EE138_HW _3 solns

EE138_HW _3 solns - EE138: Homework Assignment #3 Solutions...

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Unformatted text preview: EE138: Homework Assignment #3 Solutions (0) Fa— mowogm mm is ores-e): E5 "3:340 J1 - mi ' _ AF: Ezs E‘ E?” Lwdm 36“ Mum ZS slraéra): _ E1: ‘61:: ____L W: Mahala < was (.3): 40526} A 1‘ b9 : Wfisfilfléflfliwg 1 z.22>< 15% A E (lozeflxtoxm‘m 3/0,) ______._...__T (EDLVBRQS 0302., \n adriva J \owq Jr‘ocm 0x Wham RS (1 mad as too) 4d 2 3:360 2 0.07%;va PW>\>\<9,W\“£"t 2_ (a) Comm'fflmé a One, or mow, wed-mm 6: are, SW:wa ,, HL \mfimk/M_M ' " VMZH ca 3"" Are, doowk‘xzed CRVVJL mummy smtm bk} Jfim’. \aflKCL Odmmg "' COPsgef _ KONC, bond W gnaw-— -- o\ \JMmLo.’ a" is *Wamflawad 9mm (me, 06mm +0 amflnex in We, bowd. Cb) 'Fraa (Wad-(on; E; E: K2. 9M mmg 09.6%gi 0cm oukkowed «Them, (8 cm awe/«3&1 ' CoNimu/Lm. 31m“ N \DOWO\: E = ham?“ @935 a\\ mamas 0‘} emeer {A W30 gram” we, ELMOUQEC’L _ om tad 0?- cm warm m a ' , Nat w \Iodng 05% 'exwxgxi aw, ?€,\(iOd~LC, Pfim’mfl odbwai. ° Them, ow. &,\\O\M€C\ 0000K pwbx‘ddém avg?” bands The, widem‘wlg 01} any?) \axldS (diSCde) MAYO bfmd's am0\%\(\€x\ 0L ovflrmuum ocwrs becmse oUroms {whimd' \ass 0L3 Hma‘w Mudqu (fishy/m“ moreases. Problem #4: (a) F(E) vs. E _x M A A .0 oo .0 4:. TllllTllli’illli’rl A Fermi-Dirac Distribution F(E) O on .0 N 7 7.05 7.1 7.15 Energy (eV) 0.03 F(E) vs. E (bi 0.025 0.02 0.015 0.01 Fermi-Dirac Distribution F(E) 0.005 7.1 7. 1 1 7. 12 Energy (eV) T = 0K,F(7.11eV)= 0 T = 300 K, F(7.11 eV) ~ 0 T = 1200 K, F(7.11 eV) = 0.014 (c) Probability of an empty state = 1-F(E) T= 0K, 1-F(7.11 eV) =1 T = 300 K, 1-F(7.11eV) ~1 T = 1200 K, 1-F(7.11 eV) = 0.986 Efobmmfis; ‘ (a) M We, Pexml \eva em firm meraxf Hm "Probabdihf 0? ocvacxfim m V2. (by EF= ‘fl?’ (2&2“ ZW‘M N 2" q'L’XIO kfl/m3 Xqéflzxwfihbms x laydvm=3fl|x15€fi3 9‘/ EF : C(olaqu’E‘L‘SSY' 3MP“ x636?) 5 % TY F-(E7=O \ (C) ’F(E) * \ A — 3=Qsm \WOO') ” J‘" E—E f' EF (0 0‘ ' e“ H” ewfi?) F : o 056! T: W = L/éf'z K Mn MSW/Mm -i\.) (a) E:¢+EF’J Egh’V l? V=_E»::¢\:\EF t: * lxm‘”: — V £326“ + 50320 <9 Keri) = 1.5mm” Hi (é..CaZX(D"3L’$$ For loo mm uWrowioLejc Vadfod‘fon: b) gem: (QQZXIO’E’HIS'BMDNMS : 1.959xto"‘834 /\ ([00 MO“? ND +0 ascapa needs 80:6ng 09 a}; \@g-\— 4“ <25 + E: : C26 W + BAZADO .(g xlo“"J/e,\D= 8,971on 5 KB. M“ be EX”Efi+Eg:AE—— I AE= {fizmo’miS 0K 7w Prob\€m 1“ 7 ’1 V3=O J 7, :TT :"TT (m \wgli @ kojk 404 A (C) mm \{ has. We Nflhar «€903chin mass, We, know AHMS ‘ because, €P€adN€ Mass. is inverse“, m «hex/mil, To 'Hna- mwad'mre. 01"— E \IS. K PM: P P ...
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This note was uploaded on 01/05/2012 for the course EE 138 at UC Riverside.

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EE138_HW _3 solns - EE138: Homework Assignment #3 Solutions...

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