/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 (II) Show that the probability f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) Show that the probability for the state at the Fermi energy being occupied is exactly \(\frac{1}{2}\) , independent of temperature.

Short Answer

Expert verified
At the Fermi energy, the occupation probability is \(\frac{1}{2}\), independent of temperature.

Step by step solution

01

Understand the Fermi-Dirac Distribution

The Fermi-Dirac distribution describes the probability that an electron state with energy \(E\) is occupied at thermal equilibrium. It is given by: \[ f(E) = \frac{1}{e^{(E - \mu)/kT} + 1} \]where \(\mu\) is the chemical potential (Fermi energy at absolute zero), \(k\) is Boltzmann's constant, and \(T\) is the absolute temperature.
02

Identify the Fermi Energy Condition

At the Fermi energy level, by definition, \(E = \mu\). This means if we substitute \(E = \mu\) into the Fermi-Dirac distribution formula, we will evaluate the occupation probability at the Fermi energy.
03

Substitute Fermi Energy into the Formula

Substituting \(E = \mu\) into the Fermi-Dirac distribution gives:\[f(\mu) = \frac{1}{e^{(\mu - \mu)/kT} + 1}\]Simplifying the exponent: \[(\mu - \mu) = 0\] Thus:\[f(\mu) = \frac{1}{e^{0} + 1}\]
04

Simplify the Exponential Term

Remember that any number to the power of zero is 1, so:\[e^{0} = 1\]Thus, the formula simplifies to:\[f(\mu) = \frac{1}{1 + 1} = \frac{1}{2}\]
05

Conclude the Solution

From the calculation, you can see that the occupation probability at the Fermi energy is \(\frac{1}{2}\), independent of the temperature, because the expression reduces to \(\frac{1}{2}\) regardless of the value of \(T\). This shows that the occupation probability at the Fermi energy is always \(\frac{1}{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Fermi energy
The concept of Fermi energy is a fundamental aspect of understanding how electrons are distributed in a material at zero temperature. At absolute zero, all the lower-energy electron states up to a certain level, known as the Fermi energy, are completely filled. This is the maximum energy that an electron can have when the system is at absolute zero. Introduced by the physicist Enrico Fermi, this energy level plays a crucial role in determining the electrical properties of solids, particularly metals.

Imagine a sea of electrons constantly in motion, where each energy state can hold up to two electrons of opposite spin. At zero Kelvin ( 0°K), these states are filled up to the Fermi energy. Above this level, states are generally unoccupied because there is insufficient thermal energy to promote electrons to higher energy states.
  • Fermi energy depends on the type of atoms in the material.
  • It defines the energy boundary between filled and unfilled electron states at absolute zero.
  • This concept is pivotal in semiconductor physics and quantum mechanics.
Understanding Fermi energy helps in comprehending more complex phenomena like electrical conduction and the electronic behavior of materials.
Thermal equilibrium
Thermal equilibrium is a state where all parts of a system are at the same temperature, meaning there's no net flow of thermal energy between any parts of the system. In the context of the Fermi-Dirac distribution, which is applicable to fermions such as electrons, thermal equilibrium allows us to analyze the probability of occupancy of energy states such as the Fermi energy.

In a semiconductor or metal at thermal equilibrium, the distribution of electrons over energy states follows the Fermi-Dirac distribution function. This function statistically describes how electrons fill up available energy levels in an attempt to reach an even energy distribution.
  • No net exchange of energy occurs between system parts in thermal equilibrium.
  • Ensures the stability of the electron distribution within materials.
  • Informs how properties like conductivity change with temperature.
Understanding thermal equilibrium in quantum systems is essential for sectors like electronics, where components must remain stable under varying conditions.
Occupancy probability
Occupancy probability is a key feature of the Fermi-Dirac distribution and it describes the likelihood that an electron state will be occupied at a given temperature. Specifically, it's given by the function:\[ f(E) = \frac{1}{e^{(E - \mu)/kT} + 1} \]where \( E \) is the energy of the state, \( \mu \) represents the chemical potential (often close to Fermi energy), \( k \) is Boltzmann's constant, and \( T \) is the temperature.

When applied to the Fermi energy, the occupancy probability reflects a unique aspect: it remains exactly \( \frac{1}{2} \) regardless of temperature. This happens because when \( E = \mu \), the exponential term becomes zero:\[ e^{0} = 1 \]making the formula simplify to:\[ f(\mu) = \frac{1}{1 + 1} = \frac{1}{2} \]
  • Occupancy probability elucidates the distribution of electrons.
  • Helps predict changes in electronic properties with temperature.
  • Illustrates how systems transition between energy states efficiently.
This concept is critical for exploring phenomena such as electron transport, energy band structure, and the electronic behavior of semiconductors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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 \(E+d E . ]\)

A semiconductor diode laser emits \(1.3-\mu \mathrm{m}\) light. Assuming that the light comes from electrons and holes recombining, what is the band gap in this laser material?

(II) A very simple model of a "one-dimensional" metal consists of \(N\) electrons confined to a rigid box of width \(\ell\) We neglect the Coulomb interaction between the electrons. (a) Calculate the Fermi energy for this one-dimensional metal \(\left(E_{\mathrm{F}}=\) the energy of the most energetic electron at \right. \(T=0 \mathrm{K}\) , taking into account the Pauli exclusion principle. You can assume for simplicity that \(N\) is even. (b) What is the smallest amount of energy, \(\Delta E,\) that this \(1-\mathrm{D}\) metal can absorb? (c) Find the limit of \(\Delta E / E_{\mathrm{F}}\) for large \(N .\) What does this result say about how well metals can conduct?

7 Band Theory of Solids (I) A semiconductor is struck by light of slowly increasing frequency and begins to conduct when the wavelength of the light is 580 nm; estimate the size of the energy gap \(E_{g}\)

In the ionic salt \(\mathrm{KF}\) , the separation distance between ions is about 0.27 \(\mathrm{nm}\) . (a) Estimate the electrostatic potential energy between the ions assuming them to be point charges (magnitude 1\(e ) .\) (b) When \(\mathrm{F}^{\text { "grabs" an electron, it releases }}\) 3.41 \(\mathrm{eV}\) of energy, whereas 4.34 \(\mathrm{eV}\) is required to ionize \(\mathrm{K}\) . Find the binding energy of KF relative to free \(\mathrm{K}\) and \(\mathrm{F}\) atoms, neglecting the energy of repulsion.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.