20085ee2_1_problemset2_2008_solution

# 20085ee2_1_problemset2_2008_solution - EE2 Physics for...

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Unformatted text preview: EE2 Physics for Electrical Engineers Homework #2 Solutions Due Date: October 14th, 2008 I. 2 4 2 . 4 . 2 1 6.625 10 10 4 3.78 Therefore there are 3 states (not 4). II. 2 2 . 2 2 , 16 / / 4 16 2 1 10 √2 √2 9.11 6.625 10 10 / √2 / . But since 1eV=1.6×10-19 J, . 1.06 . 10 10.41 Therefore there are 10 states (not 11). III. 4 Normalizing to volume V: 4 2 / / 2 / / 1.06 10 / 7E+46 6E+46 5E+46 Density 4E+46 3E+46 2E+46 1E+46 0 0 0.5 1 1.5 Energy (eV) 2 2.5 IV. n=1022 cm-3=1028 m-3, m=9.11×10-31 kg, h=6.63×10-34 J-S. 3 8 2 / / / 2.71 10 1.69 0 0.5 1 Energy (eV) 1.5 2 2.5 V. EF was found from question 4 to be 1.69 eV, and at absolute zero the probability of finding particle at energy level EF is 50%, thus the following graph. 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 Energy (eV) 1.5 2 2.5 VI. Same as problem 4. VII. Maximum velocity: Average velocity: 5.78 . . 7.71 / 10 / 10 VIII. EF was found from question 4 as 1.69 eV=2.71×10-19 Joules. And , for T=300 k, plotting we get: 1.2 1 0.8 0.6 0.4 0.2 0 0 ‐0.2 0.5 1 1.5 2 2.5 ...
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