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A state \(63 \mathrm{meV}\) above the Fermi level has a probability of occupancy of \(0.090\). What is the probability of occupancy for a state 63 meV below the Fermi level?

Short Answer

Expert verified
The probability is 0.910.

Step by step solution

01

Understanding the Fermi-Dirac Distribution

The probability of occupancy of an energy state at a given energy level is described by the Fermi-Dirac distribution function, which is given by \[ f(E) = \frac{1}{e^{(E - E_F)/(k_B T)} + 1}\]where \(E\) is the energy of the state, \(E_F\) is the Fermi energy, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature.
02

Relationship Between Energies Symmetrically Positioned About the Fermi Level

For a state \(E_1\) above the Fermi level with a known occupancy probability \(f(E_1) = p\), the probability of another state \(E_2\) symmetrically positioned below the Fermi level is \(1 - p\). This is because the Fermi-Dirac distribution is symmetric about the Fermi level.
03

Compute the Probability of the Lower Energy State

The stated problem gives \(p = 0.090\) for the state 63 meV above the Fermi level. The symmetry of the Fermi-Dirac distribution indicates that the probability of occupancy for a state 63 meV below the Fermi level is \[1 - 0.090 = 0.910.\]
04

Conclusion

The probability of occupancy for the state 63 meV below the Fermi level is \(0.910\). This result relies on the symmetry property of the Fermi-Dirac distribution function.

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Key Concepts

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

Probability of Occupancy
The probability of occupancy is an important concept in statistical mechanics and quantum physics, especially when discussing how particles such as electrons are distributed among available energy states. This concept is governed by the Fermi-Dirac distribution, which calculates the likelihood that a particular energy level is occupied by a fermion, like an electron, at finite temperature.

The Fermi-Dirac distribution formula is expressed as: \[ f(E) = \frac{1}{e^{(E - E_F)/(k_B T)} + 1} \]Here,
  • \(E\) is the energy level in question.
  • \(E_F\) signifies the Fermi energy or Fermi level.
  • \(k_B\) is the Boltzmann constant, a crucial part of thermodynamic equations.
  • \(T\) stands for the absolute temperature in Kelvin.
This formula helps us determine the occupancy or the probability that a state at energy \(E\) is occupied by a fermion.

This distribution is vital for predicting the behavior of electrons in solids, especially at absolute zero temperature. At that point, all states below the Fermi level are filled, while those above are empty.
Fermi Level
In solid-state physics, the Fermi level represents a critical concept. It is the energy level at which the probability of occupancy for an electron is 50% at absolute zero temperature.

The Fermi level is crucial because it serves as the dividing line between occupied and unoccupied electron states in a material. An electron at this level contributes significantly to the electrical and thermal properties of materials:
  • At temperatures close to absolute zero, all energy levels below the Fermi level are filled with electrons.
  • As temperature rises, electrons gain energy, potentially jumping to higher energy states above the Fermi level.
  • The Fermi level helps in predicting how electrons distribute in various energy states as temperature changes.
The position of the Fermi level is influenced by the material type and its doping level. In metals, it lies within the conduction band, while in insulators and semiconductors, it's located between the valence and conduction bands.
Symmetry of Energy States
The symmetry of energy states around the Fermi level is a fascinating property of the Fermi-Dirac distribution. This symmetry implies that energy levels equidistant from the Fermi level but on opposite sides have complementary probabilities of occupancy.

For instance, if a state is above the Fermi level with a probability \(p\) to be occupied, the corresponding state at the same energy distance below the Fermi level will have a probability of \(1 - p\). This characteristic is inherent to the Fermi-Dirac function's form:\[ f(E) = \frac{1}{e^{(E - E_F)/(k_B T)} + 1} \]This function is symmetric about \(E = E_F\).

This property allows physicists and engineers to easily deduce the behavior of electrons in materials, aiding in the design of semiconductors and electronic devices. It helps in understanding how conduction occurs at various temperatures and energies, ultimately influencing the development of more efficient materials.

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Most popular questions from this chapter

A silicon-based MOSFET has a square gate \(0.50 \mu \mathrm{m}\) on edge. The insulating silicon oxide layer that separates the gate from the \(p\) -type substrate is \(0.20 \mu \mathrm{m}\) thick and has a dielectric constant of 4.5. (a) What is the equivalent gate-substrate capacitance (treating the gate as one plate and the substrate as the other plate)? (b) Approximately how many elementary charges \(e\) appear in the gate when there is a gate-source potential difference of \(1.0 \mathrm{~V}\) ?

A certain metal has \(1.70 \times 10^{28}\) conduction electrons per cubic meter. A sample of that metal has a volume of \(6.00 \times 10^{-6}\) \(\mathrm{m}^{3}\) and a temperature of \(200 \mathrm{~K}\). How many occupied states are in the energy range of \(3.20 \times 10^{-20} \mathrm{~J}\) that is centered on the energy \(4.00 \times 10^{-19} \mathrm{~J} ?\) (Caution: Avoid round-off in the exponential.)

A certain material has a molar mass of \(20.0 \mathrm{~g} / \mathrm{mol}\), a Fermi energy of \(5.00 \mathrm{eV}\), and 2 valence electrons per atom. What is the density \(\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) ?

The Fermi energy of copper is \(7.0 \mathrm{eV}\). Verify that the corresponding Fermi speed is \(1600 \mathrm{~km} / \mathrm{s}\).

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) \(4.4 \mathrm{eV},(\mathrm{b}) 5.4 \mathrm{eV},(\mathrm{c}) 5.5 \mathrm{eV},(\mathrm{d}) 5.6 \mathrm{eV}\), and \((\mathrm{e}) 6.4 \mathrm{eV} ?(\mathrm{f})\) At what temperature is the probability \(0.16\) that a state with energy \(E=5.6 \mathrm{eV}\) is occupied?

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