Q 12-12-112JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
A $\text{He}^+$ ion is in its first excited state. Its ionization energy is
Answer: (A) $13.60\ \text{eV}$
For $\text{He}^+$ ($Z = 2$), $E_n = -\dfrac{13.6Z^2}{n^2} = -\dfrac{54.4}{n^2}\ \text{eV}$. The first excited state is $n = 2$:
$$E_2 = -13.6\ \text{eV}$$
so its ionization energy is $13.60\ \text{eV}$.
Solution by Sreeraj P, M.Sc Physics