The ground state energy of a hydrogen atom is | vedclass

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(A) For a hydrogen-like species,the energy is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For $He^+$,$Z = 2$. Given $E = -3.4 \ eV$,we have $-3

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$1, 3$

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(A) For a hydrogen-like species,the energy is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$.For $He^+$,$Z = 2$. Given $E = -3.4 \ eV$,we have $-3.4 = -13.6 	imes \frac{2^2}{n^2}$,which simplifies to $n^2 = 16$,so $n = 4$.Given azimuthal quantum number $\ell = 2$,the state is $4d$.Number of angular nodes $= \ell = 2$.Number of radial nodes $= n - \ell - 1 = 4 - 2 - 1 = 1$.Since $He^+$ is a single-electron species,the nuclear charge experienced is the full nuclear charge $Z = 2e$,so statement $(4)$ is incorrect.Thus,statements $(1)$ and $(3)$ are true.

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