Answer
As n increases, the maximum L is $\sqrt{(n-1)(n)}\hbar $, which gets closer to the value $n \hbar$ postulated in the Bohr model.
Work Step by Step
The maximum orbital angular momentum for a given n occurs for $l=n-1$. We also know that $L=\sqrt{l(l+1)}\hbar$.
For n=2, $l_{max}=1$ and $L=\sqrt{l(l+1)}\hbar=\sqrt{2(2+1)}\hbar =1.41\hbar$
For n=20, $l_{max}=19$ and $L=\sqrt{l(l+1)}\hbar=\sqrt{19(19+1)}\hbar =19.49\hbar$
For n=200, $l_{max}=199$ and $L=\sqrt{l(l+1)}\hbar=\sqrt{199(199+1)}\hbar =199.5\hbar$
The trend is clear. As n increases, the maximum L is $\sqrt{(n-1)(n)}\hbar $, which gets closer to the value $n \hbar$ postulated in the Bohr model.
