- #1

- 4

- 0

**1.**

**Homework Statement**A simple model atom is composed of a point-like nucleus with charge ##+Q## and an electron charge distribution

$$

\rho(\vec{r})=-\dfrac{\left|Q\right|}{\pi a^2 r}exp(-2r/a)

$$

where ##a## is a constant. Show that the ionization energy (the energy to remove the electron to infinity) of this model is

$$

I = \frac{3}{8}\frac{Q^2}{\pi\epsilon_0a}

$$

## Homework Equations

The energy it takes to assemble a continuous charge distribution.

$$

W = \frac{\epsilon_0}{2}\int_{\textrm{all space}}E^2 dV

$$

The energy it takes to move a charged particle in an electric field from the point a to b.

$$

W = -Q\int_a^b \textbf{E}\cdot d\textbf{l}

$$

Maxwell equation for electrostatics

$$

\nabla \cdot \textbf{E} = \frac{1}{\epsilon_0}\rho(\vec{r})

$$

## The Attempt at a Solution

I think that the ionization energy is the energy it takes to assemble the charge distribution created by the electron cloud plus the energy it takes to remove the positive nucleus. If I can find and calculate the electric field created by the charge distribution at a given point ##r##, I can easily use the first and second equation in

**relevant equations**to calculate the energy it takes to assemble the charge distribution and the energy it takes to move the positive nucleus from the origin out to infinity.

Since the charge distribution is spherical symmetric I was thinking that I can use Maxwells first equation to calculate the E-field using Gauss law.

$$

\oint \textbf{E}\cdot d\textbf{a} = \frac{1}{\epsilon_0}\int \rho(\vec{r})dV

$$

From this I get that

$$

4\pi r^2 E(r) = \frac{4\pi}{\epsilon_0}\int_0^r -\dfrac{\left|Q\right|}{\pi a^2 r^\prime}exp(-2r^\prime/a) r^{\prime 2}dr^\prime

$$

I use primed r on the right hand side to differentiate my limit from my integration variable. Solving the integral on the right hand side using partial integration I end up with that the electric field is give by

$$

E(r)=-\frac{Q}{4\pi\epsilon_0 r^2 a}(a-e^{-2r/a}(2r+a))

$$

Now comes the big problem. When I'm going to calculate the work it takes to move the nucleus out to infinity I get an integral that diverges.

$$

W = -Q\int_a^b \textbf{E}\cdot d\textbf{l} = \frac{Q^2}{4\pi\epsilon_0 a}\int_\infty^0\frac{a-e^{-2r/a}(2r+a)}{r^2}dr = Diverges

$$

I don't know what I'm doing wrong and would appreciate help very much! Maybe I'm missing something? Perhaps I should assume that r<<a so I can do some kind of Taylor expansion of the exponential.