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

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Solution

$| m_{l} | = 1$
means $m_{l}$ can be $+ 1 and - 1$
So, for n = 4 six orbitals are possible and bear six electrons with $s = \frac{1}{2}$

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n = 4,

m_{l}

m_{s}

n = 4, | m_{l} | = 1

m_{s} = - \frac{1}{2}

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m_{s} = - \frac{1}{2}

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