# The time average potential of neutral hydrogen atom

1. Jan 23, 2015

### NewtonApple

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 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

since and

from product rule

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

Last edited by a moderator: May 9, 2017
2. Jan 23, 2015

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

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

$=-2e^{-\alpha r}\frac{1}{r^{3}}+\alpha e^{-\alpha r}\frac{1}{r^{2}}+e^{-\alpha r}\frac{\partial^{2}}{\partial r^{2}}\left(\frac{1}{r}\right)$

Last edited: Jan 23, 2015
3. Jan 23, 2015

### 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.