# Question f0e05

Jun 22, 2016

Here's my take on this.

#### Explanation:

I assume that you're interested in finding out how many electrons can share those two quantum numbers

$n = 4 \text{ }$ and $\text{ } {m}_{s} = - \frac{1}{2}$

The idea here is that you can use the fact that the number of orbitals each energy level can hold is given by the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{no. of orbitals} = {n}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$n$ - the principal quantum number, the quantum number that gives you the energy level on which the electron resides

In your case, the fourth energy level will have a total of

$\text{no. of orbitals} = {4}^{2} = 16$

This tells you that the fourth energy level can hold a total of $16$ orbitals. Since each orbital can hold a maximum of $2$ electrons, it follows that the maximum number of electrons that can share the quantum number $n = 4$ is

${\text{no. of e"^(-) = 2 xx "16 orbitals" = "32 e}}^{-}$

Now, half of these electrons will have spin-up, which is given by ${m}_{s} = + \frac{1}{2}$, and the other half will have spin-down, which is given by ${m}_{s} = - \frac{1}{2}$.

Since you're interested in the number of electrons that have $n = 4$ and ${m}_{s} = - \frac{1}{2}$, you will get

"no. of electrons" = "32 e"^(-)/2 = color(green)(|bar(ul(color(white)(a/a)color(black)("16 e"^(-))color(white)(a/a)|)))#