hw2 solutions

hw2 solutions - ____~____._.ti_.'§“mama___WWWM_._WL__M_...

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Unformatted text preview: ____~____._.ti_.'§“mama___WWWM_._WL__M_ MW According to classical physics, the average energy of an electron in an electmn gas at thermal equilibrium is 3k T/ 2. Determine, for T = 300 K, the average electron energy 27;; (in e__V___), average electron momentum, and the de Broglie wavelength. /2,3) 1‘ L. E f\ l # - . ~2/ 5:32, =2§a+l¢38XIO x300:- QIx/O jet/“QC .35 : 0'9” 6‘19 :3 5—4.: 1/: f: dflséfet/ ‘2‘: i _7_’ _ .— fi 5 E— QWP’ F=V2mE=/Zx£//Xll7xéZ/XIO '2! ~23“ =_-_ /. 06500 A70! 7 ,w ~59 LI 2 4625} la /. 064x104; ...._—— P‘ *9 Q, 282%: W) = @2514». —7 6,23x/0 em 625,4 .,_ ' E f .5! 2.4 (a) An electron is moving with a velocity of 2 x 106 cmfs. Determine the electron energy (in eV), momentum, and de Broglie wavelength (in A). (b) The de Broglie wavelength of an electron is 125 A. Determine the electron energy (in eV), momen— /t%\\ tum, and velocity. 4 2% In an]; i 3 2. —22 l 1 4/ E c2149: Kara—[<7] == ééfi/lxm x (2)0019: 7.322K/0 3' .l 2_ --22 _ ~ *2 .——_-» 1”: WW“: 7,)” max 2x A)": I, fax/0 [ado m/s 2 : "é _ 6' 625%” 3‘1 3 cacxm“? 3 53¢ "a (a -26 =- I m: . x10 cw. 5322/00 0 . = 363.9 A _L__;________ i a u/ ’A.: -: /23X/0 I ' -3‘f _26 A IZE‘XM ‘26 -2/ 5:; P71: ' _ (3.2mm )2”, /. 542x217 3’ 2m 2x 5/ 5Z9”: Egyr: fax, mael/ we .2 "’2" 1; é ’ m 5’3’W_;, ——- 5.82M? m/s ——- 5282x113 em/s ill x/tJ é :54 5 2.5 It is desired to produce x—ray radiation with a wavelength of l A. (0) Through what 7 ' potential voltage difference must the electron be accelerated-in vacuum so thatjt can, upon colliding with a target, generate such a photon? (Assume that all of the electron‘s energy is transferred to the photon.) (b) What is the de Broglie wavelength :3 of the electron in part (a) just before it hits the target? ' l E E .9 i IO‘W -: lo—lpm (mlxdfimflwrgk “A39 (P w on, g XML. 6,92 3 " GD: / P IE = bjéiimx x“) = {’fi‘flxm fif— J E p {of (hollow WHEY-71,40 i rZIJM/U _}b-- (X'M ) l AV __ li‘id’o’fl‘im -.,12,42 kv “‘3 //6X ’0 7‘: a ,D. WEI/2 by _[/Zx/.?Pfx/0-)5 g 5/ '7 .. ,.._———__———— -:. . 'X/0 I 14/” , an): M” filo/S l»: --—‘ :- Mfi—d— '= 0,”)(j0—“i’ln =' y’I/XIJS‘HI "W A ?.'//x29’3’x 6.40.107 (7” A“ _,_—o-—-- 077me were M‘Sr‘dh . 2,10 2.10 An electron is bound in a one—dimensional infinite potential well with a width of 100 A. Determine the electron energy levels for n = 1. 2, 3. reg“: $16 : w): o emu van; (Hue wab‘xfi 4—9135”! av. 12925an in: and/TEL LS 2040 fan/[T I '1- __ h / ' -%—:;‘+ %_U:”\/(x)]$=o '. Lha ‘-'-' Eg‘RW(KnK> : gums;pr __,_, L1,“ :9: (KwX) dzolvguang-WLM weéx E (f \ { 4‘1 2 w V 7. \ le,‘05%x Io )x Mm ,8 ’- _____________________:-__-—-—" .- thg h 2x‘Mlxro‘3‘ 100x IO M IanlO Eh: v3" ammo-3w 12‘: 3.7éx|5_ge_/V ' ' Ea: hSox ('0 ev ,2. \Orfi‘ bum 2—15; mooh‘f'l'eo' 1/14/08 11:39 AM C:\Documents and Settin s\Herman\M Documen...\HW roblem2 15.m 1 of 1 % problem 2.15 homework % problem 2,15; modified 1 2 3 4 clear, close all 5 set(0,‘defaulttextfontsize‘,12) 6 x0 : 0.6139; % 0.1582 7 8 m0 3 9.11e-3l; % kg 9 h I 6.625e—34; % J.s 10 hbar = h/(2*pi); 11 q = 1.6e—19; % Cb 12 a = 66—10; % m, 6 Angstrom 13 V0 : q*5; %[J}, 5 eV 14 15 N = 47.199; % 2*m0*a“2*V0/hbar“2 16 17 18 z = 0:0.01:1; 19 betax = sqrt(N*x0).*z; 20 wf x sqrt(xO/(1—x0))*cos{betax) +sin(betax); 21 figure 22 plot{z,wf.“2,'k—'),grid on, set(gca,'FontSize',12) 23 xlabel('x coordinate in potential well x/a') 24 ylabelt'wave function ( \psiu2/C )AZ') 25 26 gtext(['E/Vfi0 = ',num25tr(x0},', region 11']} 27 2' 15M. a: 6mm; 1/at52v.r<ar E<£é flaw: «QM stmfizsng: EI/Vo = 0.15592 aaan Eg/Va = a we. m sigma“ «for 7%.: WW fun/2077014 I." ryl‘oml’, "‘5 f 4:: ' ——w ‘ it We; wSK/NbiDWK/Wzflf Mew C, £5: £1 Oahsf'ékfll) b: E/l/o mud N= ZWéa‘Z/fi: 47:2? “5‘” MATH?) ear 5r‘mifi”) PM/C/E/VC 03%. X/O— WMJIN 714,1 mofiumwfiw £92144; affix Sogufir'a’g I11 hadrons T gnaw/77’ w ‘ - “f ‘ ‘ a . /- )- ke“ (w Nalfln’il. IS ff“ 7‘0 0” wcfranf (j frobgew 2, u)”, wadié‘ed E/VO region z 2_5 7777 ................ .. N “.7. “W wrrfimmvwr; rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr w wave function ( 01sz )2 01 _|. O 5 ......................... ...................... .2 .......................... .g ........................... ..._.._.._. ~ : : : z . 1 5 ‘ ' 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x coordinate in potential well xla 112,04 2(th low/Troy. +0 firm J 0m £6:me {s M'X2042a mdxeta (Lb/C): Hf; 53+ 9“"(mgjfi’ 5" E/VOI 05*5‘1 N: 47,2 (m/oaamfi’: oh {iw'ob‘ 93 = O forxaa V‘nod‘a‘pl'IQA 1.4 .0 oo wave function ( 1;;le )2 o '0: .0 4:. 0 0.1 0.2 0.3 0.4 0.5 0.6- 0.7 0.8 0.9 1 x coordinate in potential well x/a 4‘ bloat 4‘0 4”“; an flad‘van (9 0‘ x=Q ‘ ._ ‘. DeTeAwwfwa ‘HM +0+0~Q humben 01L enema? $911129 fun SI‘UC’OM beTweav. EC omol 136+ \21‘ QT 300 {A 129(23):, (a) fior‘ GQAE. (2.11); Maud wdromfc, 9+4ng (‘m‘i‘MCOWLvtrjfaw baud! : i IT * 3/ ‘ 3C (E3 1 _1:.3L2mh7 a” E— be (1022‘ WWW?! PM um FU§Qume_) Tim M VIM/2W 0? Women! qua‘fum $+om Faume Vo-Qqu is: E E75? ‘LT r 6.1+ th {at i N .._ (figment; ._.. ewfigxrgi’aagewah'éaa, E EU 8 3 ‘ 3 N1: (:0th &\ _____ 3E} (2w:)3é(k-U/z E—Ec=%. - edEa-dau 81' i a —3 _ ; (a) g‘ ‘ N 1 (2x LOQX‘frhm 21<O.02592<be(0‘?)}/11 3 (a em .53”? W was i ‘? v3 '5 ’2 N1, : QOIQX MA ‘: WWW- (Lo am As [at am E m b 1 3/9. '19 x J: 2 i Thug N CGaAs): (04051))(ZJ'ZMO = 3.2QMO 702w: —————w—m—';'““‘_—— _ ‘10? m. : fl .0 co .0 a) Fermi-Dirac function f(E) O In .0 N f=300K,§EF=10e§V 0.9 0.92 0.94 0.96 0.98 1 1.02 1.04 1.061., 1.08 1.1 a. (2. Us: Mawwfls p—gor H1; FUN/m" DI‘MC /% [31001001154214] {unaltth 7-6p- ’U’§. («21% E,==—/L7€ (/‘For‘ flphpaaflm? (a) T=3o K) (1,) T=3ap(j(e) frappOK 10191" 0.9-5 E/E— 5-1.1. Nat/- Mu gap of 609/4 (Ii/Egg —- an «0f 310mm owl 9:5;- L/wa Mpppw/x lo #1.: 542an gig-T47 0K _? we, 0. 35mm ( mommy?” \ Calculate the temperature at which there is a 10‘6 probability that an energy state l above the Fermi energy level is occupied by an electron. ‘ O. 357;?6/ E" 2:71;: I frobabfP/i}; Han/T state is coaufl'ed. — L E l? t 10 EF I — L ‘ hT E—EF = b7 PmOoé—O 2 FLIP/p.009) hT - 0’35}? Kev] = 0.025‘12V £4.00) A? 1*. 0,025“ g .02 ___——--""*‘o a [k] 0 Sfix 1.592(10'2?’ _. 300.3% ______________—- ...
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This note was uploaded on 04/29/2008 for the course EEC 140A taught by Professor Fink during the Winter '08 term at UC Davis.

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hw2 solutions - ____~____._.ti_.'§“mama___WWWM_._WL__M_...

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