The ratio of the energies of the hydrogen atom in its first to second excited state is :
Answer & explanation
Correct answer: option 3
First excited state i.e. second orbit (n = 2)
Second excited state i.e. third orbit (n = 3)
$∵ E=-\frac{13.6}{n^2} \Rightarrow \frac{E_2}{E_3}=\left(\frac{3}{2}\right)^2=\frac{9}{4}$