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(C) The potential energy $(PE)$ of an electron in the $n$-th orbit is given by $PE = -27.2 	imes \frac{Z^2}{n^2} \, eV$. For $He^+$,$Z = 2$ and $n =

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$-6.8$

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(C) The potential energy $(PE)$ of an electron in the $n$-th orbit is given by $PE = -27.2 	imes \frac{Z^2}{n^2} \, eV$. For $He^+$,$Z = 2$ and $n = 2$,so $PE = -27.2 	imes \frac{2^2}{2^2} = -27.2 \, eV$. This matches the given information.The energy of an electron in the $n$-th orbit of a hydrogen atom $(Z = 1)$ is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \, eV = -13.6 	imes \frac{1^2}{n^2} \, eV$.The first excited state of a hydrogen atom corresponds to $n = 2$. Thus,$E_2 = -13.6 	imes \frac{1}{2^2} = -13.6 	imes \frac{1}{4} = -3.4 \, eV$.The question asks for the double value of this energy: $2 	imes (-3.4 \, eV) = -6.8 \, eV$.

If the potential energy of an electron in the second orbit of $He^+$ is $-27.2 \, eV$,then calculate the double value of the energy of the first excited state of a hydrogen atom in $eV$.

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