An electron in a multi-electron atom has ${\mathbf{m}}_{\mathbf{l}}\mathbf{=}\mathbf{+}\mathbf{4}$ For this electron, what are (a) the value of I, (b) the smallest possible value of n, and (c) the number of possible values of ${\mathbf{m}}_{\mathbf{s}}$?

An electron in a multi-electron atom has ${\mathrm{m}}_1=+4$.

For every value of the orbital quantum number, there exist (21+1) values of magnetic quantum number ranging from -I to +I. Thus, the maximum value of this magnetic quantum number is equal to the value of the orbital quantum number. Now, for every value of the principal quantum number, there are n values of I ranging from 0 to (n-1) . Thus, the smallest value of n is given by (I+1). Now, every electron has two spin orientations, thus the value of the magnetic quantum number is determined by these spins.

Formula:

The smallest possible value of n is

n =I +1 ….. (1)

The value of $\mathcal{l}$considering the table is given by,

$$\begin{gathered} \mathrm{I}={\left({\mathrm{m}}_{\mathrm{I}}\right)}_{\max } \\ =4 \end{gathered}$$

Hence, the value is 4.

Using the above value of I=4 in equation (1), we can get the smallest possible value of n as follows:

n = 4 + 1

=5

Hence, the smallest possible value of is 5.

Using the concept, we know that there are two possible spin orientations of electron,

${\mathrm{m}}_{\mathrm{s}}=\pm \frac{1}{2}$

Hence, there are two possible values.