Week7HWSolutionsV2

The average inverse apparent mean free path is f m

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Unformatted text preview: f ⎞ 2q 2 ∫ λ ( E ) + L M ( E ) ⎜ − ∂ E0 ⎟ dE h ⎝ ⎠ ⎛h⎞ R=⎜ 2⎟ ⎝ 2q ⎠ λ ( E) 1 ⎛ ∂ f0 ⎞ ∫ λ ( E ) + L M ( E ) ⎜ − ∂ E ⎟ dE ⎝ ⎠ The corresponding ballistic resistance is: ⎛h⎞ 1 Rball = ⎜ 2 ⎟ ⎛ ∂ f0 ⎞ ⎝ 2q ⎠ ∫ M E ⎜ − ∂ E ⎟ dE ⎝ ⎠ So we can re- write the resistance as: ⎡ ⎤ ⎢ ⎥ 1 1 ⎢⎛ h ⎞ ⎥ R = Rball + ⎢⎜ 2 ⎟ − ⎥ ⎛ ∂f ⎞ ⎝ 2q ⎠ λ ( E) ⎛ ∂f ⎞ ⎢ M ( E ) ⎜ − 0 ⎟ dE ∫ M ( E ) ⎜ − 0 ⎟ dE ⎥ ∫ λ ( E) + L ⎝ ∂E ⎠ ⎝ ∂E ⎠ ⎢ ⎥ ⎣ ⎦ ⎡⎧ ⎫⎤ ⎫⎧ ⎛ ∂f ⎞ ⎢⎪ M ( E ) ⎜ − 0 ⎟ dE ⎪⎥ ⎪⎪ ∫ ⎝ ∂E ⎠ 1 ⎪⎪ ⎪⎥ ⎢ ⎪⎛ h ⎞ R = Rball + ⎢ ⎨⎜...
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