2_1_Midterm solution

2_1_Midterm solution - Winter 2010 1 p 1 10 24 = = 9.48 10...

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1. (a) 9 34 24 10 48 . 9 2 10 626 . 6 10 1 × = × × = = - - π h p k 1/m (b) m k 10 9 10 63 . 6 10 48 . 9 2 2 - × = × = = λ (c) eV J m p E e 430 . 3 10 488 . 5 2 19 * 2 = × = = - (d) Velocity s m m p v e g / 10 098 . 1 6 * × = = (e) m m p h x μ 53 10 53 . 0 10 4 10 63 . 6 4 4 30 34 = = = Δ = Δ - - - (f) Frequency Hz v v f g p 14 10 29 . 8 2 × = = = 2. L n k L nh h p n L n L = = = = = 2 2 2 The energy levels are given by 2 2 2 2 2 2 2 mL n m p E n h = = , since electrons obey Pauli principle, only 2 electrons (opposite spin) are in each level. Therefore we can occupy up to n= 6 (11/2=5 filled states + 1 half filled level. eV mL E 136 . 0 10 17 . 2 2 6 20 2 2 2 2 6 = × = = - h F E is the highest filled level at 0 K. Therefore eV E E F 136 . 0 6 = = . First excited level is 7 th level, eV mL E 185 . 0 10 96 . 2 2 7 20 2 2 2 2 7 = × = = = + = + - - e e kT E E 3.
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This note was uploaded on 05/10/2010 for the course EE 190007201 taught by Professor Bahramjalali during the Winter '10 term at UCLA.

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2_1_Midterm solution - Winter 2010 1 p 1 10 24 = = 9.48 10...

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