# Why there are nodes in orbital?

I know that because Schrödinger equations solution for probability of that region is 0, but I was looking for an intuitive explanation.

How can you intuitively explain the existence of nodes in an orbital?

• I have no way of knowing what would fit in your mind and what would not. – Ivan Neretin Aug 24 '16 at 16:21
• I don't think kids in high school have the tools to do that @Ivan. – Akshar Gandhi Aug 24 '16 at 16:25
• @ahmad The point of all this discussion is, there is another property of matter, which is not intuitive at all. The "property" $F=ma$ is intuitive because we see it everywhere. Schrodinger's equation is another such "property", which is not at all intuitive. It is this property that gives birth to those nodes. You will have to study a little bit of quantum mechanics to understand more of it. Food for thought: "Properties" like the electrostatic force and so on also apply at that scale, along with this new one. – FreezingFire Aug 24 '16 at 17:41
• en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane – permeakra Aug 24 '16 at 19:07
• @getafix Let's keep it friendly, please (re: your first comment). Thanks! – jonsca Aug 25 '16 at 0:33

Imagine a tight rope with fixed ends (like a string in a guitar). If you shake it it will vibrate. Both ends remains fixed. If you take a high frame rate video and see it in slow motion, at a fixed horizontal length (let say $x$ positions) the rope will oscillate up and down with some amplitude. It can be seen one or more intermediate $x$ positions that remains always with null amplitude (if it is in state close to a stationary one). There you have nodes in one dimensional string.
• Just to add to this, the 3D analogue of the "fixed ends" of the rope is the requirement that the wavefunction be square-integrable (aka normalisable), i.e. $\psi \to 0$ as $r \to \infty$. That would explain the presence of the radial nodes, angular nodes are a bit harder to justify imo. – orthocresol Aug 25 '16 at 4:41