Kinectic energy of a three-dimensional fermi gas of N free electrons at absolute zero

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Show that the kinectic energy of a three-dimensional fermi gas of N free electrons at absolute zero is (Mathematica code used)
u = 3/5 N Subscript[\[Epsilon], F]

Now I know total energy of N particles is this integral

u = \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(\
\[Epsilon]\ P[\[Epsilon]] \[DifferentialD]\[Epsilon]\)\)

which is made up of the density of the states and probability of the electron to occupy level with energy \[Epsilon] at temp T.

So P[\[Epsilon]] is this big horrible looking thing. My guess is that there must be a be an easy way to integrate it that comes about from absolute zero tempature because the final answer seems so nice.

Any help, would be nice. thanks!
 
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Re: kinectic energy of a three-dimensional fermi gas of N free electrons at absolute

You need to express the Fermi momentum of the system through its number of particles:

[tex]
N = \frac{V}{(2\pi)^{3}} \int_{0}^{k_{F}}{4\pi k^{2} dk}
[/tex]

The Fermi energy [itex]E_{F}[/itex] is defined as the kinetic energy of the particles with Fermi momentum

Once you had done that, the total kinetic energy of the particles is:
[tex]
(E_{\mathrm{kin}})_{\mathrm{tot}} = \frac{V}{(2\pi)^{3}} \, \int_{0}^{k_{F}}{\frac{\hbar^{2} k^{2}}{2 m} \, {4 \pi k^{2} \, dk}
[/tex]

In the final result you need to eliminate [itex]k_{F}[/itex].
 
20
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Re: kinectic energy of a three-dimensional fermi gas of N free electrons at absolute

ahhk, yeah that works. you could have explained it more but I got it in the end.
I'll fill in the gaps for anyone else who might look at this in the future.

use Ef=((khbar)^2)/2m to get the it to look right. and the big integral, you need to know that f(E) = 0 for E>Ef and f(E) = 1 for 0<E<Ef

So put the upper limit to Ef (because anything above this range in the integral equals zero) and also anything inside this range "E<Ef" aka 0 to Ef the f(E)=1, so this makes the integral really easy.

Thanks again Dickfore.
 

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