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(A) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the ground state

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First excited state energy of $He^{+}$

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(A) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For the ground state of $H$ $(Z=1, n=1)$,$E_1 = -13.6 	imes \frac{1^2}{1^2} = -13.6 	ext{ eV}$.Now,calculate the energy for the options:$A$: $He^{+}$ $(Z=2)$,first excited state $(n=2)$: $E_2 = -13.6 	imes \frac{2^2}{2^2} = -13.6 	ext{ eV}$.$B$: $Be^{3+}$ $(Z=4)$,ground state $(n=1)$: $E_1 = -13.6 	imes \frac{4^2}{1^2} = -217.6 	ext{ eV}$.$C$: $Li^{2+}$ $(Z=3)$,first excited state $(n=2)$: $E_2 = -13.6 	imes \frac{3^2}{2^2} = -30.6 	ext{ eV}$.$D$: $Li^{2+}$ $(Z=3)$,ground state $(n=1)$: $E_1 = -13.6 	imes \frac{3^2}{1^2} = -122.4 	ext{ eV}$.Thus,the ground state energy of $H$ is equal to the first excited state energy of $He^{+}$.

The ground state energy of $H$ is equal to

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