I'm currently taking Quantum Mechanics in school and I came across a table displaying the following in the given order:

  • $E_n (\text{energy levels}), g_n (\text{eigen functions})$
    • $n \ l, [m]$
  • $E_1, 1$
    • $1s, [0]$
  • $E_2, 4$
    • $2s, [0]$
    • $2p, [-1,0,1]$
  • $E_3, 9$
    • $3s, [0]$
    • $3p, [-1,0,1]$
    • $3d, [-2,-1,0,1,2]$

Now, as I understand $n$ defines the size of the orbital, $l$ the shape and $m$ the orientation.

My question is, how does the atoms shell model fit into this if one can have two electrons per orbital? Are $1s$ the same as the orbital $2s$, but just renamed for the sake of higher energy levels? Or do two similarly looking orbits exist for different energy levels? For example two "layers" of spherical electron clouds for an atom with $4$ electrons.

I just can't see how $2$, $8$, $18$ becomes $2$, $8$, $8$ as we were taught in Physics at high school? Am I missing something?

  • $\begingroup$ Don't forget that electrons have spin, which doubles the number of states. $\endgroup$
    – Qmechanic
    Nov 23 '18 at 18:22
  • $\begingroup$ @Qmechanic "two electrons per orbital" :) However, I am more interested in how one can calculate the energy contribution of single orbitals, as I suspect this might be the cause of 2, 8, 8 $\endgroup$
    – Gjert
    Nov 23 '18 at 18:24

I just can't see how 2,8,18 becomes 2,8,8 as we were taught in Physics at high school? Am I missing something?

Yes, what you are missing, is Aufbau principle.

Following this, you find the first three rows of the periodic table have size 2,8,8 even though the first three energy levels have degeneracy 2,8,18.

Note that the periodic table is organised based on Ionization Energy. The increase in ionization from one element to the next is quicker as the p-subshell is being filled up than the other subshells (see graph in link), so elements reach maximum stability (Noble gas) when the p subshell has been filled.

Putting all this together we have the following result:

Subshells are filled in the order:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p

But then this is divided into rows of the periodic table like:


2s, 2p

3s, 3p

4s, 3d, 4p

5s, 4d, 5p

So the number of elements in each row of the periodic table is:






  • $\begingroup$ Nice, I can't however see how this is derived from the Quantum Mechanical wave function? $\endgroup$
    – Gjert
    Nov 23 '18 at 18:36
  • $\begingroup$ What you have to understand is that solving quantum mechanics for the hydrogen atom does not give you the right answer. A hydrogen atom only has one electron which can be in any one of the nlm states. But atoms in the periodic table have many electrons - this makes it a different quantum problem. $\endgroup$
    – isometry
    Nov 23 '18 at 18:41
  • $\begingroup$ Okai, I assume its because of interaction between the different electrons? $\endgroup$
    – Gjert
    Nov 23 '18 at 18:44
  • 2
    $\begingroup$ Yes this is where chemists take over and talk about things like electron screening and penetration. $\endgroup$
    – isometry
    Nov 23 '18 at 18:49
  • $\begingroup$ Is it just impractical to solve the schrodinger equation for heavier atoms as a multi-body system of electrons? $\endgroup$ Nov 25 '18 at 19:29

The Hamiltonian of the hydrogen atom is separable in spherical coordinates into differential equations of r, azimuthal angle, and axial angle. There is a coordinate independent "spin" portion of the wave function as well.

Bound states of a Hamiltonian are quantizable. The states are discrete with discrete and discontinuous. Eigen values. That's n, the principle quantum number and associated with the radial wave function. The numbers m, and l are associated with the Angular coordinates via the Associated Legendre polynomials. Their product, with the spin function, give you the whole wave function in terms of the 4 quantum numbers.


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