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semisolex03 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Exercise Solutions 21 Chapter 3 Exercise Solutions E3.1 = + 1 10 sin cos α α α a a a By trial and error, α a rad = 5 305 . Now 2 5 305 2 2 mE a h = . so E ma x x x 2 2 2 2 2 34 2 31 10 2 5 305 2 5 305 1054 10 2 9.11 10 5 10 = = ( ) ( ) . . . h b g b gb g or E x J eV 2 19 6 86 10 4.29 = = . Also 2 1 2 mE a h = π so E ma x x x 1 2 2 2 2 34 2 31 10 2 2 1054 10 2 9.11 10 5 10 = = ( ) ( ) ( ) π π h . b g b gb g or E x J eV 1 19 2.41 10 150 = = . Then E E E = = 2 1 4.29 150 . or E eV = 2.79 E3.2 g E m h E E c n c ( ) = 4 2 3 2 3 π * / b g Then g m h E E dE T n C E E kT c c = + z 4 2 3 2 3 1 2 π * / / b g a f = F H I K + 4 2 2 3 3 2 3 3 2 π m h E E n c c c E E kT * / / b g a f or g m h kT T n = F H I K ( ) 4 2 2 3 3 2 3 3 2 π * / / b g which yields g x x T = ( ) F H I K 4 2 108 9.11 10 6 625 10 2 3 31 3 2 34 3 π . . / b g b g × ( ) 0 0259 16 10 19 3 2 . . / x b g which yields g x m x cm T = = 2.12 10 2.12 10 25 3 19 3 E3.3 We have g m h E E dE T p v E kT E v v = z 4 2 3 2 3 1 2 π * / / b g a f which yields g m h E E T p
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