# How many electrons are in n=3, l= 2?

Jun 12, 2017

${\text{10 e}}^{-}$

#### Explanation:

The idea here is that you must use the value of the angular momentum quantum number, $l$, which tells you the energy subshell in which an electron resides, to find the possible values of the magnetic quantum number, ${m}_{l}$.

The number of values that the magnetic quantum number can take tells you the number of orbitals that are present in a given subshell.

So, you know that the magnetic quantum number depends on the value of the angular momentum quantum number

${m}_{l} = \left\{- l , - \left(l - 1\right) , \ldots , - 1 , \textcolor{w h i t e}{-} 0 , + 1 , \ldots , + \left(l - 1\right) , + l\right\}$

In your case, you have $l = 2$, which is an accepted value for the angular momentum quantum number given the fact that the principal quantum number, $n$, is equal to $3$, so you can say that

${m}_{l} = \left\{- 2 , - 1 , \textcolor{w h i t e}{-} 0 , + 1 , + 2\right\}$

This tells you that the $d$ subshell, which is denoted by $l = 2$, holds a total of $5$ orbitals.

Since each orbital can hold a maximum of $2$ electrons, one having spin-up and one having spin-down, you can say that you have

5 color(red)(cancel(color(black)("orbitals"))) * "2 e"^(-)/(1color(red)(cancel(color(black)("orbital")))) = "10 e"^(-)

Therefore, a maximum number of $10$ electrons can share these two quantum numbers in an atom.

$n = 3 , l = 2$

These electrons are located on the third energy level, in the $3 d$ subshell.