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Chapter 41: Problem 31

In a particular crystal, the highest occupied band is full. The crystal is transparent to light of wavelengths longer than \(295 \mathrm{~nm}\) but opaque at shorter wavelengths. Calculate, in electron-volts, the gap between the highest occupied band and the next higher (empty) band for this material.

Short Answer

Expert verified
The energy gap is approximately 4.20 eV.

Step by step solution

01

Understanding the Problem

The problem asks us to find the energy gap between the highest occupied band and the next higher (empty) band in a crystal, given that the crystal is opaque to light below a wavelength of 295 nm. This implies that wavelengths shorter (greater frequency) than 295 nm have enough energy to bridge this gap.
02

Determine Frequency of the Wavelength

To find the energy of the photons that the crystal is opaque to, we first convert the given wavelength to frequency using the speed of light formula: \[ c = \lambda \times f \]Where \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\), and \(\lambda\) is 295 nm (or \(295 \times 10^{-9}\) m). Solving for \(f\) gives:\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{295 \times 10^{-9}} \approx 1.0169 \times 10^{15} \text{ Hz} \]
03

Calculate the Energy using Planck's Equation

The energy of the photons corresponding to this frequency is given by Planck's equation:\[ E = h \times f \]where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\). Substituting the frequency, we get:\[ E = 6.626 \times 10^{-34} \times 1.0169 \times 10^{15} \approx 6.73 \times 10^{-19} \text{ J} \]
04

Convert the Energy from Joules to Electron-Volts

To convert the energy from Joules to electron-volts, use the conversion factor: \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\). Thus, the energy in electron-volts is:\[ \text{Energy (eV)} = \frac{6.73 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.20 \text{ eV} \]

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Key Concepts

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

Photon Energy
Photon energy is a fundamental concept in physics and plays a crucial role in understanding the band gap in crystals. When light interacts with matter, it can be thought of as a stream of particles known as photons. Each of these photons carries a specific amount of energy, which is directly related to its frequency.
- The higher the frequency, the larger the photon energy. - The lower the frequency, the smaller the photon energy.
This relationship means that light of shorter wavelength (higher frequency) has greater photon energy. In the context of our crystal problem, only photons with energy above a certain threshold can interact with the crystal, revealing the presence of the band gap.
Wavelength Conversion
In the realm of physics and material science, converting wavelengths to frequency is a common practice when determining energy levels. Wavelength and frequency are two properties of light that are inversely related through the speed of light.
The formula \[c = \lambda \times f\] connects these two, where:
  • \(c\) is the speed of light, approximately \(3 \times 10^8\) m/s.
  • \(\lambda\) is the wavelength of light.
  • \(f\) is the frequency of light.

By converting the wavelength, typically given in nanometers (nm), to a frequency, we can then use this information to calculate the photon energy. This conversion is essential when understanding what energy photons must have to be absorbed by or pass through materials.
Planck's Equation
Planck's Equation is a simple but powerful tool used to find the energy of a photon once its frequency is known. The equation is \[E = h \times f\]where:
  • \(E\) represents the energy of the photon.
  • \(h\) is Planck's constant, roughly \(6.626 \times 10^{-34}\) J路s.
  • \(f\) is the frequency of the photon.

This equation tells us that energy is directly proportional to frequency, making it easier to determine the energy that a photon can transfer to or absorb from a material. Applying this equation to photons interacting with a crystal helps identify the energy gap significant in crystal conduction properties.
Electron-Volt Conversion
In physics, it鈥檚 often necessary to convert between different energy units to make calculations more convenient or relevant to the material being studied. Energy is often initially calculated in Joules; however, in atomic and particle physics, electron-volts (eV) are a more common unit.
To convert energy in Joules (J) to electron-volts (eV), we use the conversion factor:\[1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\]
This conversion allows scientists and students to express energy in a unit that is more manageable and contextually appropriate for phenomena on the atomic scale. It simplifies the interpretation of energy diagrams and calculations, such as identifying the band gap energy in a crystal structure.

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Most popular questions from this chapter

(a) What maximum light wavelength will excite an electron in the valence band of diamond to the conduction band? The energy gap is \(5.50 \mathrm{eV}\). (b) In what part of the electromagnetic spectrum does this wavelength lie?

What is the probability that a state \(0.0620 \mathrm{eV}\) above the Fermi energy will be occupied at (a) \(T=0 \mathrm{~K}\) and (b) \(T=320 \mathrm{~K}\) ?

What is the Fermi energy of gold (a monovalent metal with molar mass \(197 \mathrm{~g} / \mathrm{mol}\) and density \(\left.19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) ?

The Fermi energy for silver is \(5.5 \mathrm{eV} .\) At \(T=0^{\circ} \mathrm{C}\), what are the probabilities that states with the following energies are occupied: (a) \(4.4 \mathrm{eV}\), (b) \(5.4 \mathrm{eV}\), (c) \(5.5 \mathrm{eV}\), (d) \(5.6 \mathrm{eV}\), and (e) \(6.4 \mathrm{eV}\) ? (f) At what temperature is the probability \(0.16\) that a state with energy \(E=5.6 \mathrm{eV}\) is occupied?

When a photon enters the depletion zone of a \(p-n\) junction, the photon can scatter from the valence electrons there, transferring part of its energy to each electron, which then jumps to the conduction band. Thus, the photon creates electron-hole pairs. For this reason, the junctions are often used as light detectors, especially in the \(x\)-ray and gamma-ray regions of the electromagnetic spectrum. Suppose a single \(662 \mathrm{keV}\) gamma-ray photon transfers its energy to electrons in multiple scattering events inside a semiconductor with an energy gap of \(1.1 \mathrm{eV}\), until all the energy is transferred. Assuming that each electron jumps the gap from the top of the valence band to the bottom of the conduction band, find the number of electron-hole pairs created by the process.
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