# Question #0d46c

Feb 2, 2018

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

#### Explanation:

The first thing to mention here is that the coefficient added to the name of the energy subshell tells you the energy shell in which the subshell is located.

In other words, the coefficient tells you the energy level on which the electron is located inside the atom. This energy level corresponds to the principal quantum number, $n$.

In your case, you have the $\textcolor{red}{1} s$ subshell, so

$n = \textcolor{red}{1}$

Next, the letter added to the name of the subshell, which actually gives the identity of the subshell, corresponds to the angular momentum quantum number, $l$.

For a given value of $n$, the angular momentum quantum number can take the values

$l = \left\{0 , 1 , 2 , \ldots , n - 1\right\}$

$n = \textcolor{red}{1} \implies l = 0$

This corresponds to the $s$ subshell, as shown by the letter $s$ present in the name of the subshell.

Now, each subshell can hold a number of orbitals given by the number of values that the magnetic quantum number, ${m}_{l}$, can take.

For a given subshell $l$, you have

${m}_{l} = \left\{- 1 , - \left(l - 1\right) , \ldots , - 1 , 0 , 1 , \ldots , \left(l - 1\right) , l\right\}$

In this case, the $s$ subshell can hold a single orbital, the $s$ orbital, because

$l = 0 \implies {m}_{l} = 0$

Finally, the spin quantum number, ${m}_{s}$, tells you the spin of the electron inside an orbital. Each orbital can hold a maximum of two electrons of opposite spins as stated by Pauli's Exclusion Principle, so the spin quantum number can take one of two possible values, the positive value being assigned to a spin-up electron and the negative value being assigned to a spin-down electron.

${m}_{s} = \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

This means that you can put together two quantum number sets to describe an electron located in the $\textcolor{red}{1} s$ subshell.

$n = \textcolor{red}{1} , l = 0 , {m}_{l} = 0 , {m}_{s} = + \frac{1}{2}$

This electron is located in the first energy shell, in the $1 s$ subshell, in the $1 s$ orbital, and has spin-up.

$n = \textcolor{red}{1} , l = 0 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$

This electron is located in the first energy shell, in the $1 s$ subshell, in the $1 s$ orbital, and has spin-down