# Question #86f7b

Jul 8, 2017

${\text{9 e}}^{-}$

#### Explanation:

The question essentially wants you to find half of the maximum number of electrons that can reside on the third energy level, i.e. the number of orbitals present on this energy level.

As you know, the spin quantum number, ${m}_{s}$, or simply $s$, which tells you the spin of the electron, can take two possible values

• ${m}_{s} = + \frac{1}{2} \to$ the electron has spin-up
• ${m}_{2} = - \frac{1}{2} \to$ the electron ahs spin-down

According to Pauli's Exclusion Principle, each individual orbital can hold a maximum of two electrons, one having spin-up and the other spin-down.

This implies that in order to find the total number of electrons that can have ${m}_{s} = - \frac{1}{2}$, all you have to do is find the total number of electrons that reside on the third energy level and divide it by $2$.

Now, the total number of electrons that can reside in a given energy level is equal to

$\text{total no. of electrons} = 2 {n}^{2}$

Here $n$, the principal quantum number, represents the energy level on which the electron resides.

$\text{total no. of electrons} = 2 \cdot {3}^{2} = 18$
This means that the number of electrons that can have ${m}_{2} = - \frac{1}{2}$ on the third energy level is equal to
${\text{18 e"^(-)/2 = "9 e}}^{-}$
In other words, the third energy level can hold $9$ orbitals.