key041b

key041b - PHYS 320 Exam #2 on 4-7-04 (Ave. = 84.1%, = 10%)...

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PHYS 320 Exam #2 on 4-7-04 (Ave. = 84.1%, σ = 10%) 1 E = 600 eV x 400 eV Region B Region A Region C Region D L n = 5 0.2L P(x) 0.4L 0.6L 0.8L Finite Well E 1 = 2 eV E 2 = 4E 1 = 8 eV E 3 = 9E1 = 18 eV E ~ n 2 Tunneling 1. Write down the 1-D, time-independent Schrödinger Equation for the potential inside an infinite well . Identify and write down the eigenfunctions and eigenvalues for this equation. In your own words, explain why only certain energies are allowed for an electron “trapped” in an infinite well. (Ave = 82.4%) 22 2 222 2 Schrödinger Eqn.: where ( ) 0 inside the well 2 2 Eigenfunctions : ( ) si n Eigenvalues: 2 nn E Vx m x n xn xE LL mL pp ∂Ψ - = Ψ=  Ψ ==   h h Because the electron wavefunction must be zero at the boundaries of the well, only certain wavelengths will “fit” inside the well. This is analogous to standing waves forming on a string of a fixed length. The allowed wavelengths then correspond to quantized electron energies, as calculated by the de Broglie relationship. 2. The ground state energy level of a potential well is shown. Draw and label the values of the next two energy levels, and sketch the probability function of the n = 3 level. If the electron is excited to the n = 3 level, what could happen? (Ave = 88.9%) If the electron is excited to the third energy level, it can tunnel through the finite barrier on the right side and escape the well. This tunneling probability decreases exponentially with the width of the barrier. 3. An electron is in the n = 5 excited state of a 1-D infinite square well (width L). Draw the probability distribution of the electron with labeled axes and indicate the probability of finding the electron between x = 0.3L and 0.7L by shading in the appropriate area. Write down the formula necessary for calculating this
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This homework help was uploaded on 04/22/2008 for the course PHYS 320 taught by Professor Baski during the Spring '04 term at VCU.

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key041b - PHYS 320 Exam #2 on 4-7-04 (Ave. = 84.1%, = 10%)...

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