By putting the value of \(n,\) we get the value of energy.
Thus, the energy of electron in first excited state of hydrogen atom \(=-\frac{13.6}{n^{2}}=-\frac{13.6}{4}=-3.4 \mathrm{eV}\).
[ઉપયોગ: $\left.{h}=6.63 \times 10^{-34}\, {Js}, {m}_{{e}}=9.0 \times 10^{-31}\, {~kg}\right]$