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Electron in a One Dimensional Infinite Potential Well

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    An electron is confined to a narrow evacuated tube. The tube, which has length of 2m functions as a one dimensional infinite potential well.

    A: What is the energy difference between the electrons ground state and the first excitied state.

    B: What quantum number n would the energy difference between adjacent energy levels be 1eV - which is measurable, unlike the result of part A.


    2. Relevant equations

    [tex]
    E_n=\frac{\hbar^2 \pi^2 n^2}{2mL^2}
    [/tex]


    3. The attempt at a solution
    I must not be using the equation correctly, as I am getting values many many orders of magnitudes out. Is m in the equation the mass of the particle? So in this case the mass of the electron?


    [tex]
    E_1=\frac{\hbar^2 \pi^2 n^2}{2mL^2}= \frac{(1.055 \times 10^{-34})^2 \pi^2 }{2(9.109 \times 10^{-31})2^2} = 1.507 \times 10^{-38} J \\
    E_2=\frac{\hbar^2 \pi^2 n^2}{2mL^2}= \frac{(1.055 \times 10^{-34})^2 \pi^2 }{2(9.109 \times 10^{-31})} = 6.03 \times 10^{-38} J \\
    \Delta E = E_2 - E_1 = 4.523 \times 10^{-38} J = 2.82 \times 10^{-19} eV
    [/tex]

    For B: I got a quantum number of around 3.5 billion :D, so I have to be doing something wrong. Also my main question for part B was the bit where it carries on with "- which is measurable, unlike the result of A", I have no idea what they mean by that, so if anyone does I would really appreciate it.

    Thanks :)
     
    Last edited: Dec 7, 2014
  2. jcsd
  3. Dec 7, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The answer looks good. 2 meters is just an extremely wide potential well.
    An energy difference of 1 eV is measurable, an energy difference of 10-19 eV (for an electron) is not.
     
  4. Dec 7, 2014 #3
    Oh right, ok thanks. Because part B was talking about 1eV I thought my answer was way way off.

    And thanks for clarifying the wording on part B, that makes sense. Thanks :)
     
  5. Dec 7, 2014 #4
    I am really stuck on part B now, I just cant see how it can be solved. (Im using the other version of the equation with h instead of h-bar)

    [tex]
    (1.602 \times 10^{-19})=\frac{h^2n_1^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2}
    [/tex]

    So obviously the RHS has to be equal to the charge of the electron, but I just cannot see how I can solve this. Any help/advice/feedback is greatly appreciated.

    Thanks :)
     
  6. Dec 7, 2014 #5

    mfb

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    Staff: Mentor

    Don't forget the units.

    You have calculated the energy levels already, for adjacent energy levels n1 is just 1 larger than n2 (using your sign convention). This is a quadratic equation.
     
  7. Dec 7, 2014 #6
    I can t seem to get a quadratic, as the way I am doing it the n^2 cancel. What I have done is below

    [tex]
    1eV=\frac{h^2n_1^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
    1eV=\frac{h^2(n_2+1)^2}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
    1eV=\frac{h^2(n_2^2+2n_2+1)}{8mL^2} - \frac{h^2n_2^2}{8mL^2} \\
    1eV=\frac{h^2(n_2^2+2n_2+1)8mL^2-h^2n_28mL^2}{8mL^2} \\
    (1.602 \times 10^{-19})8mL^2=h^2(n_2^2+2n_2+1)8mL^2-h^2n_28mL^2 \\
    (1.602 \times 10^{-19})8mL^2=h8mL^2n_2^2+2h^28mL^2n_2+h^28ml^2-h^2n_28mL^2 \\
    (1.602 \times 10^{-19})8mL^2=2h^28mL^2n_2 + h^28mL^2\\

    n_2=\frac{(1.602 \times 10^{-19})h^28ml^2-h^28mL^2}{2h^2mL^2}
    [/tex]

    Which if I have calculated it correctly comes out to negative one half, which is impossible.

    EDIT: After using wolfram alpha, and sorting out an error, I got the result of n to be equal to the charge on the electron, which is also wrong.
     
    Last edited: Dec 7, 2014
  8. Dec 7, 2014 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Ah right, the equation becomes linear. Well, just makes it easier.

    Something with the 8mL^2 went wrong in the fourth line.
     
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