Atomic Structure - Result Question 95
####71. The maximum number of electrons that can have principal quantum number, $n=3$ and spin quantum number, $m _s=-1 / 2$, is
(2011)
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Answer:
Correct Answer: 71. $(22.8 nm)$
Solution:
- 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
\section*{| $1 L$ | $1 L$ | $1 L$ | $1 L$ | | :— | :— | :— | :— | :— | :— | :— | :— |}
Total 18 electrons, 9 electrons with $s=+\frac{1}{2}$ and 9 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 $+\frac{1}{2}$.
