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2_1_midterm08

2_1_midterm08 - EE2 Professor Harold Fetterman Feb 12 2008...

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Unformatted text preview: EE2 Professor Harold Fetterman Feb. 12, 2008 Midterm Closed Book 1a. Light of wavelength A1 falls on a metal surface. In this metal surface X1 eV are required to remove an electron. What is the kinetic energy of a) the fastest and slowest emitted photoelectrons? b) what is the stopping potential? c) What is the cut off wavelength for this metal? Explain, in a few sentences, the main point that this experiment demonstrates. 1b. We looked at commutators of operators [A,B ] .=. AB-BA. What are the values of [px , x ] and [px , y ]. Also find the value of [E , t]. What does it mean for measurements when [A, B] = 0 ? What does AEAt 2 h / 2 mean in terms of a physical system with lifetime At ? 2a.In the figure below an electron is in a two dimensional box with the ‘Px wavefunction corresponding to an energy of 4 eV. mu»! ~a/2 8/2 W. ' 1'by What is the total energy of this 2D system? If the ‘15 wavefunction was raised to an n = 3 level what would the total energy be? 2b. A energy band is approximately fitted by the expression: ) Where a is a lattice constant. A) calculate the effective mass a k=o and at the zone boundary k = n/a. Also find the value of k for minimum electron velocity. 2k2 E(k) 2 E0 (1 — e’“ 3a. Looking at the Density of states for electrons in two dimensions we were able to get expressions for the total number of occupied states as a function of energy. a) Find the value of EF as a function of n and b) the average energy <E> (at T = 0) c) Finally, find the average value of E2 ( <E2>). 3b. Using our equations for an intrinsic semiconductor and that n1 = pi . Show that Em lies below the middle of the bandgap by kT 1n (mn* /mp*)3/ 4. Also show that the product np is a function of bandgap and is equal to n12 Now taking 11 type material (Na z 0) the highest operating temperature is when the # of thermally generated carriers In is equal to 1/ 10 the number contributed by impurities m = 0.1 Na, Write an expression for Tmax . 2U?) 2 471‘ V(2m)% 5% two—dimensional ha 2v 3 g<a)de=4"’"Adsr 20’) (1/3: 8w - d VII3 N "' . I; J; L L it“v 1" t” 3‘3" d" * h’ ’ 'P 1’ 1 -h x y z x y z x 3'. 831 V 1 , 3 I = 2(1)) d? HE) = ha "W": Pad? Pa; " “1’5 H E " '5‘" e. 1:7 + l . 3k Probabiiity Offindingaparlficle ' . ~ ‘ 31 -= ~ne< WI > F a: am has ....__. 3‘?de (1)1 dynamxcalvanable assocmted operator x,y,orz —¢ x,y,orz v - 3% “*2? f<x,y,z) —~ /(x,y,z> p —+ EV ’* 3k 0 g a fig 1'61" _ _ moq‘ ‘E _ "13“!" 5“ 2(4wanfm) B 2(411Kseofi)’ 1w - ¢ + 2V P(x,£)dx = ‘P‘(x,£)‘l’(x,t) dx h’ 2 2. mobility has the dimension anal/"sec" m ' = 2;“; D a z 53:- = .3; (3f; 53%) %L M3) + v(x)v(x) = Ev(x) ' 39: 1 __ .J..§£ ‘2 2 22 f(Ec)=—§—_—§W~c(7 ) E21” pa: 2 h k a”? + 1 2 2m 2m x 1_ E __ 1 (PE dk m1) dnE 47‘ (27m); (E~— Ea)? dE h dt dk " h dis? “3? 8(1)” ‘1’ F ha (6w "T +1) 2 E E Jhap : m _ 11 r; f (1715.: p c ( c) c J": g p“ n E + g D 2—: 2111:: RT 3/2 2131:: RT 3/2 N __ °° Z d5 4: g ._ . V '— (E) -—.--—.---_—.---—- “c c 2 (”if”) J “v 2 ( r3 ) . - L 6””er + 1 o“ = —— ‘ 6 ‘ 15 d5 (E? 2 “JV 2(5) W n" 2(2fimckT)%e *(EL—w—‘C 8) NU+Ng==p0+Nd .914 +1 — )‘1‘2 Tu‘ . . ‘ “'* dxmenszon of volt __ "213,12 ‘ 2 kT % Elna“ “7:575:23 8 -EF ____ W mc L n: N¢c~( k7" ) N‘“2( h'z ) lc=mz/'L‘ "21.1.”. E~E+E 'E a , P 2 (2‘17“ kT)i —{§%§£ P + N‘ E" = "ZhZHQ/Zml'z 2' ’721'72!’8r#115 = W e = n A I ”0P0“ -~ N NI. cxp[;_. E4] h? Emu...“ = (hzflzx'ZmLI)(n; + n3 -:- 5%) kT 1 + _ ,_ .___—.._—._..—.—— . ND —- ND (1 (Eu-Es) ) N; _ NA _(£ Er) kT .... fl”???- ,‘ = —~.__r_:.__.l~_ Pi: Nve‘(£‘*fi,)/RT e + 1 1 + 6%.: '1 NC exp[ kT ] ? S "X P; ____ ‘ ‘- n‘ n 1/):(35) "w $41 + L’) 'P=Asinkr k V2717: “5'35 —E . z , ,4 (£112 1 dB: 1 dB; " h L {2 \chchng 52-“: —-.___... n da; q dz: q d3: ¢(I’ysz) = 1!),(r) zpygy) 1191(2) ...
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