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Hydrogen, Couloumbs and Quantum Mechanics

  1. Mar 13, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    In the derivation of the energy levels in the hydrogen atom one commonly assumes that the nucleus is a point charge. However, in reality the size of the nucleus is of the order of [tex] 1fm=10^−15m[/tex]. Since this is very much smaller than the typical distance of the electron from the nucleus, which is of the order of [tex] a_0 \approx 0.5 Angstoms = 0.5 *10^{-10}m [/tex], the finite size of the nucleus can be taken into account perturbatively.

    a)

    Assume that the nucleus of the hydrogen atom is a uniformly charged spherical shell of radius [tex] \delta [/tex]. According to Gauss’ law [tex] \int E \cdot dA = Q_{enclosed}/\epsilon_0[/tex], the electric field outside this shell is the same as for a point charge at the centre. However, from Gauss’ law also follows that the electric field
    inside the shell is zero. Since the force [tex]-eE_r[/tex] on the electron is given by the gradient [tex] -\partial V/\partial r [/tex] of the potential, the potential must thus be constant inside the shell and indistinguishable from the Coulomb potential outside it.

    Sketch the potential V (r), find the expression for it, and then find the perturbation [tex] \Delta [/tex]V relative to the Coulomb potential generated by a point-like nucleus.

    b)

    Use the ground-state wave function for a hydrogen atom with a point-like nucleus,

    [tex] \Phi_{100}(r,\theta, \phi) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0} [/tex]

    and calculate the shift of the ground-state energy due to the distribution of the nuclear charge overa finite spherical shell to first order in the perturbation.

    c)
    Expand the result of (b) in powers of [tex] \delta/a_0 [/tex], retain only the first non-vanishing term, and thus show that the ground-state energy shift is approximately

    [tex] \Delta E_1 \approx \frac{e^2}{6\pi\epsilon_0}\frac{\delta^2}{a_0^3} [/tex]

    2. Relevant equations

    Given on sheet:

    [tex] e^x = \Sigma \frac{x^n}{n!}[/tex]

    3. The attempt at a solution

    Okay B, I think is a normal inverse square law graph, similar to that of gravity. It goes up with constant gradient until it reaches the radius R, then decreases inverse square law curve.

    I am not sure however what to do for b). Do I need to insert the potential into a Schrödinger Equation?
     
  2. jcsd
  3. Mar 14, 2009 #2

    TFM

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    Okay, I have worked out that for b), I need to use the equation:

    [tex] \Delta E = \int^{\infty}_{-\infty}\phi^*_n \Delta V(r) \phi_n [/tex]

    Now they give you phi, but I am slightly unsure as to how I get the [tex] Delta V(r)[/tex]. I thunk this bit is also part of a).

    Any helpful suggestions of how to get this value?

    Also, I think I have the graph wrong. I t should be more of a plateau, with the sides curving downwards, as attached.
     

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