If an electron in an atom has orbital angular momentum with values limited by 3, how many values of (a)${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$and (b)${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$can the electron have? In terms of h, m, and e, what is the greatest allowed magnitude for (c)${\mathit{L}}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$and (d)${\mathit{\mu }}_{\mathbf{o}\mathbf{r}\mathbf{b}\mathbf{,}\mathbf{z}}$? (e) What is the greatest allowed magnitude for the z component of the electron’s net angular momentum (orbital plus spin)? (f) How many values (signs included) are allowed for the z component of its net angular momentum?

By using the concept of quantum numbers, we can find the number of values and greatest allowed values.

Formulae:

Number of different values of $m_l$is given by $(2l+1)$.

The angular momentum is,

$L_{orb,z}=\frac{m_{l}h}{2\pi }$

For the given value of $l,m_l$varies from $–l$to $l$. Thus, in this case, $l=3$, and the number of different values of $m_l$is

$2l+1=2\left(3\right)+1=7$

So, there are seven different values of $L_{orb,z}$.

Similarly, since ${\mu }_{orb,z}$is directly proportional to $m_l$, there are total seven different values of${\mu }_{orb,z}$.

As you know,

$L_{orb,z}=\frac{m_{l}h}{2\pi }$

So, the greatest allowed value of $L_{orb,z}$is given by

$\frac{{\left|m_l\right|}_{maximum}h}{2\pi }=\frac{3h}{2\pi }$

Greatest allowed magnitude of $L_{orb,z}$is$\frac{3h}{2\pi }$.

Since ${\mu }_{orb,z}=-m_l\mu B$, the greatest allowed value of ${\mu }_{orb,z}$is given by

${\left|m_l\right|}_{maximum}\mu B=\frac{3eh}{4\pi m_e}$

Greatest allowed magnitude of ${\mu }_{orb,z}$is $\frac{3eh}{4\pi m_e}$

The z component of the net angular momentum of the electron is given by

$\begin{array}{rcl}{L}_{net,z}& =& L_{orb,Z}+L_{s,Z}\\ & =& \frac{m_{l}h}{2\pi }+\frac{m_{s}h}{2\pi }\end{array}$

From the given value, Thus, role="math" localid="1663054632477" $m_l=3$and $m_s=\frac{1}{2}$.

$\begin{array}{rcl}{\left[L_{net,z}\right]}_{maximum}& =& \frac{\left(3+\frac{1}{2}\right)h}{2\pi }\\ & =& \frac{3.5h}{2\pi }\end{array}$

Hence. the greatest allowed magnitude for z component of the electron’s net angular momentum is$\frac{3.5h}{2\pi }$.