# What is the formula for calculating a nodal point?

Dec 29, 2017

There is none. You need the wave function for the particular orbital, which is not readily available unless the atom is hydrogen...

In the case of hydrogenic atoms, i.e. one-electron atoms, the wave functions are available in most physical chemistry textbooks up through $n = 3$.

As a simple but not undercomplicated example, consider one of the $2 p$ orbitals: The hydrogenic atom wave function for the $2 {p}_{z}$ is:

${\psi}_{2 p z} = {R}_{21} \left(r\right) {Y}_{1}^{0} \left(\theta , \phi\right)$

$= \frac{1}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{3 / 2} \sigma {e}^{- \sigma / 2} \cos \theta$,

where:

• $\sigma = Z r / {a}_{0}$.
• $Z$ is the atomic number.
• ${a}_{0} = \text{0.0529177 nm}$ is the Bohr radius.
• $r$ is the radial distance away from the nucleus.
• ${R}_{n l} \left(r\right)$ is the radial component of the wave function.
• ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ is the angular component of the wave function.

The radial nodes are where ${R}_{n l} \left(r\right) = 0$, and the angular nodes are where ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right) = 0$.

Since nonzero constants are clearly never zero, we just have to pick out the functions that have $r$ or $\theta$. That is, we just need the fact that:

${R}_{21} \left(r\right) \propto \sigma {e}^{- \sigma / 2}$

${Y}_{1}^{0} \left(\theta , \phi\right) \propto \cos \theta$

So, solving for the radial nodes, with $\sigma = Z r / {a}_{0}$:

$\frac{Z r}{{a}_{0}} {e}^{- Z r / 2 {a}_{0}} = 0$

This exponential never goes to zero except at $\infty$, at which the orbital doesn't exist, so we can eliminate it.

$\implies \frac{Z r}{{a}_{0}} = 0$

From here we can realize that there are no radial nodes in the $2 {p}_{z}$ orbital.

The only solution to this is $r = 0$, and at $r = 0$, no electrons exist because there is no electron that can fit in a radius of $\text{0 meters}$.

ANGULAR NODES?

Solving for the angular nodes,

$\cos \theta = 0$

And this only nonredundantly applies for $\theta \in \left[0 , \pi\right]$.

So, in this domain, $\cos \theta = 0$ at $\theta = \frac{\pi}{2}$, or ${90}^{\circ}$, relative to the axis of the orbital (the $z$ axis).

Furthermore, there is no $\phi$ dependence, so integration over the $\phi$ variable to generate this orbital probability density contributes only a constant of $2 \pi$:

${\text{Probability Density of 2p}}_{z}$

$= {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} {\psi}_{2 p z}^{\text{*}} {\psi}_{2 p z} {r}^{2} \mathrm{dr} \sin \theta d \theta d \phi$

$= \textcolor{red}{\frac{1}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{5} {\int}_{0}^{2 \pi} d \phi} {\int}_{0}^{\pi} {\cos}^{2} \theta \sin \theta d \theta {\int}_{0}^{\infty} {r}^{2} {e}^{- Z r / {a}_{0}} \mathrm{dr}$

$= \textcolor{red}{\frac{2 \pi}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{5}} {\int}_{0}^{\pi} {\cos}^{2} \theta \sin \theta d \theta {\int}_{0}^{\infty} {r}^{2} {e}^{- Z r / {a}_{0}} \mathrm{dr}$

Having no dependence on the value of $\phi$ for a wave function that goes to zero at a particular value of $\theta$ means that a nodal plane is generated.

Therefore, the angular node is a plane perpendicular to the $z$ axis of the $2 {p}_{z}$ orbital, and is the $x y$ plane, just like it shows in the above diagram.