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At \(T=300 \mathrm{~K},\) how far above the Fermi energy is a state for which the probability of occupation by a conduction electron is \(0.10 ?\)

Short Answer

Expert verified
The state is approximately 0.054 eV above the Fermi energy.

Step by step solution

01

Understanding the Fermi-Dirac Distribution

The probability that an energy state is occupied by an electron at temperature \( T \) is given by the Fermi-Dirac distribution formula: \[ f(E) = \frac{1}{e^{\frac{E - E_F}{kT}} + 1} \] where \( f(E) \) is the probability of occupation, \( E \) is the energy of the state, \( E_F \) is the Fermi energy, \( k \) is Boltzmann's constant, and \( T \) is the temperature in Kelvin.
02

Setting Up the Equation

Since the probability of occupation is \( 0.10 \), set \( f(E) = 0.10 \) in the Fermi-Dirac distribution equation: \[ 0.10 = \frac{1}{e^{\frac{E - E_F}{kT}} + 1} \] This can be rearranged to: \[ e^{\frac{E - E_F}{kT}} + 1 = 10 \] which simplifies to: \[ e^{\frac{E - E_F}{kT}} = 9 \]
03

Solving for E - EF

Take the natural logarithm of both sides to isolate \( E - E_F \): \[ \frac{E - E_F}{kT} = \ln(9) \] Therefore, \( E - E_F = kT \ln(9) \).
04

Calculating the Result

Substitute \( k = 8.617 \times 10^{-5} \) eV/K and \( T = 300 \) K into the equation: \[ E - E_F = (8.617 \times 10^{-5})(300) \ln(9) \] This equates to \( E - E_F \approx 0.054 \) eV, which is the energy difference above the Fermi energy.

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Key Concepts

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

Fermi energy
Fermi energy is a fundamental concept in quantum mechanics and solid-state physics. It represents the highest energy level that electrons occupy in a material at absolute zero temperature. In a metal, as temperature increases, electrons gain thermal energy and can move to energy levels higher than the Fermi energy. This is crucial for understanding the electronic properties of materials.

Think of the Fermi energy as the baseline or cap for electron energies at zero Kelvin. At this temperature, electrons fill the energy states up to the Fermi level, and no state above it is occupied. However, as we introduce heat, states above the Fermi energy become accessible as electrons gain energy and transition to these states.
  • It acts like a "cut-off" level for electrons.
  • Determines the distribution of electrons in different energy states at a given temperature.
  • Defines the electronic characteristics of different materials, like conductors, insulators, and semiconductors.
Understanding Fermi energy is crucial for predicting how electrons behave in various materials and conditions.
Boltzmann's constant
Boltzmann's constant ( \( k \) ) is a fundamental constant that links the average kinetic energy of particles in a gas with the temperature of the gas. In the context of the Fermi-Dirac distribution, it plays a key role in describing how electrons distribute themselves among available energy states at a given temperature.

Boltzmann's constant is used to convert temperature into energy units, providing a bridge between macroscopic thermodynamic quantities and quantum atomic properties. This conversion is important since energies are typically measured in electron volts (eV) in solid-state physics.
  • The value of Boltzmann's constant is approximately \( 8.617 imes 10^{-5} \) eV/K.
  • It connects temperature with energy, which is essential for analyzing processes at a microscopic level.
  • Facilitates understanding of the thermal behavior of materials and systems in thermodynamics and statistical mechanics.
Boltzmann's constant helps in calculating temperature-dependent phenomena, allowing us to predict how particles like electrons behave under different thermal conditions.
probability of occupation
The probability of occupation in the Fermi-Dirac distribution is a measure of how likely an electron is to occupy a given energy state at a specific temperature. This is governed by the Fermi-Dirac distribution function, which takes into account quantum statistics.

This concept becomes important in materials science and electronics as it helps predict electron behavior in solids. The function is a statistical way to determine the likelihood of electrons filling available states, particularly in metals, semiconductors, and superconductors.
  • At absolute zero, all states below the Fermi energy are fully occupied, and those above are empty.
  • As temperature increases, electrons are more likely to occupy higher energy states.
  • Probability values range from 0 (no chance of occupation) to 1 (certainty of occupation).
The probability of occupation is fundamental for understanding electrical conductivity, energy band structures, and thermal properties of materials.

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Most popular questions from this chapter

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 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.

The Fermi energy of aluminum is \(11.6 \mathrm{eV} ;\) its density and molar mass are \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\) and \(27.0 \mathrm{~g} / \mathrm{mol},\) respectively. From these data, determine the number of conduction electrons per atom.

The compound gallium arsenide is a commonly used semiconductor, having an energy gap \(E_{g}\) of \(1.43 \mathrm{eV}\). Its crystal structure is like that of silicon, except that half the silicon atoms are replaced by gallium atoms and half by arsenic atoms. Draw a flattened-out sketch of the gallium arsenide lattice, following the pattern of Fig. \(41-10 a\). What is the net charge of the (a) gallium and (b) arsenic ion core? (c) How many electrons per bond are there? (Hint: Consult the periodic table in Appendix G.)

The occupancy probability function (Eq. \(41-6)\) can be applied to semiconductors as well as to metals. In semiconductors the Fermi energy is close to the midpoint of the gap between the valence band and the conduction band. For germanium, the gap width is \(0.67 \mathrm{eV}\). What is the probability that (a) a state at the bottom of the conduction band is occupied and (b) a state at the top of the valence band is not occupied? Assume that \(T=290 \mathrm{~K}\). (Note: In a pure semiconductor, the Fermi energy lies symmetrically between the population of conduction electrons and the population of holes and thus is at the center of the gap. There need not be an available state at the location of the Fermi energy.)

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?

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