# Question e0f39

Aug 15, 2015

The most basic model is that of the idealized hydrogen atom. This can be generalized to other atoms, but those models have not been solved.

#### Explanation:

An atom is at it's most basic form a positively charged heavy particle (the nucleus) with negatively charged lightweight particles moving around it.

For the simplest model possible, we assume the nucleus to be so heavy, that it remains fixed in the origin. That means we don't have to take its motion into account. Now we are left with the electron. This electron moves the electric field of the charged nucleus. The nature of this field is given to us by classical electrostatics.

Lastly we ignore relativistic effects and effects caused by the spin of the electron, and we are left with just a charged particle in an electric field.

Now we identify a wavefunction with the electron $\Psi \left(\vec{r} , t\right)$. We use the model described above to write down the Schrödinger equation.
iћdel/(delt)Psi(vecr,t)=[-ћ^2/(2m_e)grad^2+V(vecr)]Psi(vecr,t)

The potential energy term $V \left(\vec{r}\right)$ can be derived from Coulombs law. The force acting on the electron is given by
$\vec{F} \left(\vec{r}\right) = - {q}^{2} / \left(4 \pi {\epsilon}_{0} | | \vec{r} | {|}^{3}\right) \vec{r}$
where $q$ is the absolute value of the charge of both the electron and the nucleus.
The potential is given by the following where $\gamma$ is a path going from infinity, where the potential is $0$, to $\vec{r}$:
$V \left(\vec{r}\right) = - {\int}_{\gamma} \vec{F} \left(\vec{s}\right) \cdot \mathrm{dv} e c s = {q}^{2} / \left(4 \pi {\epsilon}_{0}\right) {\int}_{\infty}^{r} \frac{1}{s} ^ 2 \mathrm{ds} = - {q}^{2} / \left(4 \pi {\epsilon}_{0} r\right)$.
Here we have used $r = | | \vec{r} | |$.

This gives us:
iћdel/(delt)Psi(vecr,t)=-[ћ^2/(2m_e)grad^2+q^2/(4piepsilon_0r)]Psi(vecr,t).

Fortunately for us, it is possible to determine eigenfunctions and values for the energy, that means functions $\psi \left(\vec{r}\right)$ and values $E$ of the form
-[ћ^2/(2m_e)grad^2+q^2/(4piepsilon_0r)]psi(vecr,t)=Epsi(vecr,t)#
These solutions are quite tedious to write down, so I will only do that when you ask me to, but the point is, we can solve this.

This gives us an energy spectrum for the hydrogen, plus wavefunctions belonging to each energy, or the so called orbitals of Hydrogen atom.

Unfortunately, for more complex atoms, this doesn't do the job anymore, since when you have multiple atoms, they will also exert a force on eachother. This plus of course the momentum and electron-nucleus potential term gives a lot of extra terms in the Schrödinger equation, andt until now, nobody has been able to solve it exactly. There are however ways to approximate the solution. Which I will not show here.