semisolpr02

semisolpr02 - Semiconductor Physics and Devices: Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 2 Solutions Manual Problem Solutions 9 Chapter 2 Problem Solutions 2.1 Computer plot 2.2 Computer plot 2.3 Computer plot 2.4 For problem 2.2; Phase =− = 2 π λ ω x t constant Then 2 0 ⋅− = dx dt or dx dt v p == + F H I K 2 For problem 2.3; Phase =+ = 2 x t constant Then 2 0 ⋅+ = dx dt or dx dt v p F H I K 2 2.5 Eh hc hc E ==⇒ = ν Gold: Ee V x J () 4 90 4 90 16 10 19 .. . b g So =⇒ 6625 10 3 10 4.90 16 10 34 10 19 . . xx x bg b g b g 2.54 10 5 xc m or λµ = 0254 . m Cesium: V x J 190 190 16 10 19 . b g So 654 10 34 10 19 5 . . x m b g or = 0654 . m 2.6 (a) Electron: (i) K.E. = Te V x J 11 6 1 0 19 . pm T x x −− 2 2 9.11 10 16 10 31 19 b g or pxk g m s 54 10 25 ./ h p x x 34 25 . . or = ° 12.3 A (ii) K.E. = V x J 100 17 T 2 g m s 24 h p = ° 123 . A (b) Proton: K.E. = V x J 6 1 0 19 T x x 22 1 6 7 1 0 1 6 1 0 27 19 b g or px k g m s 2.31 10 23 / h p x x 2.31 10 34 23 . or = ° 0287 . A (c) Tungsten Atom: At. Wt. = 183.92 For V x J 6 1 0 19 T = 2 = 218392 16610 27 19 . b g or k g m s 313 10 22 h p x x 34 22 . . or = ° 00212 . A (d) A 2000 kg traveling at 20 m/s: v ( ) 2000 20 or g m s 41 0 4 / h p x x 0 34 4 . or = −° 166 10 28 . xA
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 2 Solutions Manual Problem Solutions 10 2.7 Ek T avg == () 3 2 3 2 00259 . or Ee V avg = 0 01727 . Now pm E avg avg = 2 = −− 2 9.11 10 0 01727 16 10 31 19 xx bgb g .. or px k g m s avg =− 7.1 10 26 / Now λ h p x x 6625 10 7.1 10 34 26 . or = ° 933 . A 2.8 Eh hc pp p ν Now E p m e e = 2 2 and p h e e =⇒ E m h e e = F H G I K J 1 2 2 Set EE pe = and λλ = 10 Then hc m h m h p F H G I K J F H G I K J 1 2 1 2 10 2 2 which yields p h mc = 100 2 hc hc h mc mc p p == = ⋅ = 100 2 2 100 2 29 . 1110 310 100 31 8 2 b gb g So Ex J k e V 164 10 103 15 .. 2.9 (a) Em v x x 1 2 1 2 9.11 10 2 10 23 1 4 2 bg b g or J 1822 10 22 e V = 114 10 3 . Also v x x 9.11 10 2 10 31 4 b g k g m s 26 ./ Now h p x x 34 26 . . = ° 364 A (b) p hx x 125 10 34 10 . pxk g m s 53 10 26 Also v p m x x xm s = 9.11 10 582 10 26 31 4 . or vx c m s = 6 Now v x x 1 2 1 2 9.11 10 1 4 2 b g or J 154 10 21 e V = 9.64 10 3 2.10 (a) hc x === 3 10 110 34 8 10 . b gb g or J = 199 10 15 Now V x x V =⋅ ⇒ = 16 10 15 19 so Vx V k V 12.4 10 12.4 3 (b) E x x 2 2 9.11 10 31 15 602 10 23 xk g m s Then h p x x 34 23 . . = ° 011 . A
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 2 Solutions Manual Problem Solutions 11 2.11 (a) p x x == h 1054 10 10 34 6 . px k g m s =− 28 ./ (b) E hc hc p h pc = F H I K λ So ∆∆ Ecp x x () 3 10 1054 10 82 8 bg b g . or Ex J =⇒ 316 10 20 . Ee V = 0198 . 2.12 (a) p x x x h 12 10 34 10 . k g m s 878 10 26 (b) E p m x x =⋅ 1 2 1 2 510 2 26 2 29 . b g J 7.71 10 23 e V = 4.82 10 4 2.13 (a) Same as 2.12 (a), k g m s 26 (b) E p m x x 1 2 1 2 2 26 2 26 . b g J 7.71 10 26 e V = 4.82 10 7 2.14 p x x x = h 10 34 2 32 .
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This note was uploaded on 05/07/2008 for the course ECE E-302 taught by Professor Nabet during the Spring '08 term at Drexel.

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semisolpr02 - Semiconductor Physics and Devices: Basic...

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