# What are the possible values of quantum number ℓ when n =1 ?

Mar 22, 2018

$l = 0$

#### Explanation:

The relationship between the possible values of the angular momentum quantum number, $l$, and the principal quantum number, $n$, is given by

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l = \left\{0 , 1 , 2 , \ldots , n - 1\right\}}}}$

For a given quantum number $n$, the angular momentum quantum number can take $n$ possible values that range from $0$ to $n - 1$.

In your case, you have $n = 1$, which means that the angular momentum quantum number can take $1$ possible value.

$l = 0$

This tells you that the first energy shell can hold a single energy subshell, the $s$ subshell.

$n = 1 , l = 0$

The electron is located in the first energy shell, in the $1 s$ subshell.

Consequently, the first energy shell can hold a single orbital, the $1 s$ orbital, because the magnetic quantum number, ${m}_{l}$, which tells you the number of distinct orbitals present in a given subshell, can take one possible value.

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

So for $l = 0$, you have

${m}_{l} = 0 \to$ the $s$ orbital

And so

$n = 1 , l = 0 , {m}_{l} = 0$

The electron is located in the first energy level, in the $1 s$ subshell, and in the $1 s$ orbital.