To calculate Fermi energy from arbitrary dispersion relation

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Homework Statement


problem statement is attached as problem.pdf


Homework Equations



eqn are given in the pdf file

The Attempt at a Solution


I have tried in vain to connect Fermi energy with dispersion relation. I just dont have any clue ,I also tried to determine the effective number of electron but still I dont know how to get the fermi energy from that
 

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  • #2
DrDu
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I didn't check but I suppose the dispersion relation is monotonous in the range of interest (plot it!).
What is the spacing of the k values if you assume the crystal to be of length L?
How many electrons will there be in the crystal of length L?
Hence what is the maximal k value occupied?
 
  • #3
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Thanks for your reply.But I failed to understand a thing.Please be a little descriptive.
 
  • #4
I have the same problem, if can plz help.
 
  • #6
Assume that there are no additional quantum degrees of freedom (spin etc.) for the electron to have degeneracy. I also assume the problem is 1 dimensional crystal.

In that case if 1/3 of the unit cells (aka lattice spots) are occupied, the electron band is 1/3 full. If there were 1 electron in every lattice points, there wouldn't be space for electrons to fit in that band anymore (Pauli exclusion!), so they'd have to take up a different orbital (aka a different band). Since the band lattice momenta $$ka \equiv K \in -\pi \rightarrow \pi$$, that means all the states from $$ K \in -{\pi \over 3} \rightarrow {\pi \over 3}$$ are taken. The fermi energy, which is the energy of the highest state that is occupied, is the value of $$max (\epsilon(K))$$ such that $$K \in (- {\pi \over 3}, {\pi \over 3})$$, in other words, it is

$$\epsilon_F = \epsilon({\pi \over 3} = ka)$$
 
  • #7
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Density of states = Vk / Vstates in 3-dimension. Here the problem is in 1-dimension. So density of states will be ratio of lengths,
Density of states (1D)= KF / (2π/a) ×2 = 1/3
2 is for electron degeneracy accounting spins and length in 1D K-space is from -π/a to +π/a (as given in the problem |ka|≤π).
Hence, KF a = π/3. This gives Fermi energy, ε(kF ) = ε0 / 4 .
 
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Another approach is to look for ##k_F##. If the number of electrons per atom in the conduction band is 1/3, then this implies a hole in the reciprocal lattice every third site at 0 Kelvin. The Fermi momentum is associated with the shortest wave length commiserate with the periodicity of the hole distribution in the lattice.
##k_F = \frac {2\pi} {\lambda_F} = \frac {2\pi} {3a}##
##\epsilon_F = 2\epsilon_0[sin^2(\frac {k_Fa} {2}) - \frac {1} {6}sin^2(k_Fa)] = 2\epsilon_0[ \frac {1} {4} - \frac {1} {6}\frac {3} {4}] = \frac {\epsilon_0} {4}##
 

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