/*! 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 81 Most of the Sun's radiation has ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Most of the Sun's radiation has wavelengths shorter than \(1100 \mathrm{nm}\). For a solar cell to absorb all this, what energy gap ought the material have?

Short Answer

Expert verified
The material should have an energy gap of approximately 1.12 eV.

Step by step solution

01

Understanding the Problem

To absorb all radiation with wavelengths shorter than 1100 nm, the solar cell's material must have an energy gap corresponding to this wavelength or shorter. The energy gap is related to the wavelength of the absorbed light through the energy equation for photons.
02

Relating Wavelength to Energy

The energy of a photon is given by the formula: \[ E = \frac{hc}{\lambda} \]where:- \( E \) is the energy in joules (J),- \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\),- \( c \) is the speed of light \(3 \times 10^8 \text{ m/s}\), and- \( \lambda \) is the wavelength in meters.
03

Converting Wavelength to Meters

The given wavelength is in nanometers, so first, convert 1100 nm to meters using \[ 1 \text{ nm} = 10^{-9} \text{ m} \]. Therefore, \[ 1100 \text{ nm} = 1100 \times 10^{-9} \text{ m} = 1.1 \times 10^{-6} \text{ m} \].
04

Calculating the Energy Gap

Substitute the values into the energy equation to find the energy gap:\[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.1 \times 10^{-6}} \].Calculate the value:\[ E \approx 1.80 \times 10^{-19} \text{ J} \].
05

Converting Energy to Electronvolts

The energy gap is often expressed in electronvolts (eV), so convert from joules to eV using the conversion factor \[ 1 ext{ eV} = 1.602 \times 10^{-19} ext{ J} \].Thus, the energy in eV is:\[ E_{\text{eV}} = \frac{1.80 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 1.12 ext{ eV} \].

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.

Solar Cells
Solar cells are devices that convert light energy, particularly from the sun, into electrical energy. This is accomplished through a process called the photovoltaic effect. The main component of a solar cell is a semiconductor material, such as silicon, which absorbs sunlight and generates an electrical current.

The efficiency of solar cells depends largely on their ability to absorb a wide spectrum of sunlight. To do this effectively, the semiconductor material needs an energy gap that matches the energy of incoming photons from the sun. If the energy gap is too large, lower-energy photons will not be absorbed. Consequently, a solar cell material must be chosen carefully to ensure it absorbs as many photons as possible.

Solar cells are beneficial because they provide a renewable energy solution. By harnessing the sun's energy, we can reduce dependence on fossil fuels and lower environmental impact. Understanding the interaction between material properties and light absorption is crucial for developing more efficient solar cells.
Photon Energy
Photon energy is the energy carried by a single photon, which is a tiny discrete particle of light or electromagnetic radiation. It plays a vital role in the context of solar cells because it determines whether a photon can be absorbed by the solar cell material. The energy of a photon is directly related to its wavelength by the equation: \[ E = \frac{hc}{\lambda} \] where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

For solar cells, it is crucial that the energy of incoming photons exceeds the energy gap of the semiconductor material. If the photon's energy matches or exceeds this gap, the material can absorb the photon and create an electron-hole pair, which is essential for creating an electric current. The more energy a photon carries (i.e., the shorter its wavelength), the more energy it can potentially contribute to the electrical output of a solar cell.

In the context of the sun's radiation, many photons have wavelengths shorter than 1100 nm. Therefore, for a solar cell to effectively convert this radiation into electricity, its energy gap must allow for the absorption of photons with such high energies.
Wavelength Conversion
Wavelength conversion is the process of manipulating photons to change their wavelengths so that they better match the absorption properties of materials like those found in solar cells. This process can enhance the efficiency of solar cells by altering light to contain wavelengths that are more easily absorbed by the cell's semiconductor material.

The concept is particularly useful when dealing with materials that do not naturally absorb a broad spectrum of sunlight effectively. Techniques such as using coatings or layering materials with different energy gaps can facilitate wavelength conversion, redirecting or transforming photons into ranges more suited for energy absorption.

Understanding wavelength conversion is important because it maximizes the potential energy harnessed from solar radiation. By optimizing the range of absorbed wavelengths, wavelength conversion enables solar cells to convert more sunlight into usable energy, thus boosting their overall efficiency.

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

What, roughly, is the ratio of the density of molecules in an ideal gas at \(285 \mathrm{~K}\) and \(1 \mathrm{~atm}\left(\right.\) say \(\left.\mathrm{O}_{2}\right)\) to the density of free electrons (assume one per atom) in a metal (copper) also at \(285 \mathrm{~K} ?\)

A TV remote control emits IR light. If the detector on the TV set is not to react to visible light, could it make use of silicon as a "window" with its energy gap \(E_{g}=1.14 \mathrm{eV} ?\) What is the shortest-wavelength light that can strike silicon without causing electrons to jump from the valence band to the conduction band?

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

Show that the probability for the state at the Fermi energy being occupied is exactly \(\frac{1}{2}\), independent of temperature.

(II) 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 \(\left(E_{\mathrm{F}}=\) the energy of the most energetic electron at \right. \(T=0 \mathrm{K}\) , 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?

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.