# How many electrons have n = 3, m_l = 0, and m_s = +1/2?

Oct 4, 2015

Three electrons.

#### Explanation:

The idea here is that each electron in an atom has an unique set of quantum numbers, which describes exclusively that electron.

The four quantum numbers are In your case, the only quantum number that is not accounted for is the angular momentum quantum number, $l$, which can take values from $0$ to $\left(n - 1\right)$, where $n$ is the principal quantum number.

So, the question wants you to find the number of electrons that

• can be located on the third energy level, since $n = 3$;
• can be located in an orbital that has a specific orientation, since ${m}_{l} = 0$
• have spin-up, since ${m}_{s} = + \frac{1}{2}$

Start by finding how many orbitals can accomodate these electrons. SInce you know what the relationship of $l$ is to $n$, you can say that

l = 0, 1, ..., (n-1) implies l = {0; 1; 2}

For $l = 0$, which corresponds to the s-orbital, you can have one electron located in the 3s-orbital that has spin-up.

For $l = 1$, which corresponds to the 3p-orbitals, you can have an electron located in the $3 {p}_{y}$ orbital, for which ${m}_{l} = 0$, that has spin-up.

Finally, for $l = 2$, which corresponds to the 3d-orbitals, you can have an electron located in the $3 {d}_{y x}$, which would correspond to ${m}_{l} = 0$, that has spin-up.

Therefore, these quantum numbers can correspond to a total of three electrons

• $n = 3 \text{ " l=0" "m_; = 0" } {m}_{s} = + \frac{1}{2}$
• $n = 3 \text{ "l=1" "m_l = 0" } {m}_{s} = + \frac{1}{2}$
• $n = 3 \text{ "l=2" "m_l = 0" } {m}_{s} = + \frac{1}{2}$