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Electron in finite square well

  1. Oct 26, 2006 #1
    Hey,

    An electron is in a finite square well of 1 Å so the question is to find the values of the well's depth V0 that have exactly two state ?

    How to proceed with this - finding the eigenvalues En = \hbar^2\pi^2 / 2ma^2

    Thanks in advance
     
  2. jcsd
  3. Oct 26, 2006 #2

    OlderDan

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    I think you will find all you need here

    http://musr.physics.ubc.ca/~jess/p200/sq_well/sq_well.html
     
  4. Oct 26, 2006 #3
    hmmm - I have read the notes and they make some things more clear but I dont get how I can deduce V0 knowing that there should be exactly two steady states without En = Pi^2 hbar^2 n^2 /(2*m*a^2)

    Could one be a little more specific, please.
     
  5. Oct 26, 2006 #4

    OlderDan

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    As the notes say, there is always at least one bound state (I assume we are talking bound states here, since there are always infinitely many states). The well must be deep enough for there to be a second bound state, but not deep enough for three. The notes make reference to a transcendental equation for E that must be solved graphically. I think you need to work with that.
     
  6. Oct 26, 2006 #5
    Hey,

    I plot three equations than, a straight line k/k0 and a sinusoidal where I disregard the areas not allowed by the tan-function. I get a lot of intersections of k - than how to convert these to energies of V0 ?

    Hope it is okay I ask again - thanks
     
  7. Oct 26, 2006 #6

    OlderDan

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    In the notes after equation 11 there is a conversion to express things in terms of the dimensionless quantities defined as θ and θo. You can graph both sides of the resulting equation as a function of θ using different values of θo. For the LHS, graph both the +tanθ curve and the -cotθ curve. For very small θo there will be only one intersection of the RHS with the tanθ curve. If you make θo somewhat larger, the RHS curve will intersect the -cotθ curve at the horizontal axis. This is the onset of the second bound state. Increasing θo further will make the RHS curve intesect the tanθ curve again at the zero of the next positive tanθ cycle. That is the onset of the third bound state. These two larger values of θo define the limits of Vo that will give two and only two bound states. If you continue to increase θo you will get more and more intersections corresponding to an increasing number of bound states.
     
    Last edited: Oct 26, 2006
  8. Feb 2, 2009 #7
    hi.
    The problem statement, all variables and given/known data
    Seven electrons are trapped in a one dimensional infinite square well of length L. What is the ground state energy of this system as a multiple of h2 / 8mL2?

    2. Relevant equations
    Energy of a single electron in state n is n2h2 / 8mL2

    3. The attempt at a solution
    Pauli exclusion principle says all 7 must have different quantum numbers.
    starting from n = 1, we have L = 0 and mL = 0, and ms = -1/2 and 1/2, so there are two electrons in n = 1.
    For n = 2, we have two electrons for L = 0
    for L = 1, we have mL = -1, 0, 1, which means this subshell can hold 6 electrons. The remaining 3 electrons go into this subshell then.

    Final tally: 2 electrons for n = 1 and 5 electrons for n = 2.

    Total energy as a multiple of the given term then = 2*1^2 + 5*2^2 = 22.
     
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