The energy of an H atom in the n^{th} orbit i | vedclass

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Question

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

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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 species is given by the formula: $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.For a hydrogen atom $(H)$,$Z = 1$,so $E_n(H) = -13.6 	imes \frac{1^2}{n^2} = E_n$.For a singly ionized helium atom $(He^+)$,$Z = 2$,so $E_n(He^+) = -13.6 	imes \frac{2^2}{n^2} = -13.6 	imes \frac{4}{n^2}$.Substituting $E_n$ into the expression for $He^+$,we get $E_n(He^+) = 4 	imes (-13.6 	imes \frac{1^2}{n^2}) = 4E_n$.

The energy of an $H$ atom in the $n^{th}$ orbit is $E_n$. What will be the energy of the $n^{th}$ orbit of a singly ionized helium atom $(He^+)$?

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