Solving for the distribution of charge where time-averaged potential is given | Quantum Science Philippines

Solving for the distribution of charge where time-averaged potential is given

by Sim P. Bantayan, MSPhysics I, MSU-IIT

Problem 1.5

The time-averaged potential of a neutral hydrogen atom is given by $\Phi = \frac{q}{4\pi\epsilon}\frac{e^{\alpha r}}{r}(1 + \frac{\alpha r}{2})$

where q is the magnitude of the electronic charge, and $\alpha^{-1} = a_0/2, a_0$ being the Bohr radius. Find the distribution of charge( both continuous and discrete) that will give this potential and interpret your result physically.

Solution:

We can solve the distribution of charge by solving first the charge density using the Poisson’s equation $\bigtriangledown^2 \Phi= \frac{\rho}{\epsilon_0}$.

We are given a hint to find the discrete distribution of charge, meaning likely a delta function will be in our answer.

Now, we solve Poisson’s equation for  $\rho$, $\bigtriangledown^2 \Phi= \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})[\frac{e^{\alpha r}}{r} + \frac{\alpha e^{\alpha r}}{2}]$

where $\bigtriangledown^2 \equiv \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})$.

But $\bigtriangledown^2 \Phi= \frac{\rho}{\epsilon_0}$. So, $\frac{\rho}{\epsilon_0}= \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})[\frac{e^{\alpha r}}{r} + \frac{\alpha e^{\alpha r}}{2}]$

Using the product rule on the first term, we obtain $\rho = -\frac{q}{4\pi}\frac{1}{r^2}\frac{\partial}{\partial r}(e^{-\alpha r}r^2\frac{\partial}{\partial r}(\frac{1}{r}) - \alpha re^{-\alpha r} - \frac{\alpha^2 r^2}{2}e^{-\alpha r})$

Then we distribute the $\frac{1}{r^2}\frac{\partial}{\partial r}$ term and we get $\rho = -\frac{q}{4\pi}[\frac{1}{r^2}\frac{\partial}{\partial r}((e^{-\alpha r}r^2\frac{\partial}{\partial r}(\frac{1}{r})) - \frac{1}{r^2}\frac{\partial}{\partial r}(\alpha re^{-\alpha r}) - \frac{1}{r^2}\frac{\partial}{\partial r}(\frac{\alpha^2 r^2}{2}e^{-\alpha r}]$

And then using product rule, we have $\rho = -\frac{q}{4\pi}[e^{-\alpha r}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})\frac(1)(r) + \frac{\alpha}{r^2}e^{-\alpha r} + \frac{\alpha^2}{r}e^{-\alpha r} - \frac{\alpha}{r^2}e^{-\alpha r} + \frac{\alpha^3}{2}e^{-\alpha r} - \frac{\alpha^2}{r}e^{-\alpha r}]$

After the terms cancel, we are only left with $\rho = -\frac{q}{4\pi}[e^{-\alpha r}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})\frac{1}{r} + \frac{\alpha^3}{2}e^{-\alpha r} ]$

Using our delta function equation for the first term, $\rho = -\frac{q}{4\pi}[-4\pi\delta(\vec{r}) + \frac{\alpha^3}{2}e^{-\alpha r}]$,

thus $\rho = q\delta(\hat{r}) - \frac{q}{8\pi}\alpha^3e^{-\alpha r}$

Physically, this is the point charge of the proton nucleus represented by the

delta function at the center of the atom, surrounded by the negative electron

cloud.