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Homework Help: The time average potential of neutral hydrogen atom

  1. Jan 23, 2015 #1
    1. The problem statement, all variables and given/known data

    The time-averaged potential of a neutral hydrogen atom is given by


    where q is the magnitude of the electronic charge, and latex.png being the Bohr radius. Find the distribution of charge( both continuous and discrete) that will give this potential and interpret your result physically.

    2. Relevant equations


    3. The attempt at a solution


    latex.png since latex.png and latex.png

    from product rule



    now I'm stuck here no idea how to handle first term.
    Last edited by a moderator: May 9, 2017
  2. jcsd
  3. Jan 23, 2015 #2
    Ok I think I figure it out.

    lets take derivative one term at a time using product rule

    [itex]\frac{1}{r^{2}}\frac{\partial}{\partial r}\left[r^{2}e^{-\alpha r}\frac{\partial}{\partial r}\left(\frac{1}{r}\right)\right]=\frac{1}{r^{2}}\left[2re^{-\alpha r}\frac{\partial}{\partial r}\left(\frac{1}{r}\right)+r^{2}\left(-\alpha\right)e^{-\alpha r}\frac{\partial}{\partial r}\left(\frac{1}{r}\right)+r^{2}e^{-\alpha r}\frac{\partial^{2}}{\partial r^{2}}\left(\frac{1}{r}\right)\right][/itex]

    putting [itex]\frac{\partial}{\partial r}\left(\frac{1}{r}\right)=-\frac{1}{r^{2}}[/itex]

    [itex]\frac{1}{r^{2}}\left[2re^{-\alpha r}\left(-\frac{1}{r^{2}}\right)+r^{2}\left(-\alpha\right)e^{-\alpha r}\left(-\frac{1}{r^{2}}\right)+r^{2}e^{-\alpha r}\frac{\partial^{2}}{\partial r^{2}}\left(\frac{1}{r}\right)\right]=\frac{1}{r^{2}}\left[-2e^{-\alpha r}\frac{1}{r}+\alpha e^{-\alpha r}+r^{2}e^{-\alpha r}\frac{\partial^{2}}{\partial r^{2}}\left(\frac{1}{r}\right)\right][/itex]

    [itex]=-2e^{-\alpha r}\frac{1}{r^{3}}+\alpha e^{-\alpha r}\frac{1}{r^{2}}+e^{-\alpha r}\frac{\partial^{2}}{\partial r^{2}}\left(\frac{1}{r}\right)[/itex]
    Last edited: Jan 23, 2015
  4. Jan 23, 2015 #3


    User Avatar
    Education Advisor
    Gold Member

    Is it exp(alpha r) or exp(- alpha r) ?

    I have not checked your algebra. But you stopped with there still being a derivative. You are not done.
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