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    Charge distribution of the Hydrogen atom

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    The time-averaged potential of a neutral hydrogen atom is given by 4s (1 + 7) wo r where q is the magnitude of the electronic charge, and cx-' = ao/2, ao being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically.

    It is taken from "Classical Electrodynamic 3rd" by Jackson and I know it will use the Poisson Equation.

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    Please see the attached solution for the complete solution.
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    111Equation Chapter 1 Section 1
    Lets rewrite the potential as:
    (Please see the attached file)

    Poisson equation states that:
    (Please see the attached file)

    Therefore:
    (Please see the attached file)

    The Del operator in spherical coordinates is:
    (Please see the attached file)

    The Laplacian in spherical coordinates is given by:
    (Please see the attached file)

    We use the property of the Laplacian of a product:
    (Please see the attached file)

    In our case:
    (Please see the attached file)

    The functions are angular-independent. Therefore:
    (Please see the attached file)
    And:
    (Please see the attached file)

    Where we used the known relations (see appendix for proof)
    (Please see the attached file)
    For the last term:
    (Please see the attached file)
    So we get:
    (Please see the attached file)

    We know that the delta function is zero everywhere except at the origin, where so with no loss of information we can write it as:
    (Please see the attached file)

    The first term is the charge of a proton (discrete point charge) located at the origin.
    The second term is the negative charge of an electron smeared over a cloud that decays exponentially.

    Appendix: The Laplacian of 1/r
    We would like to calculate the Laplacian of
    According to (1.5) this gives:
    (Please see the attached file)

    But this is true only if , otherwise we encounter singularities.
    So let's integrate (Please see the attached file) over a volume of sphere of radius a :
    (Please see the attached file)

    Using Stokes theorem:
    (Please see the attached file)

    Where S is now the surface of the sphere.

    Now:
    (Please see the attached file)

    On the surface of the sphere
    (Please see the attached file)

    And:
    (Please see the attached file)

    Where [omega] is the solid angle.

    Plugging this into (1.15) we get:
    (Please see the attached file)

    Since the solid angle of an entire sphere is 4[pi]

    So we have:
    (Please see the attached file)

    It is zero outside of the origin, but if we integrate it over any sphere that contains the origin we get a finite value.
    We recall that by definition, and for any for any volume that contains the origin:
    (Please see the attached file)

    Therefore we can write (1.20) as:
    (Please see the attached file)

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    © BrainMass Inc. brainmass.com October 5, 2022, 1:00 am ad1c9bdddf>
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