# Total electrons with quantum number n=4 l=1 s=-1/2?

May 10, 2018

Three.

#### Explanation:

Each electron has a set of four atomic quantum numbers: [1]

• $n$, the principal quantum number,
• $l$, the orbital angular momentum quantum number,
• ${m}_{l}$, the magnetic quantum number, and
• ${m}_{s}$ or equivalently $s$, the electron spin quantum number.

By the Pauli Exclusion Principle, each electron in an atom shall have a unique set of the four quantum numbers. Therefore, the number of quantum number values combinations possible determines the number of electrons satisfying such conditions.

The question gives the value of three out of the four quantum numbers:

• $n = 4$;
• $l = 1$;
• $\textcolor{\mathrm{da} r k b l u e}{{m}_{l}}$ is yet to be determined;
• ${m}_{s} = \frac{1}{2}$

$\textcolor{\mathrm{da} r k b l u e}{{m}_{l}}$ shall be an integer between $- l$

${m}_{l} - l , - l + 1 , - l + 2 , \ldots , - 1 , 0 , 1 , 2 , \ldots , l - 1 , l$

and therefore has $l - \left(- l\right) + 1 = 2 \cdot l + 1 = \textcolor{\mathrm{da} r k b l u e}{3}$ possible values. As a result, there are a total of three possible combinations of atomic quantum numbers, and hence three electrons that satisfy the given conditions in a single atom.

Alternative Explanation
Here's an alternative approach that does not require knowledge about the connection between $l$ and ${m}_{l}$.

The $\textsf{\text{orbital}}$ angular momentum quantum number $l$ identifies sublevel the electron is located at in an atom:

• $l = \textcolor{b l u e}{0}$ corresponds to an $\textcolor{b l u e}{s}$ orbital, which contains $1$ orbit; [2]
• $l = 1$ corresponds to a $p$ orbital that holds $3$ orbits;
• $l = 2$ corresponds to a $d$ orbital which holds $5$ orbits;

and so on so forth.

The Pauli's Exclusion Principle suggests that the two electrons in a single electron orbit shall have the opposite spins, $\frac{1}{2}$ (upward-spinning) and $- \frac{1}{2}$ (downward-spinning.) However, the question is asking only for electrons that spin upwards- which have $s = \frac{1}{2}$. Also, electrons in question shall lie in the $\textcolor{\mathrm{da} r k g r e e n}{4} \textcolor{\mathrm{da} r k b l u e}{p}$ orbital- with $\textcolor{\mathrm{da} r k g r e e n}{n = 4}$ and $\textcolor{\mathrm{da} r k b l u e}{p = 1}$.

A $p$ orbital can hold at most $3$ upward-spinning electrons, giving the final answer of three.

References
[1] Kamenko, Anastasiya, et. al, "Quantum Numbers", https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Quantum_Numbers

[2] "Quantum Numbers", https://www.ck12.org/chemistry/quantum-numbers/lesson/Quantum-Numbers-CHEM/