Fermi energy condensed matter exam problem

[SMC]

Homework Statement

Hello, I am preparing a condensed matter exam and I was wondering if I could get some help on the following question from a past exam paper:

Show that for the free electron gas at zero temperature the Fermi energy is given by:

$$ε_{F}=\frac{\hbar^{2}}{2m}(3π^{2}N)^{2/3}$$

and the fermi wave number by:

$$k_{F}=(3π^{2}N)^{1/3}$$

where N is the volume density of electrons.

Homework Equations

the previous question was to derive the density of states:

$$D(ε) = \frac{V}{2π^{2}}(\frac{2m}{\hbar^{2}})^{3/2} ε^{1/2}$$

but I don't know if I have to use that or not

The Attempt at a Solution

how I would have done it is derive $k_{F}$ from $$N=\frac{VK_{F}^{3}}{3π^{2}}$$ which is obtained by saying number of electrons N is equal to volume of fermi sphere over k-space volume element $(2π/L)^{3}$.

and then state that the dispersion for a free electron gas is
$$ε(k) = \frac{\hbar^{2}k^{2}}{2m}$$ to get fermi energy.

but this method gives a result with the V present in the equation while the equations in the question don't have the V for some reason. Also I derived wavenumber first then energy while the question asks to derive energy first then wavenumber.

 

[Oxvillian]

SMC - I think the $N$ in the question is the number of particles per unit volume. Otherwise their equations aren't dimensionally correct.

 

[SMC]

yes ok that makes sense I should of noticed that, thank you Oxvillan. but I still don't understand why it asks me to derive fermi energy before wavenumber. is there a way of deriving fermi energy without deriving wavenumber first?

or maybe the questions are just in the wrong order

 

[Oxvillian]

Actually I would have done the wavenumber first too

But you can also integrate the density of states in energy space from zero to whatever the Fermi energy is and set that equal to $N$.

 

[paranormal]

Hello,

Thank you for reaching out for help with your condensed matter exam problem. The derivation you have described is correct and is commonly used to find the Fermi energy and Fermi wave number for a free electron gas at zero temperature. The reason that the equations provided in the question do not have the volume term is because the volume density of electrons (N) is already taken into account. This means that the equations are already in terms of N, and do not require the volume term.

To derive the Fermi energy, we can use the expression for the density of states, D(ε), and set it equal to the number of electrons, N. This gives us:

N = \frac{V}{2π^{2}}(\frac{2m}{\hbar^{2}})^{3/2} ε^{1/2}

From here, we can solve for ε to get the Fermi energy:

ε_{F} = \frac{\hbar^{2}}{2m}(3π^{2}N)^{2/3}

To find the Fermi wave number, we can use the relation between the Fermi energy and Fermi wave number, ε_{F} = \frac{\hbar^{2}k_{F}^{2}}{2m}. Plugging in the expression for the Fermi energy, we get:

\frac{\hbar^{2}k_{F}^{2}}{2m} = \frac{\hbar^{2}}{2m}(3π^{2}N)^{2/3}

Solving for k_{F}, we get:

k_{F} = (3π^{2}N)^{1/3}

I hope this helps with your exam preparation. Best of luck!