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Chapter 41: Problem 41

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 4.21 electron-volts (eV).

Step by step solution

01

Understand the Problem

The problem states that the crystal is transparent to light with wavelengths longer than 295 nm and opaque to shorter ones. This means that light with a wavelength of 295 nm has just enough energy to bridge the gap between the occupied and unoccupied energy bands. We need to calculate the energy of this light in electron-volts (eV).
02

Use the Energy-Wavelength Relationship

Light energy can be calculated using the formula: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \mathrm{Js}) \), \( c \) is the speed of light \( (3 \times 10^8 \mathrm{m/s}) \), and \( \lambda \) is the wavelength (295 nm or 295 x 10^{-9} meters).
03

Calculate the Energy in Joules

Substitute the values into the formula: \[E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{295 \times 10^{-9}}\]Calculate to find:\[E \approx 6.74 \times 10^{-19} \mathrm{J}\]
04

Convert Energy to Electron-Volts

The energy found is in Joules. We convert Joules to electron-volts using the conversion factor: 1 eV = \( 1.602 \times 10^{-19} \) Joules.\[E = \frac{6.74 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.21 \mathrm{eV}\]
05

Conclude the Calculation

Thus, the energy gap between the highest occupied band and the next higher empty band in the crystal is approximately 4.21 eV.

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Key Concepts

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

Energy-Wavelength Relationship
The relationship between energy and wavelength is fundamental in physics and can be understood through the behavior of light. Light behaves both as a wave and a particle, and its energy, wavelength, and frequency are connected via the equation:
\[ E = \frac{hc}{\lambda} \]This equation tells us that energy (E) is directly proportional to Planck's constant (h) and the speed of light (c), and inversely proportional to the wavelength (\( \lambda \)) of the light.
- **Planck's constant (h)**: A fundamental constant with a value of approximately \( 6.626 \times 10^{-34} \) Js.- **Speed of light (c)**: A constant value of \( 3 \times 10^8 \) meters per second.- **Wavelength (\( \lambda \))**: The distance over which the wave's shape repeats, usually measured in meters or nanometers.So, with a shorter wavelength, the energy of the light increases. In the given problem, since the crystal is transparent at 295 nm, the energy at this wavelength is just sufficient to bridge the band gap of the crystal.
Planck's Constant
Planck's constant is a key element in the field of quantum mechanics. It represents the smallest possible unit of energy and is used to calculate energy at the quantum level. This constant has a value of \( 6.626 \times 10^{-34} \) Joules seconds (Js).
Since Planck's constant is a very small number, it reflects the discrete or quantized nature of energy in quantum mechanics.
This quantized approach challenged classical physics and was revolutionary at the time of its introduction. It laid the foundation for understanding phenomena that classical physics could not explain, such as the photoelectric effect, black-body radiation, and the band theory of solids.
In the context of the energy-wavelength relationship, Planck's constant is crucial because it helps us calculate the energy carried by light waves, like the energy that allows electrons to move between energy bands within a crystal.
Electron-Volt Conversion
In physics, the electron-volt (eV) is a unit of energy that is convenient for expressing tiny energy values at the atomic or subatomic level. One electron-volt is equivalent to the energy gained by an electron when it moves through an electric potential difference of one volt.
In standard units, \( 1 \mathrm{eV} = 1.602 \times 10^{-19} \) Joules. This conversion is very useful in quantum mechanics and solid-state physics.
Using eV instead of Joules often simplifies equations and makes values easier to work with when dealing with the energy associated with atomic or electronic transitions.
For example, in the original exercise, the energy calculated from the wavelength in Joules is converted to eV to provide a more meaningful interpretation of the band gap in the context of electronic structures.

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Most popular questions from this chapter

A silicon sample is doped with atoms having donor states \(0.110 \mathrm{eV}\) below the bottom of the conduction band. (The energy gap in silicon is \(1.11 \mathrm{eV} .\) ) If each of these donor states is occupied with a probability of \(5.00 \times 10^{-5}\) at \(T=300 \mathrm{~K},\) (a) is the Fermi level above or below the top of the silicon valence band and (b) how far above or below? (c) What then is the probability that a state at the bottom of the silicon conduction band is occupied?

Copper, a monovalent metal, has molar mass \(63.54 \mathrm{~g} / \mathrm{mol}\) and density \(8.96 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the number density \(n\) of conduction electrons in copper?

In a simplified model of an undoped semiconductor, the actual distribution of energy states may be replaced by one in which there are \(N_{v}\) states in the valence band, all these states having the same energy \(E_{v},\) and \(N_{c}\) states in the conduction band, all these states having the same energy \(E_{c}\). The number of electrons in the conduction band equals the number of holes in the valence band. (a) Show that this last condition implies that $$\frac{N_{c}}{\exp \left(\Delta E_{c} / k T\right)+1}=\frac{N_{v}}{\exp \left(\Delta E_{v} / k T\right)+1}$$in which$$\Delta E_{c}=E_{c}-E_{\mathrm{F}} \quad \text { and } \quad \Delta E_{v}=-\left(E_{v}-E_{\mathrm{F}}\right)$$ (b) If the Fermi level is in the gap between the two bands and its distance from each band is large relative to \(k T,\) then the exponentials dominate in the denominators. Under these conditions, show that$$E_{\mathrm{F}}=\frac{\left(E_{c}+E_{v}\right)}{2}+\frac{k T \ln \left(N_{v} / N_{c}\right)}{2}$$and that, if \(N_{v} \approx N_{c}\), the Fermi level for the undoped semiconductor is close to the gap's center.

A certain computer chip that is about the size of a postage stamp \((2.54 \mathrm{~cm} \times 2.22 \mathrm{~cm})\) contains about 3.5 million transistors. If the transistors are square, what must be their maximum dimension? (Note: Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than \(0.7 \mu \mathrm{m}\) are now commonly and inexpensively fabricated.)

At \(1000 \mathrm{~K},\) the fraction of the conduction electrons in a metal that have energies greater than the Fermi energy is equal to the area under the curve of Fig. \(41-8 b\) beyond \(E_{\mathrm{F}}\) divided by the area under the entire curve. It is difficult to find these areas by direct integration. However, an approximation to this fraction at any temperature \(T\) is frac\(=\frac{3 k T}{2 E_{\mathrm{F}}}\) Note that \(f r a c=0\) for \(T=0 \mathrm{~K},\) just as we would expect. What is this fraction for copper at (a) \(300 \mathrm{~K}\) and (b) \(1000 \mathrm{~K}\) ? For copper, \(E_{\mathrm{F}}=7.0 \mathrm{eV} .\) (c) Check your answers by numerical integration using Eq. \(41-7\)
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