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Homework Help: Finite Well and Schrodinger's Equation

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    An electron is trapped in a finite well of width 0.5 nm and depth of 50 eV. The wavefunction is symmetric about the center of the well (x = 0.25 nm). If the electron has energy 29.66 eV and ψ(0) = 1.42 (nm)-1/2, then what is the probability for finding the particle in the left half of the well (0 < x < 0.25 nm)?


    2. Relevant equations
    Schrodinger's Equation in the classically forbidden zone: C1ekx + C2e-kx

    3. The attempt at a solution
    My idea was that because the probability density from negative infinity to .25 nm (L/2) would be .5, I could simply find p(<0) by integrating Schrodinger's equation and subtracting that from .5. However I cannot find K.

    I know the equation for K is [tex]\sqrt{2m/(h/2\Pi)^2(U-E)}[/tex] and it seems like I've been given those 4 variables. However my homework will not accept my answer for K. For reference I've been getting k = 9.25 nm^-1.
     
  2. jcsd
  3. Feb 22, 2010 #2

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    Hello yayz0rs,

    Welcome to Physics Forums.

    I'm getting a different answer for k.

    My solution to Shrodinger's equation in that region matches yours (at least involving k). There's nothing wrong with your equation for k, as far as I can tell. But still I'm getting a different value than yours when calculating the number.

    For your reference, your value for k is different from mine by close to a factor of [tex] \sqrt{2 \pi} [/tex]. Are you sure you are squaring the [tex] 2\pi [/tex] along with the h? Are you making sure to use the correct value for h (as opposed to [tex] \hbar [/tex] which as the [tex] 2 \pi [/tex] already factored in)?
     
  4. Feb 22, 2010 #3

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    By the way, your equation, as it is written, could use a little clarity by adding parenthesis or something appropriately. In my last post, above, I assume your equation really means

    [tex] k = \sqrt{\frac{2m}{\hbar^2}(U-E)} [/tex]

    where

    [tex] \hbar = \frac{h}{2\pi} [/tex]

    in which case, we both got the same equation for k.
     
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