JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Modern Physics

For a hydrogen-like ion the energy of the level $n$ is $E_n=-13.6\frac{Z^2}{n^2}\text{ eV}$. The ion ${\mathrm{Li}}^{2+}$ has $Z=3$, and its second excited state is $n=3$. Thus $E_3=-13.6\times \frac{3^2}{3^2}=-13.6\text{ eV}$, and the energy needed to free the electron from this level is $13.6\text{ eV}$. Written as $x\times 10^{-1}\text{ eV}$ this is $136\times 10^{-1}\text{ eV}$, so the value of $x$ is $136$.