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

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$4E_n$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For a hydrogen atom,the atomic number $Z = 1$,so $E_n = -13.6 	imes \frac{1^2}{n^2} = -\frac{13.6}{n^2}$.For a helium ion $(He^+)$,the atomic number $Z = 2$,so the energy $E'_{n}$ is given by: $E'_{n} = -13.6 	imes \frac{2^2}{n^2} = -13.6 	imes \frac{4}{n^2}$.Comparing the two expressions,we get: $E'_{n} = 4 	imes (-13.6 	imes \frac{1}{n^2}) = 4E_n$.Therefore,the energy of the $n^{th}$ orbit in a helium ion is $4E_n$.

If the energy of the $n^{th}$ orbit in a hydrogen atom is $E_n$,what will be the energy of the $n^{th}$ orbit in a helium ion $(He^+)$?

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