/*! 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} Q21P Show that most of the energies d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that most of the energies determined by Equation 5.64are doubly degenerate. What are the exceptional cases? Hint: Try it for N=1,2,3,4.... , to see how it goes. What are the possible values of cos(ka)in each case?

Short Answer

Expert verified

There is no degeneracy at the top and the bottom of the band whencoska=±1

Step by step solution

01

Define the Schrodinger equation

  • A differential equation for the quantum mechanical description of matter in terms of the wave-like characteristics of particles in a field. The solution has to do with the probability density of a particle in space and time.
  • The time-dependent Schrodinger equation is represented as

¾±Ä§ddtψ(t)>=HÁåœÏˆ(t)>

02

Analyze the condition

For Ka=2Ï€²Ô/Nwe have a condition on n:n=0,12,...,N-1. Because for larger n, we don't get new solutions.

We are going to write solutions for cosKafor couple of N

Ka=2Ï€²ÔN

Value of different ncorresponds to a distinct state

03

Analyze the energies

N=1,n=0coska=1Non- degenerate

N=2,n=0,1coska=1,1Non- degenerate

N=3,n=0,1,2cos(ka)=1,-12,-12First one is non -degenerate other one is degenerate

N=4,n=0,1,2,3cos(ka)=1,0,-1,0Two are non -degenerate and two are degenerate

When we have coska=±1, then we don't have degeneracy. This happens at the top and at the bottom of the band.

So, we don’t have degeneracy at the top and the bottom of the band whencos(ka)=±1

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

We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, E=p2/2m,,with the relativistic formula, E=p2c2+m2c4-mc2. Momentum is related to the wave vector in the usual way: p=hk. In particular, in the extreme relativistic limit, E≈pc=hck.

(a) Replace h2k2n Equation 5.55 by the ultra-relativistic expression, hck, and calculateEtotin this regime.

dE=h2k22mVÏ€2k2dk (5.55).

(b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc , such that gravitational collapse occurs for N>N_{C}is called the Chandrasekhar limit.

(c) At extremely high density, inverse beta decaye-+p+→n+v,converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star non relativistic ally?

(a) Find the percent error in Stirling’s approximation for z = 10 ?

(b)What is the smallest integer z such that the error is less than 1%?

(a) Figure out the electron configurations (in the notation of Equation

5.33) for the first two rows of the Periodic Table (up to neon), and check your

results against Table 5.1.

1s22s22p2(5.33).

(b) Figure out the corresponding total angular momenta, in the notation of

Equation 5.34, for the first four elements. List all the possibilities for boron,

carbon, and nitrogen.

LJ2S+1 (5.34).

(a) Suppose you put both electrons in a helium atom into the n=2state;

what would the energy of the emitted electron be?

(b) Describe (quantitatively) the spectrum of the helium ion,He+.

Find the energy at the bottom of the first allowed band, for the caseβ=10 , correct to three significant digits. For the sake of argument, assume αa=1eV.

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.