 nm) falls within the green range. So the color of the emitted light is green.
04

Calculate the minimum battery voltage required

The minimum battery voltage required to power the LED can be found using the given band gap energy. Since the band gap energy is given in electron volts, we can use the energy directly to determine the minimum required voltage. The relation between energy and voltage is given as: $$E = qV$$ where \(E\) is the energy, \(q = 1.6 \times 10^{-19} \mathrm{C}\) is the charge of an electron, and \(V\) is the voltage. We will solve for the voltage: $$V = \dfrac{E}{q} = \dfrac{2.36 \mathrm{eV}}{1.6 \times 10^{-19} \mathrm{C}} \approx 2.36 \mathrm{V}$$ The minimum battery voltage required for the LED is approximately 2.36 V.

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Most popular questions from this chapter

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