/*! 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} 56E (a) From equation (9.34) and the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) From equation (9.34) and the Fermi-Dirac distribution given in Exercise 53, obtain an expression for EF(T), the Fermi temperature for a collection of fermion oscillators, (b) Show that EFo=ε. (c) Plot EF(T)versuskBTεfrom 0tokBT6=1.5. (d) By what percent does the Fermi energy drop from its maximum T=0value when kBTrises to 25%ofε?

Short Answer

Expert verified

a) The expression for EF(T) is kBTlneEkBT-1.

b) It is proved that data-custom-editor="chemistry" EF0=E.

c) Graph is shown in the solution part.

d) The percent drop in Fermi energy is 0.462%.

Step by step solution

01

Formula used

The expression for Fermi Dirac distribution is given by,

1eE-EFkBT+1=1eEkBTEeEBTT+1

The expression for drop in Fermi energy is

∆=EF0-EFEF0×100%

02

Calculate the expression for Fermi Dirac distribution

a)

The expression for EF(T) is calculated as,

1eE-EFkBT+1=1eEkBTEeBT+1eE-EFkBT=eEkBTeEkBT-1lneE-EFkBT=lneEkBTeEkBT-1E-EFkBT=EkBT-lneEkBT-1EkBT-EFkBT=EkBT-lneEkBT-1EF=kBTlneEkBT-1

03

Calculate the expression for EF0

b)

The expression for EF0 is calculated as,

EF=kBTlneEkBT-1EF0=kBTlneEkBT=kBTEkBT=E

04

The graph of EF(T) versus kBTm

c)

The graph is

05

Calculate the percent drop in Fermi energy

d)

The expression for drop in Fermi energy is calculated as,

∆=EF0-EFEF0×100%=E-EFEF×100%=1-EEF×100%=1-kBTElneEkBT-1×100%∆=1-kBTElneEkBT-1×100%=1-0.25lne10.25-1×100%=0.462%

Therefore, the percent drop in Fermi energy is 0.462%.

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!

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 electrons’ contribution to the heat capacity of a metal is small and goes to 0as T→0. We might try to calculate it via the total internal energy, localid="1660131882505" U=∫EN(E)D(E)dE, but it is one of those integrals impossible to do in closed form, and localid="1660131274621" N(E)FDis the culprit. Still, we can explain why the heat capacity should go to zero and obtain a rough value.

(a) Starting withN(E)FDexpressed as in equation (34), show that the slope N(E)FDdEatE=EFis-1(4kBT).

(b) Based on part (a), the accompanying figure is a good approximation to N(E)FDwhen Tis small. In a normal gas, such as air, whenTis raised a little, all molecules, on average, gain a little energy, proportional to kBT. Thus, the internal energy Uincreases linearly with T, and the heat capacity, ∂U∂T, is roughly constant. Argue on the basis of the figure that in this fermion gas, as the temperature increases from 0to a small value T, while some particles gain energy of roughly kBT, not all do, and the number doing so is also roughly proportional to localid="1660131824460" T. What effect does this have on the heat capacity?

(c)Viewing the total energy increase as simply ∆U= (number of particles whose energy increases) (energy change per particle) and assuming the density of states is simply a constant Dover the entire range of particle energies, show that the heat capacity under these lowest-temperature conditions should be proportional to kBREFT. (Trying to be more precise is not really worthwhile, for the proportionality constant is subject to several corrections from effects we ignore).

Nuclear density is approximately 1017 kg/m3. (a) Treating them as a gas of fermions bound together by the (no electrostatic) "strung attraction." calculate EFfor the neutrons in lead-206 (82 protons and 124 neutrons). (b) Treating them the same way. what would EFbe for the protons? (c) In fact, the energies of the most energetic neutrons and protons, those at the Fermi energy, are essentially equal in lead-206? What has been left out of pars (a) or (b) that might account for this?

The Fermi energy in a quantum gas depends inversely on the volume, Basing your answer on Simple Chapter 5 type quantum mechanics (not such quaint notions as squeezing classical particles of finite volume into a container too small). Explain why.

Suppose we have a system of identical particles moving in just one dimension and for which the energy quantization relationship isE=bn2/3, wherebis a constant andan integer quantum number. Discuss whether the density of states should be independent ofE, an increasing function ofE, or a decreasing function ofE.

A scientifically untrained but curious friend asks, "When I walk into a room, is there a chance that all the air will be on the other side?" How do you answer this question?

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.