The maximum number of electrons that can have principal quantum number, $n=3$ and spin quantum number, $m_s=-\frac{1}{2}$, is

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Aicharya · Accepted Answer

When $n=3, l=0,1,2$ i.e., there are $3 s, 3 p$ and $3 d$ orbitals. If all these orbitals are completely occupied as

$ \text { Total } 18 \text { electrons, } 9 \text { electrons with } s=+\frac{1}{2} \text { and } 9 \text { with } s=-\frac{1}{2} . $ Alternatively In any $n$th orbit, there can be a maximum of $2 n^2$ electrons. Hence, when $n=3$, number of maximum electrons $=18$. Out of these 18 electrons, 9 can have spin $-\frac{1}{2}$ and remaining nine with spin $=+\frac{1}{2}$

The maximum number of electrons that can have principal quantum number, $n=3$ and spin quantum number,…

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