\(E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV}\)

For ground state, \(n=1\)

\(\therefore \,\,{E_1} = \frac{{ - 13.6}}{{{1^2}}}\,{\text{ = }}\,{\text{ - }}\,{\text{13}}{\text{.6}}\,{\text{eV}}\)

For first excited state, \(n=2\)

\(\therefore \quad E_{2}=\frac{-13.6}{2^{2}}=-3.4 \mathrm{eV}\)

Kinetic energy of an electron in the first excited state is

\(K=-E_{2}=3.4 \mathrm{eV}\)