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Particle in a box

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data

    hi

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 3, 2009
  2. jcsd
  3. Mar 3, 2009 #2
    Hey Chemasdf,

    Welcome to PF!

    In respect to your question, L^2: an area maybe? I hope someone will correct me if I'm wrong that this can be manipulated.

    The Bob
     
  4. Mar 3, 2009 #3

    Tom Mattson

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    You're using the formula for the 1D particle in a box. If you're talking about a 2D box then there should be 2 quantum numbers, not just one.

    Wavefunctions and energies for the 2D box are given below.

    [tex]\psi_{m,n}(x,y)=\frac{2}{\sqrt{L_xL_y}}\sin\left(\frac{m\pi x}{L_x}\right)\sin\left(\frac{n\pi y}{L_y}\right)[/tex]

    [tex]E_{m,n}=\frac{\hbar^2\pi^2}{2m}\left[\left(\frac{m}{L_x}\right)^2+\left(\frac{n}{L_y}\right)^2\right][/tex]
     
  5. Mar 3, 2009 #4
    I'm still having trouble determining the quantum numbers (n). Can someone give me a hint as to how to solve for "n". I cannot find the wavenumber without knowing the "n" which is not given. Thanks
     
  6. Mar 3, 2009 #5

    Tom Mattson

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    The ground state is [itex]n=m=1[/itex]. I would take the first excited state to be the next highest energy level.
     
  7. Mar 3, 2009 #6
    does this calculation involve any degenerate level considerations?
     
  8. Mar 4, 2009 #7

    Tom Mattson

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    Why on Earth did you delete the problem statement?
     
  9. Mar 4, 2009 #8
    maybe he found the answer? but then he should have deleted the post, not the question...o_O weird...
     
  10. Mar 4, 2009 #9

    Tom Mattson

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    Even if he did find the answer, people took the time to reply. It's disrespectful to destroy a thread like this.
     
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