Question

In an atom, total number of electrons having quantum numbers n = 4, $|m_l|$ = 1 and $m_s$ = $-\frac{1}{2}$ is ________.

Answer 1

The quantum numbers are defined as follows:
$n = 4$: Principal quantum number.
$|m_l| = 1$: Absolute value of the magnetic quantum number.
$m_s = -\frac{1}{2}$: Spin quantum number.
Step 1: Determine the possible values of $l$ and $m_l$
For $n = 4$, the possible values of the azimuthal quantum number $l$ are:
$l = 0, 1, 2, 3.$
For each $l$, the possible values of $m_l$ are as follows:
$l = 0$: $m_l = 0$.
$l = 1$: $m_l = -1, 0, +1$.
$l = 2$: $m_l = -2, -1, 0, +1, +2$.
$l = 3$: $m_l = -3, -2, -1, 0, +1, +2, +3$.
From the given condition $|m_l| = 1$, the possible values of $m_l$ are:
$m_l = -1 \text{ or } +1.$
Step 2: Count the orbitals corresponding to $n = 4$ and $|m_l| = 1$
For $l = 1$: $m_l = \pm1$ (2 orbitals).
For $l = 2$: $m_l = \pm1$ (2 orbitals).
For $l = 3$: $m_l = \pm1$ (2 orbitals).
The total number of orbitals with $|m_l| = 1$ is:
$2 + 2 + 2 = 6 \, \text{orbitals}.$
Step 3: Assign electrons with $m_s = -\frac{1}{2}$
Each orbital can hold one electron with $m_s = -\frac{1}{2}$. Thus, the total number of electrons is:
$6 \, \text{electrons}.$
Final Answer: 6.