# Question #6f111

Nov 7, 2017

$\text{9 electrons}$

#### Explanation:

The trick here is to realize that you're actually looking for the number of orbitals that are present on the third energy level in a given atom.

This is the case because, as you know, each orbital can hold a maximum of $2$ electrons of opposite spins. In other words, for each orbital, you have one electron that has a spin quantum number, ${m}_{s}$, equal to $+ \frac{1}{2}$ and one electron that has a spin quantum number equal to $- \frac{1}{2}$.

Since you're only looking for the number of electrons that have spin-up, you need to take the total number of electrons present per orbital and divide it by $2$. This, of course, gives you the number of orbitals.

As you know, the total number of orbitals present on an energy level described by a principal quantum number $n$ is given by

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{no. of orbitals} = {n}^{2}}}}$

$n = 3$

which means that the third energy level can hold a maximum of

$\text{no. of orbitals} = {3}^{2}$

$\text{no. of orbitals} = 9$

Since each orbital can hold a single electron that has ${m}_{s} = + \frac{1}{2}$, you can say that a total of $9$ electrons can have

$n = 3 , {m}_{s} = + \frac{1}{2}$

in a given atom.