EE 338_HW3_Solutions

# EE 338_HW3_Solutions - EE 338 Fall 2007 HW#3 Solutions 1)...

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Unformatted text preview: EE 338 Fall 2007 HW#3 Solutions 1) We know that E = hv and also v = For gold: = 253nm For cesium: = 653nm h . If we combine the two equations we get = hc . E 2) a.) The kinetic energy of the electron is E = 1 2 mv = 1.14meV . The momentum is simply 2 p2 p = mv or E = p = 2mE which give the same result, p = 1.82 10 -26 N s . Using 2m h this momentum, we find the de Broglie wavelength of this electron de Broglie = = 364 . p b.) From a given de Broglie wavelength, we can get the momentum h = 5.3 10 - 26 N s . From the momentum we get the energy to be p= de Broglie p2 p E= = 9.6meV . The velocity of the particle is v = = 5.8 10 4 m/sec m 2m 3) a.) In (i), atoms with 1 valence electron are located in Group I in the periodic table; Li, Na, K. In (ii), atoms with 4 valence electrons are located in Group IV; C, Si, Ge. b) Atoms in (i) are called halogens and have 7 valence electrons. In (ii) all the atoms have completed their valence shells and are called noble gasses. These elements are very inert to chemical reactions. 4) a.) To elevate a valence electron to conduction band E = hv 1.42eV v 343THz b.) c The corresponding wavelength is = = 8738 . v 9/30/2007 5) Refer to Chapter 2.3 in the course book and class notes. 6) Emin Emin The effective mass is inversely proportional to the curvature of the parabola that approximates the minimum energy point in a dispersion curve. Here the red parabolas in two figures show the curvature of the minimum energy point. The curvature of particle A * is much steeper than particle B, which means m * < m B . A 2 ...
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## This note was uploaded on 02/27/2008 for the course EE 338 taught by Professor Dapkus during the Fall '07 term at USC.

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EE 338_HW3_Solutions - EE 338 Fall 2007 HW#3 Solutions 1)...

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