Physics - Modern Physics Question with Solution | TestHub

Given, ground state energy of hydrogen atom $$E_1=-13.6\mathrm{eV}$$ Energy of electron in first excited state $$\begin{array}{l}\text{ (ie, }n=2\text{ ) }\\ E_2=-\frac{13.6}{(2{)}^2}\mathrm{eV}\end{array}$$ Therefore, excitation energy $$\begin{array}{rl}\Delta E& =E_2-E_1\\ & =-\frac{13.6}{4}-(-13.6)\\ & =-3.4+13.6=10.2\mathrm{eV}\end{array}$$