# What is the formula for calculating a nodal point?

##### 1 Answer

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

As a simple but not undercomplicated example, consider one of the

The hydrogenic atom wave function for the

#psi_(2pz) = R_(21)(r)Y_(1)^(0)(theta,phi)#

#= 1/(4sqrt(2pi)) (Z/a_0)^(3//2) sigmae^(-sigma//2)costheta# ,where:

#sigma = Zr//a_0# .#Z# is the atomic number.#a_0 = "0.0529177 nm"# is the Bohr radius.#r# is the radial distance away from the nucleus.#R_(nl)(r)# is the radial component of the wave function.#Y_(l)^(m_l)(theta,phi)# is the angular component of the wave function.

The radial nodes are where

Since nonzero constants are clearly never zero, we just have to pick out the functions that have

#R_(21)(r) prop sigmae^(-sigma//2)#

#Y_(1)^(0)(theta,phi) prop costheta#

**RADIAL NODES?**

So, solving for the radial nodes, with

#(Zr)/(a_0)e^(-Zr//2a_0) = 0#

This exponential never goes to zero except at

#=> (Zr)/(a_0) = 0#

From here we can realize that there are **no radial nodes** in the

The only solution to this is

**ANGULAR NODES?**

Solving for the angular nodes,

#costheta = 0#

And this only nonredundantly applies for

So, in this domain,

Furthermore, there is no

#"Probability Density of 2p"_z#

#= int_(0)^(2pi) int_(0)^(pi) int_(0)^(oo) psi_(2pz)^"*"psi_(2pz)r^2dr sinthetad theta d phi#

#= color(red)(1/(4sqrt(2pi)) (Z/a_0)^(5) int_(0)^(2pi) d phi) int_(0)^(pi) cos^2theta sintheta d theta int_(0)^(oo) r^2e^(-Zr//a_0)dr#

#= color(red)((2pi)/(4sqrt(2pi))(Z/a_0)^(5)) int_(0)^(pi) cos^2theta sintheta d theta int_(0)^(oo) r^2e^(-Zr//a_0)dr#

Having no dependence on the value of

Therefore, the **angular node** is a plane perpendicular to the *plane*, just like it shows in the above diagram.