Let's now consider atoms.

First, spectroscopic notation for angular-momentum states:

[table]AM value | Letter | What

0 | s | sharp

1 | p | principal

2 | d | diffuse

3 | f | faint

4 | g

5 | h[/table]

(n,L)

^{2*S+1}_{J}
n = principal quantum number

L = angular-momentum letter

S = total spin (usually on the left side)

J = total angular momentum (spin + orbit)

Back to atoms.

The first one is

**hydrogen**. The nucleus has charge Z = 1, and to cancel it out, it has 1 electron. Easy. It's in the lowest energy level, with n = 1 and lj = 0, or the 1s state. Thus, it has configuration 1s

^{1}
The next one is

**helium**. Z = # electrons = 2. What is their combined wavefunction?

For their spins, let the up states be u1 and u2 for each electron, and the down states d1 and s2. Their positions will be x1 and x2.

If their total spin is 1, their wavefunction is

X(x1,x2) * {u1*u2, 1/sqrt(2)*(u1*d2+d1*u2), d1*d2} (projected AM: +1, 0, -1)

From FD stats, X(x2,x1) = - X(x1,x2)

If their total spin is 0, their wavefunction is

X(x1,x2) * 1/sqrt(2)*(u1*d2 - d1*u2) (projected AM: 0)

From FD stats, X(x2,x1) = + X(x1,x2)

To be in their lowest energy level, both electrons must be in the 1s state. This means that their combined wavefunction must be X(x1,x2) = X

_{1s}(x1)*X

_{1s}(x2), which is, of course, symmetric. This forces their spin to 0.

Thus, helium has 2 electrons in the 1s state, making its configuration 1s

^{2}.

The next one is

**lithium**, with Z = # e's = 3. The additional electron must be in either the 2s or 2p state, because from the Pauli Exclusion Principle, it cannot be in the 1s state.

Is the third electron 2s or 2p? The higher the angular momentum, the more it avoids the nucleus, and the more the nucleus's charge gets canceled out by the inner electrons, meaning that the electron "sees" less effective charge. Thus, higher-angular-momentum states for the same principal quantum number get more energy when other electrons are present. This makes the additional electron 2s instead of 2p, making lithium 1s

^{2}*2s

^{1} or (He)*2s

^{1}.

**Beryllium** is like helium, with (He)*2s

^{2}.

**Boron** gets an electron in its 2p state, since the 2s states are all filled. It is thus (He)*2s

^{2}*2p

^{1}
**Carbon** gets another 2p electron, making it (He)*2s

^{2}*2p

^{2} However, the two 2p electrons are unpaired, with parallel spins, for a total spin of 1. This is because total spin 0 (antisymmetric) would make their combined orbital wavefunction symmetric, making the electrons tend to be near each other. An antisymmetric orbital wavefunction makes them far, and this gives them spin 1 (symmetric spins).

**Nitrogen** gets another 2p electron, with all three 2p electrons unpaired: (He)*2s

^{2}*2p

^{3}
**Oxygen** gets another 2p electron, and its spin has to pair up with another 2p electron's spin: (He)*2s

^{2}*2p

^{4}
**Fluorine** gets another 2p electron, and its spin pairs up with another 2p electron, giving 1 unpaired electron and 2 pairs of paired electrons: (He)*2s

^{2}*2p

^{5}
**Neon** gets another 2p electron, making all its 2p electrons paired: (He)*2s

^{2}*2p

^{6}
**Sodium**, (Ne)*3s

^{1}, to

**argon**, (Ne)*3s

^{2}*3p

^{6}, add electrons in the 3s and 3p states in similar fashion, because the lower ones are all occupied.

**Potassium**, (Ar)*4s

^{1} adds an electron in the 4s instead of the 3d state, because the electron gets closer to the nucleus in the 4s state than in the 3d one. After

**calcium**, (Ar)*4s

^{2}, the electrons go into the 3d shell:

**scandium**, (Ar)*4s

^{2}*3d

^{1}, to

**zinc**, (Ar)*4s

^{2}*3d

^{10}. Then the 4p electrons get added, from

**gallium**, (Ar)*4s

^{2}*3d

^{10}*4p

^{1}, to

**krypton**, (Ar)*4s

^{2}*3d

^{10}*4p

^{6}. There are a few complications along the way with the 4s and 3d electrons; in some of the elements, an electron gets moved from the 4s to the 3d state.

**Rubidium**, (Kr)*5s

^{1}, to

**xenon**, (Kr)*5s

^{2}*4d

^{10}*5p

^{6}, work in the same way, complete with similar 5s-4d complications.

After cesium, (Xe)*6s

^{1}, and barium, (Xe)*6s

^{2}, we get the filling of the 4f electron states, from

**lanthanum**, (Xe)*6s

^{2}*5d

^{1} instead of (Xe)*6s

^{2}*4f

^{1}, to

**ytterbium**, (Xe)*6s

^{2}*4f

^{14}.

**Lutetium**, (Xe)*6s

^{2}*4f

^{14}*5d

^{1}, to

**radon**, (Xe)*6s

^{2}*4f

^{14}*5d

^{10}*6p

^{6}, work like before.

**Francium**, (Rn)*7s

^{1} to

**ununoctium**, (Rn)*7s

^{2}*5f

^{14}*6d

^{10}*7p

^{6}, work in the same way.

Periodic table (electron configurations) has a nice table with which states have how many electrons for each element.

Notice that the number of electron "slots" in each state is 2*(2*lj+1), from the 2 spin states and the (2*lj+1) orbital angular-momentum states (lj = OAM). The 2 for spin states is from (2*(1/2)+1), since electrons have spin 1/2.

Also notice the "fill order" of the state "shells":

1s

2s, 2p

3s, 3p

4s, 3d, 4p

5s, 4d, 5p

6s, 4f, 5d, 6p

7s, 5f, 6d, 7p