Electrons Satisfying Specific Quantum Number Conditions in an Atom

To determine the total number of electrons in an atom satisfying the given quantum numbers: $n = 4, \quad |m_l| = 1, \quad \text{and} \quad m_s = -\frac{1}{2}$

we analyze the possible quantum states step-by-step:

1. Principal Quantum Number ($n = 4$): The possible values of the azimuthal quantum number ($l$) for $n = 4$ are: $l = 0, 1, 2, 3 \quad (\text{corresponding to } 4s, 4p, 4d, \text{ and } 4f \text{ subshells})$

2. Magnetic Quantum Number ($m_l$): The condition $|m_l| = 1$ implies that $m_l$ can take the values: $m_l = +1 \quad \text{or} \quad m_l = -1$

   Now, we check which subshells contain orbitals with $m_l = \pm 1$:

   • For $l = 0$ ($4s$ subshell): $m_l = 0$ only. No orbitals satisfy $|m_l| = 1$.
   • For $l = 1$ ($4p$ subshell): $m_l \in \{-1, 0, +1\}$. There are $2$ orbitals with $|m_l| = 1$ ($m_l = -1, +1$).
   • For $l = 2$ ($4d$ subshell): $m_l \in \{-2, -1, 0, +1, +2\}$. There are $2$ orbitals with $|m_l| = 1$ ($m_l = -1, +1$).
   • For $l = 3$ ($4f$ subshell): $m_l \in \{-3, -2, -1, 0, +1, +2, +3\}$. There are $2$ orbitals with $|m_l| = 1$ ($m_l = -1, +1$).

   Thus, the total number of orbitals with $n = 4$ and $|m_l| = 1$ is: $\text{Total orbitals} = 2 + 2 + 2 = 6$

3. Spin Quantum Number ($m_s = -1/2$): In any single orbital, there is exactly $1$ electron with a spin quantum number of $m_s = -\frac{1}{2}$.

4. Total Number of Electrons: $\text{Total electrons} = (\text{Total orbitals with } n = 4, |m_l| = 1) \times (\text{Electrons per orbital with } m_s = -1/2)$ $\text{Total electrons} = 6 \times 1 = 6$