Chapter 41 - Atomic Physics - Exercises and Problems - Page 1207: 5

$E = -0.378~eV$ $L = 3.46~\hbar$

In the $6f$ state, then $n = 6$ We can find the energy $E$: $E = -\frac{13.60~eV}{n^2}$ $E = -\frac{13.60~eV}{6^2}$ $E = -0.378~eV$ In the $6f$ state, then $l = 3$ We can find the value of $L$: $L = \sqrt{l~(l+1)}~\hbar$ $L = \sqrt{(3)~(3+1)}~\hbar$ $L = 3.46~\hbar$