Answer
$L_{\text {net }, z}=\frac{3.5 h}{2 \pi} .$
Work Step by Step
From formula the $z$ component of the net angular momentum of the electron is given by
$
L_{\mathrm{net}, z}=L_{\mathrm{orb}, z}+L_{s, z}=\frac{m_i h}{2 \pi}+\frac{m_s h}{2 \pi} .
$
For the maximum value of $L_{\text {net, } z}$ let $m_{\ell}=\left[m_{\ell}\right]_{\max }=3$ and $m_s=\frac{1}{2}$. Thus
$
L_{\text {net }, z}=\left\{3+\frac{1}{2}\right) \frac{h}{2 \pi}=\frac{3.5 h}{2 \pi} .
$
