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Unformatted text preview: ECE344: Practice questions for Midterm Exam 1: Solutions, October 13, 2009 1. Problem. a. An electron is moving with a velocity of 2 × 10 6 cm/s. Determine the electron energy in eV, its momentum, and de Broglie wavelength in nm. b. The de Broglie wavelength of another electron is 12.5 nm. Determine its energy (in eV), momentum, and velocity. Solution. Let’s express the velocity as υ = 2 × 10 4 m/s. (Note: We use ‘m’, not ‘cm’, to be consistent with the use of kg, Joules, eV, etc.) a. Then, from p = mυ , using m = 9 . 1 × 10 31 kg, we have p = 1 . 82 × 10 26 kg cm/s, or a wavevector of magnitude k = p/ ¯ h = 1 . 73 × 10 8 /m. Then, computing the energy E from E = p 2 / (2 m ) we get E = 1 . 82 × 10 22 J = 1.14 meV. Finally, we can compute the wavelength λ from λ = h/p = 3 . 63 × 10 8 m. Note that k = 2 π/λ , as it should. b. Since p = h/λ = 5 . 28 × 10 26 kg m/s, we have E = p 2 / (2 m ) = 1 . 53 × 10 22 J = 9.55 mev....
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 Spring '10
 Polizinni
 intrinsic carrier concentration, ev, Fermi level

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