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Electron in a Finite Square Well

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron in a finite square well has 6 distinct energy levels. If the finite square well is 10nm long determine:

    a) Approximate the possible values for the depth of the finite square well ##V_0##.
    b) Using a well depth value in the middle of the results obtained from part a) find the energy when the electron is in the ##n=3## state.
    c) For the ##n=3## state determine the un-normalized wave function
    d)For ##-20nm<x<20nm## draw the ##n=3## wave function
    e) For ##-20nm<x<20nm## draw the probability function

    2. Relevant equations
    ##-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)##

    3. The attempt at a solution

    Part a) of the question is somewhat tedious and involves solving transcendental equations, I essentially just followed how they solved it on this site which gave me the the possible range for the depth of the finite square well to be ##V_0=0.0237eV## to ##V_0=0.0333eV##

    Now for part b), taking the average of these gives ##V_0=0.0285eV## although I'm not sure how I can use this to find the energy of the electron in the n=3 state. In my textbook they seem to use the equation ##E_n=n^2\frac{\pi^2\hbar^2}{2mL^2}## which doesn't seem to mention the potential at all.

    plot.PNG

    From the plot I've listed the coordinates of the n=3 intersection in case that's useful. The lines are: ##y=-\mbox{cot}(x)##, ##y=\tan(x)##, and ##y=\sqrt{(\frac{8.648}{x})^2-1}## where ##8.648=\mu=L\sqrt{\frac{2m}{\hbar^2}V_0}##
     
    Last edited: Mar 20, 2016
  2. jcsd
  3. Mar 21, 2016 #2
    Well i saw your refereed treatment and i think its exact and Transcendental equations are are not a problem as it solves graphically the two relations obtained from applying the boundary conditions at the walls of the well-
    As you wish to have six possible eigen states the well depth will have a bounding value that you have to calculate and taking half does not mean thay one can get the three-
    The above calculations are well known in case of nuclear potentials also-say deutreron in a finite well - however if well depth is large compared to energies involved then approximations can be made giving simpler results for energy values.
    my advise is to follow your referred site-
     
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