Explore Atomic Orbitals 5a

Schrödinger's Equation

For a hydrogen-like atom (one nucleus and one electron), Schrödinger's Equation looks like:

(1)

\begin{align} E \psi(\mathbf{r}) = - \frac{{\hbar}^{2}}{2m} {\nabla}^{2}\psi(\mathbf{r}) + V(\mathbf{r}) \psi(\mathbf{r}) \end{align}

where $\psi(\mathbf{r})$ is the electron's wave function at position $\mathbf{r}$.

The solution to this equation is:

(2)

\begin{align} \psi(r, \theta, \phi) = \sqrt{{\left( \frac{2}{n {a}_{0}} \right)}^{3} \frac{(n - l - 1)!}{2n{[(n+l)!]}^{3}} } {e}^{-\rho/2} {\rho}^{l} {L}^{2l+1}_{n-l-1}(\rho) . {Y}^{m}_{l}(\theta, \phi) \end{align}

where:
${a}_{0}$ is the Bohr radius;
$\rho = \frac{2r}{n {a}_{0}}$;
${L}^{2l+1}_{n-l-1}(\rho)$ is a generalised Laguerre polynomial; and
${Y}^{m}_{l}(\theta, \phi)$ is a spherical harmonic function.

For the moment, there are three important bits:

• The "shape" of an orbital from the centre out can be separated from the shape of the orbital as you go around the atom.
• The spherical harmonic function determines the shape as you go around the atom, which is what we'd "see" if we looked at an atom under an ubermicroscope.
• The specific shapes are determined by the numbers $n$, $l$ and $m$, which are called quantum numbers, and (amongst other things) must always be whole numbers.

• Return to Explore Atomic Orbitals.