/*! 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} Q34P Calculate the Fermi energy for n... [FREE SOLUTION] | 91影视

91影视

Calculate the Fermi energy for noninteracting electrons in a two-dimensional infinite square well. Let 蟽 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=h2m

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=2h22mnx2lx2+ny2ly2=h2k22m,withk=nxlx,nyly

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鈥-this time a plane-and each state occupies an area 2/lxly=2/A( whereAlxly is the area of the well). Two electrons per state means

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

14k2=Nq22A,orkF=2NqA1/2=21/2

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

EF=h2kF22m=h22m2=h2m

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

EF=h2m

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 91影视!

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

a) Hund鈥檚 first rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium?

(b) Hund鈥檚 second rule says that, for a given spin, the state with the highest total orbital angular momentum (L) , consistent with overall antisymmetrization, will have the lowest energy. Why doesn鈥檛 carbon haveL=2? Note that the 鈥渢op of the ladder鈥(ML=L)is symmetric.

(c) Hund鈥檚 third rule says that if a subshell(n,l)is no more than half filled,
then the lowest energy level hasJ=lL-SI; if it is more than half filled, thenJ=L+Shas the lowest energy. Use this to resolve the boron ambiguity inProblem 5.12(b).

(d) Use Hund鈥檚 rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.12(b). Hint: Always go to the 鈥渢op of the ladder鈥 to figure out the symmetry of a state.

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

Find the average energy per free electron (Etot/Nd), as a fraction of the

Fermi energy. Answer:(3/5)EF

(a) Construct the completely anti symmetric wave function (xA,xB,xC)for three identical fermions, one in the state 5, one in the state 7,and one in the state 17

(b)Construct the completely symmetric wave function (xA,xB,xC)for three identical bosons (i) if all are in state 11(ii) if two are in state 19and another one is role="math" localid="1658224351718" 1c) one in the state 5, one in the state 7,and one in the state17

Discuss (qualitatively) the energy level scheme for helium if (a) electrons were identical bosons, and (b) if electrons were distinguishable particles (but with the same mass and charge). Pretend these 鈥渆lectrons鈥 still have spin 1/2, so the spin configurations are the singlet and the triplet.

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.