/*! 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 23 Show that, at \(T=0 \mathrm{~K}\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that, at \(T=0 \mathrm{~K}\), the average energy \(E_{\mathrm{avg}}\) of the conduction electrons in a metal is equal to \(\frac{3}{5} E_{\mathrm{F}}\). (Hint: By definition of average, \(E_{\mathrm{avg}}=(1 / n) \int E N_{\mathrm{o}}(E) d E\), where \(n\) is the number density of charge carriers.)

Short Answer

Expert verified
At T=0 K, the average energy of conduction electrons is \( \frac{3}{5} E_F \).

Step by step solution

01

Understanding the given definitions

We are given that the average energy of conduction electrons is defined as \( E_{\text{avg}} = \frac{1}{n} \int_0^{E_F} E \cdot N(E) \, dE \), where \( N(E) \) is the density of states, \( E_F \) is the Fermi energy, and \( n \) is the number density of charge carriers.
02

Define number density and density of states

At absolute zero temperature \( T = 0 \), electrons fill up to the Fermi energy \( E_F \). The number density of states is given by \( N(E) = \frac{3N}{2E_F} \left(\frac{E}{E_F}\right)^{1/2} \). And, the number density \( n \) is \( n = \int_0^{E_F} N(E) \, dE \).
03

Express the integral for number density, n

Integrating \( N(E) \) over the range from \( 0 \) to \( E_F \), we get \( n = \int_0^{E_F} \frac{3N}{2E_F} \left(\frac{E}{E_F}\right)^{1/2} \text{d}E \), solving this gives \( n = N \).
04

Calculate the average energy integral

Substitute the expression for \( N(E) \) into the integral for \( E_{\text{avg}} \): \( E_{\text{avg}} = \frac{1}{N} \int_0^{E_F} E \cdot \frac{3N}{2E_F}\left(\frac{E}{E_F}\right)^{1/2} dE \). This simplifies to \( E_{\text{avg}} = \frac{3}{2E_F} \int_0^{E_F} E^{3/2} dE \).
05

Integrate to find E_avg

Solve the integral \( \int_0^{E_F} E^{3/2} \text{d}E \). This evaluates to \( \frac{2}{5} E_F^{5/2} \). Substituting back, we get \( E_{\text{avg}} = \frac{3}{2E_F} \cdot \frac{2}{5} E_F^{5/2} \), which simplifies to \( \frac{3}{5} E_F \).
06

Confirm the result

Therefore, at \( T=0 \mathrm{~K} \), the average energy \( E_{\text{avg}} \) is equal to \( \frac{3}{5} E_{\mathrm{F}} \), proving the statement given in the problem.

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 Fermi energy, denoted as \( E_F \), is a crucial concept in understanding the behavior of electrons in a solid, particularly metals. It represents the highest energy level occupied by electrons at absolute zero temperature \( T = 0 \) K. This implies that below this temperature, electrons fill energy levels up to \( E_F \), with no electron possessing energy greater than this limit.
The significance of the Fermi energy comes from its role in determining the electrical, thermal, and optical properties of the material. It's a way to understand how electrons are distributed across energy levels in a metal. Since conduction electrons — the electrons responsible for electricity conduction — are involved in these energy levels, knowing the Fermi energy helps predict how these electrons will behave under varying conditions. This understanding helps in designing materials with specific electronic properties.
Thus, Fermi energy isn't just a theoretical construct but plays a practical role in material science and engineering, making it an essential concept in physics and chemistry.
Density of States
The density of states (DOS), denoted \( N(E) \), tells us how many electron states are available in a solid at each energy level. In terms of conduction electrons, it is critical because it describes how energy levels vary with energy changes. The DOS function plays an important role in understanding the electronic properties of materials.
The expression for density of states at absolute zero temperature is given by:
  • \( N(E) = \frac{3N}{2E_F} \left(\frac{E}{E_F}\right)^{1/2} \)
where \( N \) is the total number of electrons, and \( E_F \) is the Fermi energy.
This mathematical relationship reveals that DOS increases with energy \( E \). It indicates more states become available to electrons as they gain energy, explaining why materials have different conductive properties at different energies and temperatures.
Understanding DOS helps in predicting how electrons fill these states under varying thermal and electrical conditions, which is very important in designing electronic components.
Average Energy
Average energy \( E_{\text{avg}} \) offers insight into the typical energy level an electron within a metal might have. At \( T = 0 \ \mathrm{K} \), the exercise shows that the average energy of conduction electrons is \( \frac{3}{5} E_{\mathrm{F}} \). This value emerges from integrating the product of energy \( E \) and density of states \( N(E) \) over the range of \( 0 \) to \( E_F \).'
The formula used here is:
  • \( E_{\text{avg}} = \frac{1}{n} \int_0^{E_F} E \cdot N(E) \, dE \)
where \( n \) is the number density of electrons. The integral accounts for all energies les electron could possess when filled up to \( E_F \), averaging over them to find this typical value.
This concept is crucial because it helps in estimating how conduction electrons behave collectively within a material. It also has implications for calculating the heat capacity and electrical conductivity of metals, allowing engineers to design more efficient materials.
Number Density
Number density, denoted \( n \), refers to the number of charge carriers per unit volume in a material. In metals, these charge carriers are electrons, and their density significantly impacts a metal's ability to conduct electricity.
At \( T = 0 \ \mathrm{K} \), the step-by-step solution defines number density as:
  • \( n = \int_0^{E_F} N(E) \, dE \)
This equation integrates the density of states \( N(E) \) from \( 0 \) to the Fermi energy \( E_F \). Solving this integral provides the total number of electrons that can contribute to conduction within the metal.
Number density is crucial because it determines the free electron gas model's behavior, which predicts how electrons contribute to properties like electrical and thermal conductivity. Higher number density generally indicates better conductivity, making it a key parameter when assessing material properties.

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

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. 41-10a. 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.)

(a) Find the angle \(\theta\) hetween adjacent nearest-neighbor bonds in the silicon lattice. Recall that each silicon atom is bonded to four of its nearest neighbors. The four neighbors form a regular tetrahedron-a pyramid whose sides and base are equilateral triangles. (b) Find the bond length, given that the atoms at the corners of the tetrahedron are 388 pm apart.

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?

What is the probability that a state \(0.0620 \mathrm{eV}\) above the Fermi energy will be occupied at (a) \(T=0 \mathrm{~K}\) and (b) \(T=320 \mathrm{~K}\) ?

A potassium chloride crystal has an energy band gap of \(7.6\) eV above the topmost occupied band, which is full. Is this crystal opaque or transparent to light of wavelength \(140 \mathrm{~nm}\) ?

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.