1. We need the number of electrons possible in an atom with: $n=4, \quad |m_l|=1, \quad m_s=-\frac{1}{2}$

2. For a given principal quantum number $n=4$, the allowed values of azimuthal quantum number $l$ are: $l=0,1,2,3$

3. Now $m_l$ depends on $l$, and for each $l$: $m_l=-l,-l+1,\dots,0,\dots,l-1,l$ We need: $|m_l|=1 \Rightarrow m_l=+1 \text{ or } -1$

4. Check which $l$ values allow $m_l=\pm 1$:

   • For $l=0$: only $m_l=0$ → not allowed
   • For $l=1$: $m_l=-1,0,+1$ → allowed
   • For $l=2$: $m_l=-2,-1,0,+1,+2$ → allowed
   • For $l=3$: $m_l=-3,-2,-1,0,+1,+2,+3$ → allowed

5. So allowed $(l,m_l)$ combinations are:

   • $l=1$: $m_l=\pm 1$ → 2 combinations
   • $l=2$: $m_l=\pm 1$ → 2 combinations
   • $l=3$: $m_l=\pm 1$ → 2 combinations

   Total number of spatial states: $2+2+2=6$

6. Since $m_s=-\frac{1}{2}$ is already fixed, each such state corresponds to exactly one electron.

7. Therefore, total number of electrons is: $6$

Final Answer: $6$