The ratio of the energies of the hydrogen ato | vedclass

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the first excited state, t

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$9/4$

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(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the first excited state, the principal quantum number is $n_1 = 2$.For the second excited state, the principal quantum number is $n_2 = 3$.The energy in the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} 	ext{ eV}$.The energy in the second excited state is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} 	ext{ eV}$.The ratio of the energies is $\frac{E_2}{E_3} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.

The ratio of the energies of the hydrogen atom in its first to second excited state is

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