To calculate Fermi energy from arbitrary dispersion relation

[m1rohit]

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 don't have any clue ,I also tried to determine the effective number of electron but still I don't know how to get the fermi energy from that

 

[DrDu]

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?

 

[m1rohit]

Thanks for your reply.But I failed to understand a thing.Please be a little descriptive.

 

[sayak datta]

I have the same problem, if can please help.

 

[khandelwal deeksha]

i also have the same problem . please help me

 

[paralleltransport]

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

$$\epsilon_F = \epsilon({\pi \over 3} = ka)$$

 

[Phy4Me]

Density of states = V_k / V_states in 3-dimension. Here the problem is in 1-dimension. So density of states will be ratio of lengths,
Density of states (1D)= K_F / (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, K_F a = π/3. This gives Fermi energy, ε_{(kF )} = ε₀ / 4 .

 

[Fred Wright]

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.
$k_F = \frac {2\pi} {\lambda_F} = \frac {2\pi} {3a}$
$\epsilon_F = 2\epsilon_0[sin^2(\frac {k_Fa} {2}) - \frac {1} {6}sin^2(k_Fa)] = 2\epsilon_0[ \frac {1} {4} - \frac {1} {6}\frac {3} {4}] = \frac {\epsilon_0} {4}$