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For a certain semiconductor, the longest wavelength radiation that can be absorbed is \(1.92 \mathrm{~mm} .\) What is the energy gap in this semiconductor?

Short Answer

Expert verified
The energy gap is approximately 0.0646 meV.

Step by step solution

01

Identify Given Information

We are given the longest wavelength, \( \lambda = 1.92 \text{ mm} \), which is equivalent to \( \lambda = 1.92 \times 10^{-3} \text{ m} \). We need to find the energy gap.
02

Convert Wavelength to Meters

The wavelength is already converted to meters as \( 1.92 \times 10^{-3} \text{ m} \). This step ensures we are using consistent units for calculations.
03

Use Energy-Wavelength Relation

The energy of a photon is related to its wavelength by the equation \( E = \frac{h c}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \) and \( c \) is the speed of light \( 3 \times 10^{8} \text{ m/s} \).
04

Substitute Constants into Formula

Substitute \( h \) and \( c \) into the energy formula: \[E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.92 \times 10^{-3}}\]
05

Calculate Energy

Calculate the energy by performing the division: \[E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.92 \times 10^{-3}} = 1.035 \times 10^{-22} \text{ J}\]
06

Convert Energy to Electronvolts

Convert the energy from joules to electronvolts using the conversion factor where \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):\[E_{\text{eV}} = \frac{1.035 \times 10^{-22}}{1.602 \times 10^{-19}} \approx 0.0646 \text{ meV}\]
07

Report the Energy Gap

The energy gap of the semiconductor is approximately \( 0.0646 \text{ meV} \).

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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 within the realm of physics, especially when dealing with semiconductors. When light is absorbed by a semiconductor, it interacts with electrons, which can lead to electron excitation and conduction. This interaction is what we refer to when we talk about photon energy. The energy of a photon is inherently tied to its frequency and, inversely, its wavelength. This relationship allows us to calculate photon energy using basic principles of physics. A photon’s energy is quantified by the equation \( E = h \cdot f \), where \( E \) is the energy, \( h \) is Planck’s constant, and \( f \) is the frequency of the light. Since we often deal with wavelength in semiconductors, we can rearrange this formula using \( f = \frac{c}{\lambda} \) to relate energy directly to wavelength. The key takeaway is that shorter wavelengths correspond to higher energy photons, which are necessary to overcome the energy gap in certain semiconductors.
Planck's Constant
Planck's constant (\( h \)) is one of the cornerstones of quantum mechanics. It bridges the gap between particle physics and classical wave theories. It is a tiny constant with a value of approximately \( 6.626 \times 10^{-34} \text{ J s} \). Despite its small size, it plays a critical role in determining the energy levels in atoms and semiconductors.Planck's constant is crucial in calculating photon energy. When you know the wavelength of light, you can use Planck's constant to find how much energy a photon carries. This is particularly important in physics because it helps us understand how light interacts with materials, such as semiconductors, including how much energy is needed to excite an electron past the energy gap.
Wavelength-Energy Relation
The wavelength-energy relation is central to understanding how semiconductors interact with light. The equation \( E = \frac{h \cdot c}{\lambda} \) expresses this relationship. Here, \( E \) is the energy of the photon, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.This relation demonstrates that energy is inversely proportional to wavelength. This means the longer the wavelength, the less energy the photon will carry. Applying this to semiconductors, if a particular semiconductor can only absorb light up to a certain wavelength, this gives us a direct measure of its energy gap.In practice, by knowing the maximum wavelength that a semiconductor can absorb (as seen in the exercise with a wavelength of \( 1.92 \text{ mm} \)), we can compute the minimum photon energy necessary to bridge the energy gap. This makes the wavelength-energy relation an indispensable tool in modern physics and material science.

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Most popular questions from this chapter

(1I) For a one-dimensional potential well of width \(\ell,\) start with \(E=n^{2} \frac{h^{2}}{8 m \ell^{2}}, \quad n=1,2,3, \cdots\) and show that the number of states per unit energy interval for an electron gas is given by \(g_{\ell}(E)=\sqrt{\frac{8 m \ell^{2}}{h^{2} E}}\) Remember that there can be two electrons (spin up and spin down) for each value of \(n .[\) Hint. Write the quantum number \(n\) in terms of \(E .\) Then \(g_{t}(E)=2 d n / d E\) where \(d n\) is the number of energy levels between \(E\) and \(E+d E . ]\)

An amplifier has a voltage gain of 65 and a \(25-\mathrm{k} \Omega\) load (output) resistance. What is the peak output current through the load resistor if the input voltage is an ac signal with a peak of \(0.080 \mathrm{~V} ?\)

(II) Binding energies are often measured experimentally in kcal per mole, and then the binding energy in eV per mole- cule is calculated from that result. What is the conversion factor in going from kcal per mole to eV per molecule? What is the binding energy of KCl ( \(=4.43\) eV ) in kcal per mole?

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

8 Semiconductors and Doping (II) Suppose that a silicon semiconductor is doped with phosphorus so that one silicon atom in \(1.2 \times 10^{6}\) is replaced by a phosphorus atom. Assuming that the "extra" electron in every phosphorus atom is donated to the conduction band, by what factor is the density of conduction electrons increased? The density of silicon is 2330 \(\mathrm{kg} / \mathrm{m}^{3}\) , and the density of conduction electrons in pure silicon is about \(10^{16} \mathrm{m}^{-3}\) at room temperature.

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