Question

Why is the maximum number of (electrons, orbitals) related to each principal energy level equals `2n^2`?

Answer 1

Because of several noticed patterns:

• There are `n` orbitals in each electron "shell". For example, the `n = 2` shell has two types of orbitals, `s` and `p`.
• Of the `n` orbital types (subshells), each type (which corresponds to each `l`) has `2l + 1` number of actual orbitals.
• `l_max = n - 1`.

Because of that, at each energy level `n`, we have the following:

`n = 1`: One `1s` orbital, because `2(0) + 1 = bb(1)`

`n = 2`: One `2s`, three `2p` orbitals, because `[2(0) + 1] + [2(1) + 1] = bb(4)`.

`n = 3`: One `3s`, three `3p`, five `3d` orbitals, because `[2(0) + 1] + [2(1) + 1] + [2(2) + 1] = bb(9)`.

...etc.

As a general rule then, the total number of orbitals in an electron "shell" is:

`bb(n_"orbs") = sum_(l = 0)^(l_max) (2l + 1) = bb(sum_(l = 0)^(n - 1) (2l + 1))`

If we work this out and turn it into a simpler form:

`=> (2*0 + 1) + (2*1 + 1) + (2*2 + 1) + . . . + (2l_max + 1)`

`= (2*0) + (2*1) + (2*2) + . . . + (2l_max) + n`

`= 2(0 + 1 + 2 + 3 + . . . + l_max) + n`

Now if we realize that the sum of the natural numbers is the last number (`l_max`) plus the next number (`l_max + 1`), then the quantity divided by `2`, we have:

`=> cancel(2)([l_max*(l_max + 1)]/cancel(2)) + n`

`= l_max*(l_max + 1) + n`

Now substitute `l_max = n - 1` to get:

`= (n - 1)*((n - 1) + 1) + n`

`= n^2 - n + n`

`=> color(blue)(n_"orbs" = n^2)`

Therefore, the total number of orbitals in one quantum level is `n^2`.

Since the maximum number of electrons in each orbital is `2`, the maximum number of electrons in an entire quantum level is `color(blue)(2n^2)`.