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Q. A $He^+$ ion is in its first excited state. Its ionization energy is :

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Solution:

$E = -13.6 \frac{z^{2}}{n^{2}} eV$
As He+ is 1st excited state
$\therefore z = 2, n = 2$
$E = -13.6\, eV$
As total energy of $He^+$ in 1st excited state is $-13.6\, eV$, ionisation energy should be $+13.6\,eV$.