Given that En = −3.4 eV:
$-3.4=-\frac{13.6}{{\mathit{n}}^2}$
${\mathit{n}}^2=\frac{13.6}{3.4}=4$
$\mathit{n}=2$
For the second orbit (n = 2), the angular momentum becomes:
$\mathit{L}=\frac{2\mathit{h}}{2\mathit{\pi }}=\frac{\mathit{h}}{\mathit{\pi }}$
The angular momentum of the electron in this orbit is $\frac{\mathit{h}}{\mathit{\pi }}$