Gostota stanj in stisljivost elektronskega plina

Iz Fizika trdne snovi 2007 - 2008

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Stisljivost

Stisljivost opazovanega sistema je količina, ki pove, za koliko se mu zmanjša prostornina V, če malo povečamo tlak p, v primerjavi z začetno prostornino:

$\chi = -\frac{1}{V} \frac{\partial V}{\partial p}.$

V tem primeru bo prikladneje izračunati njeno obratno vrednost - modul stisljivosti K

$K = \frac{1}{\chi} = -V \frac{\partial p}{\partial V},$

saj bomo tlak dobili iz definicije

$p = - \frac{\partial E}{\partial V},$

torej iz povečanja energije sistema E ob zmanjšaju njegove prostornine. Energijo izračunamo tako, da seštejemo prispevke vseh delcev. Ti pri temperaturi T = 0 zasedajo stanja do Fermijeve energije ε_F:

$E = \int_0^{\varepsilon_F} g(\varepsilon) \varepsilon d\varepsilon = \frac{2}{5} \frac{\sqrt{2m^3}}{\pi^2\hbar^3} V\varepsilon_F^{5/2},$

kjer pa je tudi Fermijeva energija odvisna od prostornine. V k prostoru N delcev zaseda stanja do k_F, ki ustreza ε_F:

$N = 2 \frac{V_{kF}}{V_{k0}} = 2 \frac{4\pi k_F^3}{3} \Bigl(\frac{L}{2\pi}\Bigr)^3 = \frac{V k_F^3}{3\pi^2},$

od koder lahko izrazimo Fermijev valovni vektor in ga nadomestimo s Fermijevo energijo

$\varepsilon_F = \frac{\hbar^2 k_F^2}{2m}= \frac{\hbar^2}{2m} \Bigl(3\pi^2 \frac{N}{V}\Bigr)^{2/3}.$

Ko dobljeno vstavimo nazaj v izraz za celotno energijo, ostane

$E = \frac{3}{5}\varepsilon_F N \propto V^{-2/3}.$

Tlak sedaj dobimo z odvajanjem energije

$p = + \frac{3}{5}\frac{2}{3}\varepsilon_F\frac{N}{V} = \frac{2}{5} \varepsilon_F n \propto V^{-5/3},$

kjer je n številska gostota delcev. Stisljivostni modul sledi s še enim odvodom

$K = + V \frac{2}{5}\frac{5}{3}\varepsilon_F\frac{N}{V^2} = \frac{2}{3} \varepsilon_F n = \frac{5}{3}p.$

V dveh oz. eni razsežnosti je ob upoštevanju drugačne gostote stanj in smiselno spremenjenih definicijah tlaka in stisljivosti postopek povsem enak. Rezultati so zbrani v spodnji preglednici.

Povzetek

	3D	2D	1D
$n \$	$\frac{N}{V}$	$\frac{N}{S}$	$\frac{N}{L}$
$g(\varepsilon)$	$\frac{V\sqrt{2m^3\varepsilon}}{\pi^2 \hbar^3}$	$\frac{Sm}{\pi\hbar^2}$	$\frac{L}{\pi\hbar} \sqrt{\frac{2m}{\varepsilon}}$
$\varepsilon_F$	$\frac{\hbar^2}{2m} (3\pi^2 n)^{2/3}$	$\frac{\pi\hbar^2}{m} n$	$\frac{\pi\hbar^2}{8m} n^2$
$E \$	$\frac{3}{5} \varepsilon_F N$	$\frac{1}{2} \varepsilon_F N$	$\frac{1}{3} \varepsilon_F N$
$p \$	$\frac{2}{5} \varepsilon_F n$	$\frac{1}{2} \varepsilon_F n$	$\frac{2}{3} \varepsilon_F n$
$K \$	$\frac{5}{3} p$	$2p \$	$3p \$

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