Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
For a hydrogen atom in the $p$ state ($ l = 1 $), the possible values of the orbital angular momentum component along the z-axis, $ L_z $, are:
$$
L_z = m\hbar \quad \text{where} \quad m = 1, 0, -1
$$
Thus, the possible values of $ L_z $ are:
$$
L_z = \hbar, 0, -\hbar
$$
The spin angular momentum component along the $z$-axis, $ S_z $, for an electron is:
$$
S_z = \pm \frac{1}{2} \hbar
$$
Now we can calculate the total angular momentum component along the $z$-axis, $ J_z $, which is given by
$$
J_z = L_z + S_z=m_j\hbar
$$
By combining the possible values of $ L_z $ and $ S_z $, we get the following results as shown in the table below.
\begin{array}{|c|c|c|c|}
\hline
L_z & S_z& m_j & J_z \\
\hline
\\\hbar & +\dfrac{1}{2}\hbar & \dfrac{3}{2} & \dfrac{3}{2}\hbar \\\\\hline\\
\hbar & -\dfrac{1}{2}\hbar & \dfrac{1}{2} & \dfrac{1}{2}\hbar \\\\\hline\\
0 & +\dfrac{1}{2}\hbar& \dfrac{1}{2} & \dfrac{1}{2}\hbar \\\\\hline\\
0 & -\dfrac{1}{2}\hbar & -\dfrac{1}{2} & -\dfrac{1}{2}\hbar \\\\\hline\\
-\hbar & +\dfrac{1}{2}\hbar & -\dfrac{1}{2} & -\dfrac{1}{2}\hbar \\\\\hline\\
-\hbar & -\dfrac{1}{2}\hbar & -\dfrac{3}{2}& -\dfrac{3}{2}\hbar \\\\\hline
\hline
\end{array}
$$\color{blue}{\bf [b]}$$
Values for $ j = \frac{1}{2} $ and $ j = \frac{3}{2} $
For $ j = \frac{1}{2} $, the values of $ m_j $ can be:
$$
m_j = \pm \frac{1}{2}
$$
Since $ j $ can only change in integer increments, the possible values of $ J_z $ for $ j = \frac{1}{2} $ are:
$$
\boxed{J_z = \frac{1}{2} \hbar, -\frac{1}{2} \hbar}
$$
For $ j = \frac{3}{2} $, the values of $ m_j $ can be:
$$
m_j = \pm \frac{3}{2}, \pm \frac{1}{2}
$$
Thus, the possible values of $ J_z $ for $ j = \frac{3}{2} $ are:
$$
\boxed{J_z = \frac{3}{2} \hbar, \frac{1}{2} \hbar, -\frac{1}{2} \hbar, -\frac{3}{2} \hbar}
$$
These results show how the total angular momentum component $ J_z $ can take on different values depending on the combination of the orbital and spin angular momentum components.
