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The width of a finite potential well

  1. Nov 6, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron is enclosed in a potential well, whose walls are ##V_0 = 8.0eV## high. If the energy of the ground state is ##E = 0.50eV##, approximate the width of the well.

    Answer: ##0.72nm##

    2. Relevant equations
    For an electron in a potential well, whose energy is less than the height of the potential walls, outside the well the time-independent Schrödinger equation becomes:
    \begin{equation}
    \Psi ''(x) = \frac{2m}{\hbar^2}(V_0 - E) \Psi (x) = \alpha^2 \Psi (x),
    \end{equation}
    where
    \begin{equation}
    \alpha^2 = \frac{2m}{\hbar} (V_0 - E) > 0
    \end{equation}
    Inside the well, where ##V(x) = 0##, it is the familiar:
    \begin{equation}
    \Psi ''(x) = -k^2 \Psi (x), k^2 = \frac{2mE}{\hbar^2}
    \end{equation}
    By doing a whole bunch of math (by requiring that ##\Psi## be continuously differentiable at the potential walls and solving the resulting linear system with the assumption that it has more than one solution, meaning ##det() = 0##), we and up with the result
    \begin{equation}
    \frac{\sin (kL)}{\cos (kL)} = \tan (kL) = \frac{\alpha}{k}.
    \end{equation}
    Substituting ##\alpha## from ##(2)## and ##k## from ##(3)##, we get
    \begin{equation}
    \tan (kL) = \sqrt{\frac{V_0 - E}{E}}
    \end{equation}
    3. The attempt at a solution
    Now my idea was to use ##(5)## to solve for ##L## as follows:
    [tex]
    \tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\
    \iff\\
    kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\
    \iff\\
    L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}})

    = \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\

    = \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\
    \arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\

    = 2.084 751 \cdot 10^{-8}m,
    [/tex]
    which is a bit off. Any idea what I'm doing wrong?
     
    Last edited: Nov 6, 2016
  2. jcsd
  3. Nov 6, 2016 #2

    Simon Bridge

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  4. Nov 6, 2016 #3

    TSny

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    I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

    Also, make sure your calculator is in the radian mode for taking the arctan.
     
  5. Nov 7, 2016 #4
    Yeah, it was the radian thing, and the fact that the well goes from -L to L. The online material for our book (Tipler, Modern Physics, itself glosses over this part, the mathematics in particular) actually uses the letter ##a## in place of ##L##, but I figured I could just change the place of the origin without doing anything to make up for it, so to speak.

    Thanks again.
     
    Last edited: Nov 7, 2016
  6. Nov 7, 2016 #5
    Apparently there are even and odd solutions. The online material for our book states, that the ##-\cot(ka)## actually corresponds to the odd case.
     
  7. Nov 7, 2016 #6

    Simon Bridge

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    Yes?
     
  8. Nov 9, 2016 #7
    I as just noting that the link (the part of the page) you provided used ##-\cot## instead of ##\tan##. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
     
  9. Nov 9, 2016 #8

    Simon Bridge

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    Yes? Where are you up to in your solution?
     
  10. Nov 13, 2016 #9
    Gah, sorry. I just fell of the map for a week, didn't I? I basically forgot to use radians in my calculation and also forgot, that the formula applies to a potential well from ##-a## to ##a##, not ##0## to ##L##.

    Should I have posted my final numerical answer in this thread? I guess that would have been in good housekeeping...

    The answer I got was ##\approx 0.73nm##.
     
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