Week1HWSolutions

# 1 and write the answer as i1 w ece 656 3 2 m 2

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Unformatted text preview: ntegral ∞ I1 = ∫W EC 2 m* ( E − EC ) π ( E − EF ) 1+ e 2 m* ( E − EC ) ( E − E ) dE ≈ ∫ W eF π E ∞ 1 k BT C k BT dE but it is easier to recognize that “non- degenerate” means E F << E or η F << 0 and that F 1/ 2 (η F ) → exp (η F ) for η F << 0 so we can use the result of prob. 1) and write the answer as I1 ≈ W ECE- 656 3/ 2 m* 2π ( k BT ) exp (η F ) 2 Fall 2013 Mark Lundstrom 8/24/2013 ECE 656 Homework 1: Week 1 (continued) 3) For still more practice, work out this integral: ∞ ⎛ ∂f ⎞ I 2 = ∫ M ( E ) ⎜ − 0 ⎟ dE , ⎝ ∂E ⎠ E C where M ( E ) is as given in problem 1). Solution: From the form of the Fermi function, we see that ⎛ ∂ f0 ⎞ ⎛ ∂ f0 ⎞ ⎜ − ∂E ⎟ = ⎜ + ∂E ⎟ ⎝ ⎠⎝ F ⎠ so the integral becomes ∞ ⎛ ∂f ⎞ I 2 = ∫ M ( E ) ⎜ + 0 ⎟ dE . ⎝ ∂ EF ⎠ EC Since we are integrating with respect to energy, not Fermi energy, we can move the derivative outside of the integral to write ∞ ∞ ∞ ∂ 1 ∂ 1∂ I2 = ∫ M ( E ) f0 ( E ) dE = k BT ∂( E k T ) E∫ M ( E ) f0 ( E ) dE = k BT ∂η F E∫ M ( E ) f0 ( E ) dE ∂ EF E F B C C C The integral can be recognized as the one we worked out in prob. 1), so ⎫ m*k BT 2π ∂ 3/ 2 m* 2π 1∂ 1 ∂⎧ ⎪ ⎪ I2 = I= F (η ) ( k BT ) F 1/2 (η F )⎬ = W ⎨W k BT ∂η...
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## This document was uploaded on 01/15/2014.

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