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quiz5solution - E = 2 h 2 w 2 mL 2 = 2 1.055" 10 34 J $...

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EE 439 Name_______________________________________ Quiz 5 – Nov. 6, 2009 Put your final answers on this sheet and attach any additional sheets behind. You must include your work to get full credit. A finite-height quantum well has barriers that are 0.5-eV high and a well width of 1 nm. Find the first quantized energy level of the well. The characteristic equation for finding the energies of the even-symmetry states of a finite-height quantum well: w tan w = R 2 " w 2 where w = mEL 2 2 h 2 and R = mU o L 2 2 h 2 . For the given well configuration, R = 9.11 " 10 # 31 kg ( ) 0.5 eV ( ) 1.6 " 10 # 19 J / eV ( ) 10 # 9 m ( ) 2 2 1.055 " 10 # 34 J $ s ( ) 2 = 1.81 Inserting this value into the characteristic equation, w tan w = 3.27 " w 2 . Use some type of iteration on this transcendental equation (calculator equation solver, enlightened guessing, or whatever), leads to a value of w = 0.99. Then we can find the corresponding energy:
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Unformatted text preview: E = 2 h 2 w 2 mL 2 = 2 1.055 " 10 # 34 J $ s ( ) 2 0.99 ( ) 2 9.11 " 10 # 31 kg ( ) 10 # 9 ( ) 2 = 2.39 " 10 # 20 J = 0.15 eV Potentially useful equations and data: Square-barrier tunneling: T " 4 E U o # E ( ) U o 2 exp # $ L ( ) ; = 2 m U o # E ( ) h 2 Energy levels in an infinitely deep quantum well: E n = n 2 " 2 h 2 2 mL 2 p = h E = h # "# = c h = 6.63x10 –34 J·s = 4.14x10 –15 eV·s; h = 1.055x10 –34 J·s = 6.595x10 –16 eV·s (Planck’s constant) c = 3.0x10 8 m/s (speed of light in vacuum) k = 1.38x10 –23 J/K = 8.617x10 –5 eV/K (Boltzmann’s constant, not wave number) m = 9.11x10 –31 kg (mass of an electron) q = 1.6x10 –19 C (charge of an electron) ε o = 8.85x10 –12 F/m (free-space permittivity) 1 eV = 1.6x10 –19 J...
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quiz5solution - E = 2 h 2 w 2 mL 2 = 2 1.055" 10 34 J $...

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