# Is it true that for a value of n (n-principal quantum number), the value of m_l (m_l-magnetic quantum number) is equal to n^2? If it is true, can you please explain this to me? Thanks.

Jul 6, 2015

The actual value, no. The number of values, yes.

#### Explanation:

The relationship between the principal quantum number, $n$, and the magnetic quantum number, ${m}_{l}$, actually goes through the angular momentum quantum number, $l$.

The angular momentum quantum number can take any value that ranges from $0$ to $n - 1$

$l = 0 , 1 , 2 , \ldots , \left(n - 1\right)$

The magnetic quantum number can take any value that ranges from $- l$ to $l$

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

If you take into account the possible values of $l$, the magnetic quantum number can thus have values that range from

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

So, for example, if $n = 2$, ${m}_{l}$ can be equal to

${m}_{l} = 0$ $\to$ for $l = 0$

and

${m}_{l} = \left\{- 1 , 0 , 1\right\}$ $\to$ for $l = 1$

As you can see, the individual values of ${m}_{l}$ don't even come close to the value of ${n}^{2}$.

However, the total number of values ${m}_{l}$ can take, if you take into account the values of $l$, is indeed equal to ${n}^{2}$.

The magnetic quantum number actually tells you how many orbitals a subshell has. In the above example, if $n = 2$, you get

$n = 2$, $l = 0$, ${m}_{l} = 0$ $\to$ one 2s-orbital;

{:(n=2, l=1, m_l = -1->2p_x), (n=2, l=1, m_l = 0->2p_y), (n=2, l=1, m_l = 1->2p_z) :}} $\to$ three 3p-orbitals.

This means that the second energy level has a total of two subshells, the 2s-subshell and the 2p-subshell.

The number of orbitals each subshell contains is given by ${m}_{l}$. The 2s-subshell contains one orbital, since you only have one possible value for ${m}_{l}$.

The 2p-subshell contains 3 orbitals, each denoted by a different value of ${m}_{l}$.

The total number of orbitals in the second energy level is 4, which is equal to ${n}^{2}$.

If you try this with $n = 3$, you'll find that you get a total number of $9$ orbitals

• one 3s-orbital;
• three 3p-orbitals;
• five 3d-orbitals.