Question:
$\mathrm{AHe}^{+}$ion is in its first excited state. Its ionization energy is:
Correct Option: , 3
Solution:
(3) $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$
For $\mathrm{He}^{+}, E_{2}=\frac{-13.6(2)^{2}}{2^{2}}=-13.60 \mathrm{eV}$
Ionization energy $=0-E_{2}=13.60 \mathrm{eV}$
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