Calculate the Fermi energy for noninteracting electrons in a two-dimensional infinite square well. Let σ be the number of free electrons per unit area.

Short Answer

Expert verified

The Fermi energy for electrons in a two-dimensional infinite square well is

EF=πh2σm

Step by step solution

01

Definition of Fermi energy of electron

The greatest energy that an electron may hold at 0K is known as the Fermi energy.

Equation 5.50

Enxny=π2h22mnx2lx2+ny2ly2=h2k22m,withk=πnxlx,πnyly

02

Calculating the Fermi energy for electrons in a two-dimensional infinite square well

Each state is represented by an intersection on a grid in k-space”-this time a plane-and each state occupies an area π2/lxly=π2/A( whereAlxly is the area of the well). Two electrons per state means

Enxnynz=h22mnx2lx2+ny2ly2+nz2lz2=h2k22m …(5.50).

14πk2=Nq2π2A,orkF=2πNqA1/2=2πσ1/2

where σNq/Ais the number of free electrons per unit area.

EF=h2kF22m=h22m2πσ=πh2σm

Thus the Fermi energy for electrons in a two-dimensional infinite square well is

EF=πh2σm

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 Vaia!

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

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?

Chlorine has two naturally occurring isotopes,CI35and CI37. Show that

the vibrational spectrum of HCIshould consist of closely spaced doublets,

with a splitting given by v=7.51×10-4v, where v is the frequency of the

emitted photon. Hint: Think of it as a harmonic oscillator, with ω=k/μ, where

μis the reduced mass (Equation 5.8 ) and k is presumably the same for both isotopes.

Typically, the interaction potential depends only on the vectorr=r1-r2between

the two particles. In that case the Schrodinger equation seperates, if we change variables from r1,r2andR=(m1r1+m2r2)I(m1+m2)(the center of mass).

(a)Show that

localid="1655976113066" r1=R+(μ/m1)r,r2=R-(μ/m)r,and1=(μ/m)R+r,2=(μ/m1)R-r,where

localid="1655976264171" μ=m1m2m1+m2(5.15).

is the reduced mass of the system

(b) Show that the (time-independent) Schrödinger equation (5.7) becomes

-h22m112ψ-h22m222ψ+=-h22(m1+m2)R2ψ-h22μr2ψ+V(r)ψ=.(5.7).

(c) Separate the variables, letting ψ(R,r)=ψR(R)ψr(r)Note that ψRsatisfies the

one-particle Schrödinger equation, with the total mass (m1+m2)in place of m, potential zero, and energy ERwhile ψrsatisfies the one-particle Schrödinger equation with the reduced mass in place of m, potential V(r) , and energy localid="1655977092786" Er. The total energy is the sum: E=ER+Er. What this tells us is that the center of mass moves like a free particle, and the relative motion (that is, the motion of particle 2 with respect to particle 1) is the same as if we had a single particle with the reduced mass, subject to the potential V. Exactly the same decomposition occurs in classical mechanics; it reduces the two-body problem to an equivalent one-body problem.

Suppose you had three particles, one in stateψa(x), one in stateψb(x), and one in stateψc(x). Assuming ψa,ψb, andψc are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row isψa(x1),ψb(x1),ψc(x1) , etc., whese second row isψa(x2),ψb(x2),ψc(x2) , etc., and so on (this device works for any number of particles).

Certain cold stars (called white dwarfs) are stabilized against gravitational collapse by the degeneracy pressure of their electrons (Equation 5.57). Assuming constant density, the radius R of such an object can be calculated as follows:

P=23EtotV=23h2kF510π2m=(3π2)2/3h25mp5/3(5.57)

(a) Write the total electron energy (Equation 5.56) in terms of the radius, the number of nucleons (protons and neutrons) N, the number of electrons per nucleon d, and the mass of the electron m. Beware: In this problem we are recycling the letters N and d for a slightly different purpose than in the text.

Etot=h2V2π2m0kFK4dk=h2kF5V10π2m=h2(3π2Nd)5/310π2mV-2/3(5.56)

(b) Look up, or calculate, the gravitational energy of a uniformly dense sphere. Express your answer in terms of G (the constant of universal gravitation), R, N, and M (the mass of a nucleon). Note that the gravitational energy is negative.

(c) Find the radius for which the total energy, (a) plus (b), is a minimum.

R=(9π4)2/3h2d5/3GmM2N1/3

(Note that the radius decreases as the total mass increases!) Put in the actual numbers, for everything except , using d=1/2 (actually, decreases a bit as the atomic number increases, but this is close enough for our purposes). Answer:

(d) Determine the radius, in kilometers, of a white dwarf with the mass of the sun.

(e) Determine the Fermi energy, in electron volts, for the white dwarf in (d), and compare it with the rest energy of an electron. Note that this system is getting dangerously relativistic (seeProblem 5.36).

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.

Sign-up for free