# If n = 2, l = 1, and m_l = 0 is desired for a given atom, how many quantum states are allowed for any such electron(s) in this atom?

Apr 7, 2017

Well, by the Pauli Exclusion Principle (no two electrons can share all four of the same quantum numbers), if we specify three out of four quantum numbers, we are stuck with only the quantum number ${m}_{s}$ as a free choice, i.e. we can only choose the spin.

• $n = 2$ corresponds to the second energy level, i.e. you either have $2 s$ or $2 p$ orbitals. We don't know yet.
• $l = 1$ corresponds to $p$ orbitals, so we are in a $2 p$ orbital, but we don't know which of the three we are in, yet.
• ${m}_{l} = 0$ by default corresponds to the $2 {p}_{z}$ atomic orbital, i.e. the $z$-directional $p$ orbital rather than the $x$- or $y$-directional $p$ orbital.

So, we are in a $2 {p}_{z}$ orbital, and thus choose which spin the electron can have (up/down). Obviously, we only have two choices, then, so we can only have:

(1) $\left(n , l , {m}_{l} , {m}_{s}\right) = \left(2 , 1 , 0 , - \frac{1}{2}\right)$

(2) $\left(n , l , {m}_{l} , {m}_{s}\right) = \left(2 , 1 , 0 , + \frac{1}{2}\right)$