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  4. The energy of He+ ion in the first excited state will be

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Q. The energy of $He^{+}$ ion in the first excited state will be

NTA AbhyasNTA Abhyas 2020Atoms

Solution:

$E=-\left(13.6\right)\frac{n^{2}}{Z^{2}}eV$
$E=-13.6\frac{2^{2}}{2^{2}}=-13.6eV$