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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the ground sta

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$9: 4$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6 	ext{ eV}}{n^2}$.For the ground state,$n = 1$.The first excited state corresponds to $n = 2$.The second excited state corresponds to $n = 3$.We need the ratio of energies in the first excited state $(E_2)$ and the second excited state $(E_3)$.$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} 	ext{ eV}$.$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} 	ext{ eV}$.The ratio of the energy in the first excited state to the energy in the second excited state is $\frac{E_2}{E_3} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.Thus,the ratio is $9: 4$.

The ratio of the energies of the electron in the hydrogen atom in the first and second excited states is

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