Angular Momentum of 0 for Electrons in Ground Energy State

[pivoxa15]

How can there be an angular momentum of 0 for electrons in the ground energy state? It is moving around the nucleus just like it is when the electrons are in the 2s state. But in that case there is angular momentum.

 

[ZapperZ]

How do you know it is "moving"?

Zz.

 

[malawi_glenn]

Does not s equals $$l = 0$$ ?

And have you regarded the quantum mechanical view of the electron "moving" around the nucleus? What does the quantum numbers $$l$$ represent?

 

[George Jones]

Let me think out loud for a bit.

Suppose that a classical object of mass $m$ is subject to an inwardly directed central force of magnitude $A/r^2$. If the object is in a circular orbit of radius $R$, then the magnitude of its angular momentum is $L = mvR$. Also, Newton's second law gives

$$m \frac{v^2}{R} = \frac{A}{R^2},$$

which leads to

$$v = \sqrt{\frac{A}{m}} \frac{1}{\sqrt{R}}.$$

Use in this in the expression for angular momentum:

$$L = \sqrt{Am} \sqrt{R}.$$

Thus, classically, arbitrarily small values of $L$ are allowed and as an orbit's radius goes to zero, so too does the magnitude of the angular moment.

In this limit, the preferred axis "disappars". Quantum mechanically, $L = 0$ states are spherically symmetric.

As I say, I'm thinking out loud, so don't take anything I've written seriously.

 

[pivoxa15]

How do you know it is "moving"?

Zz.

Good point. It could be stationary clouds of some sort. However at higher energetic states, |L| is not zero so does it imply it is 'moving'?

Does not s equals $$l = 0$$ ?

Thats right. l=0 when in the s state.

What does the quantum numbers $$l$$ represent?

Good question. What does the quantum numbers represent? A direct answer is numerals to track the solution of the Schrodinger equation for a hyrdogenic system. And when the classical analogies of quantum operators are applied to the SE, out comes these numerals that we assigned. The angular momentum operator spits out these L numbers which we have to intepret it as angular momentum.

 

[MidasDeWolf]

Good question! However, your last remark contains an error. Namely, all S-states have zero angular momentum! This is very counter intuitive but also an experimental fact...

The problem is that we are used to think classically... However, on the atomic level experiments show that electrons behave wave-like. In order to understand your problem you must first accept this view...

Now, if you solve Schrodinger's wave equation then you can obtain so-called stationairy solutions. Simplified they represent standing waves in a Coulomb potential (the attractive force between proton and electron). The ground state and all other S-states then represent standing waves in the radial direction without an angular dependence. Because of this spherical symmetry there is not angular preference, and as a consequence no angular momentum.

I hope this brings you closer to your final answer!

Midas

 

[ZapperZ]

Good point. It could be stationary clouds of some sort. However at higher energetic states, |L| is not zero so does it imply it is 'moving'?

If it is "moving", especially in the geometry that has been described for the p,d,f,etc. orbitals, then it should radiate, no? But it doesn't.

So where do you think is the source of the problem here? Think about it. We have already addressed some of this in the FAQ.

Zz.

 

[pivoxa15]

If it is "moving", especially in the geometry that has been described for the p,d,f,etc. orbitals, then it should radiate, no? But it doesn't.

So where do you think is the source of the problem here? Think about it. We have already addressed some of this in the FAQ.

Zz.

I am starting to think that its best not to think about a picture of the atom at all and just do the maths and get experimental results because nothing seems to make sense. The electron is moving but there is no paths. That is all I can infer. The electron can't be stationary for if it was then I will know its exact location so HUP is violated. There can't be a paths because then we will know its position and velocity instantaneously.