How do you explain the fact that the angular momentum of an electron in the s-orbital is zero?

Question

I know it's something beyond the domain of classical mechanics but what exactly is it?

If it's got to do with the geometry of the s-orbital as Bamboozel puts it, can someone explain the same in the case of the p and d-orbitals where angular momentum is sqrt2*h/2pi and 2*sqrt3*h/2pi

Answer 1

Well, since we are trying to "be careful" here, let's also make sure we be extra careful in saying that the atomic orbitals somehow implies a "velocity" or speed of anything. It doesn't. By saying such things, we are already implicitly implying a well-defined charged particle moving around. You don't have such things until a position measurement is done. Before then, an electron in an s-orbital, for example, has no well-defined position and identity. Rather, based on the wavefunction alone, it is "spread out" in a uniform sphere around the nucleus. So the electron is everywhere simultaneously (which is connected to the Schrodinger Cat-type puzzlement - another illustration that things in QM are interconnected). This is how we get an angular momentum of zero for the s-orbital - from the geometry of the orbital itself.

This is another illustration where our social language can cause many confusion in trying to describe things that have no linguistic equivalent. As soon as we say "electron moves in an orbit", a whole range of implications kick in. We automatically imply that there is this well-defined object that we can track along the way and moving in a well-defined trajectory. QM implies no such thing, at least as far as atomic orbitals are concerned. We have seen a whole zoo of evidence where an "electron" can simultaneously spread itself into many locations to produce unclassical effects (bonding-antibonding bands, etc.) .

Answer 2

I suppose you are asking about why the azimuthal quantum number for s orbital is zeo.
so let me try the answer.
azimuthal quantum number l is related to angular momentum by the equation L= h' * (l*(l+1) )e1/2. where l is a non negative integer.

and azimuthal quatum no. governs the ellipticity of the probability of cloud and number of planar nodes going through the nucleus.
and for s orbital, whuich is sperical, ellepticcity is zero, and hence there is no probability to find the electron cloud at necleus.
so the number of nodes for s orbital is zero.

you can get more information in the wikipedia site.

Answer 3

an s-orbital surrounds the nucleus like a sphere. an orbital is where the electron is likely to be found. when s = 0, then the electron is not likely to be found at the nucleus. only p-orbitals, d-orbitals, f-orbitals have nodes (the flower petal model) and may be found at the nucleus.

Answer 4

It's the Mexicans' fault, actually.

Answer 5

Dont worry. Its all relative.