For ionising an excited hydrogen atom,the req | vedclass

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Question

(A) The ionization energy $(IE)$ of an atom is the minimum energy required to remove an electron from the atom to infinity. For the ground state $(n=1

Vedclass · Accepted Answer

$3.4$ or less

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(A) The ionization energy $(IE)$ of an atom is the minimum energy required to remove an electron from the atom to infinity. For the ground state $(n=1)$,the ionization energy is $IE_1 = E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$. The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$. For an excited state,$n \ge 2$. For the first excited state $(n=2)$,$IE_2 = |E_2| = \frac{13.6}{2^2} = 3.4 \ eV$. For the second excited state $(n=3)$,$IE_3 = |E_3| = \frac{13.6}{3^2} = 1.51 \ eV$. As $n$ increases,the ionization energy decreases. Therefore,for any excited state $(n \ge 2)$,the required energy will be $3.4 \ eV$ or less.

For ionising an excited hydrogen atom,the required energy in $eV$ will be -

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