What will be the ionization energy in eV for | vedclass

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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 $(n=1)$,$E_1 =

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$3.4$ or less

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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 $(n=1)$,$E_1 = -13.6 \, eV$.For excited states,$n \geq 2$.For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \, eV$.Ionization energy is the energy required to remove the electron from its current state to $n = \infty$ (where $E = 0$).For $n=2$,the ionization energy is $0 - (-3.4) = 3.4 \, eV$.For higher excited states $(n > 2)$,the energy $E_n$ becomes closer to $0$,so the ionization energy $(0 - E_n)$ will be less than $3.4 \, eV$.Therefore,the ionization energy for an excited hydrogen atom is $3.4 \, eV$ or less.

What will be the ionization energy in $eV$ for an excited hydrogen atom?

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