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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$.The sec

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$.The second excited state corresponds to $n = 3$.The energy of the electron in the second excited state is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.The energy required to ionize the atom from this state is the energy needed to bring the electron to $n = \infty$,which is $E_{\infty} - E_3 = 0 - (-1.51) = 1.51 \ eV$.

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to ionize a hydrogen atom in its second excited state is ........... $eV$.

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