Electronic Specific Heat constant

[Sturnn17]

Homework Statement

Hi all. This isn't homework per say, more lack of understanding of something when reading around the notes.My problem is in trying to derive the electronic specific heat constant [itex]γ[/itex] in 2 dimensions.

Homework Equations

I know the general formula for specific heat is [itex] c^{el} = {\frac{π^2 g(ε_F) k^2 T}{3}} [/itex]

I also derived the density of states in 2D (g(ε)) to be [itex] {\frac{m^*}{ ħ^2 π}} [/itex]

And that [itex] c^{el} = γ~T [/itex]

The Attempt at a Solution

I know that I could trivially combine my equations together to give [itex] c^{el} ={\frac{π m^* k^2 T}{3ħ^2}} [/itex]

However I am confused to what this actually represents, is this the contribution to the specific heat by each electron?

I am looking for [itex]γ[/itex] in units of [itex] J~mol^{-1}~K^{-2} [/itex] or something similar so I would think I need to introduce avogadros number into my equation somehow but I have failed so far.

I tried to include it by working out n (the number of electrons per unit area)

[itex] n = ~^{ε_F}_{0}\int g(ε) dε =~ ^{ε_F}_{0}\int{\frac{m^*}{ ħ^2 π}} dε = {\frac{m^*}{ ħ^2 π}}~ε_F [/itex]

[itex] πk^{2}_{F}= {\frac{N}{ 2}}({\frac{2π}{ L}})^2 [/itex] => [itex] k_F= (2nπ)^{1/2} [/itex]

[itex] ε_F= {\frac{ħ^2 (2nπ)}{ 2m^*}} = {\frac{ħ^2 nπ}{m^*}}[/itex]

[itex] n = {\frac{m^*}{ ħ^2 π}}~{\frac{ħ^2 nπ}{m^*}}= n... [/itex] This evidently wasn't very helpful.

Thanks for taking time to read my post and for any input you may have.

 

[mark726]

Hi there,

It looks like you're on the right track with your attempt at a solution. The equation you have derived, c^{el} = {\frac{π m^* k^2 T}{3ħ^2}}, represents the contribution to the specific heat by each electron in 2D. To convert this to units of J~mol^{-1}~K^{-2}, you can use the fact that the Avogadro's number, N_A, is equal to 6.022 x 10^23 mol^{-1}. So, to get the specific heat in terms of moles, you can simply multiply your equation by N_A. This would give you c^{el} = {\frac{N_A π m^* k^2 T}{3ħ^2}}.

However, if you want to express the specific heat per unit area (in units of J~m^{-2}~K^{-2}), you can use your expression for the number of electrons per unit area, n, and multiply your equation by n. This would give you c^{el} = {\frac{π m^* k^2 T}{3ħ^2}}~n.

I hope this helps clarify things for you. Let me know if you have any further questions.