The time average potential of neutral hydrogen atom

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Homework Statement


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

latex.png


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.

Homework Equations


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latex.png


The Attempt at a Solution


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latex.png


latex.png
since
latex.png
and
latex.png


from product rule

latex.png


latex.png


now I'm stuck here no idea how to handle first term.
 
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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]
 
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