Given that the ground state ionization energy of a hydrogen atom is $+13.6 \mathrm{~eV}$, the exact

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Given that the ground state ionization energy of a hydrogen atom is $+13.6 \mathrm{~eV}$, the exact energy of the electron when residing in the $4^{	ext{th}}$ excited state ($n=5$) of a $\mathrm{He}^+$ ion is

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(b) : The energy of an electron in a hydrogen-like ion is $E_n = -13.6 	imes \frac{Z^2}{n^2} \mathrm{~eV}$. For a Helium ion ($\mathrm{He}^+$), the atomic number $Z = 2$. The $4^{	ext{th}}$ excited state corresponds to the principal quantum number $n = 5$. Substituting these values into the formula yields $E_5 = -13.6 	imes \frac{2^2}{5^2} = -13.6 	imes \frac{4}{25} = \frac{-54.4}{25} = -2.176 \mathrm{~eV}$.