Calculate the energy of a $H{e}^{+}ion$ in its first excited state solution

Solution: The energy is $E_0=\frac{-Rhc{L}^2}{n^2}=-\frac{\left(13.6eV\right)Z^2}{n^2}$
For a $H{e}^{+}ion,Z=2$ and the for the first excited state $n=2$ so that the energy of $H{e}^{+}$ ion in the first excited state is $-13.6eV$