Atomic 'magic' numbers.

center o bass

I'm currently reading about nuclear and particle physics in B.R.Martins 'An introduction to nuclear and particle physics'.

In chapter 7.3 he introduces the shell model of the nucleus and draws an analogy to the periodic table and how we think of it being constructed of progressively placing more and more electrons in 'orbitals' given by the quantum numbers n,l and ml.

From the study of the hydrogen atom we know that the energy levels are degenerate with 2n^2. Martin further notes that if a shell or a subshell is filled then

[tex]\sum m_s = \sum m_l = 0[/tex]
which implies that
[tex]\vec L = \vec S = 0 = \vec J = \vec L + \vec S.[/tex]
He then states that 'For any atom with a closed shell or a closed sub-shell, the electrons are paired off and thus no valence electrons are avaiable. Such atoms are therefore chemically inert. It is straight forward to work out the atomic numbers at which this occurs. These are

[tex]Z = 2,10,18,36,54.[/tex]'

But I do not understand how these numbers are obtained. If one follows the 'hydrogen model' I would think that the numbers at which we have a closed shell or subshell would be

[tex] Z = 2,4,10,12,18,28, \ldots[/tex]

corresponding to the electron configurations

[tex] 1s^2, 1s^2 2s^2, 1s^2 2s^2 2p^6,1s^2 2s^2 2p^63s^2, \ldots[/tex]

What have I missunderstood here?

mfb

The strong interaction in a nucleus is a bit different from the hydrogen atom. There, electrons have a potential given by other particles (the nucleus), with an 1/r^2-law. This is not true for the strong interaction, where the interaction is short-ranged and given by other nucleons in the same volume. As a result, the energy levels depend heavily on the quantum states, and you get a nice mess with different magic numbers.

center o bass

But how, then, can it be that

?
Can these be calculated within a theoretical framework?

mfb

You can work out these numbers if you know the energy levels.
Calculating these energy levels (numerically) is not so easy, but it is possible.