Explanation:
Energy of the electron in the \({n^{\text {th }}}\) orbit is \({E_{n}=\dfrac{E_{1}}{n^{2}}=\dfrac{-13.6}{n^{2}} {eV}}\)
For \({n=2, E_{2}=\dfrac{-13.6}{4} {eV}=-3.4 {eV}}\)
Angular momentum of the electron in the \({n^{\text {th }}}\) orbit is
\(L = mvr = \frac{{nh}}{{2\pi }} \Rightarrow L = \frac{{2h}}{{2\pi }} = \frac{h}{\pi }\,\,\,\,\,\,\,\,\,\,(n = 2)\).
So correct option is (1)