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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. Lets express the velocity as = 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 = m , 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 from = h/p = 3 . 63 10 8 m. Note that k = 2 / , as it should. b. Since p = h/ = 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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This note was uploaded on 09/13/2010 for the course ECE ECE344 taught by Professor Polizinni during the Spring '10 term at University of Massachusetts Boston.
 Spring '10
 Polizinni

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