Physics - Modern Physics Question with Solution | TestHub

Energy E of an atom with principal quantum number $\mathrm{n}$ is given by $E=\frac{-13.6}{{\mathrm{n}}^2}{Z}^2$ for first excited state $\mathrm{n}=2$ and for ${\mathrm{He}}^{+}\mathrm{Z}=2$ $$\begin{array}{rl}\Rightarrow \mathrm{E}& =\frac{-136\times (2{)}^2}{(2{)}^2}\\ & =-13.6\mathrm{eV}\end{array}$$