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Evaluate the integrals (Equation5.108 and 5.109) for the case of identical fermions at absolute zero. Compare your results with equations 5.43 and5.45. (Note for electrons there is an extra factor of 2 in Equations 5.108 and 5.109. to account for the spin degeneracy.)

Short Answer

Expert verified

The integrals for the case of identical fermions at absolute zero are

Etot=V20π2mh3(2mEF)5/2

Step by step solution

01

Definition of Fermi energy

A notion in quantum physics is called Fermi Energy. The Fermi energy is the value of the Fermi level at absolute zero temperature. The sea of fermions, in which no particle can live, is a clear indicator of it.

02

Evaluating the integrals

Equation 5.108 N=v2π20k2n()dk,, where n()is given as (T0)by Eq. 5.104. So N=V2π20kmaxk2dk=V2π2k3max3,wherekmax is given byh2k2max2m=μ(0)=EFkmax=2mEFh

Compare Eq. 5.43, which says

EF=h22m(3π2NqV)2/3,or(2mEF)3/2h3=3π2NqV,N=V3π2qh3(2mEF)3/2.

Here q = 1, and Eq. 5.108 needs an extra factor of 2 on the right, to account for spin, so the two formulas agree.

Equation 5.109

Etot=Vh24π2m0kmaxk4dk=Vh24π2mk5max5Etot=V20π2mh3(2mEF)5/2

Compare Eq. 5.45, which says Etot=Vh210π2mk5max.Again, Eq. 5.109 for electrons has an extra factor of 2, so the two agree.

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Most popular questions from this chapter

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.

The density of copper is8.96g/cm3,and its atomic weight is63.5g/mole

(a) Calculate the Fermi energy for copper (Equation 5.43). Assume d = 1, and give your answer in electron volts.

EF=ħ22m3ρπ22/3 (5.43).

(b) What is the corresponding electron velocity? Hint: SetEF=1/2mv2Is it safe to assume the electrons in copper are nonrelativistic?

(c) At what temperature would the characteristic thermal energyrole="math" localid="1656065555994" (kBT,wherekBkBis the Boltzmann constant and T is the Kelvin temperature) equal the Fermi energy, for copper? Comment: This is called the Fermi temperature,TF

. As long as the actual temperature is substantially below the Fermi temperature, the material can be regarded as “cold,” with most of the electrons in the lowest accessible state. Since the melting point of copper is 1356 K, solid copper is always cold.

(d) Calculate the degeneracy pressure (Equation 5.46) of copper, in the electron gas model.

P=23EtotV=23ħ2kF510π2m=3π22/3ħ25mρ5/3

Obtain equation 5.76 by induction. The combinatorial question is this: How many different ways you can put N identical balls into d baskets (never mind the subscriptfor this problem). You could stick all of them into the third basket, or all but one in the second basket and one in the fifth, or two in the first and three in the third and all the rest in the seventh, etc. Work it out explicitly for the casesN=1,N=2,N=3andN=4; by that stage you should be able to deduce the general formula.

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?

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

E1=-m2h2e24π'2=-13.6eV(4.77).

(b) Find the separation in wavelength between the red Balmer lines n=3n=2for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line n=2n=1?.

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