FYS2160 Oblig 8. Lars Kristian Henriksen Universitetet i Oslo November 13, 2015

Transcription

1 FYS216 Oblig 8 Lars Krisian Henriksen Universiee i Oslo November 13, 215

2 FERMIGASSER 2 a) Oppgave 1 In he n-vecor space, he densiy of saes D(n x, n y, n z ) equals 2 as every possible sae can be occupied by one spin up paricle and one spin down paricle. In he n-space, his will change o D(n) = πn 2 hrough D(n x, n y, n z )dn x dn y dn z = D(n x, n y, n z ) 4π 8 n2 dn = πn 2 dn Since ε pc = π hc L (n2 x + n 2 y + n 2 z) 1/2 in he n-vecor space, in he n space he energy ε π hc L (n)1/2 = an where a = cπ h L. To make he ransiion o he ε space o find D(ε) we use he following: dε dn = d dε an = a dn = dn a and ha i follows from ε = an ha n = ε. From his, we can derive he following a D(n)dn = πn 2 dn = π( ε dε )2 a a = D(ε)dε We now see ha D(ε) = π( ε a )2 1 a = π a 3 ε2 like we waned o show.

3 FERMIGASSER 3 b) As earlier we use ε = an bu since we wan o look a he fermi-energy we use ε F = an F where n F is he maximum lengh of he n-vecor in n space. We solve N = f(ε F, µ, T )D(ε)dε The facors in his inegral is known (assuming T=). The fermi-dirac disribuion is 1 up o ε and over. D(ε) was calculaed earlier. From his i is easy o show ha N = n F = 3 3N π f(ε F, µ, T )D(ε)dε = π 3 n3 F In his ask i is needed o solve for n f for he use in ε F = an F. Noe ha L = V 1/3. This expression hen expands o ε F = an F = a 3 3N π = cπ h ( 3N V 1/3 π ( 3n = cπ h π where n = N/V. This is he Fermi energy of a gas wih N elecrons, and he expression we waned o find. ) 1/3 ) 1/3

4 FERMIGASSER 4 c) To find he oal energy a zero emperaure i is needed o add an energy o every sae/paricle N. This relaes o he inegral U = ε F f(ε, µ, T ) D(ε) ε dε wih he upper inegral limi se o he Fermi energy level, as he energy above his equals zero (assuming T=). This inegral is easily solved as we know he differen componens from earlier calculaions U = = ε F ε F = πε4 F 4a 3 f(ε, µ, T ) D(ε) ε dε 1 π a 3 ε2 εdε Cleaning his up and insering he previous expression for a produces U 3 4 Nε F which is wha we waned o find.

5 FERMIGASSER 5 a) I use ha D(ε) = 3N 2ε 3/2 F change he inegral variable in N = Oppgave 2 ε 1/2 and f(ε, µ, T ) = 1 ( ). The nex sep is o ε µ 1 + exp kt f(ε, µ, T )D(ε)dε from ε o x. Firs is observed ha ε = ε F x dε = ε F dx f(ε F, µ, T ) dε hen ranslaes o f(x, c, ) dx by: (using = kt/ε F, c = µ/ε F and x = ε/ε F ) 1 ( ) dε ε µ 1 + exp kt ε F ( xεf cε F exp ε F ) dx = + 1 ε F ( x c exp ) dx + 1 N hen ranslaes o N = = f(x, c, )D(ε)dx 3N 2ε 3/2 F = 3 2 N ε 1/2 ε F ( ) dx x c exp + 1 ε 1/2 F ε 1/2 1 ( x c exp = 3 ( ) x 1/2 2 N ε 1/2 1 ( ε x c exp 1 = 3 2 which is wha we waned o show. x 1/2 ( ) dx x c exp + 1 ) dx + 1 ) dx + 1

6 FERMIGASSER 6 b) The fac ha = means ha kt/ε F =. This ranslaes o a emperaure equal o zero (boh ε F and k are fixed values). This means ha he Fermi-Dirac disribuion curve has a sharp edge a ε F so ha ε F = µ, he chemical poenial equals he Fermi energy level. c) 1 c mo.5.5 c Figure 1. The dimensionless chemical poenial µ as a funcion of he dimensionless emperaure The inegral in a) is solved numerically hrough combining and c values ha makes he inegral equal 3/2. These values of wih he corresponding c-value is ploed in he figure above.

7 FERMIGASSER 7 d) 3.5 u() = U(T )/ε F mo 3 u = U/εF Figure 2. The dimensionless energy u as a funcion of he dimensionless poenial c

8 FERMIGASSER 8 e) 1.6 Varmekapasie/k B mo sni 1.4 CV/kB sni Figure 3. Hea capaciy C V as a funcion of he dimensionless emperaure. We know ha C V = u = U(T )/ε F = U(T ). From his i is observed ha kt/ε F kt he plo is acually he same as he acual C V plo. Through he usage of dimensionless variables we obain acual hea capaciy.

9 FERMIGASSER 9 The code used for his a 1 clear all; 2 3 = linspace(.1,2,35); 4 5 c = linspace( 2.5,1.1,2); 6 7 lis = zeros(lengh(c),1); 8 flis = zeros(lengh(),1); 9 1 for j= 1:lengh() for i=2:lengh(c) F (sqr(x)./ (exp((x c(i))./(j)) + 1)); lis(i,:) = c(i 1); 17 if inegral(f,,inf) 2./3. 18 break 19 end 2 21 end flis(j,:) = lis(i); end 26 %% Plo 27 figure('paperposiion',[ ], 'Color',[1 1 1]) 29 3 axes('linewidh',2.,'fonsize',12); 31 plo(,flis,'linewidh',2.) 32 xlim('auo') xlabel('','fonsize',14) 35 ylabel('c','fonsize',14) 36 ile('c mo ', 'FonWeigh','bold','FonSize',14,'Inerpreer','laex') %% 4 prin('oppg2c',' depsc'); %% Oppg d 43 uin = zeros(lengh(),1); 44 for i=1:lengh() 45 uu (3./2.* (sqr(x)./ (exp((x flis(i)')./(i)) + 1)).* x); 46 uin(i) = inegral(uu,,inf); 47 end 48 %% Plo 49 figure('paperposiion',[ ],... 5 'Color',[1 1 1]) axes('linewidh',2.,'fonsize',12);

10 FERMIGASSER 1 53 plo(,uin,'linewidh',2.) 54 xlim('auo') xlabel('$$$$','fonsize',14,'inerpreer','laex') 57 ylabel('$$u=u/\varepsilon_f$$','fonsize',14,'inerpreer','laex') 58 ile('$$u() = U(T)/\varepsilon_F$$ mo $$$$', 'FonWeigh','bold','FonSize',14,'Inerpreer','laex') 6 61 %% 62 prin('oppg2d',' depsc'); 63 %% Oppg e % CV = (uin(2.:end) uin(1.:end 1.))./ ((2.:end)'... (1.:end 1.)'); 66 CV = diff(uin)./diff(') 67 sni = ((2.:end) + (1.:end 1.))./2.; %% Plo 7 figure('paperposiion',[ ], 'Color',[1 1 1]) axes('linewidh',2.,'fonsize',12); 74 plo(sni',cv,'linewidh',2.) 75 xlim('auo') xlabel('$$$$ sni','fonsize',14,'inerpreer','laex') 78 ylabel('$$c_v*k_b$$','fonsize',14,'inerpreer','laex') 79 ile('varmekapasie*$$k_b$$ mo $$$$ sni',... 8 'FonWeigh','bold','FonSize',14,'Inerpreer','laex') %% 83 prin('oppg2e',' depsc');