The time average potential of neutral hydrogen atom

[NewtonApple]

Homework Statement

[/B]
The time-averaged potential of a neutral hydrogen atom is given by

where q is the magnitude of the electronic charge, and

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

[/B]

The Attempt at a Solution

[/B]

since

and

from product rule

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

 

[NewtonApple]

Ok I think I figure it out.

lets take derivative one term at a time using product rule

\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]

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

\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]

=-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)

 

[DEvens]

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.