Physics - Modern Physics Question with Solution | TestHub

The energy levels of the hydrogen atom are given by $E_n=-\frac{13.6}{n^2}$ eV,

where $n$ is the principal quantum number.

The most portable value of resonance energy corresponds to the transition from the ground state

($n=1$) to the first excited state ($n=2$).

The energy difference is

$E_2-E_1=-13.6(\frac{1}{2^2}-\frac{1}{1^2})=-13.6(\frac{1}{4}-1)=-13.6(-\frac{3}{4})=10.2$ eV.

Therefore, the answer is 10.2 eV.