/*! 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 At what wavelength will an LED r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At what wavelength will an LED radiate if made from a material with an energy gap \(E_{\mathrm{g}}=1.6 \mathrm{eV} ?\)

Short Answer

Expert verified
The LED will radiate at a wavelength of 775 nm.

Step by step solution

01

Understand the Relationship between Energy and Wavelength

The wavelength at which an LED radiates is related to the energy gap, \( E_g \), of the material through the equation \( E = \frac{hc}{\lambda} \), where \( E \) is the energy in joules, \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ J} \cdot \text{s} \)), \( c \) is the speed of light (\(3.00 \times 10^8 \text{ m/s} \)), and \( \lambda \) is the wavelength in meters. Our goal is to solve this equation for \( \lambda \).
02

Convert Energy Gap from Electronvolts to Joules

Given the energy gap \( E_g = 1.6 \text{ eV} \), we convert it to joules using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Thus, \( E = 1.6 \times 1.602 \times 10^{-19} \text{ J} = 2.5632 \times 10^{-19} \text{ J} \).
03

Solve for Wavelength

Using the formula \( E = \frac{hc}{\lambda} \), we rearrange to find \( \lambda = \frac{hc}{E} \). Substituting the values we have: \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}, c = 3.00 \times 10^8 \text{ m/s}, \text{ and } E = 2.5632 \times 10^{-19} \text{ J} \), we get \( \lambda = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{2.5632 \times 10^{-19}} \).
04

Calculate the Wavelength

Carrying out the calculation gives \( \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{2.5632 \times 10^{-19}} = 7.75 \times 10^{-7} \text{ m} \). Converting this to nanometers gives \( 775 \text{ nm} \), since \( 1 \text{ m} = 10^9 \text{ nm} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Gap
The energy gap, often denoted as \(E_g\), is a fundamental property of a material that determines the energy needed to move an electron from the valence band to the conduction band. This energy transition is what produces light in a light-emitting diode (LED).
An LED's color, corresponding to its wavelength of emitted light, is directly linked to the size of the energy gap. The larger the energy gap, the shorter the wavelength and higher the energy of the light. Conversely, a smaller energy gap means a longer wavelength and lower energy light.
In our exercise, the energy gap is given as \(1.6\text{ eV}\). This refers to the energy barrier that electrons need to overcome to emit light. By understanding this concept, we know the energy involved which is crucial for calculating the wavelength.
Planck's Constant
Planck's constant is a fundamental constant denoted as \(h\), with a value of \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\). It plays a critical role when dealing with quantum mechanics and the relationship between energy and frequency of electromagnetic waves.
In the context of LEDs, Planck's constant helps us determine the energy of photons emitted by the semiconductor material. The equation \(E = \frac{hc}{\lambda}\) demonstrates how Planck's constant directly relates energy \(E\) to wavelength \(\lambda\), by incorporating the speed of light \(c\). Using Planck's constant allows us to calculate precise energy levels corresponding to specific wavelengths of light emitted by an LED.
Wavelength Conversion
When calculating the wavelength of radiation emitted from an LED, we often start from an energy value expressed in joules and convert it to wavelength using the equation \(\lambda = \frac{hc}{E}\). The result typically comes in meters, which we then convert to nanometers to match common LED wavelength specifications.
In simpler terms, one meter equals \(10^9\) nanometers, so multiplying by this factor converts the LED's wavelength from meters to nanometers. This is crucial since most practical applications, such as lighting and electronics involving LEDs, require the wavelength to be specified in terms of nanometers.
Electronvolt to Joule Conversion
In physics, different energy levels are measured in various units, but converting these energies to a standard unit like joules can simplify calculations. An electronvolt (eV) is a unit commonly used to express energy on a smaller scale, suitable for atomic and subatomic particles.
When converting the energy gap of an LED from electronvolts to joules, we use the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). This conversion is vital as it allows us to use standard equations involving Planck's constant and the speed of light, which are defined in joules. For our exercise, converting the energy gap of \(1.6 \text{ eV}\) to \(2.5632 \times 10^{-19} \text{ J}\) enabled precise calculation of the LED's emission wavelength.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(II) (a) For copper at room temperature \((T=293 \mathrm{K})\) , calculate the Fermi factor, Eq. \(14,\) for an electron with energy 0.12 \(\mathrm{eV}\) above the Fermi energy. This represents the probability that this state is occupied. Is this reasonable? (b) What is the probability that a state 0.12 \(\mathrm{eV}\) below the Fermi energy is occupied? (c) What is the probability that the state in part \((b)\) is unoccupied? \(f(E)=\frac{1}{e^{\left(E-E_{\mathrm{F}}\right) / k T}+1}\)

The energy gap between valence and conduction bands in germanium is \(0.72 \mathrm{eV}\). What range of wavelengths can a photon have to excite an electron from the top of the valence band into the conduction band?

The energy gap \(E_{\mathrm{g}}\) in germanium is \(0.72 \mathrm{eV}\). When used as a photon detector, roughly how many electrons can be made to jump from the valence to the conduction band by the passage of a \(730-\mathrm{keV}\) photon that loses all its energy in this fashion?

A very simple model of a "one-dimensional" metal consists of \(N\) electrons confined to a rigid box of width \(\ell\). We neglect the Coulomb interaction between the electrons. (a) Calculate the Fermi energy for this one-dimensional metal \(\left(E_{\mathrm{F}}=\right.\) the energy of the most energetic electron at \(T=0 \mathrm{~K}\) ), taking into account the Pauli exclusion principle. You can assume for simplicity that \(N\) is even. \((b)\) What is the smallest amount of energy, \(\Delta E,\) that this \(1-\mathrm{D}\) metal can absorb? \((c)\) Find the limit of \(\Delta E / E_{\mathrm{F}}\) for large \(N\). What does this result say about how well metals can conduct?

A strip of silicon \(1.8 \mathrm{~cm}\) wide and \(1.0 \mathrm{~mm}\) thick is immersed in a magnetic field of strength \(1.3 \mathrm{~T}\) perpendicular to the strip (Fig. \(40-47\) ). When a current of \(0.28 \mathrm{~mA}\) is run through the strip, there is a resulting Hall effect voltage of \(18 \mathrm{mV}\) across the strip (Section \(27-8\) ). How many electrons per silicon atom are in the conduction band? The density of silicon is \(2330 \mathrm{~kg} / \mathrm{m}^{3}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.