Physics - Modern Physics Question with Solution | TestHub

$\mathrm{E}=\frac{-\left(13.6\right){\mathrm{z}}^2}{{\mathrm{n}}^2}$
$=\frac{-\left(13.6\right){\left(2\right)}^2}{2^2}=-13.60\mathrm{ }\mathrm{e}\mathrm{V}$
Total energy of$\mathrm{H}{\mathrm{e}}^{+}$ion is in its first excited state is$-13.60\mathrm{ }\mathrm{e}\mathrm{V}$. Hence, its ionization energy will be$13.60\mathrm{ }\mathrm{e}\mathrm{V}$