My textbook says that for a $d$-electron, the orbital angular momentum is: $$ \frac{\sqrt{6}h}{2\pi} $$ But since: $$ mvr = \frac{nh}{2\pi}$$ $$ n \in N$$ where $N$ is the set of natural numbers, how is the value of $n$ an irrational number?
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asked Jul 10, 2015 at 7:56
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5$\begingroup$ First, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Secondly, you confused the orbital angular momentum quantum number $l$ with the actual value of the orbital angular momentum $L$. $\endgroup$– WildcatJul 10, 2015 at 8:23
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2$\begingroup$ And surely you made a mistake in your first equation: it should be $h$ instead of $n$ here, of course. Or perhaps it was a mistake (misprint) in the book which confused you. $\endgroup$– WildcatJul 10, 2015 at 8:36
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So, first, as I said in my comment, you confused the orbital angular momentum quantum number $l$ with the principal quantum number $n$. Both can take only integer values (positive and non-negative respectively), but for totally unrelated reasons.
Secondly, the orbital angular momentum itself is related to its quantum number $l$ by the following equation, $$ \hat{L}^2 \Psi = l (l + 1) \hbar^2 \Psi \, , $$ where $\hbar^2{l(l+1)}$ is the square of the magnitude of an electron orbital angular momentum. Thus, while the orbital angular momentum quantum number $l$ is always a non-negative integer, the magnitude of an electron orbital angular momentum $\sqrt{l(l+1)} \hbar$ can take irrational values.
For instance, for a d-electron, $l=2$, and thus, the magnitude of its orbital angular momentum is $\sqrt{2(2+1)} \hbar = \sqrt{6} \hbar = \sqrt{6} h / 2 \pi$.
