Question

How many electrons have quantum numbers `(n,m_s) = (3,-1/2)`?

Answer 1

Nine.

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`underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")`
`" "" "" "" "" "" "3d`

`ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))`
`underbrace(" "" "" "" "" "" "" "" ")`
`" "" "" "3p`
`ul(darr color(white)(uarr))`
`3s`

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Well, `l` runs from `0` to `n - 1`, i.e. `l_(max) = n-1`. Thus, we have access to `l = 0,1, . . . (3-1)`, which are the `s,p,` and `d` orbitals.

For each `l`, the degeneracy of the orbital set is `2l+1`, so `l` and the magnetic quantum number `m_l` are and correspond to...

• `l = 0`, `m_l = {0}`,
  number of `s` orbitals = `2(0) + 1 = bb1`

• `l = 1`, `m_l = {-1,0,+1}`,
  number of `p` orbitals = `2(1) + 1 = bb3`

• `l = 2`, `m_l = {-2,-1,0,+1,+2}`,
  number of `d` orbitals = `2(2) + 1 = bb5`

(recalling that `m_l = {-l, . . . , 0, . . . , +l}`)

Hence, `1+3+5 = bb9` orbitals of `n = 3`.

The spin quantum number of `-1/2` designates spin-down, and by the Pauli Exclusion Principle, one can only have one kind of electron spin per electron in each orbital. In other words, either

`ul(darr color(white)(uarr))` or `ul(darr uarr)`

are allowed (the choice of spin for the first electron is arbitrary), but

`ul(darr darr)` or `ul(uarr uarr)`

is not.

Therefore, the total number of configurations for one spin-down electron in each possible orbital is the same number as the total number of orbitals.

In other words, there are `bb(9)` such electrons that can have `m_s = -1/2` and `n = 3`, occupying the `bb(9)` total `n = 3` orbitals with `m_l = {0}, {-1,0,+1}, {-2,-1,0,+1,+2}`.