How many nodal surfaces are in a 2s orbital and 3px orbital?

[RPN]

Hello,
New to chemistry need help with this question and I don't understand what a nodal surface is
How mnay nodal surfaces are there for
a) a 2s orbital
b) a 3px orbital
If anyone can help me that would be great.
Thank you

 

[dextercioby]

Here's a link.

http://www.chemistry.nmsu.edu/studntres/chem111/resources/notes/atomic_orbitals.html

And this is trully awesome

http://www.shef.ac.uk/chemistry/orbitron/AOs/2s/index.html


Daniel.

 

[RPN]

Thanks for the quick reply!
So I went to the links and I think these are the answers
a) a 2s orbital has 1 nodal surface
b) a 3px orbital has 2 nodal surfaces
Is that right?
Thanks for the help!

 

[dextercioby]

Yes,it is correct.

Daniel.

 

[RPN]

Thank you so much. I really appreicate all the help provided here.

 

[Gokul43201]

Dexter, the Orbitron site is really neat.

Just a few additional points here :

1. The nodal points come from the radial part of the wavefunction.

2. A neat little observation (at least to me) : We know that $R_{1s}$ decays exponentially from the origin and hence has no nodes. 2s is orthogonal to 1s. So, if 1s is positive everywhere and the integral of the product is zero (orthogonality), then 2s must have a positive and a negative region. In other words, 2s must have at least one node.

3. $\psi _{n,l,m}$ has n-l-1 nodes in $R_{n,l}$ which are not at the origin, and one node at the origin for p,d and f. For $\psi _{n,l,m}$, the node at the origin gives rise to l nodal surfaces (planes or cones) passing through the origin (hence the s orbitals have no nodal surfaces passing through the origin).

The upshot of point 3 is that there are (n-l-1) + l = n-1 nodal surfaces, totally.