# Question #7c3c0

Dec 8, 2016

Here's what I got.

#### Explanation:

I'm assuming that you're supposed to figure out how many electrons can share the two quantum numbers given to you

$n = 4 \text{ }$ and $\text{ } l = 0$

As you know, we use four quantum numbers to describe the location and spin of an electron in an atom. The principal quantum number, $n$, describes the energy level on which the electron resides. In other words, the value of $n$ tells you the energy shell on which you can find the electron.

In your case, $n = 4$ means that the electron is located on the fourth energy level.

The angular momentum quantum number, $l$, describes the subshell in which the electron resides. More specifically, you have

• $l = 0 \to$ the s subshell
• $l = 1 \to$ the p subshell
• $l = 2 \to$ the d subshell
• $l = 3 \to$ the f subshell

In your case, $l = 0$ means that the electron is located in the s subshell.

Now, the maximum number of electrons that can occupy the s subshell, regardless of the energy level on which they're located, is equal to $2$.

This is the case because the s subshell can only hold one orbital, as given by the magnetic quantum number, ${m}_{l}$, which for $l = 0$ is equal to $0$.

Moreover, each individual orbital can hold a maximum of $2$ electrons, one having spin-up, or ${m}_{s} = + \frac{1}{2}$, and the other having spin-down, or ${m}_{s} = - \frac{1}{2}$.

Therefore, you can say that a maximum of $2$ electrons can share the quantum numbers

$n = 4 \text{ }$ and $\text{ } l = 0$