QJEEM23Ph85ca: JEE Main 2023 Physics - Class 12 JEEM23Ph85ca Solution: The energy of He + ion in its first excited state is, (The…

The energy levels of a one-electron ion can be described by the formula:

$$E_n=-\frac{Z^2}{n^2}\times E_0$$

where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13.6 eV).

For the $${\mathrm{He}}^{+}$$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:

$$E_2=-\frac{2^2}{2^2}\times (-13.6\mathrm{eV})=-13.6\mathrm{eV}$$

So, the energy of the $${\mathrm{He}}^{+}$$ ion in its first excited state is $$-13.6\mathrm{eV}$$.