semisolpr02

# semisolpr02 - Semiconductor Physics and Devices: Basic...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 12

semisolpr02 - Semiconductor Physics and Devices: Basic...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online