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

May 26, 2017

Nine.

$\underline{\downarrow \textcolor{w h i t e}{\uparrow}} \text{ "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" } \underline{\downarrow \textcolor{w h i t e}{\uparrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" }}$
$\text{ "" "" "" "" "" } 3 d$

$\underline{\downarrow \textcolor{w h i t e}{\uparrow}} \text{ "ul(darr color(white)(uarr))" } \underline{\downarrow \textcolor{w h i t e}{\uparrow}}$
$\underbrace{\text{ "" "" "" "" "" "" "" }}$
$\text{ "" "" } 3 p$
$\underline{\downarrow \textcolor{w h i t e}{\uparrow}}$
$3 s$

Well, $l$ runs from $0$ to $n - 1$, i.e. ${l}_{\max} = n - 1$. Thus, we have access to $l = 0 , 1 , . . . \left(3 - 1\right)$, which are the $s , p ,$ and $d$ orbitals.

For each $l$, the degeneracy of the orbital set is $2 l + 1$, so $l$ and the magnetic quantum number ${m}_{l}$ are and correspond to...

• $l = 0$, ${m}_{l} = \left\{0\right\}$,
number of $s$ orbitals = $2 \left(0\right) + 1 = \boldsymbol{1}$

• $l = 1$, ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$,
number of $p$ orbitals = $2 \left(1\right) + 1 = \boldsymbol{3}$

• $l = 2$, ${m}_{l} = \left\{- 2 , - 1 , 0 , + 1 , + 2\right\}$,
number of $d$ orbitals = $2 \left(2\right) + 1 = \boldsymbol{5}$

(recalling that ${m}_{l} = \left\{- l , . . . , 0 , . . . , + l\right\}$)

Hence, $1 + 3 + 5 = \boldsymbol{9}$ orbitals of $n = 3$.

The spin quantum number of $- \frac{1}{2}$ 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

$\underline{\downarrow \textcolor{w h i t e}{\uparrow}}$ or $\underline{\downarrow \uparrow}$

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

$\underline{\downarrow \downarrow}$ or $\underline{\uparrow \uparrow}$

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 $\boldsymbol{9}$ such electrons that can have ${m}_{s} = - \frac{1}{2}$ and $n = 3$, occupying the $\boldsymbol{9}$ total $n = 3$ orbitals with ${m}_{l} = \left\{0\right\} , \left\{- 1 , 0 , + 1\right\} , \left\{- 2 , - 1 , 0 , + 1 , + 2\right\}$.