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

(A) The total number of electrons in a shell with principal quantum number $n$ is $2n^{2}$.
For $n=4$,the total number of electrons is $2(4)^{2} = 32$.
In a neutral atom with a filled shell,half of the electrons have spin $m_{s}=+\frac{1}{2}$ and half have spin $m_{s}=-\frac{1}{2}$.
Therefore,the number of electrons with $n=4$ and $m_{s}=-\frac{1}{2}$ is $\frac{32}{2} = 16$.
$(b)$ The quantum numbers $n=3$ and $l=0$ correspond to the $3s$ orbital.
Since an $s$ orbital can hold a maximum of $2$ electrons,the number of electrons with $n=3$ and $l=0$ is $2$.