alleryFor principal quantum number $n$, the orbital quantum number $l$ takes values
A$\pm 1/2$
B$1, 2, \ldots, n$
C$0, 1, 2, \ldots, (n-1)$
D$-n, -n+1, \ldots, n$
Answer & Solution
Correct answer: C. $0, 1, 2, \ldots, (n-1)$
$l$ goes from $0$ to $n-1$.
So $n=1$ has only $l=0$ (s)
$n=2$ has $l=0$ (s) and $l=1$ (p)
$n=3$ has s, p, d; etc.
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