Chapter 41 - Atomic Physics - Exercises and Problems - Page 1206: 3

$L = 4.47~\hbar$

We can find $n$: $E_n = -\frac{13.60~eV}{n^2} = -0.544~eV$ $n^2 = \frac{13.60}{0.544}$ $n = \sqrt{\frac{13.60}{0.544}}$ $n = 5$ When $n = 5$, the maximum possible value of $l$ is $l = 4$ We can find the angular momentum $L$ when $l = 4$: $L = \sqrt{l~(l+1)}~\hbar$ $L = \sqrt{(4)~(4+1)}~\hbar$ $L = 4.47~\hbar$