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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 π λ ω = dx dt or dx dt v p = = + F H I K ω λ π 2 For problem 2.3; Phase = + = 2 π λ ω x t constant Then 2 0 π λ ω + = dx dt or dx dt v p = = − F H I K ω λ π 2 2.5 E h hc hc E = = = ν λ λ Gold: E eV x J = = ( ) 4 90 4 90 16 10 19 . . . b g So λ = ( ) 6 625 10 3 10 4.90 16 10 34 10 19 . . x x x b gb g b g 2.54 10 5 x cm or λ µ = 0 254 . m Cesium: E eV x J = = ( ) 190 190 16 10 19 . . . b g So λ = ( ) 6 625 10 3 10 190 16 10 6 54 10 34 10 19 5 . . . . x x x x cm b gb g b g or λ µ = 0 654 . m 2.6 (a) Electron: (i) K.E. = = = T eV x J 1 16 10 19 . p mT x x = = 2 2 9.11 10 16 10 31 19 b gb g . or p x kg m s = 5 4 10 25 . / λ = = h p x x 6 625 10 5 4 10 34 25 . . or λ = ° 12.3 A (ii) K.E. = = = T eV x J 100 16 10 17 . p mT = 2 p x kg m s = 5 4 10 24 . / λ = h p λ = ° 123 . A (b) Proton: K.E. = = = T eV x J 1 16 10 19 . p mT x x = = 2 2 167 10 16 10 27 19 . . b gb g or p x kg m s = 2.31 10 23 / λ = = h p x x 6 625 10 2.31 10 34 23 . or λ = ° 0 287 . A (c) Tungsten Atom: At. Wt. = 183.92 For T eV x J = = 1 16 10 19 . p mT = 2 = ( ) 2 183 92 166 10 16 10 27 19 . . . x x b gb g or p x kg m s = = 313 10 22 . / λ = = h p x x 6 625 10 313 10 34 22 . . or λ = ° 0 0212 . A (d) A 2000 kg traveling at 20 m/s: p mv = = ( )( ) 2000 20 or p x kg m s = 4 10 4 / λ = = h p x x 6 625 10 4 10 34 4 . or λ = ° 166 10 28 . x A
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 2 Solutions Manual Problem Solutions 10 2.7 E kT avg = = ( ) 3 2 3 2 0 0259 . or E eV avg = 0 01727 . Now p mE avg avg = 2 = ( ) 2 9.11 10 0 01727 16 10 31 19 x x b g b g . . or p x kg m s avg = 7.1 10 26 / Now λ = = h p x x 6 625 10 7.1 10 34 26 . or λ = ° 93 3 . A 2.8 E h hc p p p = = ν λ Now E p m e e = 2 2 and p h e e = λ E m h e e = F H G I K J 1 2 2 λ Set E E p e = and λ λ p e = 10 Then hc m h m h p e p λ λ λ = = F H G I K J F H G I K J 1 2 1 2 10 2 2 which yields λ p h mc = 100 2 E E hc hc h mc mc p p = = = = λ 100 2 2 100 2 = 2 9.11 10 3 10 100 31 8 2 x x b gb g So E x J keV = = 164 10 10 3 15 . . 2.9 (a) E mv x x = = 1 2 1 2 9.11 10 2 10 2 31 4 2 b gb g or E x J = 1822 10 22 . E x eV = 114 10 3 . Also p mv x x = = 9.11 10 2 10 31 4 b gb g p x kg m s = 1822 10 26 . / Now λ = = h p x x 6 625 10 1822 10 34 26 . . λ = ° 364 A (b) p h x x = = λ 6 625 10 125 10 34 10 . p x kg m s = 5 3 10 26 . / Also v p m x x x m s = = = 5 3 10 9.11 10 582 10 26 31 4 . . / or v x cm s = 582 10 6 . / Now E mv x x = = 1 2 1 2 9.11 10 582 10 2 31 4 2 b gb g . or E x J = 154 10 21 . E x eV = 9.64 10 3 2.10 (a) E h hc x x x = = = ν λ 6 625 10 3 10 1 10 34 8 10 . b gb g or E x J = 199 10 15 . Now E e V x x V = = 199 10 16 10 15 19 . . b g so V x V kV = = 12.4 10 12.4 3 (b) p mE x x = = 2 2 9.11 10 199 10 31 15 b gb g . = 6 02 10 23 . / x kg m s Then λ = = h p x x 6 625 10 6 02 10 34 23 . . λ = ° 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 .
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