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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=V202mh3(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=v220k2n()dk,, where n()is given as (T0)by Eq. 5.104. So N=V220kmaxk2dk=V22k3max3,wherekmax is given byh2k2max2m=(0)=EFkmax=2mEFh

Compare Eq. 5.43, which says

EF=h22m(32NqV)2/3,or(2mEF)3/2h3=32NqV,N=V32qh3(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=Vh242m0kmaxk4dk=Vh242mk5max5Etot=V202mh3(2mEF)5/2

Compare Eq. 5.45, which says Etot=Vh2102mk5max.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

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.

Derive the Stefan-Boltzmann formula for the total energy density in blackbody radiation

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(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,
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(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.

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?

Suppose we use delta function wells, instead of spikes (i.e., switch the sign ofin Equation 5.57). Analyze this case, constructing the analog to Figure 5.6. this requires no new calculation, for the positive energy solutions (except that is now negative; use =-1.5 for the graph), but you do need to work out the negative energy solutions (letk-2mE/handZ-ka,forE<0) and , for). How many states are there in the first allowed band?

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