CHAPTER 5 FREE ELECTRON THEORY

Size: px

Start display at page:

Download "CHAPTER 5 FREE ELECTRON THEORY"

Jeffery Wheeler
5 years ago
Views:

1 CHAPTER 5 REE ELECTRON THEORY

2 r Elctron Thory Many solids conduct lctricity. Thr ar lctrons that ar not bound to atos but ar abl to ov through th whol crystal. Conducting solids fall into two ain classs; tals and siconductors. ρ ( ) tals ; RT Ω and incrass by th addition of sall aounts of ipurity. Th rsistivity norally dcrass onotonically with dcrasing tpratur. ρ( RT ) ρ( RT ) pur siconductor tal and can b rducd by th addition of sall aounts of ipurity. Siconductors tnd to bco insulators at low T.

3 Why obil lctrons appar in so solids and othrs? Whn th intractions btwn lctrons ar considrd this bcos a vry difficult qustion to answr. Th coon physical proprtis of tals; Grat physical strngth High dnsity Good lctrical and thral conductivity, tc. This chaptr will calculat ths coon proprtis of tals using th assuption that conduction lctrons xist and consist of all valnc lctrons fro all th tals; thus tallic Na, Mg and Al will b assud to hav 1, and 3 obil lctrons pr ato rspctivly. A sipl thory of fr lctron odl which works rarkably wll will b dscribd to xplain ths proprtis of tals.

4 Why obil lctrons appar in so solids and not othrs? According to fr lctron odl (EM), th valanc lctrons ar rsponsibl for th conduction of lctricity, and for this rason ths lctrons ar trd conduction lctrons. Na 11 1s s p 6 3s 1 Valanc lctron (loosly bound) Cor lctrons This valanc lctron, which occupis th third atoic shll, is th lctron which is rsponsibl chical proprtis of Na.

5 Whn w bring Na atos togthr to for a Na tal, Na tal Na has a BCC structur and th distanc btwn narst nighbours is 3.7 A Th radius of th third shll in Na is 1.9 A Solid stat of Na atos ovrlap slightly. ro this obsrvation it follows that a valanc lctron is no longr attachd to a particular ion, but blongs to both nighbouring ions at th sa ti.

6 A valanc lctron rally blongs to th whol crystal, sinc it can ov radily fro on ion to its nighbour, and thn th nighbour s nighbour, and so on. This obil lctron bcos a conduction lctron in a solid Th roval of th valanc lctrons lavs a positivly chargd ion. Th charg dnsity associatd th positiv ion cors is sprad uniforly throughout th tal so that th lctrons ov in a constant lctrostatic potntial. All th dtails of th crystal structur is lost whn this assunption is ad. According to EM this potntial is takn as zro and th rpulsiv forc btwn conduction lctrons ar also ignord.

7 Thrfor, ths conduction lctrons can b considrd as oving indpndntly in a squar wll of finit dpth and th dgs of wll corrsponds to th dgs of th sapl. Considr a tal with a shap of cub with dg lngth of L, Ψ and E can b found by solving Schrödingr quation V L/ 0 L/ h = E ψ ψ Sinc, V =0 By ans of priodic boundary conditions Ψ s ar running wavs. ψ ( x + L, y + L, z + L) = ψ ( x, y, z)

8 Th solutions of Schrödingr quations ar plan wavs, rr 1 ikr 1 x y z ψ ( x, y, z) = = V V i( k x+ k y + k z) Noralization constant whr V is th volu of th cub, V=L 3 Na = pλ π Na = p k π whr, k = λ π π k = p = p Na L So th wav vctor ust satisfy π k x = p L k = π y q π k L z = r L ; ; whr p, q, r taking any intgr valus; +v, -v or zro.

9 Th wav function Ψ(x,y,z) corrsponds to an nrgy of E = h k = h + + E ( k k k ) x y z th ontu of p =h ( k x, k y, k z) Enrgy is copltly kintic 1 v k = h v k =h p =h k

10 W know that th nubr of allowd k valus insid a sphrical shll of k-spac of radius k of Vk g( k) dk = dk, π whr g(k) is th dnsity of stats pr unit agnitud of k.

11 Th nubr of allowd stats pr unit nrgy rang? Each k stat rprsnts two possibl lctron stats, on for spin up, th othr is spin down. = g( E) = g( k) dk de g( E) de g( k) dk E = h k de k dk = h E k = h V g( E ) = dk g( k) π kk E h hde k g( E) = V ( ) E 3 π h 3/ 1/

12 Ground stat of th fr lctron gas Elctrons ar frions (s=±1/) and oby Pauli xclusion principl; ach stat can accoodat only on lctron. Th lowst-nrgy stat of N fr lctrons is thrfor obtaind by filling th N stats of lowst nrgy.

13 Thus all stats ar filld up to an nrgy E, known as ri nrgy, obtaind by intgrating dnsity of stats btwn 0 and E, should qual N. Hnc Rbr g( E) E E = V ( ) E 3 π h 3/ 1/ V V N = g( E) de = ( ) E de = ( E ) 3 3 π h 3π h 0 0 3/ 1/ 3/ Solv for E (ri nrgy); E h 3 π N = V /3

14 Th occupid stats ar insid th ri sphr in k-spac shown blow; radius is ri wav nubr k. k z E ri surfac E=E h 3 π N = V k E = h /3 k ro ths two quation k can b found as, k 1/3 y 3π N k = V k x Th surfac of th ri sphr rprsnt th boundary btwn occupid and unoccupid k stats at absolut zro for th fr lctron gas.

15 Typical valus ay b obtaind by using onovalnt potassiu tal as an xapl; for potassiu th atoic dnsity and hnc th valanc lctron dnsity N/V is 1.40x so that 19 E = J =.1 V k = A 1 ri (dgnracy) Tpratur T by E 4 T = = K k B E = k BT

16 It is only at a tpratur of this ordr that th particls in a classical gas can attain (gain) kintic nrgis as high as E. Only at tpraturs abov T will th fr lctron gas bhav lik a classical gas. ri ontu P =h k P = V P V = = s 6 1 Ths ar th ontu and th vlocity valus of th lctrons at th stats on th ri surfac of th ri sphr. So, ri Sphr plays iportant rol on th bhaviour of tals.

17 Typical valus of onovalnt potassiu tal; /3 h 3 π N E = =.1 V V π 1/3 N 3 k = = A V 1 P V = = s 6 1 E 4 T = = K k B

19 Th fr lctron gas at finit tpratur At a tpratur T th probability of occupation of an lctron stat of nrgy E is givn by th ri distribution function f = + 1 D ( E E )/ k BT 1 ri distribution function dtrins th probability of finding an lctron at th nrgy E.

20 ri unction at T=0 and at a finit tpratur = + 1 f 0.5 D ( )/ B 1 E E k T f D =? At 0 K f D (E,T) i. E<E 1 f = =1 1 D ( )/ B 1 + E E k T ii. E>E f = 1 D ( )/ B 1 + = E E k T E<E E E>E E 0

21 ri-dirac distribution function at various tpraturs,

22 Nubr of lctrons pr unit nrgy rang according to th fr lctron odl? Th shadd ara shows th chang in distribution btwn absolut zro and a finit tpratur. n(e,t) g(e) T>0 T=0 n(e,t) nubr of fr lctrons pr unit nrgy rang is just th ara undr n(e,t) graph. n( ET, ) = g( E) fd ( ET, ) E E

23 ri-dirac distribution function is a sytric function; at finit tpraturs, th sa nubr of lvls blow E is ptid and sa nubr of lvls abov E ar filld by lctrons. n(e,t) g(e) T=0 T>0 E E

24 Hat capacity of th fr lctron gas ro th diagra of n(e,t) th chang in th distribution of lctrons can b rsbld into triangls of hight 1/g(E ) and a bas of k B T so 1/g(E )k B T lctrons incrasd thir nrgy by k B T. n(e,t) T>0 T=0 g(e) Th diffrnc in thral nrgy fro th valu at T=0 K 1 ET ( ) E(0) g( E )( k BT ) E E

25 Diffrntiating with rspct to T givs th hat capacity at constant volu, E T C v = = g( E ) k B T N = E g( E ) 3 3 N 3N g( E ) = E = k T B 3N k T C g( E ) k T k T = = v B B B 3 T C v = Nk B T Hat capacity of r lctron gas

26 Transport Proprtis of Conduction Elctrons ri-dirac distribution function dscribs th bhaviour of lctrons only at quilibriu. If thr is an applid fild (E or B) or a tpratur gradint th transport cofficint of thral and lctrical conductivitis ust b considrd. Transport cofficints σ,elctrical conductivity K,Thral conductivity

27 Total hat capacity at low tpraturs C = γt + βt 3 Elctronic Hat capacity Lattic Hat Capacity whr γ and β ar constants and thy can b found drawing C v /T as a function of T

28 Equation of otion of an lctron with an applid lctric and agntic fild. r dv E v B dt = ur r ur This is just Nwton s law for particls of ass and charg (-). Th us of th classical quation of otion of a particl to dscrib th bhaviour of lctrons in plan wav stats, which xtnd throughout th crystal. A particl-lik ntity can b obtaind by suprposing th plan wav stats to for a wavpackt.

29 Th vlocity of th wavpackt is th group vlocity of th wavs. Thus r ur r dω 1 de hk p v = r = r = = dk h dk E = hω = p = hk h k So on can us quation of dv/dt r r dv v ur r ur + = E v B dt τ (*) τ = an fr ti btwn collisions. An lctron loss all its nrgy in ti τ

30 In th absnc of a agntic fild, th applid E rsults a constant acclration but this will not caus a continuous incras in currnt. Sinc lctrons suffr collisions with phonons lctrons r v Th additional tr caus th vlocity v to τ dcay xponntially with a ti constant whn τ th applid E is rovd.

31 Th Elctrical Conductivty In th prsnc of DC fild only, q.(*) has th stady stat solution v r τ = ur E a constant of proportionality (obility) τ µ Mobility for = lctron Mobility dtrins how fast th charg carrirs ov with an E.

32 Elctrical currnt dnsity, J J = n( v ) v r τ = ur E Whr n is th lctron dnsity and v is drift vlocity. Hnc n = N V ur ur J = n τ E n τ σ = Elctrical conductivity Oh s law ur ur J =σ E Elctrical Rsistivity and Rsistanc ρ = 1 σ R = ρl A

33 Collisions In a prfct crystal; th collisions of lctrons ar with thrally xcitd lattic vibrations (scattring of an lctron by a phonon). This lctron-phonon scattring givs a tpratur dpndnt τ ( T ph ) collision ti which tnds to infinity as T 0. In ral tal, th lctrons also collid with ipurity atos, vacancis and othr iprfctions, this rsult in a finit scattring ti vn at T=0. τ 0

34 Th total scattring rat for a slightly iprfct crystal at finit tpratur; = + τ τ ph ( T ) τ 0 Du to phonon Du to iprfctions So th total rsistivity ρ, ρ = = + = ρ ( ) I T + ρ n τ n τ ( T ) n τ ph 0 0 Idal rsistivity Rsidual rsistivity This is known as Matthisn s rul and illustratd in following figur for sodiu spcin of diffrnt purity.

35 Rsidual rsistanc ratio Rsidual rsistanc ratio = roo tp. rsistivity/ rsidual rsistivity 6 and it can b as high as 10 for highly purifid singl crystals. ipur pur Tpratur

36 σ Collision ti σ ( RT ) =.0 x10 ( Ω ) sodiu 7 1 τ can b found by taking σ rsidual = 5.3 x10 ( Ω ) purna = σ n =.7 x τ =.6 x10 s n x10 s at RT at T=0 6 v = x / s Taking ; and l = v τ l( RT ) = 9 n lt ( = 0) = 77 µ Ths an fr paths ar uch longr than th intratoic distancs, confiring that th fr lctrons do not collid with th atos thslvs.

37 Thral conductivity, K Du to th hat tranport by th conduction lctrons K K tals non tals Elctrons coing fro a hottr rgion of th tal carry or thral nrgy than thos fro a coolr rgion, rsulting in a nt flow of hat. Th thral conductivity 1 K = C v l C 3 V V whr is th spcific hat pr unit volu l is th an spd of lctrons rsponsibl for thral conductivity sinc only lctron stats within about k BT of ε chang thir occupation as th tpratur varis. v l = v τ is th an fr path; and ri nrgy 1 1 π N T π nk T τ K = C v = k ( ) = B V τ B ε τ V T whr 1 ε = v π T C v = Nk B T

38 Widann-ranz law σ = n τ K = π nk Tτ B Th ratio of th lctrical and thral conductivitis is indpndnt of th lctron gas paratrs; 3 Lorntz nubr K π k B = =.45 x10 W Ω K σ T 3 8 K L = =.3 x 10 W ΩK σt 8 or coppr at 0 C

Exam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator.

Exam 1. It is important that you clearly show your work and mark the final answer clearly, closed book, closed notes, no calculator. Exam N a m : _ S O L U T I O N P U I D : I n s t r u c t i o n s : It is important that you clarly show your work and mark th final answr clarly, closd book, closd nots, no calculator. T i m : h o u r