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Rectangular finite potential well problem

  1. Nov 27, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    An electron enters in a finite rectangular potential well of length 4 angstroms. When the entering electrons have a kinetic energy of 0.7 eV they can travel through the region without having any reflection. Use this information to calculate the depth of the potential well.


    2. Relevant equations
    Not sure because I don't know if the kinetic energy of the electron is greater or lesser than the potential well.
    I know that inside the well, [itex]\Psi _{II}(x)= C \sin (kx)+ D \cos (kx)[/itex]. I also know the form of the wave function outside the well (thanks to https://www.physicsforums.com/showthread.php?t=540406&page=3) but I don't have the normalization, etc. I tried to search in google, wikipedia and hyperphysics and can't find the whole detailed solution.
    I was thinking of calculating the flux of probability (I need the exact wavefunctions, normalized) in order to calculate the coefficient of transmission.

    3. The attempt at a solution
    I'm confused. I don't know how to tackle the problem, too many unknowns to me.
    Any tip is appreciated.
     
  2. jcsd
  3. Nov 27, 2011 #2

    fluidistic

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    Ah I think I just realized how to solve the problem.
    The electron has a greater kinetic energy that the potential well.
    The coefficient of transmission is the one given by a barrier of potential when the energy of the particle exceed the potential barrier.
    It is worth [itex]T=\frac{1}{1+ \frac{V_0 ^2 \sin ^2 (k_1 a)}{4E(E-V_0)}}[/itex] where [itex]k_1 = \sqrt { \frac{2m(V_0-E)}{\hbar ^2}}[/itex]. But unfortunately [itex]k_1[/itex] is complex.
    So I set T=1 and I try to solve for [itex]V_0[/itex]... but this doesn't go well. This gives me [itex]V_0=0[/itex].
    By the way I used the formula given in http://en.wikipedia.org/wiki/Rectangular_potential_barrier#Analysis_of_the_obtained_expressions.
    Does someone understand better than me what's going on?
     
  4. Nov 28, 2011 #3

    fluidistic

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    Hmm, I've watched the first part of and he reaches a slightly different result than wikipedia for the transmission coefficient.
    His [itex]k_1[/itex] is worth [itex]\sqrt {\frac{2m (E-V_0)}{\hbar ^2} }[/itex].
    So for my exercise, I must solve for [itex]V_0[/itex] and [itex]n=1[/itex] in the expression a [itex]\sqrt {\frac{2m (E-V_0)}{\hbar ^2}}=n \pi[/itex]. I've checked out the units, all seems to work.
    When I plug and chug the given values for the problem, I reach that [itex]V_0\approx -1.65 eV[/itex] or [itex]-1.6 eV[/itex] if I round correctly.
    Can someone check out my work? That would after all make wikipedia "slightly wrong" for a formula.
     
    Last edited by a moderator: Sep 25, 2014
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