EE338_HW3_Solutions

EE338_HW3_Solutions - EE 338 Spring 2008 HW#3 Solutions 1 a...

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2/11/2008 1 EE 338 Spring 2008 HW#3 Solutions 1 3 a) 11 ( ) 0.27 1 1 b) ( ) ( 3 ) 0.05 c) ( ) ( 6 ) 2.47 10 FF F Ek TE kT F F EE k T fE e e k T k T +− ⎡⎤ ⎢⎥ ⎣⎦ =+ == = + + = = × Probability of finding a hole will be 1 3 a) [1 ( )] 1 1 0.27 1 1 b) [1 ( 3 )] 0.05 c) [1 ( 6 )] 2.47 10 F kT F F T fE k T e e k T k T −− Δ= =− = + + −−= ×
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2/11/2008 2 2 State being occupied by an electron: () f EE + Δ State being empty (occupied by a hole): 1( ) f −Δ 1 1e x p 1 (i) x p F FF fE E E kT E kT = +Δ − ⎡⎤ + ⎢⎥ ⎣⎦ = Δ + 1 ) 1 x p 1 1 x p 1 exp 1 x p exp exp 1 exp exp 1 (ii) x p F E E kT E kT E kT E kT kT kT kT kT E kT −− Δ = −Δ − + =− + +− = + ⎛⎞ Δ ⎜⎟ Δ + ⎝⎠ = Δ + Since (i) and (ii) are the same we proved the relation.
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2/11/2008 3 3 1 () 1e x p For , we will get exp 1. Then we can ignore the 1 in the denominator: F F F fE EE kT EE k T kT = ⎡⎤ + ⎢⎥ ⎣⎦ ±± 1 e x p exp F F F fE E k T kT kT −− −≈ = ± 4 a) From p.74 of the book 3 * 2 3 4( 2 ) () () e x p ( ) ( ) exp exp exp
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EE338_HW3_Solutions - EE 338 Spring 2008 HW#3 Solutions 1 a...

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