A hydrogen atom makes a transition from n=2 to n=1 and emits a photon. This photon strikes a doubly ionized lithium atom (Z=3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is.........

Energy of emitted photon

$E=\left[\frac{1}{1^2}-\frac{1}{2^2}\right]\times 13.6\mathrm{eV}=\frac{3}{4}\times 13.6\mathrm{eV}$

Energy required to completely remove the electron from excited state of quantum number 'n' of doubly ionized lithium,

$E^I=\frac{13.6Z^2}{n^2}\mathrm{eV}=\frac{13.6\times 9}{n^2}\mathrm{eV}$

$\text{ As }E\ge E^I\Rightarrow \frac{3}{4}\times 13.6\ge \frac{13.6\times 9}{n^2}$

$n^2\ge 3\times 4\Rightarrow n\ge \sqrt{12}=3.5$

$\therefore$ Least quantum number for the excited state = 4