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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

Before the discovery of the neutron, one theory of the nucleus proposed that the nucleus contains protons and electrons. For example, the helium-4 nucleus would contain 4 protons and 2 electrons instead of - as we now know to be true- 2 protons and 2 neutrons. (a) Assuming that the electron moves at nonrelativistic speeds, find the ground-state energy in mega-electron- volts of an electron confined to a one-dimensional box of length $5.0 \mathrm{fm}\( (the approximate diameter of the \)^{4} \mathrm{He}$ nucleus). (The electron actually does move at relativistic speeds. See Problem \(80 .)\) (b) What can you conclude about the electron-proton model of the nucleus? The binding energy of the \(^{4} \mathrm{He}\) nucleus - the energy that would have to be supplied to break the nucleus into its constituent particles-is about \(28 \mathrm{MeV} .\) (c) Repeat (a) for a neutron confined to the nucleus (instead of an electron). Compare your result with (a) and comment on the viability of the proton-neutron theory relative to the electron-proton theory.
An electron in a one-dimensional box has ground-state energy 0.010 eV. (a) What is the length of the box? (b) Sketch the wave functions for the lowest three energy states of the electron. (c) What is the wavelength of the electron in its second excited state \((n=3) ?\) (d) The electron is in its ground state when it absorbs a photon of wavelength \(15.5 \mu \mathrm{m}\). Find the wavelengths of the photon(s) that could be emitted by the electron subsequently.
(a) Show that the ground-state energy of the hydrogen atom can be written \(E_{1}=-k e^{2} /\left(2 a_{0}\right),\) where \(a_{0}\) is the Bohr radius. (b) Explain why, according to classical physics, an electron with energy \(E_{1}\) could never be found at a distance greater than \(2 a_{0}\) from the nucleus.
A double-slit interference experiment is performed with 2.0-ev photons. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (i.e., the spacing between maxima is the same)?
What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) \(1.0 \mathrm{eV}\) (b) $1.0 \mathrm{keV} .
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