The energy of a hydrogen atom in the n^{th} o | vedclass

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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: $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For a hy

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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: $E_n = -13.6 \frac{Z^2}{n^2} 	ext{ eV}$.For a hydrogen atom,the atomic number $Z = 1$,so $E_n(H) = -13.6 \frac{1^2}{n^2} = E_n$.For a singly ionized helium atom $(He^+)$,the atomic number $Z = 2$.The energy in the $n^{th}$ orbit for $He^+$ is $E_n(He^+) = -13.6 \frac{2^2}{n^2} = 4 	imes (-13.6 \frac{1^2}{n^2}) = 4E_n$.Therefore,the energy is $4E_n$.

The energy of a hydrogen atom in the $n^{th}$ orbit is $E_n$. What will be the energy in the $n^{th}$ orbit of a singly ionized helium atom?

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