How many electrons in an atom can have the following quantum numbers?
$(a)$ $n = 4, m_s = -\frac{1}{2}$
$(b)$ $n = 3, l = 0$

(A) For a given principal quantum number $n$,the total number of electrons is given by $2n^2$.
For $n = 4$,total electrons $= 2(4)^2 = 2 \times 16 = 32$.
In any shell,half of the electrons have $m_s = +\frac{1}{2}$ and half have $m_s = -\frac{1}{2}$.
Therefore,the number of electrons with $m_s = -\frac{1}{2}$ is $32 / 2 = 16$.
$(b)$ For $n = 3$ and $l = 0$,the orbital is $3s$.
$A$ single orbital can hold a maximum of $2$ electrons.
Therefore,the number of electrons is $2$.