Physics - Modern Physics Question with Solution | TestHub

For hydrogen like atoms like ${\mathrm{He}}^{+},{\mathrm{Li}}^{++}$etc, the energy in an $n^{\text{th }}$ orbit is given by $E_n=\frac{-Z^{2}13.6}{n^2}\mathrm{eV}$ For ${\mathrm{He}}^{+},Z=2$ $\therefore E_1=\frac{-(2{)}^{2}13.6}{(1{)}^2}\mathrm{eV}=-54.4\mathrm{eV}$