In an atom, the total number of electrons having the quantum numbers $n = 4, | m_{l} | = 1$ and $m_{s} = - \frac{1}{2}$ , is

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Solution

$| m_{l} | = 1$ thus, the magnetic quantum number $= + 1$ and $- 1$
$(m_{s}) = - \frac{1}{2}$
$n = 4$ can have suborbitals.
$\begin{matrix} l & m_{l} & m_{s} 4 s & 0 & 0 & + \frac{1}{2} - \frac{1}{2} 4 p & 1 & - 1 - - -, 0, + 1 - - - & + \frac{1}{2} - \frac{1}{2} 4 d & 2 & - 2, - 1 - - -, 0, + 1 - - -, + 2 & + \frac{1}{2} - \frac{1}{2} 4 f & 3 & - 3, - 2, - 1 - - -, 0, + 1 - - - + 2, + 3 & + \frac{1}{2} - \frac{1}{2} \end{matrix}$
Thus underlined electrons with $(m_{l} = - 1, + 1)$ and spin $- (\frac{1}{2})$ are six.

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