Filling Atomic Shells

Every electron in an atom is fully described by four quantum numbers: the principal quantum number $n$ , the orbital angular momentum quantum number $\ell$ , the magnetic quantum number $m_\ell$ , and the spin projection $m_s$ . The Pauli exclusion principle forbids two electrons from sharing the same set of all four. Counting how many distinct combinations are available for a given $n$ tells us the capacity of an atomic shell — the maximum number of electrons it can hold.

The quantum numbers in brief

For a given $n$ , the allowed values of $\ell$ are

$\ell = 0, 1, 2, \ldots, n-1.$

For each $\ell$ , the magnetic quantum number $m_\ell$ takes the $2\ell + 1$ values

$m_\ell = -\ell,\; -\ell+1,\; \ldots,\; \ell-1,\; \ell.$

Finally, the spin projection $m_s$ takes exactly two values, $+\tfrac{1}{2}$ and $-\tfrac{1}{2}$ , regardless of the other quantum numbers. Together, each $\ell$ value defines a subshell of capacity

$N_\ell = 2\,(2\ell + 1).$

The factor $2\ell+1$ counts the $m_\ell$ states; the factor of $2$ accounts for the two spin orientations.

Counting the capacity of a shell

The $n$ -th shell contains all subshells from $\ell = 0$ to $\ell = n-1$ . Its total electron capacity is the sum

$N_n = \sum_{\ell=0}^{n-1} 2\,(2\ell + 1) = 2 \sum_{\ell=0}^{n-1} (2\ell + 1).$

The inner sum is the sum of the first $n$ odd integers, which equals $n^2$ :

$\sum_{\ell=0}^{n-1} (2\ell + 1) = 1 + 3 + 5 + \cdots + (2n-1) = n^2.$

Therefore

$\boxed{N_n = 2n^2.}$

This compact formula is a direct consequence of the quantum number rules and the Pauli exclusion principle. It governs the period lengths of the periodic table.

Subshell notation and the periodic table

Chemists label subshells by letter: $\ell = 0$ is called s, $\ell = 1$ is p, $\ell = 2$ is d, and $\ell = 3$ is f. The capacity of each subshell is:

| Subshell | $\ell$ | $m_\ell$ values | Capacity $2(2\ell+1)$ | |----------|--------|-----------------|----------------------| | s | 0 | $0$ | 2 | | p | 1 | $-1, 0, +1$ | 6 | | d | 2 | $-2, -1, 0, +1, +2$ | 10 | | f | 3 | $-3, \ldots, +3$ | 14 |

The structure of the periodic table follows directly: the first period has $2 \times 1^2 = 2$ elements (H and He), the second and third periods each accommodate $2 \times 2^2 = 8$ elements (though the filling order for higher $n$ mixes shells due to electron-electron repulsion), and the fourth and fifth periods each span 18 elements when the $3d$ and $4d$ subshells are included.

Worked example: the $n = 2$ shell

For $n = 2$ , the allowed subshells are $\ell = 0$ (the 2s) and $\ell = 1$ (the 2p):

2s: $2(2 \times 0 + 1) = 2$ electrons.
2p: $2(2 \times 1 + 1) = 6$ electrons.
Total: $2 + 6 = 8 = 2 \times 2^2$ , consistent with the formula.

The eight elements in the second period of the periodic table (lithium through neon) correspond to filling these eight states.

Try it

This is a numerical exercise — your code should return a number. Compute the total number of electrons that can fill the $n = 3$ shell by summing the capacity of each subshell ( $\ell = 0$ , $\ell = 1$ , and $\ell = 2$ ). Each subshell $\ell$ holds $2(2\ell + 1)$ electrons.

Run your code to see the quantum state.