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Finite Spherical Well

  1. Nov 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is placed in a finite spherical well:

    [tex]V(r) = \{^{-V_{0}, if r \leq a;}_{0, if r > a.}[/tex]

    Find the ground state, by solving the radial equation with [tex]l = 0[/tex]. Show that there is no bound state if [tex]V_{0}a^{2} < \pi^{2}\hbar^{2}/8m[/tex].


    2. Relevant equations
    [tex]\frac{d}{dr}(r^{2}\frac{dR}{dr}) - \frac{2mr^{2}}{\hbar^{2}}[V(r) - E]R = l(l + 1)R[/tex].


    3. The attempt at a solution
    For [tex]r \leq a[/tex]
    [tex]\frac{d}{dr}(r^{2}\frac{dR}{dr}) - \frac{2mr^{2}}{\hbar^{2}}[V(r) - E]R = 0[/tex] [tex]\Rightarrow[/tex][tex]

    [tex]2r\frac{dR}{dr} + \frac{2mr^{2}}{\hbar^{2}}(V_{0} + E)[/tex] [tex]\Rightarrow[/tex]

    [tex]\frac{dR}{dr} = \frac{-mr}{\hbar^{2}}(V_{0} + E)[/tex] [tex]\Rightarrow[/tex]

    [tex]R = \frac{-m}{\hbar^{2}}\int r(V_{0} + E) dr[/tex]

    But I'm not sure about the second part to show that there is no bound state with the given conditions.
     
  2. jcsd
  3. Nov 15, 2007 #2
    Ok, so we can say:

    [tex]V_{0} < \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}[/tex]

    When we let [tex]n = \frac{1}{2}[/tex] or [tex]n^{2} = \frac{1}{4}[/tex]

    And if there is no bound state, then the ground state of energy must be greater than the potential. And in the ground state we have [tex]n=1[/tex] right?

    So we have:

    [tex]E_{n0} = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}[/tex].

    Which means:

    [tex]V_{0} < E_{n0}[/tex] [tex]\Rightarrow[/tex]

    [tex]\frac{\pi^{2} \hbar^{2}}{8ma^{2}} < \frac{\pi^{2} \hbar^{2}}{2ma^{2}}[/tex].

    Is that what they're looking for?
     
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