# Find the ionization energy of a simple model atom

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1. Jun 8, 2017

### Turbotanten

1. The problem statement, all variables and given/known data
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
$$I = \frac{3}{8}\frac{Q^2}{\pi\epsilon_0a}$$
2. Relevant 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
$$\nabla \cdot \textbf{E} = \frac{1}{\epsilon_0}\rho(\vec{r})$$
3. 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.

2. Jun 10, 2017

### rude man

That's not how ionization energy is defined, both in the problem statement and in general. So you're off on a tangent already.
good except you're forgetting something ...
plus what you forgot. Where did the 4π on the rhs come from?
Bad move - see top of my commentary.

Hint: how much energy to remove 1 electron to ∞ in the presence of the charge distribution consisting of the protons & electrons? You made a good move to consider Gauss, stick with it.

Last edited: Jun 10, 2017
3. Jun 10, 2017

### rude man

Upon further examination I don't like the way the problem is stated. What is "the" electron? The problem seems to suggest an atom with many protons and electrons - enough to enable a continuous charge distribution expression for ρ, then it doesn't state where the ionized electron is first located. It can't be located at the proton since that would imply infinite ionization energy.

So maybe we should assume one proton and one electron effectively situated at some mean position r0 which could be computed from the given charge density function. The answer does not include the electronic charge (1.6e-19C) which it must unless Q is the electronic charge.

I think the problem statement needs to be clarified.