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HW_Problems_3_solution

# HW_Problems_3_solution - HW_Problems_3_soln 1 The avg...

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HW_Problems_3_soln 1. The avg energy of an electron: 0 () E EPEdE ⟨ ⟩= P(E) – Probability of finding the e with energy of E 3 2 3 0 2 2 exp( ) E E E dE kT kT π ⇒⟨ ⟩= ( 1) 0 Using functions: ( ) 1 ( )= 2 (x+1)=x (x) 3 gives 2 tx x e t dt This E kT −− Γ Γ= Γ ΓΓ 2. The effective density of states 33 0 22 2 31 19 34 2 25 3 19 3 2( ) ( ) 2 at 300 , 26 9.1 10 (300 ) 2( ) (0.26 1.6 10 ) 2 (1.05 10 ) 2.56 10 2.56 10 c c m N kT k kT meV kg Nk J JS m cm = = × = ×× ×⋅ = ×= × The effective density of states at 77k 3 18 3 2 77 (77 ) (300 )( ) 3.32 10 300 cc N k N k cm = = × If * 0 0.1 mm = 3 * 19 17 3 2 0 3 * 18 17 3 2 0 (300 ) 2.56 10 ( ) 8.09 10 (77 ) 3.32 10 ( ) 1.05 10 c c m N k cm m m N k cm m = × = ×

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3. at 0 : F Ek 2 2 2 3 * 2 34 2 2 24 3 3 31 21 2 (3 ) 2 (1.05 10 ) (3 10 ) 2(0.067 9.1 10 ) 8.3 10 5.18 10 F En m Js m kg J eV π −− = ×⋅ = ×× = ×= × 17 3 Using J-D approx with
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HW_Problems_3_solution - HW_Problems_3_soln 1 The avg...

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